Notation by work
Paper VIII: Packet Operator Algorithms, Exact Compilation, and GPU Realization
All notation · Glossary · Source coverage
700 symbol spellings in 2339 mathematical passages across 48 selected source files. Subscripts and superscripts have their own entries alongside their base symbols.
The same symbol can have several meanings here. A passage classified as a definition may define another part of its expression. Read the linked context before assigning a meaning.
All 700 entries are available below.
*Definition passage · 92 passages
Search aliases: *
Definition passage: Block diagonal plus low rank operator · Read passage · 01_manuscript.tex, line 303
[Block diagonal plus low rank operator] A block diagonal plus low rank operator has the form A=D+UV^*, where D=⊕_α∈ ID_α, U,V∈ C^n× r. The abbreviation BDLR will be used below.
Exact mathematical passage
A=D+UV^*,
+Definition passage · 286 passages
Search aliases: +
Definition passage: Block diagonal plus low rank operator · Read passage · 01_manuscript.tex, line 303
[Block diagonal plus low rank operator] A block diagonal plus low rank operator has the form A=D+UV^*, where D=⊕_α∈ ID_α, U,V∈ C^n× r. The abbreviation BDLR will be used below.
Exact mathematical passage
A=D+UV^*,
=Definition passage · 592 passages
Search aliases: =
Definition passage: Packet layout · Read passage · 01_manuscript.tex, line 62
[Packet layout] A packet layout is a finite family of finite dimensional Hilbert spaces P=(X_α)_α∈ I with direct sum X=⊕_α∈ IX_α. Write n_α=dim X_α, n=∑_α∈ In_α. The coordinate inclusion and projection are ι_α:X_α→ X and π_α:X→ X_α. The ambient coordinate projector is P_α=ι_απ_α.
Exact mathematical passage
\Pcal=(X_\alpha)_{\alpha\in I}
ADefinition passage · 309 passages
Search aliases: A
Definition passage: Packet operator representation · Read passage · 01_manuscript.tex, line 91
acket operator representation] A finite packet operator representation consists of IR=( P,E,(B_βα)_(β,α)∈ E), where E⊆ I× I and B_βα:X_α⟶ X_β. Local blocks are identified with their ambient extensions by inclusion and projection. In particular, the notation P_β AP_α denotes either the ambient block or its restriction X_α→ X_β according to the displayed domain; ι_β B_βαπ_α is the corresponding ambient extension. Its evaluation is eval(IR) = ∑_(β,α)∈ E ι_β B_βαπ_α, where ι_β:X_β→ X is the coordina…
Exact mathematical passage
P_\beta AP_\alpha
A^{m-1}Definition passage · 2 passages
Search aliases: A^m-1
Definition passage: Krylov space · Read passage · 01_manuscript.tex, line 1096
[Krylov space] For a linear operator A, a vector v, and a positive integer m, define K_m(A,v) = span {v,Av,…,A^m-1v}.
Exact mathematical passage
\Kcal_m(A,v) = \operatorname{span} \{v,Av,\ldots,A^{m-1}v\}.
\absDefinition passage · 3 passages
Search aliases: abs
Definition passage: Decision gap · Read passage · 01_manuscript.tex, line 1694
[Decision gap] At an internal node v, let g_v = absK_λ(v)-S_λ(v). This is the gap between the keep and split scores after the child subproblems have been solved.
Exact mathematical passage
g_v = \abs{K_\lambda(v)-S_\lambda(v)}.
BDefinition passage · 179 passages
Search aliases: B
Definition passage: Packet operator representation · Read passage · 01_manuscript.tex, line 80
[Packet operator representation] A finite packet operator representation consists of IR=( P,E,(B_βα)_(β,α)∈ E), where E⊆ I× I and B_βα:X_α⟶ X_β. Local blocks are identified with their ambient extensions by inclusion and projection. In particular, the notation P_β AP_α denotes either the ambient block or its restriction X_α→ X_β according to the displayed domain; ι_β B_βαπ_α is the corresponding ambient extension. Its evaluation is eval(IR) = ∑_(β,α)∈ E ι_β B_βαπ_α, where ι_β:X_β→ X is the coordi…
Exact mathematical passage
\IR=(\Pcal,E,(B_{\beta\alpha})_{(\beta,\alpha)\in E}),
B_{\beta\alpha}Definition passage · 8 passages
Search aliases: B_βα beta alpha
Definition passage: Packet operator representation · Read passage · 01_manuscript.tex, line 80
[Packet operator representation] A finite packet operator representation consists of IR=( P,E,(B_βα)_(β,α)∈ E), where E⊆ I× I and B_βα:X_α⟶ X_β. Local blocks are identified with their ambient extensions by inclusion and projection. In particular, the notation P_β AP_α denotes either the ambient block or its restriction X_α→ X_β according to the displayed domain; ι_β B_βαπ_α is the corresponding ambient extension. Its evaluation is eval(IR) = ∑_(β,α)∈ E ι_β B_βαπ_α, where ι_β:X_β→ X is the coordi…
Exact mathematical passage
\IR=(\Pcal,E,(B_{\beta\alpha})_{(\beta,\alpha)\in E}),
\bytesDefinition passage · 1 passage
Search aliases: bytes operatorname
Definition passage: Arithmetic intensity · Read passage · 01_manuscript.tex, line 389
[Arithmetic intensity] When the modeled byte traffic is strictly positive, the modeled arithmetic intensity is I = fracflopbytes.
Exact mathematical passage
\Ical = \frac{\flop}{\bytes}.
cDefinition passage · 107 passages
Search aliases: c
Definition passage: Discrete precision option · Read passage · 01_manuscript.tex, line 1422
[Discrete precision option] For edge e, a discrete precision option k has a measured or modeled cost c_e,k and a nonnegative integer error weight w_e,k. Exactly one option must be selected on every edge.
Exact mathematical passage
c_{e,k}
c_{e,k}Definition passage · 1 passage
Search aliases: c_e,k
Definition passage: Discrete precision option · Read passage · 01_manuscript.tex, line 1422
[Discrete precision option] For edge e, a discrete precision option k has a measured or modeled cost c_e,k and a nonnegative integer error weight w_e,k. Exactly one option must be selected on every edge.
Exact mathematical passage
c_{e,k}
\CCDefinition passage · 12 passages
Search aliases: CC C mathbb
Definition passage: Block diagonal plus low rank operator · Read passage · 01_manuscript.tex, line 307
[Block diagonal plus low rank operator] A block diagonal plus low rank operator has the form A=D+UV^*, where D=⊕_α∈ ID_α, U,V∈ C^n× r. The abbreviation BDLR will be used below.
Exact mathematical passage
D=\bigoplus_{\alpha\in I}D_\alpha, \qquad U,V\in\CC^{n\times r}.
\CC^{n\times r}Definition passage · 3 passages
Search aliases: CC^n× r C^n× r CC times mathbb times
Definition passage: Block diagonal plus low rank operator · Read passage · 01_manuscript.tex, line 307
[Block diagonal plus low rank operator] A block diagonal plus low rank operator has the form A=D+UV^*, where D=⊕_α∈ ID_α, U,V∈ C^n× r. The abbreviation BDLR will be used below.
Exact mathematical passage
D=\bigoplus_{\alpha\in I}D_\alpha, \qquad U,V\in\CC^{n\times r}.
DDefinition passage · 158 passages
Search aliases: D
Definition passage: Block diagonal plus low rank operator · Read passage · 01_manuscript.tex, line 303
[Block diagonal plus low rank operator] A block diagonal plus low rank operator has the form A=D+UV^*, where D=⊕_α∈ ID_α, U,V∈ C^n× r. The abbreviation BDLR will be used below.
Exact mathematical passage
A=D+UV^*,
D_\alphaDefinition passage · 3 passages
Search aliases: D_α alpha
Definition passage: Block diagonal plus low rank operator · Read passage · 01_manuscript.tex, line 307
[Block diagonal plus low rank operator] A block diagonal plus low rank operator has the form A=D+UV^*, where D=⊕_α∈ ID_α, U,V∈ C^n× r. The abbreviation BDLR will be used below.
Exact mathematical passage
D=\bigoplus_{\alpha\in I}D_\alpha, \qquad U,V\in\CC^{n\times r}.
\dimDefinition passage · 6 passages
Search aliases: dim
Definition passage: Packet layout · Read passage · 01_manuscript.tex, line 70
[Packet layout] A packet layout is a finite family of finite dimensional Hilbert spaces P=(X_α)_α∈ I with direct sum X=⊕_α∈ IX_α. Write n_α=dim X_α, n=∑_α∈ In_α. The coordinate inclusion and projection are ι_α:X_α→ X and π_α:X→ X_α. The ambient coordinate projector is P_α=ι_απ_α.
Exact mathematical passage
n_\alpha=\dim X_\alpha, \qquad n=\sum_{\alpha\in I}n_\alpha.
EDefinition passage · 77 passages
Search aliases: E
Definition passage: Packet operator representation · Read passage · 01_manuscript.tex, line 80
[Packet operator representation] A finite packet operator representation consists of IR=( P,E,(B_βα)_(β,α)∈ E), where E⊆ I× I and B_βα:X_α⟶ X_β. Local blocks are identified with their ambient extensions by inclusion and projection. In particular, the notation P_β AP_α denotes either the ambient block or its restriction X_α→ X_β according to the displayed domain; ι_β B_βαπ_α is the corresponding ambient extension. Its evaluation is eval(IR) = ∑_(β,α)∈ E ι_β B_βαπ_α, where ι_β:X_β→ X is the coordi…
Exact mathematical passage
\IR=(\Pcal,E,(B_{\beta\alpha})_{(\beta,\alpha)\in E}),
eDefinition passage · 120 passages
Search aliases: e
Definition passage: Discrete precision option · Read passage · 01_manuscript.tex, line 1422
[Discrete precision option] For edge e, a discrete precision option k has a measured or modeled cost c_e,k and a nonnegative integer error weight w_e,k. Exactly one option must be selected on every edge.
Exact mathematical passage
e
\operatorname{eval}Definition passage · 1 passage
Search aliases: eval operatorname
Definition passage: Packet operator representation · Read passage · 01_manuscript.tex, line 92
B_βα:X_α⟶ X_β. Local blocks are identified with their ambient extensions by inclusion and projection. In particular, the notation P_β AP_α denotes either the ambient block or its restriction X_α→ X_β according to the displayed domain; ι_β B_βαπ_α is the corresponding ambient extension. Its evaluation is eval(IR) = ∑_(β,α)∈ E ι_β B_βαπ_α, where ι_β:X_β→ X is the coordinate inclusion.
Exact mathematical passage
\operatorname{eval}(\IR) = \sum_{(\beta,\alpha)\in E} \iota_\beta B_{\beta\alpha}\pi_\alpha,
\flopDefinition passage · 4 passages
Search aliases: flop operatorname
Definition passage: Arithmetic intensity · Read passage · 01_manuscript.tex, line 389
[Arithmetic intensity] When the modeled byte traffic is strictly positive, the modeled arithmetic intensity is I = fracflopbytes.
Exact mathematical passage
\Ical = \frac{\flop}{\bytes}.
GDefinition passage · 34 passages
Search aliases: G
Definition passage: Support graph · Read passage · 01_manuscript.tex, line 138
[Support graph] The support graph G_A has vertex set I. It has a directed edge α⟶β when P_β AP_α≠0.
Exact mathematical passage
G_A
gDefinition passage · 20 passages
Search aliases: g
Definition passage: Decision gap · Read passage · 01_manuscript.tex, line 1694
[Decision gap] At an internal node v, let g_v = absK_λ(v)-S_λ(v). This is the gap between the keep and split scores after the child subproblems have been solved.
Exact mathematical passage
g_v = \abs{K_\lambda(v)-S_\lambda(v)}.
G_ADefinition passage · 3 passages
Search aliases: G_A
Definition passage: Support graph · Read passage · 01_manuscript.tex, line 138
[Support graph] The support graph G_A has vertex set I. It has a directed edge α⟶β when P_β AP_α≠0.
Exact mathematical passage
G_A
g_vDefinition passage · 5 passages
Search aliases: g_v
Definition passage: Decision gap · Read passage · 01_manuscript.tex, line 1694
[Decision gap] At an internal node v, let g_v = absK_λ(v)-S_λ(v). This is the gap between the keep and split scores after the child subproblems have been solved.
Exact mathematical passage
g_v = \abs{K_\lambda(v)-S_\lambda(v)}.
IDefinition passage · 67 passages
Search aliases: I
Definition passage: Packet layout · Read passage · 01_manuscript.tex, line 62
[Packet layout] A packet layout is a finite family of finite dimensional Hilbert spaces P=(X_α)_α∈ I with direct sum X=⊕_α∈ IX_α. Write n_α=dim X_α, n=∑_α∈ In_α. The coordinate inclusion and projection are ι_α:X_α→ X and π_α:X→ X_α. The ambient coordinate projector is P_α=ι_απ_α.
Exact mathematical passage
\Pcal=(X_\alpha)_{\alpha\in I}
\IcalDefinition passage · 2 passages
Search aliases: Ical I mathcal
Definition passage: Arithmetic intensity · Read passage · 01_manuscript.tex, line 389
[Arithmetic intensity] When the modeled byte traffic is strictly positive, the modeled arithmetic intensity is I = fracflopbytes.
Exact mathematical passage
\Ical = \frac{\flop}{\bytes}.
\IRDefinition passage · 2 passages
Search aliases: IR mathsf
Definition passage: Packet operator representation · Read passage · 01_manuscript.tex, line 80
[Packet operator representation] A finite packet operator representation consists of IR=( P,E,(B_βα)_(β,α)∈ E), where E⊆ I× I and B_βα:X_α⟶ X_β. Local blocks are identified with their ambient extensions by inclusion and projection. In particular, the notation P_β AP_α denotes either the ambient block or its restriction X_α→ X_β according to the displayed domain; ι_β B_βαπ_α is the corresponding ambient extension. Its evaluation is eval(IR) = ∑_(β,α)∈ E ι_β B_βαπ_α, where ι_β:X_β→ X is the coordi…
Exact mathematical passage
\IR=(\Pcal,E,(B_{\beta\alpha})_{(\beta,\alpha)\in E}),
KDefinition passage · 57 passages
Search aliases: K
Definition passage: Decision gap · Read passage · 01_manuscript.tex, line 1694
[Decision gap] At an internal node v, let g_v = absK_λ(v)-S_λ(v). This is the gap between the keep and split scores after the child subproblems have been solved.
Exact mathematical passage
g_v = \abs{K_\lambda(v)-S_\lambda(v)}.
kDefinition passage · 137 passages
Search aliases: k
Definition passage: Discrete precision option · Read passage · 01_manuscript.tex, line 1422
[Discrete precision option] For edge e, a discrete precision option k has a measured or modeled cost c_e,k and a nonnegative integer error weight w_e,k. Exactly one option must be selected on every edge.
Exact mathematical passage
k
K_\lambdaDefinition passage · 5 passages
Search aliases: K_λ lambda
Definition passage: Decision gap · Read passage · 01_manuscript.tex, line 1694
[Decision gap] At an internal node v, let g_v = absK_λ(v)-S_λ(v). This is the gap between the keep and split scores after the child subproblems have been solved.
Exact mathematical passage
g_v = \abs{K_\lambda(v)-S_\lambda(v)}.
\KcalDefinition passage · 3 passages
Search aliases: Kcal K mathcal
Definition passage: Krylov space · Read passage · 01_manuscript.tex, line 1096
[Krylov space] For a linear operator A, a vector v, and a positive integer m, define K_m(A,v) = span {v,Av,…,A^m-1v}.
Exact mathematical passage
\Kcal_m(A,v) = \operatorname{span} \{v,Av,\ldots,A^{m-1}v\}.
\Kcal_mDefinition passage · 2 passages
Search aliases: Kcal_m K_m Kcal mathcal
Definition passage: Krylov space · Read passage · 01_manuscript.tex, line 1096
[Krylov space] For a linear operator A, a vector v, and a positive integer m, define K_m(A,v) = span {v,Av,…,A^m-1v}.
Exact mathematical passage
\Kcal_m(A,v) = \operatorname{span} \{v,Av,\ldots,A^{m-1}v\}.
mDefinition passage · 94 passages
Search aliases: m
Definition passage: Krylov space · Read passage · 01_manuscript.tex, line 1095
[Krylov space] For a linear operator A, a vector v, and a positive integer m, define K_m(A,v) = span {v,Av,…,A^m-1v}.
Exact mathematical passage
m
nDefinition passage · 151 passages
Search aliases: n
Definition passage: Packet layout · Read passage · 01_manuscript.tex, line 70
[Packet layout] A packet layout is a finite family of finite dimensional Hilbert spaces P=(X_α)_α∈ I with direct sum X=⊕_α∈ IX_α. Write n_α=dim X_α, n=∑_α∈ In_α. The coordinate inclusion and projection are ι_α:X_α→ X and π_α:X→ X_α. The ambient coordinate projector is P_α=ι_απ_α.
Exact mathematical passage
n_\alpha=\dim X_\alpha, \qquad n=\sum_{\alpha\in I}n_\alpha.
n_\alphaDefinition passage · 8 passages
Search aliases: n_α alpha
Definition passage: Packet layout · Read passage · 01_manuscript.tex, line 70
[Packet layout] A packet layout is a finite family of finite dimensional Hilbert spaces P=(X_α)_α∈ I with direct sum X=⊕_α∈ IX_α. Write n_α=dim X_α, n=∑_α∈ In_α. The coordinate inclusion and projection are ι_α:X_α→ X and π_α:X→ X_α. The ambient coordinate projector is P_α=ι_απ_α.
Exact mathematical passage
n_\alpha=\dim X_\alpha, \qquad n=\sum_{\alpha\in I}n_\alpha.
PDefinition passage · 93 passages
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Definition passage: Packet layout · Read passage · 01_manuscript.tex, line 75
def:packet_layout A packet layout is a finite family of finite dimensional Hilbert spaces P=(X_α)_α∈ I with direct sum X=⊕_α∈ IX_α. Write n_α=dim X_α, n=∑_α∈ In_α. The coordinate inclusion and projection are ι_α:X_α→ X and π_α:X→ X_α. The ambient coordinate projector is P_α=ι_απ_α.
Exact mathematical passage
P_\alpha=\iota_\alpha\pi_\alpha
P_\alphaDefinition passage · 24 passages
Search aliases: P_α alpha
Definition passage: Packet layout · Read passage · 01_manuscript.tex, line 75
def:packet_layout A packet layout is a finite family of finite dimensional Hilbert spaces P=(X_α)_α∈ I with direct sum X=⊕_α∈ IX_α. Write n_α=dim X_α, n=∑_α∈ In_α. The coordinate inclusion and projection are ι_α:X_α→ X and π_α:X→ X_α. The ambient coordinate projector is P_α=ι_απ_α.
Exact mathematical passage
P_\alpha=\iota_\alpha\pi_\alpha
P_\betaDefinition passage · 14 passages
Search aliases: P_β beta
Definition passage: Packet operator representation · Read passage · 01_manuscript.tex, line 91
acket operator representation] A finite packet operator representation consists of IR=( P,E,(B_βα)_(β,α)∈ E), where E⊆ I× I and B_βα:X_α⟶ X_β. Local blocks are identified with their ambient extensions by inclusion and projection. In particular, the notation P_β AP_α denotes either the ambient block or its restriction X_α→ X_β according to the displayed domain; ι_β B_βαπ_α is the corresponding ambient extension. Its evaluation is eval(IR) = ∑_(β,α)∈ E ι_β B_βαπ_α, where ι_β:X_β→ X is the coordina…
Exact mathematical passage
P_\beta AP_\alpha
\PcalDefinition passage · 5 passages
Search aliases: Pcal P mathcal
Definition passage: Packet layout · Read passage · 01_manuscript.tex, line 62
[Packet layout] A packet layout is a finite family of finite dimensional Hilbert spaces P=(X_α)_α∈ I with direct sum X=⊕_α∈ IX_α. Write n_α=dim X_α, n=∑_α∈ In_α. The coordinate inclusion and projection are ι_α:X_α→ X and π_α:X→ X_α. The ambient coordinate projector is P_α=ι_απ_α.
Exact mathematical passage
\Pcal=(X_\alpha)_{\alpha\in I}
rDefinition passage · 102 passages
Search aliases: r
Definition passage: Block diagonal plus low rank operator · Read passage · 01_manuscript.tex, line 307
[Block diagonal plus low rank operator] A block diagonal plus low rank operator has the form A=D+UV^*, where D=⊕_α∈ ID_α, U,V∈ C^n× r. The abbreviation BDLR will be used below.
Exact mathematical passage
D=\bigoplus_{\alpha\in I}D_\alpha, \qquad U,V\in\CC^{n\times r}.
SDefinition passage · 45 passages
Search aliases: S
Definition passage: Decision gap · Read passage · 01_manuscript.tex, line 1694
[Decision gap] At an internal node v, let g_v = absK_λ(v)-S_λ(v). This is the gap between the keep and split scores after the child subproblems have been solved.
Exact mathematical passage
g_v = \abs{K_\lambda(v)-S_\lambda(v)}.
sDefinition passage · 68 passages
Search aliases: s
Definition passage: One pass streaming model · Read passage · 01_manuscript.tex, line 362
[One pass streaming model] Fix a scalar size s bytes. For every active block, the one pass streaming model reads the block once, reads its source packet once, and reads and writes its target packet once. It does not assume cache reuse across edges.
Exact mathematical passage
s
S_\lambdaDefinition passage · 4 passages
Search aliases: S_λ lambda
Definition passage: Decision gap · Read passage · 01_manuscript.tex, line 1694
[Decision gap] At an internal node v, let g_v = absK_λ(v)-S_λ(v). This is the gap between the keep and split scores after the child subproblems have been solved.
Exact mathematical passage
g_v = \abs{K_\lambda(v)-S_\lambda(v)}.
\operatorname{span}Definition passage · 2 passages
Search aliases: span operatorname
Definition passage: Krylov space · Read passage · 01_manuscript.tex, line 1096
[Krylov space] For a linear operator A, a vector v, and a positive integer m, define K_m(A,v) = span {v,Av,…,A^m-1v}.
Exact mathematical passage
\Kcal_m(A,v) = \operatorname{span} \{v,Av,\ldots,A^{m-1}v\}.
UDefinition passage · 43 passages
Search aliases: U
Definition passage: Block diagonal plus low rank operator · Read passage · 01_manuscript.tex, line 303
[Block diagonal plus low rank operator] A block diagonal plus low rank operator has the form A=D+UV^*, where D=⊕_α∈ ID_α, U,V∈ C^n× r. The abbreviation BDLR will be used below.
Exact mathematical passage
A=D+UV^*,
VDefinition passage · 54 passages
Search aliases: V
Definition passage: Block diagonal plus low rank operator · Read passage · 01_manuscript.tex, line 303
[Block diagonal plus low rank operator] A block diagonal plus low rank operator has the form A=D+UV^*, where D=⊕_α∈ ID_α, U,V∈ C^n× r. The abbreviation BDLR will be used below.
Exact mathematical passage
A=D+UV^*,
vDefinition passage · 154 passages
Search aliases: v
Definition passage: Krylov space · Read passage · 01_manuscript.tex, line 1095
[Krylov space] For a linear operator A, a vector v, and a positive integer m, define K_m(A,v) = span {v,Av,…,A^m-1v}.
Exact mathematical passage
v
V^*Definition passage · 20 passages
Search aliases: V^*
Definition passage: Block diagonal plus low rank operator · Read passage · 01_manuscript.tex, line 303
[Block diagonal plus low rank operator] A block diagonal plus low rank operator has the form A=D+UV^*, where D=⊕_α∈ ID_α, U,V∈ C^n× r. The abbreviation BDLR will be used below.
Exact mathematical passage
A=D+UV^*,
wDefinition passage · 51 passages
Search aliases: w
Definition passage: Discrete precision option · Read passage · 01_manuscript.tex, line 1422
[Discrete precision option] For edge e, a discrete precision option k has a measured or modeled cost c_e,k and a nonnegative integer error weight w_e,k. Exactly one option must be selected on every edge.
Exact mathematical passage
w_{e,k}
w_{e,k}Definition passage · 1 passage
Search aliases: w_e,k
Definition passage: Discrete precision option · Read passage · 01_manuscript.tex, line 1422
[Discrete precision option] For edge e, a discrete precision option k has a measured or modeled cost c_e,k and a nonnegative integer error weight w_e,k. Exactly one option must be selected on every edge.
Exact mathematical passage
w_{e,k}
XDefinition passage · 41 passages
Search aliases: X
Definition passage: Packet layout · Read passage · 01_manuscript.tex, line 62
[Packet layout] A packet layout is a finite family of finite dimensional Hilbert spaces P=(X_α)_α∈ I with direct sum X=⊕_α∈ IX_α. Write n_α=dim X_α, n=∑_α∈ In_α. The coordinate inclusion and projection are ι_α:X_α→ X and π_α:X→ X_α. The ambient coordinate projector is P_α=ι_απ_α.
Exact mathematical passage
\Pcal=(X_\alpha)_{\alpha\in I}
X_\alphaDefinition passage · 8 passages
Search aliases: X_α alpha
Definition passage: Packet layout · Read passage · 01_manuscript.tex, line 62
[Packet layout] A packet layout is a finite family of finite dimensional Hilbert spaces P=(X_α)_α∈ I with direct sum X=⊕_α∈ IX_α. Write n_α=dim X_α, n=∑_α∈ In_α. The coordinate inclusion and projection are ι_α:X_α→ X and π_α:X→ X_α. The ambient coordinate projector is P_α=ι_απ_α.
Exact mathematical passage
\Pcal=(X_\alpha)_{\alpha\in I}
X_\betaDefinition passage · 3 passages
Search aliases: X_β beta
Definition passage: Packet operator representation · Read passage · 01_manuscript.tex, line 88
[Packet operator representation] A finite packet operator representation consists of IR=( P,E,(B_βα)_(β,α)∈ E), where E⊆ I× I and B_βα:X_α⟶ X_β. Local blocks are identified with their ambient extensions by inclusion and projection. In particular, the notation P_β AP_α denotes either the ambient block or its restriction X_α→ X_β according to the displayed domain; ι_β B_βαπ_α is the corresponding ambient extension. Its evaluation is eval(IR) = ∑_(β,α)∈ E ι_β B_βαπ_α, where ι_β:X_β→ X is the coordi…
Exact mathematical passage
B_{\beta\alpha}:X_\alpha\longrightarrow X_\beta.
\timesDefinition passage · 50 passages
Search aliases: × times
Definition passage: Packet operator representation · Read passage · 01_manuscript.tex, line 84
[Packet operator representation] A finite packet operator representation consists of IR=( P,E,(B_βα)_(β,α)∈ E), where E⊆ I× I and B_βα:X_α⟶ X_β. Local blocks are identified with their ambient extensions by inclusion and projection. In particular, the notation P_β AP_α denotes either the ambient block or its restriction X_α→ X_β according to the displayed domain; ι_β B_βαπ_α is the corresponding ambient extension. Its evaluation is eval(IR) = ∑_(β,α)∈ E ι_β B_βαπ_α, where ι_β:X_β→ X is the coordi…
Exact mathematical passage
E\subseteq I\times I
\alphaDefinition passage · 84 passages
Search aliases: α alpha
Definition passage: Packet layout · Read passage · 01_manuscript.tex, line 62
[Packet layout] A packet layout is a finite family of finite dimensional Hilbert spaces P=(X_α)_α∈ I with direct sum X=⊕_α∈ IX_α. Write n_α=dim X_α, n=∑_α∈ In_α. The coordinate inclusion and projection are ι_α:X_α→ X and π_α:X→ X_α. The ambient coordinate projector is P_α=ι_απ_α.
Exact mathematical passage
\Pcal=(X_\alpha)_{\alpha\in I}
\betaDefinition passage · 51 passages
Search aliases: β beta
Definition passage: Packet operator representation · Read passage · 01_manuscript.tex, line 80
[Packet operator representation] A finite packet operator representation consists of IR=( P,E,(B_βα)_(β,α)∈ E), where E⊆ I× I and B_βα:X_α⟶ X_β. Local blocks are identified with their ambient extensions by inclusion and projection. In particular, the notation P_β AP_α denotes either the ambient block or its restriction X_α→ X_β according to the displayed domain; ι_β B_βαπ_α is the corresponding ambient extension. Its evaluation is eval(IR) = ∑_(β,α)∈ E ι_β B_βαπ_α, where ι_β:X_β→ X is the coordi…
Exact mathematical passage
\IR=(\Pcal,E,(B_{\beta\alpha})_{(\beta,\alpha)\in E}),
\epsilonDefinition passage · 23 passages
Search aliases: ϵ epsilon
Definition passage: Candidate representation tree · Read passage · 01_manuscript.tex, line 1505
[Candidate representation tree] A candidate representation tree is a finite rooted tree. Every node v has κ_v≥0, ϵ_v≥0, representing the cost and error of keeping v as one leaf. Every internal node also has a split overhead σ_v≥0. An admissible pruning contains exactly one selected node on every root to terminal path.
Exact mathematical passage
\kappa_v\geq0, \qquad \epsilon_v\geq0,
\epsilon_vDefinition passage · 12 passages
Search aliases: ϵ_v epsilon
Definition passage: Candidate representation tree · Read passage · 01_manuscript.tex, line 1505
[Candidate representation tree] A candidate representation tree is a finite rooted tree. Every node v has κ_v≥0, ϵ_v≥0, representing the cost and error of keeping v as one leaf. Every internal node also has a split overhead σ_v≥0. An admissible pruning contains exactly one selected node on every root to terminal path.
Exact mathematical passage
\kappa_v\geq0, \qquad \epsilon_v\geq0,
\iotaDefinition passage · 7 passages
Search aliases: ι iota
Definition passage: Packet layout · Read passage · 01_manuscript.tex, line 75
[Packet layout] A packet layout is a finite family of finite dimensional Hilbert spaces P=(X_α)_α∈ I with direct sum X=⊕_α∈ IX_α. Write n_α=dim X_α, n=∑_α∈ In_α. The coordinate inclusion and projection are ι_α:X_α→ X and π_α:X→ X_α. The ambient coordinate projector is P_α=ι_απ_α.
Exact mathematical passage
\iota_\alpha:X_\alpha\to X
\iota_\alphaDefinition passage · 3 passages
Search aliases: ι_α iota alpha
Definition passage: Packet layout · Read passage · 01_manuscript.tex, line 75
[Packet layout] A packet layout is a finite family of finite dimensional Hilbert spaces P=(X_α)_α∈ I with direct sum X=⊕_α∈ IX_α. Write n_α=dim X_α, n=∑_α∈ In_α. The coordinate inclusion and projection are ι_α:X_α→ X and π_α:X→ X_α. The ambient coordinate projector is P_α=ι_απ_α.
Exact mathematical passage
\iota_\alpha:X_\alpha\to X
\iota_\betaDefinition passage · 4 passages
Search aliases: ι_β iota beta
Definition passage: Packet operator representation · Read passage · 01_manuscript.tex, line 91
f IR=( P,E,(B_βα)_(β,α)∈ E), where E⊆ I× I and B_βα:X_α⟶ X_β. Local blocks are identified with their ambient extensions by inclusion and projection. In particular, the notation P_β AP_α denotes either the ambient block or its restriction X_α→ X_β according to the displayed domain; ι_β B_βαπ_α is the corresponding ambient extension. Its evaluation is eval(IR) = ∑_(β,α)∈ E ι_β B_βαπ_α, where ι_β:X_β→ X is the coordinate inclusion.
Exact mathematical passage
\iota_\beta B_{\beta\alpha}\pi_\alpha
\kappaDefinition passage · 32 passages
Search aliases: κ kappa
Definition passage: Candidate representation tree · Read passage · 01_manuscript.tex, line 1505
[Candidate representation tree] A candidate representation tree is a finite rooted tree. Every node v has κ_v≥0, ϵ_v≥0, representing the cost and error of keeping v as one leaf. Every internal node also has a split overhead σ_v≥0. An admissible pruning contains exactly one selected node on every root to terminal path.
Exact mathematical passage
\kappa_v\geq0, \qquad \epsilon_v\geq0,
\kappa_vDefinition passage · 16 passages
Search aliases: κ_v kappa
Definition passage: Candidate representation tree · Read passage · 01_manuscript.tex, line 1505
[Candidate representation tree] A candidate representation tree is a finite rooted tree. Every node v has κ_v≥0, ϵ_v≥0, representing the cost and error of keeping v as one leaf. Every internal node also has a split overhead σ_v≥0. An admissible pruning contains exactly one selected node on every root to terminal path.
Exact mathematical passage
\kappa_v\geq0, \qquad \epsilon_v\geq0,
\lambdaDefinition passage · 77 passages
Search aliases: λ lambda
Definition passage: Decision gap · Read passage · 01_manuscript.tex, line 1694
[Decision gap] At an internal node v, let g_v = absK_λ(v)-S_λ(v). This is the gap between the keep and split scores after the child subproblems have been solved.
Exact mathematical passage
g_v = \abs{K_\lambda(v)-S_\lambda(v)}.
\piDefinition passage · 11 passages
Search aliases: π pi
Definition passage: Packet layout · Read passage · 01_manuscript.tex, line 75
gindefinition[Packet layout] A packet layout is a finite family of finite dimensional Hilbert spaces P=(X_α)_α∈ I with direct sum X=⊕_α∈ IX_α. Write n_α=dim X_α, n=∑_α∈ In_α. The coordinate inclusion and projection are ι_α:X_α→ X and π_α:X→ X_α. The ambient coordinate projector is P_α=ι_απ_α.
Exact mathematical passage
\pi_\alpha:X\to X_\alpha
\pi_\alphaDefinition passage · 5 passages
Search aliases: π_α pi alpha
Definition passage: Packet layout · Read passage · 01_manuscript.tex, line 75
gindefinition[Packet layout] A packet layout is a finite family of finite dimensional Hilbert spaces P=(X_α)_α∈ I with direct sum X=⊕_α∈ IX_α. Write n_α=dim X_α, n=∑_α∈ In_α. The coordinate inclusion and projection are ι_α:X_α→ X and π_α:X→ X_α. The ambient coordinate projector is P_α=ι_απ_α.
Exact mathematical passage
\pi_\alpha:X\to X_\alpha
\sigmaDefinition passage · 31 passages
Search aliases: σ sigma
Definition passage: Candidate representation tree · Read passage · 01_manuscript.tex, line 1510
[Candidate representation tree] A candidate representation tree is a finite rooted tree. Every node v has κ_v≥0, ϵ_v≥0, representing the cost and error of keeping v as one leaf. Every internal node also has a split overhead σ_v≥0. An admissible pruning contains exactly one selected node on every root to terminal path.
Exact mathematical passage
\sigma_v\geq0
\sigma_vDefinition passage · 12 passages
Search aliases: σ_v sigma
Definition passage: Candidate representation tree · Read passage · 01_manuscript.tex, line 1510
[Candidate representation tree] A candidate representation tree is a finite rooted tree. Every node v has κ_v≥0, ϵ_v≥0, representing the cost and error of keeping v as one leaf. Every internal node also has a split overhead σ_v≥0. An admissible pruning contains exactly one selected node on every root to terminal path.
Exact mathematical passage
\sigma_v\geq0
\ldotsDefinition passage · 19 passages
Search aliases: … ldots
Definition passage: Krylov space · Read passage · 01_manuscript.tex, line 1096
[Krylov space] For a linear operator A, a vector v, and a positive integer m, define K_m(A,v) = span {v,Av,…,A^m-1v}.
Exact mathematical passage
\Kcal_m(A,v) = \operatorname{span} \{v,Av,\ldots,A^{m-1}v\}.
\toDefinition passage · 40 passages
Search aliases: → to
Definition passage: Packet layout · Read passage · 01_manuscript.tex, line 75
[Packet layout] A packet layout is a finite family of finite dimensional Hilbert spaces P=(X_α)_α∈ I with direct sum X=⊕_α∈ IX_α. Write n_α=dim X_α, n=∑_α∈ In_α. The coordinate inclusion and projection are ι_α:X_α→ X and π_α:X→ X_α. The ambient coordinate projector is P_α=ι_απ_α.
Exact mathematical passage
\iota_\alpha:X_\alpha\to X
\inDefinition passage · 61 passages
Search aliases: ∈ in
Definition passage: Packet layout · Read passage · 01_manuscript.tex, line 62
[Packet layout] A packet layout is a finite family of finite dimensional Hilbert spaces P=(X_α)_α∈ I with direct sum X=⊕_α∈ IX_α. Write n_α=dim X_α, n=∑_α∈ In_α. The coordinate inclusion and projection are ι_α:X_α→ X and π_α:X→ X_α. The ambient coordinate projector is P_α=ι_απ_α.
Exact mathematical passage
\Pcal=(X_\alpha)_{\alpha\in I}
\sumDefinition passage · 125 passages
Search aliases: ∑ sum
Definition passage: Packet layout · Read passage · 01_manuscript.tex, line 70
[Packet layout] A packet layout is a finite family of finite dimensional Hilbert spaces P=(X_α)_α∈ I with direct sum X=⊕_α∈ IX_α. Write n_α=dim X_α, n=∑_α∈ In_α. The coordinate inclusion and projection are ι_α:X_α→ X and π_α:X→ X_α. The ambient coordinate projector is P_α=ι_απ_α.
Exact mathematical passage
n_\alpha=\dim X_\alpha, \qquad n=\sum_{\alpha\in I}n_\alpha.
\sum_{(\beta,\alpha)\in E}Definition passage · 6 passages
Search aliases: ∑_(β,α)∈ E sum beta alpha in
Definition passage: Packet operator representation · Read passage · 01_manuscript.tex, line 92
B_βα:X_α⟶ X_β. Local blocks are identified with their ambient extensions by inclusion and projection. In particular, the notation P_β AP_α denotes either the ambient block or its restriction X_α→ X_β according to the displayed domain; ι_β B_βαπ_α is the corresponding ambient extension. Its evaluation is eval(IR) = ∑_(β,α)∈ E ι_β B_βαπ_α, where ι_β:X_β→ X is the coordinate inclusion.
Exact mathematical passage
\operatorname{eval}(\IR) = \sum_{(\beta,\alpha)\in E} \iota_\beta B_{\beta\alpha}\pi_\alpha,
\sum_{\alpha\in I}Definition passage · 2 passages
Search aliases: ∑_α∈ I sum alpha in
Definition passage: Packet layout · Read passage · 01_manuscript.tex, line 70
[Packet layout] A packet layout is a finite family of finite dimensional Hilbert spaces P=(X_α)_α∈ I with direct sum X=⊕_α∈ IX_α. Write n_α=dim X_α, n=∑_α∈ In_α. The coordinate inclusion and projection are ι_α:X_α→ X and π_α:X→ X_α. The ambient coordinate projector is P_α=ι_απ_α.
Exact mathematical passage
n_\alpha=\dim X_\alpha, \qquad n=\sum_{\alpha\in I}n_\alpha.
\neqDefinition passage · 4 passages
Search aliases: ≠ neq
Definition passage: Support graph · Read passage · 01_manuscript.tex, line 142
[Support graph] The support graph G_A has vertex set I. It has a directed edge α⟶β when P_β AP_α≠0.
Exact mathematical passage
P_\beta AP_\alpha\neq0
\geqDefinition passage · 32 passages
Search aliases: ≥ geq
Definition passage: Candidate representation tree · Read passage · 01_manuscript.tex, line 1505
[Candidate representation tree] A candidate representation tree is a finite rooted tree. Every node v has κ_v≥0, ϵ_v≥0, representing the cost and error of keeping v as one leaf. Every internal node also has a split overhead σ_v≥0. An admissible pruning contains exactly one selected node on every root to terminal path.
Exact mathematical passage
\kappa_v\geq0, \qquad \epsilon_v\geq0,
\subseteqDefinition passage · 10 passages
Search aliases: ⊆ subseteq
Definition passage: Packet operator representation · Read passage · 01_manuscript.tex, line 84
[Packet operator representation] A finite packet operator representation consists of IR=( P,E,(B_βα)_(β,α)∈ E), where E⊆ I× I and B_βα:X_α⟶ X_β. Local blocks are identified with their ambient extensions by inclusion and projection. In particular, the notation P_β AP_α denotes either the ambient block or its restriction X_α→ X_β according to the displayed domain; ι_β B_βαπ_α is the corresponding ambient extension. Its evaluation is eval(IR) = ∑_(β,α)∈ E ι_β B_βαπ_α, where ι_β:X_β→ X is the coordi…
Exact mathematical passage
E\subseteq I\times I
\bigoplusDefinition passage · 8 passages
Search aliases: ⊕ bigoplus
Definition passage: Packet layout · Read passage · 01_manuscript.tex, line 66
[Packet layout] A packet layout is a finite family of finite dimensional Hilbert spaces P=(X_α)_α∈ I with direct sum X=⊕_α∈ IX_α. Write n_α=dim X_α, n=∑_α∈ In_α. The coordinate inclusion and projection are ι_α:X_α→ X and π_α:X→ X_α. The ambient coordinate projector is P_α=ι_απ_α.
Exact mathematical passage
X=\bigoplus_{\alpha\in I}X_\alpha.
\bigoplus_{\alpha\in I}Definition passage · 3 passages
Search aliases: ⊕_α∈ I bigoplus alpha in
Definition passage: Packet layout · Read passage · 01_manuscript.tex, line 66
[Packet layout] A packet layout is a finite family of finite dimensional Hilbert spaces P=(X_α)_α∈ I with direct sum X=⊕_α∈ IX_α. Write n_α=dim X_α, n=∑_α∈ In_α. The coordinate inclusion and projection are ι_α:X_α→ X and π_α:X→ X_α. The ambient coordinate projector is P_α=ι_απ_α.
Exact mathematical passage
X=\bigoplus_{\alpha\in I}X_\alpha.
\longrightarrowDefinition passage · 6 passages
Search aliases: ⟶ longrightarrow
Definition passage: Packet operator representation · Read passage · 01_manuscript.tex, line 88
[Packet operator representation] A finite packet operator representation consists of IR=( P,E,(B_βα)_(β,α)∈ E), where E⊆ I× I and B_βα:X_α⟶ X_β. Local blocks are identified with their ambient extensions by inclusion and projection. In particular, the notation P_β AP_α denotes either the ambient block or its restriction X_α→ X_β according to the displayed domain; ι_β B_βαπ_α is the corresponding ambient extension. Its evaluation is eval(IR) = ∑_(β,α)∈ E ι_β B_βαπ_α, where ι_β:X_β→ X is the coordi…
Exact mathematical passage
B_{\beta\alpha}:X_\alpha\longrightarrow X_\beta.
A_0Introduction passage · 7 passages
Search aliases: A_0
Introduction passage · Read passage · 02_supplement_04.tex, line 91
The represented matrix is defined recursively using the actual Kronecker product: M_0=[1], M_k+1=A_0⊗ M_k(A_1,…,A_k). The algorithm is defined independently. On each slice with fixed first coordinate, it recursively contracts the trailing modes. It then applies A_0 to the resulting first-coordinate fibers: y_i,u=∑_j=0^d-1(A_0)_ij Apply_k(A_1,…,A_k;x_j,·)_u. Induction, the Kronecker entry formula, and finite-sum distributivity prove Apply_k(A;x)=M_k(A)x. Thus correctness is not inferred merely fr…
Exact mathematical passage
M_0=[1],\qquad M_{k+1}=A_0\otimes M_k(A_1,\ldots,A_k).
A_1Introduction passage · 10 passages
Search aliases: A_1
Introduction passage · Read passage · 02_supplement_04.tex, line 91
The represented matrix is defined recursively using the actual Kronecker product: M_0=[1], M_k+1=A_0⊗ M_k(A_1,…,A_k). The algorithm is defined independently. On each slice with fixed first coordinate, it recursively contracts the trailing modes. It then applies A_0 to the resulting first-coordinate fibers: y_i,u=∑_j=0^d-1(A_0)_ij Apply_k(A_1,…,A_k;x_j,·)_u. Induction, the Kronecker entry formula, and finite-sum distributivity prove Apply_k(A;x)=M_k(A)x. Thus correctness is not inferred merely fr…
Exact mathematical passage
M_0=[1],\qquad M_{k+1}=A_0\otimes M_k(A_1,\ldots,A_k).
A_kIntroduction passage · 4 passages
Search aliases: A_k
Introduction passage · Read passage · 02_supplement_04.tex, line 91
The represented matrix is defined recursively using the actual Kronecker product: M_0=[1], M_k+1=A_0⊗ M_k(A_1,…,A_k). The algorithm is defined independently. On each slice with fixed first coordinate, it recursively contracts the trailing modes. It then applies A_0 to the resulting first-coordinate fibers: y_i,u=∑_j=0^d-1(A_0)_ij Apply_k(A_1,…,A_k;x_j,·)_u. Induction, the Kronecker entry formula, and finite-sum distributivity prove Apply_k(A;x)=M_k(A)x. Thus correctness is not inferred merely fr…
Exact mathematical passage
M_0=[1],\qquad M_{k+1}=A_0\otimes M_k(A_1,\ldots,A_k).
A_{\gamma\beta}Introduction passage · 1 passage
Search aliases: A_γβ gamma beta
Introduction passage · Read passage · 01_manuscript.tex, line 154
[Block composition] Let A,B∈End(X). Then (AB)_γα = ∑_β∈ I A_γβB_βα. Consequently, supp(AB) ⊆ supp(A)∘supp(B), where relational composition is defined by (γ,α)∈ E∘ F when some β satisfies (γ,β)∈ E, (β,α)∈ F.
Exact mathematical passage
\boxed{ (AB)_{\gamma\alpha} = \sum_{\beta\in I} A_{\gamma\beta}B_{\beta\alpha}. }
\operatorname{Anc}Introduction passage · 2 passages
Search aliases: Anc operatorname
Introduction passage · Read passage · 02_supplement_05.tex, line 8
For each vertex, define its strict ancestor set Anc(v)={u:there is a positive-length path uleadsto v}. For every edge u→ v, the proof establishes Anc(u)subsetneqAnc(v). The inclusion follows by appending the edge. It is strict because u is an ancestor of v but cannot be its own strict ancestor in an acyclic graph. Cardinality therefore decreases when following predecessors. This constructs a well-founded relation on which the level recursion is defined: ℓ(v)=1+max_u→ vℓ(u), max∅=0.
Exact mathematical passage
\operatorname{Anc}(v)=\{u:\text{there is a positive-length path }u\leadsto v\}.
\operatorname{Apply}Introduction passage · 2 passages
Search aliases: Apply operatorname
Introduction passage · Read passage · 02_supplement_04.tex, line 95
The represented matrix is defined recursively using the actual Kronecker product: M_0=[1], M_k+1=A_0⊗ M_k(A_1,…,A_k). The algorithm is defined independently. On each slice with fixed first coordinate, it recursively contracts the trailing modes. It then applies A_0 to the resulting first-coordinate fibers: y_i,u=∑_j=0^d-1(A_0)_ij Apply_k(A_1,…,A_k;x_j,·)_u. Induction, the Kronecker entry formula, and finite-sum distributivity prove Apply_k(A;x)=M_k(A)x. Thus correctness is not inferred merely fr…
Exact mathematical passage
y_{i,u}=\sum_{j=0}^{d-1}(A_0)_{ij}\, \operatorname{Apply}_{k}(A_1,\ldots,A_k;x_{j,\cdot})_u.
\operatorname{Apply}_kIntroduction passage · 1 passage
Search aliases: Apply_k operatorname
Introduction passage · Read passage · 02_supplement_04.tex, line 100
otimes M_k(A_1,…,A_k). The algorithm is defined independently. On each slice with fixed first coordinate, it recursively contracts the trailing modes. It then applies A_0 to the resulting first-coordinate fibers: y_i,u=∑_j=0^d-1(A_0)_ij Apply_k(A_1,…,A_k;x_j,·)_u. Induction, the Kronecker entry formula, and finite-sum distributivity prove Apply_k(A;x)=M_k(A)x. Thus correctness is not inferred merely from agreeing dimension or storage formulas.
Exact mathematical passage
\boxed{\operatorname{Apply}_k(A;x)=M_k(A)x.}
\operatorname{Apply}_{k}Introduction passage · 1 passage
Search aliases: Apply_k operatorname
Introduction passage · Read passage · 02_supplement_04.tex, line 95
The represented matrix is defined recursively using the actual Kronecker product: M_0=[1], M_k+1=A_0⊗ M_k(A_1,…,A_k). The algorithm is defined independently. On each slice with fixed first coordinate, it recursively contracts the trailing modes. It then applies A_0 to the resulting first-coordinate fibers: y_i,u=∑_j=0^d-1(A_0)_ij Apply_k(A_1,…,A_k;x_j,·)_u. Induction, the Kronecker entry formula, and finite-sum distributivity prove Apply_k(A;x)=M_k(A)x. Thus correctness is not inferred merely fr…
Exact mathematical passage
y_{i,u}=\sum_{j=0}^{d-1}(A_0)_{ij}\, \operatorname{Apply}_{k}(A_1,\ldots,A_k;x_{j,\cdot})_u.
dIntroduction passage · 54 passages
Search aliases: d
Introduction passage · Read passage · 02_supplement_04.tex, line 95
The represented matrix is defined recursively using the actual Kronecker product: M_0=[1], M_k+1=A_0⊗ M_k(A_1,…,A_k). The algorithm is defined independently. On each slice with fixed first coordinate, it recursively contracts the trailing modes. It then applies A_0 to the resulting first-coordinate fibers: y_i,u=∑_j=0^d-1(A_0)_ij Apply_k(A_1,…,A_k;x_j,·)_u. Induction, the Kronecker entry formula, and finite-sum distributivity prove Apply_k(A;x)=M_k(A)x. Thus correctness is not inferred merely fr…
Exact mathematical passage
y_{i,u}=\sum_{j=0}^{d-1}(A_0)_{ij}\, \operatorname{Apply}_{k}(A_1,\ldots,A_k;x_{j,\cdot})_u.
D_vIntroduction passage · 4 passages
Search aliases: D_v
Introduction passage · Read passage · 02_supplement_05.tex, line 42
Forward substitution with an actual stage evaluator. textitSource item 6.5; module TriangularSolve Use the graph and levels constructed on the previous page. A vertex v carries an arbitrary vector space X_v over a field. The spaces may have different dimensions. Let D_v:X_voverset∼⟶X_v, T_vu:X_u⟶ X_v be an invertible diagonal map and linear transfer maps. The operator being solved is defined by its actual block action: (Ax)_v=D_vx_v+∑_u→ vT_vux_u. No solution, solve equation or precomputed topol…
Exact mathematical passage
D_v:X_v\overset{\sim}{\longrightarrow}X_v, \qquad T_{vu}:X_u\longrightarrow X_v
\EndIntroduction passage · 2 passages
Search aliases: End operatorname
Introduction passage · Read passage · 01_manuscript.tex, line 153
[Block composition] Let A,B∈End(X). Then (AB)_γα = ∑_β∈ I A_γβB_βα. Consequently, supp(AB) ⊆ supp(A)∘supp(B), where relational composition is defined by (γ,α)∈ E∘ F when some β satisfies (γ,β)∈ E, (β,α)∈ F.
Exact mathematical passage
A,B\in\End(X)
FIntroduction passage · 75 passages
Search aliases: F
Introduction passage · Read passage · 01_manuscript.tex, line 169
[Block composition] Let A,B∈End(X). Then (AB)_γα = ∑_β∈ I A_γβB_βα. Consequently, supp(AB) ⊆ supp(A)∘supp(B), where relational composition is defined by (γ,α)∈ E∘ F when some β satisfies (γ,β)∈ E, (β,α)∈ F.
Exact mathematical passage
(\gamma,\alpha)\in E\circ F
HIntroduction passage · 29 passages
Search aliases: H
Introduction passage · Read passage · 02_supplement_04.tex, line 125
Walsh butterfly and exact addition count. textitSource item 4.7; module WalshTransform The two-point matrix is constructed as H_2=1 · 1 1 · -1, and the full family is defined by H_1=[1] and H_2N=H_2⊗ H_N =H_N · H_N
Exact mathematical passage
H_2=\begin{pmatrix}1&1\\1&-1\end{pmatrix},
H_1Introduction passage · 4 passages
Search aliases: H_1
Introduction passage · Read passage · 02_supplement_04.tex, line 128
1 · -1, and the full family is defined by H_1=[1] and H_2N=H_2⊗ H_N =H_N · H_N
Exact mathematical passage
H_1=[1]
H_2Introduction passage · 2 passages
Search aliases: H_2
Introduction passage · Read passage · 02_supplement_04.tex, line 125
Walsh butterfly and exact addition count. textitSource item 4.7; module WalshTransform The two-point matrix is constructed as H_2=1 · 1 1 · -1, and the full family is defined by H_1=[1] and H_2N=H_2⊗ H_N =H_N · H_N
Exact mathematical passage
H_2=\begin{pmatrix}1&1\\1&-1\end{pmatrix},
H_{2N}Introduction passage · 3 passages
Search aliases: H_2N
Introduction passage · Read passage · 02_supplement_04.tex, line 129
1 · -1, and the full family is defined by H_1=[1] and H_2N=H_2⊗ H_N =H_N · H_N H_N · -H_N. The two-point orthogonality identity is checked explicitly. Kronecker multiplication and induction then prove H_NH_N^T=NI, N=2^k. The ordinary transpose is sufficient because all constructed entries are real, even though the scalar field is C.
Exact mathematical passage
H_{2N}=H_2\otimes H_N =\begin{pmatrix}H_N&H_N\\H_N&-H_N\end{pmatrix}.
H_NIntroduction passage · 6 passages
Search aliases: H_N
Introduction passage · Read passage · 02_supplement_04.tex, line 129
1 · -1, and the full family is defined by H_1=[1] and H_2N=H_2⊗ H_N =H_N · H_N H_N · -H_N. The two-point orthogonality identity is checked explicitly. Kronecker multiplication and induction then prove H_NH_N^T=NI, N=2^k. The ordinary transpose is sufficient because all constructed entries are real, even though the scalar field is C.
Exact mathematical passage
H_{2N}=H_2\otimes H_N =\begin{pmatrix}H_N&H_N\\H_N&-H_N\end{pmatrix}.
iIntroduction passage · 60 passages
Search aliases: i
Introduction passage · Read passage · 02_supplement_04.tex, line 95
The represented matrix is defined recursively using the actual Kronecker product: M_0=[1], M_k+1=A_0⊗ M_k(A_1,…,A_k). The algorithm is defined independently. On each slice with fixed first coordinate, it recursively contracts the trailing modes. It then applies A_0 to the resulting first-coordinate fibers: y_i,u=∑_j=0^d-1(A_0)_ij Apply_k(A_1,…,A_k;x_j,·)_u. Induction, the Kronecker entry formula, and finite-sum distributivity prove Apply_k(A;x)=M_k(A)x. Thus correctness is not inferred merely fr…
Exact mathematical passage
y_{i,u}=\sum_{j=0}^{d-1}(A_0)_{ij}\, \operatorname{Apply}_{k}(A_1,\ldots,A_k;x_{j,\cdot})_u.
jIntroduction passage · 135 passages
Search aliases: j
Introduction passage · Read passage · 02_supplement_04.tex, line 95
The represented matrix is defined recursively using the actual Kronecker product: M_0=[1], M_k+1=A_0⊗ M_k(A_1,…,A_k). The algorithm is defined independently. On each slice with fixed first coordinate, it recursively contracts the trailing modes. It then applies A_0 to the resulting first-coordinate fibers: y_i,u=∑_j=0^d-1(A_0)_ij Apply_k(A_1,…,A_k;x_j,·)_u. Induction, the Kronecker entry formula, and finite-sum distributivity prove Apply_k(A;x)=M_k(A)x. Thus correctness is not inferred merely fr…
Exact mathematical passage
y_{i,u}=\sum_{j=0}^{d-1}(A_0)_{ij}\, \operatorname{Apply}_{k}(A_1,\ldots,A_k;x_{j,\cdot})_u.
\leadstoIntroduction passage · 6 passages
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Introduction passage · Read passage · 02_supplement_05.tex, line 8
For each vertex, define its strict ancestor set Anc(v)={u:there is a positive-length path uleadsto v}. For every edge u→ v, the proof establishes Anc(u)subsetneqAnc(v). The inclusion follows by appending the edge. It is strict because u is an ancestor of v but cannot be its own strict ancestor in an acyclic graph. Cardinality therefore decreases when following predecessors. This constructs a well-founded relation on which the level recursion is defined: ℓ(v)=1+max_u→ vℓ(u), max∅=0.
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\operatorname{Anc}(v)=\{u:\text{there is a positive-length path }u\leadsto v\}.
MIntroduction passage · 34 passages
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Introduction passage · Read passage · 02_supplement_04.tex, line 91
The represented matrix is defined recursively using the actual Kronecker product: M_0=[1], M_k+1=A_0⊗ M_k(A_1,…,A_k). The algorithm is defined independently. On each slice with fixed first coordinate, it recursively contracts the trailing modes. It then applies A_0 to the resulting first-coordinate fibers: y_i,u=∑_j=0^d-1(A_0)_ij Apply_k(A_1,…,A_k;x_j,·)_u. Induction, the Kronecker entry formula, and finite-sum distributivity prove Apply_k(A;x)=M_k(A)x. Thus correctness is not inferred merely fr…
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M_0=[1],\qquad M_{k+1}=A_0\otimes M_k(A_1,\ldots,A_k).
M_0Introduction passage · 1 passage
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Introduction passage · Read passage · 02_supplement_04.tex, line 91
The represented matrix is defined recursively using the actual Kronecker product: M_0=[1], M_k+1=A_0⊗ M_k(A_1,…,A_k). The algorithm is defined independently. On each slice with fixed first coordinate, it recursively contracts the trailing modes. It then applies A_0 to the resulting first-coordinate fibers: y_i,u=∑_j=0^d-1(A_0)_ij Apply_k(A_1,…,A_k;x_j,·)_u. Induction, the Kronecker entry formula, and finite-sum distributivity prove Apply_k(A;x)=M_k(A)x. Thus correctness is not inferred merely fr…
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M_0=[1],\qquad M_{k+1}=A_0\otimes M_k(A_1,\ldots,A_k).
M_kIntroduction passage · 5 passages
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Introduction passage · Read passage · 02_supplement_04.tex, line 91
The represented matrix is defined recursively using the actual Kronecker product: M_0=[1], M_k+1=A_0⊗ M_k(A_1,…,A_k). The algorithm is defined independently. On each slice with fixed first coordinate, it recursively contracts the trailing modes. It then applies A_0 to the resulting first-coordinate fibers: y_i,u=∑_j=0^d-1(A_0)_ij Apply_k(A_1,…,A_k;x_j,·)_u. Induction, the Kronecker entry formula, and finite-sum distributivity prove Apply_k(A;x)=M_k(A)x. Thus correctness is not inferred merely fr…
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M_0=[1],\qquad M_{k+1}=A_0\otimes M_k(A_1,\ldots,A_k).
M_{k+1}Introduction passage · 1 passage
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Introduction passage · Read passage · 02_supplement_04.tex, line 91
The represented matrix is defined recursively using the actual Kronecker product: M_0=[1], M_k+1=A_0⊗ M_k(A_1,…,A_k). The algorithm is defined independently. On each slice with fixed first coordinate, it recursively contracts the trailing modes. It then applies A_0 to the resulting first-coordinate fibers: y_i,u=∑_j=0^d-1(A_0)_ij Apply_k(A_1,…,A_k;x_j,·)_u. Induction, the Kronecker entry formula, and finite-sum distributivity prove Apply_k(A;x)=M_k(A)x. Thus correctness is not inferred merely fr…
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M_0=[1],\qquad M_{k+1}=A_0\otimes M_k(A_1,\ldots,A_k).
\maxIntroduction passage · 11 passages
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Introduction passage · Read passage · 02_supplement_05.tex, line 16
r every edge u→ v, the proof establishes Anc(u)subsetneqAnc(v). The inclusion follows by appending the edge. It is strict because u is an ancestor of v but cannot be its own strict ancestor in an acyclic graph. Cardinality therefore decreases when following predecessors. This constructs a well-founded relation on which the level recursion is defined: ℓ(v)=1+max_u→ vℓ(u), max∅=0.
Exact mathematical passage
\boxed{\ell(v)=1+\max_{u\to v}\ell(u),\qquad \max\varnothing=0.}
\max_{u\to v}Introduction passage · 1 passage
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Introduction passage · Read passage · 02_supplement_05.tex, line 16
r every edge u→ v, the proof establishes Anc(u)subsetneqAnc(v). The inclusion follows by appending the edge. It is strict because u is an ancestor of v but cannot be its own strict ancestor in an acyclic graph. Cardinality therefore decreases when following predecessors. This constructs a well-founded relation on which the level recursion is defined: ℓ(v)=1+max_u→ vℓ(u), max∅=0.
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\boxed{\ell(v)=1+\max_{u\to v}\ell(u),\qquad \max\varnothing=0.}
NIntroduction passage · 26 passages
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Introduction passage · Read passage · 02_supplement_04.tex, line 129
1 · -1, and the full family is defined by H_1=[1] and H_2N=H_2⊗ H_N =H_N · H_N H_N · -H_N. The two-point orthogonality identity is checked explicitly. Kronecker multiplication and induction then prove H_NH_N^T=NI, N=2^k. The ordinary transpose is sufficient because all constructed entries are real, even though the scalar field is C.
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H_{2N}=H_2\otimes H_N =\begin{pmatrix}H_N&H_N\\H_N&-H_N\end{pmatrix}.
\oversetIntroduction passage · 1 passage
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Introduction passage · Read passage · 02_supplement_05.tex, line 42
Forward substitution with an actual stage evaluator. textitSource item 6.5; module TriangularSolve Use the graph and levels constructed on the previous page. A vertex v carries an arbitrary vector space X_v over a field. The spaces may have different dimensions. Let D_v:X_voverset∼⟶X_v, T_vu:X_u⟶ X_v be an invertible diagonal map and linear transfer maps. The operator being solved is defined by its actual block action: (Ax)_v=D_vx_v+∑_u→ vT_vux_u. No solution, solve equation or precomputed topol…
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D_v:X_v\overset{\sim}{\longrightarrow}X_v, \qquad T_{vu}:X_u\longrightarrow X_v
r_1Introduction passage · 7 passages
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Introduction passage · extended_pages.tex, line 59
(r_1,r_2)=(1,3), the aggregated objectives are J_rm keep=4+2+3·7=27, J_rm split=2+8+3·2=16. The exact shared optimum is split. Operator 1 alone prefers keep on its per-use costs, while operator 2 prefers split. Choosing independently and trying to combine those decisions would not solve the declared one-pruning problem. headA hand-computable decision threshold Fix r_1=1 and vary r_2=r≥0. Keeping costs 6+7r and splitting costs 10+2r. Splitting is strictly cheaper when r>4/5; its first integer reu…
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r_1=1
r_2Introduction passage · 7 passages
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Introduction passage · extended_pages.tex, line 59
, the aggregated objectives are J_rm keep=4+2+3·7=27, J_rm split=2+8+3·2=16. The exact shared optimum is split. Operator 1 alone prefers keep on its per-use costs, while operator 2 prefers split. Choosing independently and trying to combine those decisions would not solve the declared one-pruning problem. headA hand-computable decision threshold Fix r_1=1 and vary r_2=r≥0. Keeping costs 6+7r and splitting costs 10+2r. Splitting is strictly cheaper when r>4/5; its first integer reuse win is r=1. …
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r_2=r\ge0
\subsetneqIntroduction passage · 1 passage
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Introduction passage · Read passage · 02_supplement_05.tex, line 12
For each vertex, define its strict ancestor set Anc(v)={u:there is a positive-length path uleadsto v}. For every edge u→ v, the proof establishes Anc(u)subsetneqAnc(v). The inclusion follows by appending the edge. It is strict because u is an ancestor of v but cannot be its own strict ancestor in an acyclic graph. Cardinality therefore decreases when following predecessors. This constructs a well-founded relation on which the level recursion is defined: ℓ(v)=1+max_u→ vℓ(u), max∅=0.
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\operatorname{Anc}(u)\subsetneq\operatorname{Anc}(v).
\suppIntroduction passage · 5 passages
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Introduction passage · Read passage · 01_manuscript.tex, line 163
[Block composition] Let A,B∈End(X). Then (AB)_γα = ∑_β∈ I A_γβB_βα. Consequently, supp(AB) ⊆ supp(A)∘supp(B), where relational composition is defined by (γ,α)∈ E∘ F when some β satisfies (γ,β)∈ E, (β,α)∈ F.
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\supp(AB) \subseteq \supp(A)\circ\supp(B),
TIntroduction passage · 167 passages
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Introduction passage · pages.tex, line 5
ctored operator, and a compiled execution plan are different objects. They may denote the same linear map, but they need not store the same data, return the same output, or incur the same costs. This guide makes those distinctions visible through calculations rather than a pipeline of boxes. vfigthree_views headThree models to carry through the paper The first is a six-dimensional operator A=D+uv^T. Every entry of A is nonzero, yet three 2×2 diagonal blocks and two thin vectors suffice to apply …
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A=D+uv^T
T_{vu}Introduction passage · 4 passages
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Introduction passage · Read passage · 02_supplement_05.tex, line 42
Forward substitution with an actual stage evaluator. textitSource item 6.5; module TriangularSolve Use the graph and levels constructed on the previous page. A vertex v carries an arbitrary vector space X_v over a field. The spaces may have different dimensions. Let D_v:X_voverset∼⟶X_v, T_vu:X_u⟶ X_v be an invertible diagonal map and linear transfer maps. The operator being solved is defined by its actual block action: (Ax)_v=D_vx_v+∑_u→ vT_vux_u. No solution, solve equation or precomputed topol…
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D_v:X_v\overset{\sim}{\longrightarrow}X_v, \qquad T_{vu}:X_u\longrightarrow X_v
uIntroduction passage · 84 passages
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Introduction passage · pages.tex, line 5
ctored operator, and a compiled execution plan are different objects. They may denote the same linear map, but they need not store the same data, return the same output, or incur the same costs. This guide makes those distinctions visible through calculations rather than a pipeline of boxes. vfigthree_views headThree models to carry through the paper The first is a six-dimensional operator A=D+uv^T. Every entry of A is nonzero, yet three 2×2 diagonal blocks and two thin vectors suffice to apply …
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A=D+uv^T
v^TIntroduction passage · 17 passages
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Introduction passage · pages.tex, line 5
ctored operator, and a compiled execution plan are different objects. They may denote the same linear map, but they need not store the same data, return the same output, or incur the same costs. This guide makes those distinctions visible through calculations rather than a pipeline of boxes. vfigthree_views headThree models to carry through the paper The first is a six-dimensional operator A=D+uv^T. Every entry of A is nonzero, yet three 2×2 diagonal blocks and two thin vectors suffice to apply …
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A=D+uv^T
xIntroduction passage · 95 passages
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Introduction passage · Read passage · 02_supplement_04.tex, line 95
The represented matrix is defined recursively using the actual Kronecker product: M_0=[1], M_k+1=A_0⊗ M_k(A_1,…,A_k). The algorithm is defined independently. On each slice with fixed first coordinate, it recursively contracts the trailing modes. It then applies A_0 to the resulting first-coordinate fibers: y_i,u=∑_j=0^d-1(A_0)_ij Apply_k(A_1,…,A_k;x_j,·)_u. Induction, the Kronecker entry formula, and finite-sum distributivity prove Apply_k(A;x)=M_k(A)x. Thus correctness is not inferred merely fr…
Exact mathematical passage
y_{i,u}=\sum_{j=0}^{d-1}(A_0)_{ij}\, \operatorname{Apply}_{k}(A_1,\ldots,A_k;x_{j,\cdot})_u.
x_{j,\cdot}Introduction passage · 1 passage
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Introduction passage · Read passage · 02_supplement_04.tex, line 95
The represented matrix is defined recursively using the actual Kronecker product: M_0=[1], M_k+1=A_0⊗ M_k(A_1,…,A_k). The algorithm is defined independently. On each slice with fixed first coordinate, it recursively contracts the trailing modes. It then applies A_0 to the resulting first-coordinate fibers: y_i,u=∑_j=0^d-1(A_0)_ij Apply_k(A_1,…,A_k;x_j,·)_u. Induction, the Kronecker entry formula, and finite-sum distributivity prove Apply_k(A;x)=M_k(A)x. Thus correctness is not inferred merely fr…
Exact mathematical passage
y_{i,u}=\sum_{j=0}^{d-1}(A_0)_{ij}\, \operatorname{Apply}_{k}(A_1,\ldots,A_k;x_{j,\cdot})_u.
X_uIntroduction passage · 1 passage
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Introduction passage · Read passage · 02_supplement_05.tex, line 42
Forward substitution with an actual stage evaluator. textitSource item 6.5; module TriangularSolve Use the graph and levels constructed on the previous page. A vertex v carries an arbitrary vector space X_v over a field. The spaces may have different dimensions. Let D_v:X_voverset∼⟶X_v, T_vu:X_u⟶ X_v be an invertible diagonal map and linear transfer maps. The operator being solved is defined by its actual block action: (Ax)_v=D_vx_v+∑_u→ vT_vux_u. No solution, solve equation or precomputed topol…
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D_v:X_v\overset{\sim}{\longrightarrow}X_v, \qquad T_{vu}:X_u\longrightarrow X_v
x_uIntroduction passage · 2 passages
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Introduction passage · Read passage · 02_supplement_05.tex, line 47
iangularSolve Use the graph and levels constructed on the previous page. A vertex v carries an arbitrary vector space X_v over a field. The spaces may have different dimensions. Let D_v:X_voverset∼⟶X_v, T_vu:X_u⟶ X_v be an invertible diagonal map and linear transfer maps. The operator being solved is defined by its actual block action: (Ax)_v=D_vx_v+∑_u→ vT_vux_u. No solution, solve equation or precomputed topological certificate is assumed.
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(Ax)_v=D_vx_v+\sum_{u\to v}T_{vu}x_u.
X_vIntroduction passage · 2 passages
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Introduction passage · Read passage · 02_supplement_05.tex, line 41
Forward substitution with an actual stage evaluator. textitSource item 6.5; module TriangularSolve Use the graph and levels constructed on the previous page. A vertex v carries an arbitrary vector space X_v over a field. The spaces may have different dimensions. Let D_v:X_voverset∼⟶X_v, T_vu:X_u⟶ X_v be an invertible diagonal map and linear transfer maps. The operator being solved is defined by its actual block action: (Ax)_v=D_vx_v+∑_u→ vT_vux_u. No solution, solve equation or precomputed topol…
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X_v
x_vIntroduction passage · 3 passages
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Introduction passage · Read passage · 02_supplement_05.tex, line 47
iangularSolve Use the graph and levels constructed on the previous page. A vertex v carries an arbitrary vector space X_v over a field. The spaces may have different dimensions. Let D_v:X_voverset∼⟶X_v, T_vu:X_u⟶ X_v be an invertible diagonal map and linear transfer maps. The operator being solved is defined by its actual block action: (Ax)_v=D_vx_v+∑_u→ vT_vux_u. No solution, solve equation or precomputed topological certificate is assumed.
Exact mathematical passage
(Ax)_v=D_vx_v+\sum_{u\to v}T_{vu}x_u.
yIntroduction passage · 18 passages
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Introduction passage · Read passage · 02_supplement_04.tex, line 95
The represented matrix is defined recursively using the actual Kronecker product: M_0=[1], M_k+1=A_0⊗ M_k(A_1,…,A_k). The algorithm is defined independently. On each slice with fixed first coordinate, it recursively contracts the trailing modes. It then applies A_0 to the resulting first-coordinate fibers: y_i,u=∑_j=0^d-1(A_0)_ij Apply_k(A_1,…,A_k;x_j,·)_u. Induction, the Kronecker entry formula, and finite-sum distributivity prove Apply_k(A;x)=M_k(A)x. Thus correctness is not inferred merely fr…
Exact mathematical passage
y_{i,u}=\sum_{j=0}^{d-1}(A_0)_{ij}\, \operatorname{Apply}_{k}(A_1,\ldots,A_k;x_{j,\cdot})_u.
y_{i,u}Introduction passage · 1 passage
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Introduction passage · Read passage · 02_supplement_04.tex, line 95
The represented matrix is defined recursively using the actual Kronecker product: M_0=[1], M_k+1=A_0⊗ M_k(A_1,…,A_k). The algorithm is defined independently. On each slice with fixed first coordinate, it recursively contracts the trailing modes. It then applies A_0 to the resulting first-coordinate fibers: y_i,u=∑_j=0^d-1(A_0)_ij Apply_k(A_1,…,A_k;x_j,·)_u. Induction, the Kronecker entry formula, and finite-sum distributivity prove Apply_k(A;x)=M_k(A)x. Thus correctness is not inferred merely fr…
Exact mathematical passage
y_{i,u}=\sum_{j=0}^{d-1}(A_0)_{ij}\, \operatorname{Apply}_{k}(A_1,\ldots,A_k;x_{j,\cdot})_u.
\cdotIntroduction passage · 6 passages
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Introduction passage · Read passage · 02_supplement_04.tex, line 95
The represented matrix is defined recursively using the actual Kronecker product: M_0=[1], M_k+1=A_0⊗ M_k(A_1,…,A_k). The algorithm is defined independently. On each slice with fixed first coordinate, it recursively contracts the trailing modes. It then applies A_0 to the resulting first-coordinate fibers: y_i,u=∑_j=0^d-1(A_0)_ij Apply_k(A_1,…,A_k;x_j,·)_u. Induction, the Kronecker entry formula, and finite-sum distributivity prove Apply_k(A;x)=M_k(A)x. Thus correctness is not inferred merely fr…
Exact mathematical passage
y_{i,u}=\sum_{j=0}^{d-1}(A_0)_{ij}\, \operatorname{Apply}_{k}(A_1,\ldots,A_k;x_{j,\cdot})_u.
\gammaIntroduction passage · 10 passages
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Introduction passage · Read passage · 01_manuscript.tex, line 154
[Block composition] Let A,B∈End(X). Then (AB)_γα = ∑_β∈ I A_γβB_βα. Consequently, supp(AB) ⊆ supp(A)∘supp(B), where relational composition is defined by (γ,α)∈ E∘ F when some β satisfies (γ,β)∈ E, (β,α)∈ F.
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\boxed{ (AB)_{\gamma\alpha} = \sum_{\beta\in I} A_{\gamma\beta}B_{\beta\alpha}. }
\ellIntroduction passage · 11 passages
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Introduction passage · Read passage · 02_supplement_05.tex, line 16
r every edge u→ v, the proof establishes Anc(u)subsetneqAnc(v). The inclusion follows by appending the edge. It is strict because u is an ancestor of v but cannot be its own strict ancestor in an acyclic graph. Cardinality therefore decreases when following predecessors. This constructs a well-founded relation on which the level recursion is defined: ℓ(v)=1+max_u→ vℓ(u), max∅=0.
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\boxed{\ell(v)=1+\max_{u\to v}\ell(u),\qquad \max\varnothing=0.}
\varnothingIntroduction passage · 2 passages
Search aliases: ∅ varnothing
Introduction passage · Read passage · 02_supplement_05.tex, line 16
r every edge u→ v, the proof establishes Anc(u)subsetneqAnc(v). The inclusion follows by appending the edge. It is strict because u is an ancestor of v but cannot be its own strict ancestor in an acyclic graph. Cardinality therefore decreases when following predecessors. This constructs a well-founded relation on which the level recursion is defined: ℓ(v)=1+max_u→ vℓ(u), max∅=0.
Exact mathematical passage
\boxed{\ell(v)=1+\max_{u\to v}\ell(u),\qquad \max\varnothing=0.}
\sum_{j=0}^{d-1}Introduction passage · 1 passage
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Introduction passage · Read passage · 02_supplement_04.tex, line 95
The represented matrix is defined recursively using the actual Kronecker product: M_0=[1], M_k+1=A_0⊗ M_k(A_1,…,A_k). The algorithm is defined independently. On each slice with fixed first coordinate, it recursively contracts the trailing modes. It then applies A_0 to the resulting first-coordinate fibers: y_i,u=∑_j=0^d-1(A_0)_ij Apply_k(A_1,…,A_k;x_j,·)_u. Induction, the Kronecker entry formula, and finite-sum distributivity prove Apply_k(A;x)=M_k(A)x. Thus correctness is not inferred merely fr…
Exact mathematical passage
y_{i,u}=\sum_{j=0}^{d-1}(A_0)_{ij}\, \operatorname{Apply}_{k}(A_1,\ldots,A_k;x_{j,\cdot})_u.
\sum_{u\to v}Introduction passage · 3 passages
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Introduction passage · Read passage · 02_supplement_05.tex, line 47
iangularSolve Use the graph and levels constructed on the previous page. A vertex v carries an arbitrary vector space X_v over a field. The spaces may have different dimensions. Let D_v:X_voverset∼⟶X_v, T_vu:X_u⟶ X_v be an invertible diagonal map and linear transfer maps. The operator being solved is defined by its actual block action: (Ax)_v=D_vx_v+∑_u→ vT_vux_u. No solution, solve equation or precomputed topological certificate is assumed.
Exact mathematical passage
(Ax)_v=D_vx_v+\sum_{u\to v}T_{vu}x_u.
\sum_{\beta\in I}Introduction passage · 1 passage
Search aliases: ∑_β∈ I sum beta in
Introduction passage · Read passage · 01_manuscript.tex, line 154
[Block composition] Let A,B∈End(X). Then (AB)_γα = ∑_β∈ I A_γβB_βα. Consequently, supp(AB) ⊆ supp(A)∘supp(B), where relational composition is defined by (γ,α)∈ E∘ F when some β satisfies (γ,β)∈ E, (β,α)∈ F.
Exact mathematical passage
\boxed{ (AB)_{\gamma\alpha} = \sum_{\beta\in I} A_{\gamma\beta}B_{\beta\alpha}. }
\circIntroduction passage · 5 passages
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Introduction passage · Read passage · 01_manuscript.tex, line 163
[Block composition] Let A,B∈End(X). Then (AB)_γα = ∑_β∈ I A_γβB_βα. Consequently, supp(AB) ⊆ supp(A)∘supp(B), where relational composition is defined by (γ,α)∈ E∘ F when some β satisfies (γ,β)∈ E, (β,α)∈ F.
Exact mathematical passage
\supp(AB) \subseteq \supp(A)\circ\supp(B),
\simIntroduction passage · 1 passage
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Introduction passage · Read passage · 02_supplement_05.tex, line 42
Forward substitution with an actual stage evaluator. textitSource item 6.5; module TriangularSolve Use the graph and levels constructed on the previous page. A vertex v carries an arbitrary vector space X_v over a field. The spaces may have different dimensions. Let D_v:X_voverset∼⟶X_v, T_vu:X_u⟶ X_v be an invertible diagonal map and linear transfer maps. The operator being solved is defined by its actual block action: (Ax)_v=D_vx_v+∑_u→ vT_vux_u. No solution, solve equation or precomputed topol…
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D_v:X_v\overset{\sim}{\longrightarrow}X_v, \qquad T_{vu}:X_u\longrightarrow X_v
\geIntroduction passage · 33 passages
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Introduction passage · extended_pages.tex, line 59
, the aggregated objectives are J_rm keep=4+2+3·7=27, J_rm split=2+8+3·2=16. The exact shared optimum is split. Operator 1 alone prefers keep on its per-use costs, while operator 2 prefers split. Choosing independently and trying to combine those decisions would not solve the declared one-pruning problem. headA hand-computable decision threshold Fix r_1=1 and vary r_2=r≥0. Keeping costs 6+7r and splitting costs 10+2r. Splitting is strictly cheaper when r>4/5; its first integer reuse win is r=1. …
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r_2=r\ge0
\otimesIntroduction passage · 11 passages
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Introduction passage · Read passage · 02_supplement_04.tex, line 91
The represented matrix is defined recursively using the actual Kronecker product: M_0=[1], M_k+1=A_0⊗ M_k(A_1,…,A_k). The algorithm is defined independently. On each slice with fixed first coordinate, it recursively contracts the trailing modes. It then applies A_0 to the resulting first-coordinate fibers: y_i,u=∑_j=0^d-1(A_0)_ij Apply_k(A_1,…,A_k;x_j,·)_u. Induction, the Kronecker entry formula, and finite-sum distributivity prove Apply_k(A;x)=M_k(A)x. Thus correctness is not inferred merely fr…
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M_0=[1],\qquad M_{k+1}=A_0\otimes M_k(A_1,\ldots,A_k).
\mathbf1Use in source · 2 passages
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Use in source · block_error.tex, line 47
nter,font=scriptsize] at (10.1,3.0575) 3; path[draw=black,fill=black!12,line width=.6pt] (11.456667,0.85) rectangle (12.41,3.144967); node[font=small,align=center,font=scriptsize] at (11.933333,0.55) Frobenius; node[font=small,align=center,font=scriptsize] at (11.933333,3.364967) 2 sqrt3; node[font=small,align=center,font=scriptsize] at (7,0.1) E _F^2=12/10^4, while E1=(3/100)1.;
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E\mathbf1=(3/100)\mathbf1
aUse in source · 103 passages
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Use in source · pages.tex, line 142
vpage9The cross term has somewhere to goCLOSED LOW-RANK PRODUCT ALGEBRA Let A=D_A+u_av_a^T be the running operator. Take D_B=diag(1,2,3,4,5,6), u_b=(1,0,1,0,1,0)^T, and v_b=(0,1,0,1,0,1)^T. Define B=D_B+u_bv_b^T. U_C=[D_Au_b+u_a(v_a^Tu_b) u_a], V_C=[v_b D_B^Tv_a]. vfigbdlr_product headWhy there are only two columns Expanding AB gives three correction terms, but two share the right factor v_b^T: D_Au_bv_b^T+u_a(v_a^Tu_b)v_b^T =(D_Au_b+u_a(v_a^Tu_b))v_b^T. The remaining term is u_av_a^TD_B. Thus A…
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A=D_A+u_av_a^T
A^{(1)}Use in source · 1 passage
Search aliases: A^(1)
Use in source · Read passage · 01_manuscript.tex, line 1579
[Shared compilation for an operator family] Let operators A^(1),…,A^(m) use the same candidate tree. Give operator j a nonnegative reuse weight r_j. Replace the local keep cost and error by κ_v = κ_v^rm setup + ∑_j=1^m r_jκ_v^(j) and ϵ_v = ∑_j=1^m r_jϵ_v^(j). Then [source reference] gives the exact optimal shared pruning for the additive weighted objective.
Exact mathematical passage
A^{(1)},\ldots,A^{(m)}
A^{(m)}Use in source · 1 passage
Search aliases: A^(m)
Use in source · Read passage · 01_manuscript.tex, line 1579
[Shared compilation for an operator family] Let operators A^(1),…,A^(m) use the same candidate tree. Give operator j a nonnegative reuse weight r_j. Replace the local keep cost and error by κ_v = κ_v^rm setup + ∑_j=1^m r_jκ_v^(j) and ϵ_v = ∑_j=1^m r_jϵ_v^(j). Then [source reference] gives the exact optimal shared pruning for the additive weighted objective.
Exact mathematical passage
A^{(1)},\ldots,A^{(m)}
A^*Use in source · 9 passages
Search aliases: A^*
Use in source · extended_pages.tex, line 8
0 · i, G=diag(5,2). Their product is a genuinely complex rectangular operator: A=Q_LGQ_R^*=mat3 · 0 4 i · 0 0 · -2 i, A^*A=diag(25,4). vfigcomplex_core headThe small calculation exposes a wrong convention The first column of Q_L has Hermitian squared length 9/25+16/25=1. Its bilinear square without conjugation is 9/25-16/25=-7/25. Thus Q_L^TQ_L is not the identity. The portable negative control replaces the adjoint by transpose and rejects the resulting isometry claim.
Exact mathematical passage
A=Q_LGQ_R^*=\mat{3&0\\4\mathrm i&0\\0&-2\mathrm i}, \qquad A^*A=\diag(25,4).
A^{-1}Use in source · 15 passages
Search aliases: A^-1
Use in source · extended_pages.tex, line 27
0 · 1 · -1 · 1. The two routes from coordinate zero to coordinate three cancel, so K^2=0. Consequently A^-1=I-K=mat1 · 0 · 0 · 0 -1 · 1 · 0 · 0 -1 · 0 · 1 · 0 0 · -1 · 1 · 1. vfig[.52]inverse_closure headWhat the closure permits The directed transitive closure includes 0→3, because it records existence of a route rather than the signed sum of its coefficients. Nevertheless (A^-1)_30=0. Reachability is therefore an upper envelope for inverse support, just as it is for product support.
Exact mathematical passage
A^{-1}=I-K=\mat{1&0&0&0\\-1&1&0&0\\-1&0&1&0\\0&-1&1&1}.
A^2Use in source · 2 passages
Search aliases: A^2
Use in source · Read passage · 02_supplement_01.tex, line 77
Cancellation control. Lean verifies the rational example A=1 · 1 -1 · -1, A^2=0. The coordinate block A_00 is nonzero, and a length-two active walk from zero to zero exists. Nevertheless the corresponding block of A^2 is zero. This is an exact counterexample to replacing support inclusion by equality.
Exact mathematical passage
A=\begin{bmatrix}1&1\\-1&-1\end{bmatrix},\qquad A^2=0.
A^{h+1}Use in source · 1 passage
Search aliases: A^h+1
Use in source · Read passage · 02_supplement_01.tex, line 60
Nilpotence certificate. If no walk of length h+1 exists between any pair of packets, then A^h+1=0. Every block vanishes, and reconstruction forces the whole operator to vanish. A second theorem derives the required absence of walks from a supplied level map ℓ:I⟶{0,…,h}, A_bc≠0 Longrightarrow ℓ(c)<ℓ(b). It proves that a length-n walk increases the level by at least n. This certifies the acyclic case with a bounded height. This block does not implement or verify an algorithm that constructs a topo…
Exact mathematical passage
\boxed{A^{h+1}=0.}
A^jUse in source · 12 passages
Search aliases: A^j
Use in source · pages.tex, line 292
With A=diag(1,2), the vectors v and Av=(1,2)^T are independent. Two polynomial samples now distinguish the packet labels. Alternatively, allowing the packet projectors as extra address operators separates the coordinates even when A=I_2. Both changes enlarge the addressing power, but in different ways. headWhen invariance is only approximate If A=D+K with D packet diagonal, then A^j-D^j ≤ jM^j-1 K , M=max{ A , D }. This controls departure from the decoupled iterate. It is not a claim that the di…
Exact mathematical passage
\|A^j-D^j\|\le jM^{j-1}\|K\|, \qquad M=\max\{\|A\|,\|D\|\}.
A^{j-1-k}Use in source · 1 passage
Search aliases: A^j-1-k
Use in source · Read passage · 01_manuscript.tex, line 1169
Use the telescoping identity A^j-D^j = ∑_k=0^j-1 A^j-1-kKD^k. Each term has norm at most M^j-1normK. The vector D^jv lies in the direct sum of local Krylov spaces.
Exact mathematical passage
A^j-D^j = \sum_{k=0}^{j-1} A^{j-1-k}KD^k.
A^mUse in source · 1 passage
Search aliases: A^m
Use in source · Read passage · 01_manuscript.tex, line 952
[Path support of powers] For every positive integer m, P_γ A^mP_α is the sum over directed walks (vertices may repeat) α=α_0 ⟶ α_1 ⟶⋯⟶ α_m=γ in G_A, with contribution A_α_mα_m-1 ⋯ A_α_1α_0. If no such path exists, the block vanishes.
Exact mathematical passage
P_\gamma A^mP_\alpha
A^nUse in source · 2 passages
Search aliases: A^n
Use in source · Read passage · 02_supplement_01.tex, line 54
F_n+1(b,a) · =∑_c: A_bc≠0A_bcF_n(c,a). The checked identity is P_bA^nP_a=F_n(b,a) (n≥0). Its proof is induction using the exact composition theorem, not an assumed support recurrence. When no active walk exists, the corresponding contribution sum is zero, and hence the actual block of A^n vanishes.
Exact mathematical passage
\boxed{P_bA^nP_a=F_n(b,a)\qquad(n\ge0).}
A^TUse in source · 15 passages
Search aliases: A^T
Use in source · pages.tex, line 107
vpage7A transpose, not an adjointKRONECKER ACTION AND ARRAY ORDER With column-major vectorization, the tensor identity is (A⊗ B)vec(X)=vec(BXA^T). Here the transpose records index order, not a Hermitian inner product. Use the explicitly complex matrix A=mat1 · i
Exact mathematical passage
(A\otimes B)\operatorname{vec}(X)=\operatorname{vec}(BXA^T).
a_0Use in source · 6 passages
Search aliases: a_0
Use in source · pages.tex, line 387
ildren. Keeping r costs one with error six; splitting it adds overhead one. Keeping either a or b costs three with error two. Splitting a costs four with zero error, and splitting b costs five with zero error. vfigtree_frontier The five complete prunings have (cost,error) pairs (1,6), (7,4), (9,2), (8,2), (10,0). At budget two the unique cheapest pruning is {a_0,a_1,b}, costing eight. It is Pareto optimal, but never minimizes cost+λerr for any λ≥0. headSee the incompatible inequalities To beat t…
Exact mathematical passage
\{a_0,a_1,b\}
A_{00}Use in source · 1 passage
Search aliases: A_00
Use in source · Read passage · 02_supplement_01.tex, line 80
-1 · -1, A^2=0. The coordinate block A_00 is nonzero, and a length-two active walk from zero to zero exists. Nevertheless the corresponding block of A^2 is zero. This is an exact counterexample to replacing support inclusion by equality.
Exact mathematical passage
A_{00}
a_1Use in source · 6 passages
Search aliases: a_1
Use in source · pages.tex, line 387
ildren. Keeping r costs one with error six; splitting it adds overhead one. Keeping either a or b costs three with error two. Splitting a costs four with zero error, and splitting b costs five with zero error. vfigtree_frontier The five complete prunings have (cost,error) pairs (1,6), (7,4), (9,2), (8,2), (10,0). At budget two the unique cheapest pruning is {a_0,a_1,b}, costing eight. It is Pareto optimal, but never minimizes cost+λerr for any λ≥0. headSee the incompatible inequalities To beat t…
Exact mathematical passage
\{a_0,a_1,b\}
a_2Use in source · 4 passages
Search aliases: a_2
Use in source · Read passage · monograph_examples.tex, line 47
A budget can select a point that no Lagrangian selects. Consider a rooted candidate tree. The root r has children a,b, and a has children a_1,a_2. Split overheads are zero. Each retained node has the following declared cost and integer error: lrrrrr Node · r · a · b · a_1 · a_2
Exact mathematical passage
a_1,a_2
a_3Use in source · 1 passage
Search aliases: a_3
Use in source · error_chain.tex, line 25
iptsize] at (10.5,2.92) frac45; path[draw=black,fill=black!12,line width=.6pt] (11.733333,0.9) rectangle (12.6,1.0125); node[font=small,align=center,font=scriptsize] at (12.166667,0.6) 3; node[font=small,align=center,font=scriptsize] at (12.166667,1.2325) frac120; node[font=small,align=center,font=scriptsize] at (7,0.05) widehat p_j=a_jwidehat p_j-1+e_j, with (a_1,a_2,a_3)=(2,3,4).;
Exact mathematical passage
(a_1,a_2,a_3)=(2,3,4)
A_{ba}Use in source · 6 passages
Search aliases: A_ba
Use in source · Read passage · 02_supplement_01.tex, line 10
0 · b≠ a, ∑_aP_a=1. Write A_ba=P_bAP_a and let E_A={(b,a):A_ba≠0}. These are exact algebraic zeros, not values discarded by a floating-point threshold.
Exact mathematical passage
A_{ba}=P_bAP_a
A_{bc}Use in source · 4 passages
Search aliases: A_bc
Use in source · Read passage · 02_supplement_01.tex, line 23
Composition. Inserting the projector identity between factors yields (AB)_ba=∑_c A_bcB_ca. If the left side is nonzero, at least one intermediate c has both A_bc≠0 and B_ca≠0. Consequently supp(AB)⊆supp(A)∘supp(B). No equality of supports is asserted.
Exact mathematical passage
\boxed{(AB)_{ba}=\sum_c A_{bc}B_{ca}.}
A_eUse in source · 7 passages
Search aliases: A_e
Use in source · Read passage · 01_manuscript.tex, line 1256
[Packet perturbation certificate] Let A= ∑_e∈ EA_e, widehat A = ∑_e∈ E(A_e+E_e), where edge e starts at packet α(e), and assume A_e=A_eP_α(e), E_e=E_eP_α(e). Then norm(widehat A-A)x ≤ ∑_e∈ E normE_e normP_α(e)x. If normE_e ≤ ε_enormA_e, then norm(widehat A-A)x ≤ ∑_e∈ E ε_enormA_e normP_α(e)x.
Exact mathematical passage
A= \sum_{e\in E}A_e, \qquad \widehat A = \sum_{e\in E}(A_e+E_e),
A_{HH}Use in source · 1 passage
Search aliases: A_HH
Use in source · Read passage · 01_manuscript.tex, line 1071
[Schur support inclusion] Split the packets into retained indices R and hidden indices H. Suppose the hidden block A_HH is invertible. The retained Schur complement is S = A_RR - A_RHA_HH^-1A_HR. A retained block (β,α) can be active only if it was already active in A_RR, or if there is a directed walk from α through hidden packets to β. The internal hidden path may have length zero, so entrance and exit may use the same hidden packet.
Exact mathematical passage
A_{HH}
A_{HH}^{-1}Use in source · 6 passages
Search aliases: A_HH^-1
Use in source · pages.tex, line 266
1 · 1 · 2. The third coordinate is hidden. Its invertible block is [2], so eliminating it produces S=A_RR-A_RHA_HH^-1A_HR =mat3/2 · -1/2 -1/2 · 3/2. vfigschur_fill headThe hidden route has a sign The retained coordinates had no direct coupling, yet both touch the hidden coordinate. The resulting correction couples them by -1/2. This is fill: a new active retained block produced by an eliminated route.
Exact mathematical passage
S=A_{RR}-A_{RH}A_{HH}^{-1}A_{HR} =\mat{3/2&-1/2\\-1/2&3/2}.
A_{HR}Use in source · 7 passages
Search aliases: A_HR
Use in source · pages.tex, line 266
1 · 1 · 2. The third coordinate is hidden. Its invertible block is [2], so eliminating it produces S=A_RR-A_RHA_HH^-1A_HR =mat3/2 · -1/2 -1/2 · 3/2. vfigschur_fill headThe hidden route has a sign The retained coordinates had no direct coupling, yet both touch the hidden coordinate. The resulting correction couples them by -1/2. This is fill: a new active retained block produced by an eliminated route.
Exact mathematical passage
S=A_{RR}-A_{RH}A_{HH}^{-1}A_{HR} =\mat{3/2&-1/2\\-1/2&3/2}.
A_iUse in source · 6 passages
Search aliases: A_i
Use in source · Read passage · 01_manuscript.tex, line 864
[Product chain perturbation] Let A_1,…,A_m be exact factors and let widehat A_j=A_j+E_j. Then widehat A_m⋯widehat A_1 - A_m⋯ A_1 ≤ ∑_j=1^m ( ∏_i=j+1^mnormwidehat A_i ) normE_j ( ∏_i=1^j-1normA_i ). Empty products are one.
Exact mathematical passage
\boxed{ \left\| \widehat A_m\cdots\widehat A_1 - A_m\cdots A_1 \right\| \leq \sum_{j=1}^m \left( \prod_{i=j+1}^m\norm{\widehat A_i} \right) \norm{E_j} \left( \prod_{i=1}^{j-1}\norm{A_i} \right). }
a_iUse in source · 6 passages
Search aliases: a_i
Use in source · Read passage · 02_supplement_06.tex, line 137
For any nonnegative decreasing sequence a_i, weights with these properties satisfy ∑_i a_iw_i≥∑_i>ka_i. This finite weighted inequality is itself proved. For k<d, use μ=a_k+1 and compare the summands to tail weights v_i=0 for i≤ k and v_i=1 otherwise: a_iv_i+μ(w_i-v_i)≤ a_iw_i. For a retained coordinate the inequality uses a_i≥μ and w_i≥0. For a tail coordinate it uses a_i≤μ and w_i≤1. Summation and ∑ w_i≥∑ v_i give the result.
Exact mathematical passage
a_i
A_i^jUse in source · 1 passage
Search aliases: A_i^j
Use in source · Read passage · 02_supplement_03.tex, line 97
Exact and approximate packet Krylov spaces. textitSource items 7.2 and 7.3 Write K_m(A,v)=span{v,Av,…,A^m-1v}. For a complete finite projector family P_i, suppose A commutes with every P_i. Let A_i=P_iAP_i. Lean proves the actual orbit identity A^jv=∑_i A_i^jP_iv and therefore K_m(A,v)⊆ ⊕_i K_m(A_i,P_iv). The case j=0 uses ∑_iP_i=I. The zeroth local power is the ambient identity, applied to P_iv, rather than an incorrectly substituted local projector power.
Exact mathematical passage
A^jv=\sum_i A_i^jP_iv
A_jUse in source · 13 passages
Search aliases: A_j
Use in source · Read passage · 01_manuscript.tex, line 716
[Rank growth in a product chain] Let A_j=D_j+U_jV_j^*, j=1,…,m, share one diagonal packet layout. Before recompression, the product A_m⋯ A_1 has a BDLR representation whose correction rank is at most ∑_j=1^m r_j.
Exact mathematical passage
A_j=D_j+U_jV_j^*, \qquad j=1,\ldots,m,
a_jUse in source · 3 passages
Search aliases: a_j
Use in source · pages.tex, line 217
vpage13Errors enter at different placesFACTOR PERTURBATION VERSUS RECOMPRESSION Let the exact scalar factors be (2,3,4) and start from p_0=1. If additive errors (1/10,1/5,1/20) enter after each partial multiplication, then widehat p_j=a_jwidehat p_j-1+e_j, widehat p_3-p_3=12e_1+4e_2+e_3=frac4120. vfigerror_chain headThe same local numbers, a different model If instead the factors themselves are perturbed, the computed product is (4+e_3)(3+e_2)(2+e_1). Its error is 402/125, not 41/20. The factor-…
Exact mathematical passage
\widehat p_j=a_j\widehat p_{j-1}+e_j,\qquad \widehat p_3-p_3=12e_1+4e_2+e_3=\frac{41}{20}.
A_{j+1}Use in source · 3 passages
Search aliases: A_j+1
Use in source · Read passage · 01_manuscript.tex, line 937
Set D_j=widehat P_j-P_j. Then D_j=A_jD_j-1+E_j, D_0=0. Repeated substitution gives D_m = ∑_j=1^m A_mA_m-1⋯ A_j+1E_j. Taking norms yields the result.
Exact mathematical passage
D_m = \sum_{j=1}^{m} A_mA_{m-1}\cdots A_{j+1}E_j.
A_{j,e}Use in source · 1 passage
Search aliases: A_j,e
Use in source · Read passage · 02_supplement_03.tex, line 75
Packetwise mixed precision. At each stage let A_j=∑_e A_j,e, widehat A_j=∑_e(A_j,e+Δ_j,e), Δ_j,e=Δ_j,eP_j,a(e). The previous packet certificate supplies the uniform operator bound normwidehat A_j-A_j≤ δ_j:=∑_enormΔ_j,enormP_j,a(e). The new theorem substitutes these δ_j into the actual product-chain estimate. The resulting vector error is at most the operator certificate times normx. A vector-specific packet bound is not silently reused as an operator norm.
Exact mathematical passage
A_j=\sum_e A_{j,e},\qquad \widehat A_j=\sum_e(A_{j,e}+\Delta_{j,e}),\qquad \Delta_{j,e}=\Delta_{j,e}P_{j,a(e)}.
A_{j-1}Use in source · 1 passage
Search aliases: A_j-1
Use in source · Read passage · 01_manuscript.tex, line 887
Use the telescoping identity widehat A_m⋯widehat A_1 -A_m⋯ A_1 = ∑_j=1^m widehat A_m⋯widehat A_j+1 E_j A_j-1⋯ A_1. Take norms term by term.
Exact mathematical passage
\widehat A_m\cdots\widehat A_1 -A_m\cdots A_1 = \sum_{j=1}^m \widehat A_m\cdots\widehat A_{j+1} E_j A_{j-1}\cdots A_1.
A_{j\ell}Use in source · 1 passage
Search aliases: A_jℓ ell
Use in source · Read passage · 01_manuscript.tex, line 606
The (i,j) entry of BXA^T is ∑_k=1^n∑_ℓ=1^m B_ikX_kℓA_jℓ. This is the coordinate formula for (A⊗ B)vec(X). Multiplying first by B and then by A^T gives the stated count.
Exact mathematical passage
\sum_{k=1}^{n}\sum_{\ell=1}^{m} B_{ik}X_{k\ell}A_{j\ell}.
A_{k+1}Use in source · 1 passage
Search aliases: A_k+1
Use in source · Read passage · 02_supplement_04.tex, line 148
Exact arithmetic count. Each pair of slices forms one sum and one difference at every trailing coordinate. The event type therefore gives A_0=0, A_k+1=2A_k+2^k+1. The closed form and its logarithmic re-expression are both checked: A_k=k2^k=Nlog_2N. The logarithm in the formal count is the natural-number binary logarithm, with its power-of-two identity proved by Mathlib. The case N=1 has zero arithmetic events.
Exact mathematical passage
A_0=0,\qquad A_{k+1}=2A_k+2^{k+1}.
a_{k+1}Use in source · 1 passage
Search aliases: a_k+1
Use in source · Read passage · 02_supplement_06.tex, line 141
For any nonnegative decreasing sequence a_i, weights with these properties satisfy ∑_i a_iw_i≥∑_i>ka_i. This finite weighted inequality is itself proved. For k<d, use μ=a_k+1 and compare the summands to tail weights v_i=0 for i≤ k and v_i=1 otherwise: a_iv_i+μ(w_i-v_i)≤ a_iw_i. For a retained coordinate the inequality uses a_i≥μ and w_i≥0. For a tail coordinate it uses a_i≤μ and w_i≤1. Summation and ∑ w_i≥∑ v_i give the result.
Exact mathematical passage
\mu=a_{k+1}
A_{k-1}Use in source · 1 passage
Search aliases: A_k-1
Use in source · Read passage · 02_supplement_04.tex, line 88
Arbitrary tensor factors. textitSource item 4.6; module StructuredTensor Let A_0,…,A_k-1 be square factors of order d. Tensor coordinates are represented by a nested product of k copies of {0,…,d-1}. The empty product is a singleton. Lean proves that the coordinate type has cardinality d^k.
Exact mathematical passage
A_0,\ldots,A_{k-1}
A_mUse in source · 7 passages
Search aliases: A_m
Use in source · Read passage · 01_manuscript.tex, line 722
[Rank growth in a product chain] Let A_j=D_j+U_jV_j^*, j=1,…,m, share one diagonal packet layout. Before recompression, the product A_m⋯ A_1 has a BDLR representation whose correction rank is at most ∑_j=1^m r_j.
Exact mathematical passage
A_m\cdots A_1
A_{m-1}Use in source · 1 passage
Search aliases: A_m-1
Use in source · Read passage · 01_manuscript.tex, line 937
Set D_j=widehat P_j-P_j. Then D_j=A_jD_j-1+E_j, D_0=0. Repeated substitution gives D_m = ∑_j=1^m A_mA_m-1⋯ A_j+1E_j. Taking norms yields the result.
Exact mathematical passage
D_m = \sum_{j=1}^{m} A_mA_{m-1}\cdots A_{j+1}E_j.
A_{RH}Use in source · 7 passages
Search aliases: A_RH
Use in source · pages.tex, line 266
1 · 1 · 2. The third coordinate is hidden. Its invertible block is [2], so eliminating it produces S=A_RR-A_RHA_HH^-1A_HR =mat3/2 · -1/2 -1/2 · 3/2. vfigschur_fill headThe hidden route has a sign The retained coordinates had no direct coupling, yet both touch the hidden coordinate. The resulting correction couples them by -1/2. This is fill: a new active retained block produced by an eliminated route.
Exact mathematical passage
S=A_{RR}-A_{RH}A_{HH}^{-1}A_{HR} =\mat{3/2&-1/2\\-1/2&3/2}.
A_{RR}Use in source · 6 passages
Search aliases: A_RR
Use in source · pages.tex, line 266
1 · 1 · 2. The third coordinate is hidden. Its invertible block is [2], so eliminating it produces S=A_RR-A_RHA_HH^-1A_HR =mat3/2 · -1/2 -1/2 · 3/2. vfigschur_fill headThe hidden route has a sign The retained coordinates had no direct coupling, yet both touch the hidden coordinate. The resulting correction couples them by -1/2. This is fill: a new active retained block produced by an eliminated route.
Exact mathematical passage
S=A_{RR}-A_{RH}A_{HH}^{-1}A_{HR} =\mat{3/2&-1/2\\-1/2&3/2}.
A_\alphaUse in source · 2 passages
Search aliases: A_α alpha
Use in source · Read passage · 01_manuscript.tex, line 1122
[Invariant packet Krylov decomposition] Assume [A,P_α]=0 for every packet. If v=∑_α v_α, v_α=P_α v, then A^jv = ∑_α A_α^jv_α for every j≥0, where A_α=P_α AP_α. Hence K_m(A,v) ⊆ ⊕_α K_m(A_α,v_α).
Exact mathematical passage
A_\alpha=P_\alpha AP_\alpha.
A_\alpha^jUse in source · 1 passage
Search aliases: A_α^j alpha
Use in source · Read passage · 01_manuscript.tex, line 1116
[Invariant packet Krylov decomposition] Assume [A,P_α]=0 for every packet. If v=∑_α v_α, v_α=P_α v, then A^jv = ∑_α A_α^jv_α for every j≥0, where A_α=P_α AP_α. Hence K_m(A,v) ⊆ ⊕_α K_m(A_α,v_α).
Exact mathematical passage
A^jv = \sum_\alpha A_\alpha^jv_\alpha
A_{\alpha_1\alpha_0}Use in source · 1 passage
Search aliases: A_α_1α_0 alpha alpha
Use in source · Read passage · 01_manuscript.tex, line 964
[Path support of powers] For every positive integer m, P_γ A^mP_α is the sum over directed walks (vertices may repeat) α=α_0 ⟶ α_1 ⟶⋯⟶ α_m=γ in G_A, with contribution A_α_mα_m-1 ⋯ A_α_1α_0. If no such path exists, the block vanishes.
Exact mathematical passage
A_{\alpha_m\alpha_{m-1}} \cdots A_{\alpha_1\alpha_0}.
A_{\alpha_m\alpha_{m-1}}Use in source · 1 passage
Search aliases: A_α_mα_m-1 alpha alpha
Use in source · Read passage · 01_manuscript.tex, line 964
[Path support of powers] For every positive integer m, P_γ A^mP_α is the sum over directed walks (vertices may repeat) α=α_0 ⟶ α_1 ⟶⋯⟶ α_m=γ in G_A, with contribution A_α_mα_m-1 ⋯ A_α_1α_0. If no such path exists, the block vanishes.
Exact mathematical passage
A_{\alpha_m\alpha_{m-1}} \cdots A_{\alpha_1\alpha_0}.
A_{\beta\alpha}Use in source · 7 passages
Search aliases: A_βα beta alpha
Use in source · pages.tex, line 20
A_βα · A block map · What are its source and target dimensions?
Exact mathematical passage
A_{\beta\alpha}
A_{\beta\alpha}^*Use in source · 1 passage
Search aliases: A_βα^* beta alpha
Use in source · Read passage · 01_manuscript.tex, line 193
[Adjoint and tensor semantics] The adjoint representation has blocks (A^*)_αβ=A_βα^*. If A and B are packet resolved on X and Y, then (A⊗ B)_(β,δ),(α,γ) = A_βα⊗ B_δγ.
Exact mathematical passage
(A^*)_{\alpha\beta}=A_{\beta\alpha}^*.
\mathcal BUse in source · 3 passages
Search aliases: B mathcal
Use in source · Read passage · 02_supplement_05.tex, line 117
Exact banded addresses and modeled work. textitSource item 4.1; module BandedCount For p packets and half-bandwidth w, the actual address set is B_p,w ={(i,j)∈{0,…,p-1}^2: i≤ j+w and j≤ i+w}. The predicate is the integer version of |i-j|≤ w. The count is first defined as a double sum of indicators, then identified with the cardinality of this subtype. It is not defined by the target closed form.
Exact mathematical passage
\mathcal B_{p,w} =\{(i,j)\in\{0,\ldots,p-1\}^2: i\leq j+w\ \text{and}\ j\leq i+w\}.
bUse in source · 148 passages
Search aliases: b
Use in source · pages.tex, line 27
F_v(b) · A budget table · Does b mean exact weight, or an upper bound?
Exact mathematical passage
F_v(b)
B^{-1}Use in source · 1 passage
Search aliases: B^-1
Use in source · Read passage · 02_supplement_05.tex, line 208
A direct inverse bound. If C=(I+E)^-1 and norm E_2≤η<1, then C=I-EC. Consequently (1-η)norm C_2≤ 1, normI+E_2≤ 1+η. The constructed normalization is invertible because S and A are. Combining the estimates gives the actual condition number bound κ_2(B):=norm B_2normB^-1_2 ≤frac1+η1-η. This norm argument does not require a separately assumed Hermitian perturbation. Nonnegative η follows from the norm bound. Nonemptiness is explicit because the condition-number and norm-of-identity convention uses …
Exact mathematical passage
\boxed{\kappa_2(B):=\norm B_2\norm{B^{-1}}_2 \leq\frac{1+\eta}{1-\eta}.}
b^2Use in source · 4 passages
Search aliases: b^2
Use in source · Read passage · 01_manuscript.tex, line 298
There are n/b active diagonal blocks. Each contains b^2 entries.
Exact mathematical passage
b^2
B_{\mathrm F}^{2}Use in source · 1 passage
Search aliases: B_ F^2 mathrm
Use in source · Read passage · 02_supplement_06.tex, line 66
For the Frobenius norm, trace cyclicity proves left and right isometric invariance. The proof also establishes the projection identity normB-QQ^*B_ F^2 =norm B_ F^2-normQ^*B_ F^2. Nonnegativity of the left side gives left compression contractivity; the right version follows by adjoints. No norm-preservation or contractivity hypothesis is inserted.
Exact mathematical passage
\norm{B-QQ^*B}_{\mathrm F}^{2} =\norm B_{\mathrm F}^{2}-\norm{Q^*B}_{\mathrm F}^{2}.
B_2Use in source · 1 passage
Search aliases: B_2
Use in source · Read passage · 02_supplement_05.tex, line 208
A direct inverse bound. If C=(I+E)^-1 and norm E_2≤η<1, then C=I-EC. Consequently (1-η)norm C_2≤ 1, normI+E_2≤ 1+η. The constructed normalization is invertible because S and A are. Combining the estimates gives the actual condition number bound κ_2(B):=norm B_2normB^-1_2 ≤frac1+η1-η. This norm argument does not require a separately assumed Hermitian perturbation. Nonnegative η follows from the norm bound. Nonemptiness is explicit because the condition-number and norm-of-identity convention uses …
Exact mathematical passage
\boxed{\kappa_2(B):=\norm B_2\norm{B^{-1}}_2 \leq\frac{1+\eta}{1-\eta}.}
B_{ca}Use in source · 2 passages
Search aliases: B_ca
Use in source · Read passage · 02_supplement_01.tex, line 23
Composition. Inserting the projector identity between factors yields (AB)_ba=∑_c A_bcB_ca. If the left side is nonzero, at least one intermediate c has both A_bc≠0 and B_ca≠0. Consequently supp(AB)⊆supp(A)∘supp(B). No equality of supports is asserted.
Exact mathematical passage
\boxed{(AB)_{ba}=\sum_c A_{bc}B_{ca}.}
B_{dc}Use in source · 1 passage
Search aliases: B_dc
Use in source · Read passage · 02_supplement_01.tex, line 33
Adjoint and tensor products. With an involution and self-adjoint projectors, the formalization proves (A_ba)^*=(A^*)_ab. For actual finite matrices and actual Kronecker products, it proves (P_b⊗ Q_d)(A⊗ B)(P_a⊗ Q_c) =A_ba⊗ B_dc. The self-adjointness hypothesis is explicit in the adjoint theorem. The tensor theorem is not a statement about an uninterpreted tensor symbol.
Exact mathematical passage
(P_b\otimes Q_d)(A\otimes B)(P_a\otimes Q_c) =A_{ba}\otimes B_{dc}.
b_iUse in source · 1 passage
Search aliases: b_i
Use in source · Read passage · 02_supplement_03.tex, line 168
Keeping node v costs κ_v and incurs integer error w_v. Splitting costs overhead σ_v in addition to the selected child cuts. The exact-weight recurrence is F_v(b)=min{ l κ_v if b=w_v, [2pt] displaystyleσ_v+ min_∑_i b_i=b∑_iF_v_i(b_i) if v is internal. . Nonexistent alternatives are infeasible. The child sum is implemented by the proved finite min-plus convolution, not by assuming an optimal allocation.
Exact mathematical passage
F_v(b)=\min\left\{ \begin{array}{l} \kappa_v\quad\text{if }b=w_v,\\[2pt] \displaystyle\sigma_v+ \min_{\sum_i b_i=b}\sum_iF_{v_i}(b_i)\quad\text{if }v\text{ is internal}. \end{array}\right.
B_{ik}Use in source · 1 passage
Search aliases: B_ik
Use in source · Read passage · 01_manuscript.tex, line 606
The (i,j) entry of BXA^T is ∑_k=1^n∑_ℓ=1^m B_ikX_kℓA_jℓ. This is the coordinate formula for (A⊗ B)vec(X). Multiplying first by B and then by A^T gives the stated count.
Exact mathematical passage
\sum_{k=1}^{n}\sum_{\ell=1}^{m} B_{ik}X_{k\ell}A_{j\ell}.
\mathcal B_{p,w}Use in source · 3 passages
Search aliases: B_p,w mathcal
Use in source · Read passage · 02_supplement_05.tex, line 117
Exact banded addresses and modeled work. textitSource item 4.1; module BandedCount For p packets and half-bandwidth w, the actual address set is B_p,w ={(i,j)∈{0,…,p-1}^2: i≤ j+w and j≤ i+w}. The predicate is the integer version of |i-j|≤ w. The count is first defined as a double sum of indicators, then identified with the cardinality of this subtype. It is not defined by the target closed form.
Exact mathematical passage
\mathcal B_{p,w} =\{(i,j)\in\{0,\ldots,p-1\}^2: i\leq j+w\ \text{and}\ j\leq i+w\}.
B_{\rm stream}Use in source · 3 passages
Search aliases: B_rm stream rm
Use in source · pages.tex, line 412
blocks accumulate into one target, the reduction or accumulation policy is additional work. Equal matrix dimensions alone do not remove that dependency. Different accumulation types also define different shape classes under the paper's convention. headCount the data separately For three diagonal 2×2 blocks, eight-byte scalars, and q right-hand sides, the stated one-pass model moves B_rm stream=8·3(4+2q+4q)=96+144q bytes. It gives 240 bytes at q=1 and 672 bytes at q=4. It deliberately assumes no …
Exact mathematical passage
B_{\rm stream}=8\cdot3(4+2q+4q)=96+144q\text{ bytes}.
B_{\mathrm{stream}}Use in source · 1 passage
Search aliases: B_stream mathrm
Use in source · Read passage · 02_supplement_02.tex, line 184
Streaming. The explicit event type counts one block read, one source read, and a target read plus write for every edge. With scalar size s, its byte count is B_stream=s∑_e(m_en_e+n_eq+2m_eq). No cross-edge cache reuse is assumed. Sparse index metadata and temporary workspace remain separate.
Exact mathematical passage
B_{\mathrm{stream}}=s\sum_e(m_en_e+n_eq+2m_eq).
b_vUse in source · 2 passages
Search aliases: b_v
Use in source · Read passage · 02_supplement_05.tex, line 53
Constructed solution. Well-founded recursion defines x_v=D_v^-1 (b_v-∑_u→ vT_vux_u). Applying D_v proves the original equation at v. If another vector satisfies all equations, induction over predecessors makes each previously used coordinate identical, and invertibility of D_v forces agreement at v. Thus the construction gives existence and uniqueness, and the block operator is bijective.
Exact mathematical passage
\boxed{x_v=D_v^{-1}\!\left(b_v-\sum_{u\to v}T_{vu}x_u\right).}
b_\betaUse in source · 1 passage
Search aliases: b_β beta
Use in source · Read passage · 01_manuscript.tex, line 1026
The equation in packet β is D_β x_β = b_β - ∑_α→βK_βαx_α. Every predecessor of β lies in an earlier level. Once those values are available, x_β is obtained from the local solve with D_β. Vertices in the same level have no directed edge between them, so their solves are independent.
Exact mathematical passage
D_\beta x_\beta = b_\beta - \sum_{\alpha\to\beta}K_{\beta\alpha}x_\alpha.
B_{\delta\gamma}Use in source · 2 passages
Search aliases: B_δγ delta gamma
Use in source · Read passage · 01_manuscript.tex, line 197
[Adjoint and tensor semantics] The adjoint representation has blocks (A^*)_αβ=A_βα^*. If A and B are packet resolved on X and Y, then (A⊗ B)_(β,δ),(α,γ) = A_βα⊗ B_δγ.
Exact mathematical passage
(A\otimes B)_{(\beta,\delta),(\alpha,\gamma)} = A_{\beta\alpha}\otimes B_{\delta\gamma}.
B_\nuUse in source · 1 passage
Search aliases: B_ν nu
Use in source · Read passage · 02_supplement_06.tex, line 54
ubsection:002textitModule SmallCoreNorms; arbitrary ambient competitors For rectangular isometries Q,V define L(G)=QGV^*, C(B)=Q^*BV, Q^*Q=I, V^*V=I. The following statements are proved for the actual matrix rank and actual matrix norms: C( L(G))=G, rank L(G)=rank G, rank C(B)≤rank B, norm L(G)_ν=normG_ν, norm C(B)_ν≤norm B_ν, ν∈{2, F}. For the spectral norm, the proof uses the Gram matrix identity (QG)^*(QG)=G^*G and the C^*-norm square identity, together with right-side adjoint symmetry. Compr…
Exact mathematical passage
\boxed{\norm{\mathcal L(G)}_\nu=\norm{G}_\nu, \qquad \norm{\mathcal C(B)}_\nu\leq\norm B_\nu, \quad \nu\in\{2,\mathrm F\}.}
\bar\kappaUse in source · 2 passages
Search aliases: barκ bar kappa
Use in source · Read passage · 02_supplement_05.tex, line 222
tκ-1sqrtκ+1. Monotonicity of square root and positive denominators prove that q is monotone. Lean also proves κ_2(B)≥1, q(κ)≥0 for κ≥1, and q(κ)<1. Thus q(κ_2(B))≤ q (frac1+η1-η). For 1≤κ and e_0≥0, an explicit classical premise e_m≤2q(κ)^m e_0 and a certified upper bound κ≤barκ yield e_m≤2q(barκ)^m e_0.
Exact mathematical passage
\kappa\leq\bar\kappa
\bmodUse in source · 2 passages
Search aliases: bmod
Use in source · pages.tex, line 73
vpage5Four coefficients, four exact frequenciesCIRCULANT MULTIPLICATION Take c=(2,1,0,-1)^T and define C(c)_jk=c_j-kbmod4. The unitary Fourier matrix is F_jk= i^-jk/2. Its normalization is visible in the exact diagonalization: C(c)=F^*diag(2Fc)F, 2Fc=(2,2-2 i,2,2+2 i)^T. vfigcirculant headFollow an actual input For x=(1,2,-1,0)^T, dense multiplication gives (0,6,0,-2)^T. Transforming x, multiplying by the four displayed eigenvalues, and applying F^* gives exactly the same vector. Every entry bel…
Exact mathematical passage
C(c)_{jk}=c_{j-k\bmod4}
CUse in source · 49 passages
Search aliases: C
Use in source · pages.tex, line 73
vpage5Four coefficients, four exact frequenciesCIRCULANT MULTIPLICATION Take c=(2,1,0,-1)^T and define C(c)_jk=c_j-kbmod4. The unitary Fourier matrix is F_jk= i^-jk/2. Its normalization is visible in the exact diagonalization: C(c)=F^*diag(2Fc)F, 2Fc=(2,2-2 i,2,2+2 i)^T. vfigcirculant headFollow an actual input For x=(1,2,-1,0)^T, dense multiplication gives (0,6,0,-2)^T. Transforming x, multiplying by the four displayed eigenvalues, and applying F^* gives exactly the same vector. Every entry bel…
Exact mathematical passage
C(c)_{jk}=c_{j-k\bmod4}
\CUse in source · 7 passages
Search aliases: C mathbb
Use in source · Read passage · 02_supplement_04.tex, line 73
Kronecker application. For A∈ C^m× m, B∈ C^n× n and X∈ C^n× m, Lean proves (A⊗ B)vec(X)=vec(BXA^T). The formal vectorization is explicitly column-major: its pair coordinate (j,i) stores X_ij. The right factor is the ordinary transpose, not the conjugate transpose. This distinction is essential over C.
Exact mathematical passage
A\in\C^{m\times m}
\mathbb CUse in source · 1 passage
Search aliases: C mathbb
Use in source · Read passage · 02_supplement_03.tex, line 6
Low-rank factors through a product chain. textitSource item 5.2 Let A_j=D_j+U_jV_j^*, U_j,V_j∈ C^n× r_j, with all D_j diagonal in one packet layout. Set P_m=A_m⋯ A_1, D^(m)=D_m⋯ D_1. The formalization constructs thin matrices L_m,R_m such that P_m=D^(m)+L_mR_m^*. Their column index type is a recursively constructed disjoint sum. Its cardinality is proved to equal ∑_j=1^m r_j.
Exact mathematical passage
A_j=D_j+U_jV_j^*,\qquad U_j,V_j\in\mathbb C^{n\times r_j},
\mathcal CUse in source · 4 passages
Search aliases: C mathcal
Use in source · Read passage · 02_supplement_06.tex, line 44
Why a small-core optimum is a global optimum. textitModule SmallCoreNorms; arbitrary ambient competitors For rectangular isometries Q,V define L(G)=QGV^*, C(B)=Q^*BV, Q^*Q=I, V^*V=I. The following statements are proved for the actual matrix rank and actual matrix norms: C( L(G))=G, rank L(G)=rank G, rank C(B)≤rank B, norm L(G)_ν=normG_ν, norm C(B)_ν≤norm B_ν, ν∈{2, F}. For the spectral norm, the proof uses the Gram matrix identity (QG)^*(QG)=G^*G and the C^*-norm square identity, together with r…
Exact mathematical passage
\mathcal L(G)=QGV^*,\qquad \mathcal C(B)=Q^*BV, \qquad Q^*Q=I,\quad V^*V=I.
C^*Use in source · 1 passage
Search aliases: C^*
Use in source · Read passage · 02_supplement_06.tex, line 63
ents are proved for the actual matrix rank and actual matrix norms: C( L(G))=G, rank L(G)=rank G, rank C(B)≤rank B, norm L(G)_ν=normG_ν, norm C(B)_ν≤norm B_ν, ν∈{2, F}. For the spectral norm, the proof uses the Gram matrix identity (QG)^*(QG)=G^*G and the C^*-norm square identity, together with right-side adjoint symmetry. Compression follows from submultiplicativity and the isometry norm bound, including empty spaces.
Exact mathematical passage
C^*
\C^{m\times m}Use in source · 1 passage
Search aliases: C^m× m C times mathbb times
Use in source · Read passage · 02_supplement_04.tex, line 73
Kronecker application. For A∈ C^m× m, B∈ C^n× n and X∈ C^n× m, Lean proves (A⊗ B)vec(X)=vec(BXA^T). The formal vectorization is explicitly column-major: its pair coordinate (j,i) stores X_ij. The right factor is the ordinary transpose, not the conjugate transpose. This distinction is essential over C.
Exact mathematical passage
A\in\C^{m\times m}
\C^{m\times t}Use in source · 1 passage
Search aliases: C^m× t C times mathbb times
Use in source · Read passage · 02_supplement_06.tex, line 6
Compact factors from actual column spaces. textitModules CompactRangeFactors and SmallCoreTransfer Let L∈ C^m× t, R∈ C^n× t, A=LR^*. Use the ordinary Euclidean inner products on the coordinate spaces. Define S_L=range(L), r_L=dim S_L, S_R=range(R), r_R=dim S_R. The implementation constructs an orthonormal basis of each actual range subspace. Let Q_L and Q_R be the matrices of the corresponding isometric inclusions. Express the range-restricted input maps in these bases to obtain R_L,R_R.
Exact mathematical passage
L\in\C^{m\times t},\qquad R\in\C^{n\times t},\qquad A=LR^*.
\C^{n\times m}Use in source · 1 passage
Search aliases: C^n× m C times mathbb times
Use in source · Read passage · 02_supplement_04.tex, line 73
Kronecker application. For A∈ C^m× m, B∈ C^n× n and X∈ C^n× m, Lean proves (A⊗ B)vec(X)=vec(BXA^T). The formal vectorization is explicitly column-major: its pair coordinate (j,i) stores X_ij. The right factor is the ordinary transpose, not the conjugate transpose. This distinction is essential over C.
Exact mathematical passage
X\in\C^{n\times m}
\C^{n\times n}Use in source · 1 passage
Search aliases: C^n× n C times mathbb times
Use in source · Read passage · 02_supplement_04.tex, line 73
Kronecker application. For A∈ C^m× m, B∈ C^n× n and X∈ C^n× m, Lean proves (A⊗ B)vec(X)=vec(BXA^T). The formal vectorization is explicitly column-major: its pair coordinate (j,i) stores X_ij. The right factor is the ordinary transpose, not the conjugate transpose. This distinction is essential over C.
Exact mathematical passage
B\in\C^{n\times n}
\mathbb C^{n\times r_j}Use in source · 1 passage
Search aliases: C^n× r_j mathbb times
Use in source · Read passage · 02_supplement_03.tex, line 6
Low-rank factors through a product chain. textitSource item 5.2 Let A_j=D_j+U_jV_j^*, U_j,V_j∈ C^n× r_j, with all D_j diagonal in one packet layout. Set P_m=A_m⋯ A_1, D^(m)=D_m⋯ D_1. The formalization constructs thin matrices L_m,R_m such that P_m=D^(m)+L_mR_m^*. Their column index type is a recursively constructed disjoint sum. Its cardinality is proved to equal ∑_j=1^m r_j.
Exact mathematical passage
A_j=D_j+U_jV_j^*,\qquad U_j,V_j\in\mathbb C^{n\times r_j},
\C^{n\times t}Use in source · 1 passage
Search aliases: C^n× t C times mathbb times
Use in source · Read passage · 02_supplement_06.tex, line 6
Compact factors from actual column spaces. textitModules CompactRangeFactors and SmallCoreTransfer Let L∈ C^m× t, R∈ C^n× t, A=LR^*. Use the ordinary Euclidean inner products on the coordinate spaces. Define S_L=range(L), r_L=dim S_L, S_R=range(R), r_R=dim S_R. The implementation constructs an orthonormal basis of each actual range subspace. Let Q_L and Q_R be the matrices of the corresponding isometric inclusions. Express the range-restricted input maps in these bases to obtain R_L,R_R.
Exact mathematical passage
L\in\C^{m\times t},\qquad R\in\C^{n\times t},\qquad A=LR^*.
\C^{r_L\times r_R}Use in source · 1 passage
Search aliases: C^r_L× r_R C times mathbb times
Use in source · Read passage · 02_supplement_06.tex, line 27
Define the small core by actual multiplication, G=R_LR_R^*∈ C^r_L× r_R. Finite matrix associativity and adjoint identities give A=Q_LGQ_R^*, rank G=rank A. The rank identity follows from both multiplication inequalities and the left inverse obtained by compression. It is not inferred from agreeing dimensions.
Exact mathematical passage
G=R_LR_R^*\in\C^{r_L\times r_R}.
c_0Use in source · 5 passages
Search aliases: c_0
Use in source · Read passage · 01_manuscript.tex, line 565
[Toeplitz circulant embedding] Let n≥1 be an integer and T∈ C^n× n be Toeplitz, with first column (c_0,c_1,…,c_n-1)^T and first row (c_0,r_1,…,r_n-1). Set g = (c_0,c_1,…,c_n-1,0,r_n-1,…,r_1)^T ∈ C^2n and append n zeros to x∈ C^n. The first n entries of the circulant product C(g)(x,0)^T equal Tx.
Exact mathematical passage
(c_0,c_1,\ldots,c_{n-1})^T
c_1Use in source · 2 passages
Search aliases: c_1
Use in source · Read passage · 01_manuscript.tex, line 565
[Toeplitz circulant embedding] Let n≥1 be an integer and T∈ C^n× n be Toeplitz, with first column (c_0,c_1,…,c_n-1)^T and first row (c_0,r_1,…,r_n-1). Set g = (c_0,c_1,…,c_n-1,0,r_n-1,…,r_1)^T ∈ C^2n and append n zeros to x∈ C^n. The first n entries of the circulant product C(g)(x,0)^T equal Tx.
Exact mathematical passage
(c_0,c_1,\ldots,c_{n-1})^T
C_2Use in source · 1 passage
Search aliases: C_2
Use in source · Read passage · 02_supplement_05.tex, line 203
A direct inverse bound. If C=(I+E)^-1 and norm E_2≤η<1, then C=I-EC. Consequently (1-η)norm C_2≤ 1, normI+E_2≤ 1+η. The constructed normalization is invertible because S and A are. Combining the estimates gives the actual condition number bound κ_2(B):=norm B_2normB^-1_2 ≤frac1+η1-η. This norm argument does not require a separately assumed Hermitian perturbation. Nonnegative η follows from the norm bound. Nonemptiness is explicit because the condition-number and norm-of-identity convention uses …
Exact mathematical passage
(1-\eta)\norm C_2\leq 1, \qquad \norm{I+E}_2\leq 1+\eta.
c_eUse in source · 34 passages
Search aliases: c_e
Use in source · pages.tex, line 342
vpage20Spend error where it is cheapestCONTINUOUS AND CLIPPED PRECISION Take sensitivities s=(1,2,4), cost coefficients c=(1,1,2), and budget τ=1. Minimizing ∑_ec_elog(1/ϵ_e) subject to ∑_es_eϵ_e=1 gives ϵ_e=fracc_e4s_e, ϵ=(1/4,1/8,1/8). vfigprecision headThe budget is weighted The three contributions to the error allowance are (1/4,1/4,1/2). Equal cost coefficients give equal weighted contributions, not equal tolerances. The third edge has twice the cost coefficient of the second, which offsets…
Exact mathematical passage
\sum_ec_e\log(1/\epsilon_e)
c_fUse in source · 2 passages
Search aliases: c_f
Use in source · Read passage · 01_manuscript.tex, line 1346
[Continuous precision allocation] Let E be a finite nonempty edge set, let s_e>0 be sensitivities and c_e>0 be costs, and let τ>0. Consider min_ε_e>0 ∑_e c_elogfrac1ε_e subject to ∑_es_eε_e=τ. The unique minimizer is ε_e = fracτ c_es_e∑_fc_f.
Exact mathematical passage
\boxed{ \eps_e = \frac{\tau c_e}{s_e\sum_fc_f}. }
c_{i-j}Use in source · 1 passage
Search aliases: c_i-j
Use in source · Read passage · 02_supplement_04.tex, line 48
Toeplitz padding and column-major vectorization. textitSource items 4.4 and 4.5 Toeplitz embedding. For n>0, define T_ij= c_i-j, · j≤ i, r_j-i, · i<j. The diagonal is supplied by c_0; the coordinate r_0 is unused. The constructed generator of length 2n is g=(c_0,…,c_n-1,0,r_n-1,…,r_1), and the padding map is the actual finite-index sum eliminator that appends n zeros.
Exact mathematical passage
T_{ij}= \begin{cases} c_{i-j},&j\le i,\\ r_{j-i},&i<j. \end{cases}
c_jUse in source · 2 passages
Search aliases: c_j
Use in source · Read passage · 01_manuscript.tex, line 978
[Polynomial support] If p(z)=∑_j=0^dc_jz^j, then supp(p(A)) is contained in the union of the support relations of A^j for 0≤ j≤ d.
Exact mathematical passage
p(z)=\sum_{j=0}^dc_jz^j,
c_{j,k}Use in source · 1 passage
Search aliases: c_j,k
Use in source · Read passage · 01_manuscript.tex, line 1438
ct discrete precision program] Let the total integer error budget be B. Define F_j(b) as the minimum cost for assigning precisions to the first j edges with total weight exactly b. Then F_0(0)=0, F_0(b)=+∞ for b≠0. Set F_j-1(u)=+∞ whenever u<0 or the state is unreachable. Then F_j(b) = min_k { F_j-1(b-w_j,k)+c_j,k }. The minimum of F_m(b) over 0≤ b≤ B is the exact minimum cost for the quantized model.
Exact mathematical passage
\boxed{ F_j(b) = \min_k \left\{ F_{j-1}(b-w_{j,k})+c_{j,k} \right\}. }
c_{j-k}Use in source · 5 passages
Search aliases: c_j-k
Use in source · pages.tex, line 94
4 · 3 · 2, x=(1,2,-1)^T. The diagonal pattern is Toeplitz, not circulant. To use a circular convolution, embed it into dimension six with first column g=(2,3,4,0,5,-1)^T and pad the input with three zeros. vfigtoeplitz headRead the wrapped indices For output index j<3, an input index k≤ j uses the first-column coefficient c_j-k. An input index k>j wraps to the stored first-row coefficient r_k-j. The padded input positions contribute nothing. Keeping the first three outputs therefore gives Tx=(-5…
Exact mathematical passage
c_{j-k}
c_{j-k\bmod4}Use in source · 2 passages
Search aliases: c_j-kbmod4 bmod
Use in source · pages.tex, line 73
vpage5Four coefficients, four exact frequenciesCIRCULANT MULTIPLICATION Take c=(2,1,0,-1)^T and define C(c)_jk=c_j-kbmod4. The unitary Fourier matrix is F_jk= i^-jk/2. Its normalization is visible in the exact diagonalization: C(c)=F^*diag(2Fc)F, 2Fc=(2,2-2 i,2,2+2 i)^T. vfigcirculant headFollow an actual input For x=(1,2,-1,0)^T, dense multiplication gives (0,6,0,-2)^T. Transforming x, multiplying by the four displayed eigenvalues, and applying F^* gives exactly the same vector. Every entry bel…
Exact mathematical passage
C(c)_{jk}=c_{j-k\bmod4}
c_{n-1}Use in source · 3 passages
Search aliases: c_n-1
Use in source · Read passage · 01_manuscript.tex, line 565
[Toeplitz circulant embedding] Let n≥1 be an integer and T∈ C^n× n be Toeplitz, with first column (c_0,c_1,…,c_n-1)^T and first row (c_0,r_1,…,r_n-1). Set g = (c_0,c_1,…,c_n-1,0,r_n-1,…,r_1)^T ∈ C^2n and append n zeros to x∈ C^n. The first n entries of the circulant product C(g)(x,0)^T equal Tx.
Exact mathematical passage
(c_0,c_1,\ldots,c_{n-1})^T
c_rUse in source · 1 passage
Search aliases: c_r
Use in source · Read passage · 01_manuscript.tex, line 548
For each frequency k, let v^(k)_j = frac1sqrt n exp(frac2π i jkn). Then (C(c)v^(k))_j = ∑_r=0^n-1c_rv^(k)_j-r = ( ∑_r=0^n-1 c_rexp(-frac2π i rkn) )v^(k)_j. The eigenvalue is sqrt n(Fc)_k. The vectors v^(k) are the columns of F^*, which gives the formula.
Exact mathematical passage
(C(c)v^{(k)})_j = \sum_{r=0}^{n-1}c_rv^{(k)}_{j-r} = \left( \sum_{r=0}^{n-1} c_r\exp\left(-\frac{2\pi\ii rk}{n}\right) \right)v^{(k)}_j.
\CC^{2n}Use in source · 1 passage
Search aliases: CC^2n C^2n CC mathbb
Use in source · Read passage · 01_manuscript.tex, line 573
[Toeplitz circulant embedding] Let n≥1 be an integer and T∈ C^n× n be Toeplitz, with first column (c_0,c_1,…,c_n-1)^T and first row (c_0,r_1,…,r_n-1). Set g = (c_0,c_1,…,c_n-1,0,r_n-1,…,r_1)^T ∈ C^2n and append n zeros to x∈ C^n. The first n entries of the circulant product C(g)(x,0)^T equal Tx.
Exact mathematical passage
g = (c_0,c_1,\ldots,c_{n-1},0,r_{n-1},\ldots,r_1)^T \in\CC^{2n}
\CC^{m\times m}Use in source · 1 passage
Search aliases: CC^m× m C^m× m CC times mathbb times
Use in source · Read passage · 01_manuscript.tex, line 587
[Kronecker modewise application] Let A∈ C^m× m, B∈ C^n× n, and X∈ C^n× m. With column major vectorization, (A⊗ B)vec(X) = vec(BXA^T). The right side has modeled cost 2n^2m+2nm^2 and stores n^2+m^2 factor entries instead of n^2m^2 entries.
Exact mathematical passage
A\in\CC^{m\times m}
\CC^{m\times r}Use in source · 1 passage
Search aliases: CC^m× r C^m× r CC times mathbb times
Use in source · Read passage · 01_manuscript.tex, line 805
[Small core recompression] Let L∈ C^m× r, R∈ C^n× r. Take compact QR factorizations L=Q_LR_L, R=Q_RR_R, where the columns of Q_L and Q_R are orthonormal. Form the small core G=R_LR_R^* and let G=WΣ Z^* be an ordered thin singular value decomposition, with nonnegative nonincreasing singular values. For an integer k≥0, retain min{k,d} columns, where d is the number of supplied thin singular directions; tail singular values beyond d are zero. Define L_k=Q_LW_kΣ_k^1/2, R_k=Q_RZ_kΣ_k^1/2. Then L_kR_k…
Exact mathematical passage
L\in\CC^{m\times r}, \qquad R\in\CC^{n\times r}.
\CC^nUse in source · 2 passages
Search aliases: CC^n C^n CC mathbb
Use in source · Read passage · 01_manuscript.tex, line 517
[Circulant spectral representation] Let n≥1 be an integer and c∈ C^n, with indices taken modulo n, and define C(c)_jk=c_j-k. Let F be the unitary discrete Fourier matrix F_jk = frac1sqrt n exp(-frac2π i jkn). Then C(c) = F^*diag(sqrt n Fc)F. The operator is determined by n scalars and can be applied through two Fourier transforms and one pointwise product.
Exact mathematical passage
c\in\CC^n
\CC^{n\times m}Use in source · 1 passage
Search aliases: CC^n× m C^n× m CC times mathbb times
Use in source · Read passage · 01_manuscript.tex, line 587
[Kronecker modewise application] Let A∈ C^m× m, B∈ C^n× n, and X∈ C^n× m. With column major vectorization, (A⊗ B)vec(X) = vec(BXA^T). The right side has modeled cost 2n^2m+2nm^2 and stores n^2+m^2 factor entries instead of n^2m^2 entries.
Exact mathematical passage
X\in\CC^{n\times m}
\CC^{n\times n}Use in source · 2 passages
Search aliases: CC^n× n C^n× n CC times mathbb times
Use in source · Read passage · 01_manuscript.tex, line 564
[Toeplitz circulant embedding] Let n≥1 be an integer and T∈ C^n× n be Toeplitz, with first column (c_0,c_1,…,c_n-1)^T and first row (c_0,r_1,…,r_n-1). Set g = (c_0,c_1,…,c_n-1,0,r_n-1,…,r_1)^T ∈ C^2n and append n zeros to x∈ C^n. The first n entries of the circulant product C(g)(x,0)^T equal Tx.
Exact mathematical passage
T\in\CC^{n\times n}
\mathrm{child}Use in source · 1 passage
Search aliases: child mathrm
Use in source · Read passage · 02_supplement_02.tex, line 20
Recurrence. For terminal nodes, F_λ(v)=κ_v+λϵ_v. For internal nodes, F_λ(v)=min{κ_v+λϵ_v, σ_v+∑_w∈child(v)F_λ(w)}. The checked result has two parts. First, F_λ(v)≤ J_λ(L) for every legal cut of the subtree. Second, the constructed function codecompile returns a legal cut with objective exactly F_λ(v). Thus the minimum is attained, not merely bounded below.
Exact mathematical passage
\boxed{F_\lambda(v)=\min\left\{\kappa_v+\lambda\epsilon_v, \;\sigma_v+\sum_{w\in\mathrm{child}(v)}F_\lambda(w)\right\}.}
\operatorname{child}Use in source · 1 passage
Search aliases: child operatorname
Use in source · Read passage · 01_manuscript.tex, line 1543
[Exact Lagrangian packet compiler] Fix λ≥0 and define J_λ( L) = cost( L) + λerr( L). For every node v, set K_λ(v) = κ_v+λϵ_v and S_λ(v) = σ_v + ∑_w∈child(v)F_λ(w). For a terminal node, set F_λ(v)=K_λ(v). For an internal node, the recurrence F_λ(v) = min{K_λ(v),S_λ(v)} computes the exact minimum Lagrangian objective over every admissible pruning of the subtree rooted at v.
Exact mathematical passage
S_\lambda(v) = \sigma_v + \sum_{w\in\operatorname{child}(v)}F_\lambda(w).
\codeUse in source · 1 passage
Search aliases: code
Use in source · Read passage · 02_supplement_03.tex, line 136
For a predicate S on costs, the common minimum specification means: ll codenone: · no cost satisfies S, codesome(v): · S(v) and v≤ z whenever S(z). Both attainment and minimality are included. Optional minimum and optional addition are proved to implement union and independent sum of feasible sets.
Exact mathematical passage
\begin{array}{ll} \code{none}: & \text{no cost satisfies }S,\\ \code{some}(v): & S(v)\text{ and }v\leq z\text{ whenever }S(z). \end{array}
\operatorname{compressed}Use in source · 1 passage
Search aliases: compressed operatorname
Use in source · Read passage · 02_supplement_02.tex, line 203
Contracts, grouping, and amortization. textitSource items 2.9, 10.2, 10.4--10.6, and 11.3 Output conversion. A decoded run returns the actual pair (decode(compressed(x)), t_run(x)+t_decode(compressed(x))). Under the declared output equality, Lean proves that this returns the same explicit result and charges the conversion time. It also proves the downstream composition identity when compressed output is consumed directly. Omitting a positive decode cost strictly understates the modeled explicit-…
Exact mathematical passage
\bigl(\operatorname{decode}(\operatorname{compressed}(x)), \;t_{\mathrm{run}}(x)+t_{\mathrm{decode}}(\operatorname{compressed}(x))\bigr).
\condUse in source · 2 passages
Search aliases: cond operatorname
Use in source · Read passage · 01_manuscript.tex, line 1192
[Block Jacobi condition bound] Let A=D+K act on a nonempty finite coordinate space and be Hermitian positive definite with positive definite block diagonal part D. Here D^-1/2 is the actual inverse of the positive square root of D. Suppose E=D^-1/2KD^-1/2 is Hermitian and normE_2≤η<1. Then cond_2(D^-1/2AD^-1/2) ≤ frac1+η1-η.
Exact mathematical passage
\boxed{ \cond_2(D^{-1/2}AD^{-1/2}) \leq \frac{1+\eta}{1-\eta}. }
\cond_2Use in source · 2 passages
Search aliases: cond_2 cond operatorname
Use in source · Read passage · 01_manuscript.tex, line 1192
[Block Jacobi condition bound] Let A=D+K act on a nonempty finite coordinate space and be Hermitian positive definite with positive definite block diagonal part D. Here D^-1/2 is the actual inverse of the positive square root of D. Suppose E=D^-1/2KD^-1/2 is Hermitian and normE_2≤η<1. Then cond_2(D^-1/2AD^-1/2) ≤ frac1+η1-η.
Exact mathematical passage
\boxed{ \cond_2(D^{-1/2}AD^{-1/2}) \leq \frac{1+\eta}{1-\eta}. }
\coprodUse in source · 1 passage
Search aliases: coprod
Use in source · Read passage · 02_supplement_02.tex, line 161
Explicit storage and operation models. textitSource items 3.1, 3.2, 3.4, 3.6, 3.8--3.10, and 4.2 For a finite collection of rectangular blocks, the scalar addresses are actual elements of coprod_e∈ E(Fin(m_e)×Fin(n_e)). The corresponding operation addresses include a right-hand-side index and one of two scalar-operation tags. Lean proves their cardinalities rather than merely defining a formula named ``cost.''
Exact mathematical passage
\coprod_{e\in E}\bigl(\operatorname{Fin}(m_e)\times\operatorname{Fin}(n_e)\bigr).
\coprod_{e\in E}Use in source · 1 passage
Search aliases: coprod_e∈ E coprod in
Use in source · Read passage · 02_supplement_02.tex, line 161
Explicit storage and operation models. textitSource items 3.1, 3.2, 3.4, 3.6, 3.8--3.10, and 4.2 For a finite collection of rectangular blocks, the scalar addresses are actual elements of coprod_e∈ E(Fin(m_e)×Fin(n_e)). The corresponding operation addresses include a right-hand-side index and one of two scalar-operation tags. Lean proves their cardinalities rather than merely defining a formula named ``cost.''
Exact mathematical passage
\coprod_{e\in E}\bigl(\operatorname{Fin}(m_e)\times\operatorname{Fin}(n_e)\bigr).
\costUse in source · 12 passages
Search aliases: cost operatorname
Use in source · pages.tex, line 387
ne. Keeping either a or b costs three with error two. Splitting a costs four with zero error, and splitting b costs five with zero error. vfigtree_frontier The five complete prunings have (cost,error) pairs (1,6), (7,4), (9,2), (8,2), (10,0). At budget two the unique cheapest pruning is {a_0,a_1,b}, costing eight. It is Pareto optimal, but never minimizes cost+λerr for any λ≥0. headSee the incompatible inequalities To beat the root, that pruning would require 8+2λ≤1+6λ, hence λ≥7/4. To beat the …
Exact mathematical passage
\cost+\lambda\err
D^{(m)}Use in source · 4 passages
Search aliases: D^(m)
Use in source · Read passage · 02_supplement_03.tex, line 10
Low-rank factors through a product chain. textitSource item 5.2 Let A_j=D_j+U_jV_j^*, U_j,V_j∈ C^n× r_j, with all D_j diagonal in one packet layout. Set P_m=A_m⋯ A_1, D^(m)=D_m⋯ D_1. The formalization constructs thin matrices L_m,R_m such that P_m=D^(m)+L_mR_m^*. Their column index type is a recursively constructed disjoint sum. Its cardinality is proved to equal ∑_j=1^m r_j.
Exact mathematical passage
P_m=A_m\cdots A_1,\qquad D^{(m)}=D_m\cdots D_1.
D^{(m-1)}Use in source · 2 passages
Search aliases: D^(m-1)
Use in source · Read passage · 02_supplement_03.tex, line 19
Construction. Suppose P_m-1=D^(m-1)+L_m-1R_m-1^*. The new factors are L_m · =[D_mL_m-1+U_m(V_m^*L_m-1) U_m],
Exact mathematical passage
P_{m-1}=D^{(m-1)}+L_{m-1}R_{m-1}^*
D^{-1}Use in source · 25 passages
Search aliases: D^-1
Use in source · pages.tex, line 162
-1 · 3. Hence D^-1u=u/4 and S=1+u^TD^-1u=frac52. The Woodbury identity reduces the entire inverse to A^-1=D^-1-D^-1uS^-1u^TD^-1 =D^-1-frac140uu^T. vfigwoodbury headTwo hypotheses, not one Invertibility of D alone does not suffice. For D=I_2, u=e_1, and v=-e_1, the update is diag(0,1). The small matrix S is zero, exactly where the correction formula would require an invalid division.
Exact mathematical passage
D^{-1}u=u/4
D^{-1/2}Use in source · 7 passages
Search aliases: D^-1/2
Use in source · pages.tex, line 301
vpage18Measure coupling after normalizationBLOCK JACOBI AS A GEOMETRIC CHANGE Let A=mat4 · 1 1 · 1, D=diag(4,1), B=D^-1/2=diag(1/2,1). Then BAB=I+E=mat1 · 1/2
Exact mathematical passage
A=\mat{4&1\\1&1},\quad D=\diag(4,1),\quad B=D^{-1/2}=\diag(1/2,1).
d^2Use in source · 3 passages
Search aliases: d^2
Use in source · Read passage · 01_manuscript.tex, line 620
[Equal tensor factors] For integers k,d≥0, a product of k square factors of order d, acting on a tensor with d^k entries, can be applied by k modewise contractions, reusing intermediate slices, with modeled cost 2k d^k+1 and factor storage kd^2.
Exact mathematical passage
kd^2
D^jUse in source · 7 passages
Search aliases: D^j
Use in source · pages.tex, line 292
With A=diag(1,2), the vectors v and Av=(1,2)^T are independent. Two polynomial samples now distinguish the packet labels. Alternatively, allowing the packet projectors as extra address operators separates the coordinates even when A=I_2. Both changes enlarge the addressing power, but in different ways. headWhen invariance is only approximate If A=D+K with D packet diagonal, then A^j-D^j ≤ jM^j-1 K , M=max{ A , D }. This controls departure from the decoupled iterate. It is not a claim that the di…
Exact mathematical passage
\|A^j-D^j\|\le jM^{j-1}\|K\|, \qquad M=\max\{\|A\|,\|D\|\}.
D^kUse in source · 1 passage
Search aliases: D^k
Use in source · Read passage · 01_manuscript.tex, line 1169
Use the telescoping identity A^j-D^j = ∑_k=0^j-1 A^j-1-kKD^k. Each term has norm at most M^j-1normK. The vector D^jv lies in the direct sum of local Krylov spaces.
Exact mathematical passage
A^j-D^j = \sum_{k=0}^{j-1} A^{j-1-k}KD^k.
d^kUse in source · 2 passages
Search aliases: d^k
Use in source · Read passage · 01_manuscript.tex, line 614
[Equal tensor factors] For integers k,d≥0, a product of k square factors of order d, acting on a tensor with d^k entries, can be applied by k modewise contractions, reusing intermediate slices, with modeled cost 2k d^k+1 and factor storage kd^2.
Exact mathematical passage
d^k
d^{k+1}Use in source · 3 passages
Search aliases: d^k+1
Use in source · Read passage · 01_manuscript.tex, line 615
[Equal tensor factors] For integers k,d≥0, a product of k square factors of order d, acting on a tensor with d^k entries, can be applied by k modewise contractions, reusing intermediate slices, with modeled cost 2k d^k+1 and factor storage kd^2.
Exact mathematical passage
\boxed{ 2k d^{k+1} }
d^{k+2}Use in source · 1 passage
Search aliases: d^k+2
Use in source · Read passage · 02_supplement_04.tex, line 106
Operation events. The event type for k+1 modes is a disjoint union of d copies of the trailing event type and the events for the final fiber contraction. Its cardinality obeys W_0=0, W_k+1=dW_k+2d^k+2. Lean proves the closed form W_k=2kd^k+1. The factor-address type has cardinality k d^2. These statements include k=0 and d=0. With no factors, the action is the identity on the singleton coordinate type and incurs zero modeled arithmetic. With d=0 and at least one factor, the coordinate type is em…
Exact mathematical passage
W_0=0,\qquad W_{k+1}=dW_k+2d^{k+2}.
d^{k-1}Use in source · 1 passage
Search aliases: d^k-1
Use in source · Read passage · 01_manuscript.tex, line 624
For k=0 there are no contractions and both modeled work and factor storage are zero. For k≥1, one mode has d^k-1 fibers of length d. Multiplying each fiber by a d by d matrix costs 2d^2, so one mode costs 2d^k+1. Sum over the k modes.
Exact mathematical passage
d^{k-1}
D_0Use in source · 1 passage
Search aliases: D_0
Use in source · Read passage · 01_manuscript.tex, line 931
Set D_j=widehat P_j-P_j. Then D_j=A_jD_j-1+E_j, D_0=0. Repeated substitution gives D_m = ∑_j=1^m A_mA_m-1⋯ A_j+1E_j. Taking norms yields the result.
Exact mathematical passage
D_j=A_jD_{j-1}+E_j, \qquad D_0=0.
D_1Use in source · 1 passage
Search aliases: D_1
Use in source · Read passage · 02_supplement_03.tex, line 10
Low-rank factors through a product chain. textitSource item 5.2 Let A_j=D_j+U_jV_j^*, U_j,V_j∈ C^n× r_j, with all D_j diagonal in one packet layout. Set P_m=A_m⋯ A_1, D^(m)=D_m⋯ D_1. The formalization constructs thin matrices L_m,R_m such that P_m=D^(m)+L_mR_m^*. Their column index type is a recursively constructed disjoint sum. Its cardinality is proved to equal ∑_j=1^m r_j.
Exact mathematical passage
P_m=A_m\cdots A_1,\qquad D^{(m)}=D_m\cdots D_1.
D_AUse in source · 23 passages
Search aliases: D_A
Use in source · pages.tex, line 142
vpage9The cross term has somewhere to goCLOSED LOW-RANK PRODUCT ALGEBRA Let A=D_A+u_av_a^T be the running operator. Take D_B=diag(1,2,3,4,5,6), u_b=(1,0,1,0,1,0)^T, and v_b=(0,1,0,1,0,1)^T. Define B=D_B+u_bv_b^T. U_C=[D_Au_b+u_a(v_a^Tu_b) u_a], V_C=[v_b D_B^Tv_a]. vfigbdlr_product headWhy there are only two columns Expanding AB gives three correction terms, but two share the right factor v_b^T: D_Au_bv_b^T+u_a(v_a^Tu_b)v_b^T =(D_Au_b+u_a(v_a^Tu_b))v_b^T. The remaining term is u_av_a^TD_B. Thus A…
Exact mathematical passage
A=D_A+u_av_a^T
D_BUse in source · 21 passages
Search aliases: D_B
Use in source · pages.tex, line 142
vpage9The cross term has somewhere to goCLOSED LOW-RANK PRODUCT ALGEBRA Let A=D_A+u_av_a^T be the running operator. Take D_B=diag(1,2,3,4,5,6), u_b=(1,0,1,0,1,0)^T, and v_b=(0,1,0,1,0,1)^T. Define B=D_B+u_bv_b^T. U_C=[D_Au_b+u_a(v_a^Tu_b) u_a], V_C=[v_b D_B^Tv_a]. vfigbdlr_product headWhy there are only two columns Expanding AB gives three correction terms, but two share the right factor v_b^T: D_Au_bv_b^T+u_a(v_a^Tu_b)v_b^T =(D_Au_b+u_a(v_a^Tu_b))v_b^T. The remaining term is u_av_a^TD_B. Thus A…
Exact mathematical passage
D_B=\diag(1,2,3,4,5,6)
D_B^*Use in source · 3 passages
Search aliases: D_B^*
Use in source · Read passage · 01_manuscript.tex, line 687
[Exact BDLR product] Let A=D_A+U_AV_A^*, B=D_B+U_BV_B^* use the same diagonal packet layout. Then AB=D_AD_B+U_CV_C^*, where U_C = D_AU_B+U_A(V_A^*U_B) · U_A and V_C = V_B · D_B^*V_A . The unrecompressed correction rank is at most r_A+r_B.
Exact mathematical passage
V_C = \begin{bmatrix} V_B&D_B^*V_A \end{bmatrix}.
D_B^TUse in source · 2 passages
Search aliases: D_B^T
Use in source · pages.tex, line 143
vpage9The cross term has somewhere to goCLOSED LOW-RANK PRODUCT ALGEBRA Let A=D_A+u_av_a^T be the running operator. Take D_B=diag(1,2,3,4,5,6), u_b=(1,0,1,0,1,0)^T, and v_b=(0,1,0,1,0,1)^T. Define B=D_B+u_bv_b^T. U_C=[D_Au_b+u_a(v_a^Tu_b) u_a], V_C=[v_b D_B^Tv_a]. vfigbdlr_product headWhy there are only two columns Expanding AB gives three correction terms, but two share the right factor v_b^T: D_Au_bv_b^T+u_a(v_a^Tu_b)v_b^T =(D_Au_b+u_a(v_a^Tu_b))v_b^T. The remaining term is u_av_a^TD_B. Thus A…
Exact mathematical passage
U_C=[D_Au_b+u_a(v_a^Tu_b)\ \ u_a],\qquad V_C=[v_b\ \ D_B^Tv_a].
D_jUse in source · 5 passages
Search aliases: D_j
Use in source · Read passage · 01_manuscript.tex, line 716
[Rank growth in a product chain] Let A_j=D_j+U_jV_j^*, j=1,…,m, share one diagonal packet layout. Before recompression, the product A_m⋯ A_1 has a BDLR representation whose correction rank is at most ∑_j=1^m r_j.
Exact mathematical passage
A_j=D_j+U_jV_j^*, \qquad j=1,\ldots,m,
D_{j-1}Use in source · 1 passage
Search aliases: D_j-1
Use in source · Read passage · 01_manuscript.tex, line 931
Set D_j=widehat P_j-P_j. Then D_j=A_jD_j-1+E_j, D_0=0. Repeated substitution gives D_m = ∑_j=1^m A_mA_m-1⋯ A_j+1E_j. Taking norms yields the result.
Exact mathematical passage
D_j=A_jD_{j-1}+E_j, \qquad D_0=0.
D_kUse in source · 7 passages
Search aliases: D_k
Use in source · Read passage · 02_supplement_06.tex, line 88
Spectral optimality from a kernel witness. textitModules DiagonalTruncation and SpectralOptimality Let σ_1≥⋯≥σ_d≥0 and D=diag(σ_1,…,σ_d), D_k=diag(σ_1,…,σ_min(k,d),0,…,0). Lean uses indices beginning at zero; the formulas on this page use the conventional one-based notation. The rank of D_k is proved to be at most k, even when retained singular values are zero.
Exact mathematical passage
D=\diag(\sigma_1,\ldots,\sigma_d),\qquad D_k=\diag(\sigma_1,\ldots,\sigma_{\min(k,d)},0,\ldots,0).
D_mUse in source · 3 passages
Search aliases: D_m
Use in source · Read passage · 01_manuscript.tex, line 937
Set D_j=widehat P_j-P_j. Then D_j=A_jD_j-1+E_j, D_0=0. Repeated substitution gives D_m = ∑_j=1^m A_mA_m-1⋯ A_j+1E_j. Taking norms yields the result.
Exact mathematical passage
D_m = \sum_{j=1}^{m} A_mA_{m-1}\cdots A_{j+1}E_j.
D_v^{-1}Use in source · 2 passages
Search aliases: D_v^-1
Use in source · Read passage · 02_supplement_05.tex, line 53
Constructed solution. Well-founded recursion defines x_v=D_v^-1 (b_v-∑_u→ vT_vux_u). Applying D_v proves the original equation at v. If another vector satisfies all equations, induction over predecessors makes each previously used coordinate identical, and invertibility of D_v forces agreement at v. Thus the construction gives existence and uniqueness, and the block operator is bijective.
Exact mathematical passage
\boxed{x_v=D_v^{-1}\!\left(b_v-\sum_{u\to v}T_{vu}x_u\right).}
D_\betaUse in source · 2 passages
Search aliases: D_β beta
Use in source · Read passage · 01_manuscript.tex, line 1026
The equation in packet β is D_β x_β = b_β - ∑_α→βK_βαx_α. Every predecessor of β lies in an earlier level. Once those values are available, x_β is obtained from the local solve with D_β. Vertices in the same level have no directed edge between them, so their solves are independent.
Exact mathematical passage
D_\beta x_\beta = b_\beta - \sum_{\alpha\to\beta}K_{\beta\alpha}x_\alpha.
\mathrm{decode}Use in source · 1 passage
Search aliases: decode mathrm
Use in source · Read passage · 02_supplement_02.tex, line 203
Contracts, grouping, and amortization. textitSource items 2.9, 10.2, 10.4--10.6, and 11.3 Output conversion. A decoded run returns the actual pair (decode(compressed(x)), t_run(x)+t_decode(compressed(x))). Under the declared output equality, Lean proves that this returns the same explicit result and charges the conversion time. It also proves the downstream composition identity when compressed output is consumed directly. Omitting a positive decode cost strictly understates the modeled explicit-…
Exact mathematical passage
\bigl(\operatorname{decode}(\operatorname{compressed}(x)), \;t_{\mathrm{run}}(x)+t_{\mathrm{decode}}(\operatorname{compressed}(x))\bigr).
\operatorname{decode}Use in source · 3 passages
Search aliases: decode operatorname
Use in source · Read passage · 01_manuscript.tex, line 231
[Contract conversion] Suppose one algorithm returns a compressed representation R and another returns an explicit object Y. A direct timing comparison is valid only if decode(R)=Y and the time required by decode is included, or if the downstream task consumes R directly.
Exact mathematical passage
\operatorname{decode}(R)=Y
\detUse in source · 2 passages
Search aliases: det
Use in source · pages.tex, line 175
For the running model, the checker multiplies the proposed inverse by A in both orders and obtains the identity. It also checks the determinant relation det A=det Ddet S. These tests verify the actual factors and sign of the correction, not just a numerical solve residual. headConditioning survives the reduction A small S can still be ill-conditioned. Reducing its dimension does not bound S^-1 . The manuscript's inverse estimate keeps that term explicit, along with the norms of D^-1u and v^TD^-1…
Exact mathematical passage
\det A=\det D\det S
\operatorname{DFT}Use in source · 3 passages
Search aliases: DFT operatorname
Use in source · Read passage · 02_supplement_04.tex, line 22
Normalization and inversion. The proof identifies matrix application with the ordinary counting-measure discrete Fourier transform: Fx=n^-1/2DFT(x). It separately identifies F^*x with the appropriately scaled inverse transform. Mathlib's checked Fourier inversion theorem and the proved scalar normalization yield both identities F^*F=I, FF^*=I. No unitarity assumption is inserted.
Exact mathematical passage
Fx=n^{-1/2}\operatorname{DFT}(x).
\diagUse in source · 22 passages
Search aliases: diag operatorname
Use in source · extended_pages.tex, line 3
vpage29A complex range needs an adjointTHE INNER PRODUCT IS PART OF THE CONTRACT The earlier compact example is real. Over C, an isometry means Q^*Q=I, with conjugation. Take the exact Gaussian-rational factors Q_L=mat3/5 · 0 4 i/5 · 0 0 · 1, Q_R=mat1 · 0 0 · i, G=diag(5,2). Their product is a genuinely complex rectangular operator: A=Q_LGQ_R^*=mat3 · 0
Exact mathematical passage
Q_L=\mat{3/5&0\\4\mathrm i/5&0\\0&1},\qquad Q_R=\mat{1&0\\0&\mathrm i},\qquad G=\diag(5,2).
\operatorname{diag}Use in source · 4 passages
Search aliases: diag operatorname
Use in source · Read passage · 02_supplement_04.tex, line 36
Convolution and reconstruction. An explicit bijective change of index proves DFT(C(c)x)_k =DFT(c)_kDFT(x)_k. This gives the matrix intertwining relation FC(c)=diag(DFTc)F. Multiplication by the proved inverse reconstructs the circulant: C(c)=F^*diag(sqrt n Fc)F. The two-transform application formula is also checked for every input vector. The result includes n=1 and arbitrary complex coefficients. Hermiticity is not required.
Exact mathematical passage
FC(c)=\operatorname{diag}(\operatorname{DFT}c)F
\distUse in source · 1 passage
Search aliases: dist operatorname
Use in source · Read passage · 01_manuscript.tex, line 1160
[Approximate packet Krylov bound] Write A=D+K, where D is packet diagonal. For every positive integer j, normA^j-D^j ≤ jM^j-1normK, where M=max{normA,normD}. Consequently, dist(A^jv,⊕_α K_j+1(D_α,v_α)) ≤ jM^j-1normKnormv.
Exact mathematical passage
\dist(A^jv,\bigoplus_\alpha\Kcal_{j+1}(D_\alpha,v_\alpha)) \leq jM^{j-1}\norm{K}\norm{v}.
\operatorname{dist}Use in source · 1 passage
Search aliases: dist operatorname
Use in source · Read passage · 02_supplement_03.tex, line 124
Define the comparison space L_j+1=⊕_i K_j+1(P_iDP_i,P_iv). Packet diagonality is first shown to imply commutation. The exact theorem then proves D^jv∈ L_j+1. Using that vector as a witness in the infimum distance gives dist(A^jv, L_j+1) ≤ jM^j-1normKnormv. This statement uses Lean's actual metric infimum distance to the subspace. It does not assume that the comparison vector lies there. The block-Jacobi condition bound and the preconditioned conjugate-gradient corollary are not completed by thes…
Exact mathematical passage
\boxed{\operatorname{dist}(A^jv,\mathcal L_{j+1}) \leq jM^{j-1}\norm{K}\norm{v}.}
e_0Use in source · 3 passages
Search aliases: e_0
Use in source · Read passage · 02_supplement_05.tex, line 222
The preconditioning corollary. Define q(κ)=fracsqrtκ-1sqrtκ+1. Monotonicity of square root and positive denominators prove that q is monotone. Lean also proves κ_2(B)≥1, q(κ)≥0 for κ≥1, and q(κ)<1. Thus q(κ_2(B))≤ q (frac1+η1-η). For 1≤κ and e_0≥0, an explicit classical premise e_m≤2q(κ)^m e_0 and a certified upper bound κ≤barκ yield e_m≤2q(barκ)^m e_0.
Exact mathematical passage
e_0\geq0
e_1Use in source · 9 passages
Search aliases: e_1
Use in source · pages.tex, line 173
-1 · 3. Hence D^-1u=u/4 and S=1+u^TD^-1u=frac52. The Woodbury identity reduces the entire inverse to A^-1=D^-1-D^-1uS^-1u^TD^-1 =D^-1-frac140uu^T. vfigwoodbury headTwo hypotheses, not one Invertibility of D alone does not suffice. For D=I_2, u=e_1, and v=-e_1, the update is diag(0,1). The small matrix S is zero, exactly where the correction formula would require an invalid division.
Exact mathematical passage
u=e_1
e_1^TUse in source · 1 passage
Search aliases: e_1^T
Use in source · woodbury.tex, line 15
(1.65,2.1) rectangle (2.5,2.95); node[font=small,align=center,font=small] at (2.075,2.525) frac38; node[font=small,align=center] at (6.25,3.5) D^-1u=tfrac14 u; node[font=small,align=center] at (6.25,2.25) S=1+u^TD^-1u=tfrac52; node[font=small,align=center] at (6.25,0.9) A^-1=D^-1-tfrac140uu^T; node[font=small,align=center,font=small] at (11.45,4.25) I-e_1e_1^T; path[draw=black!30,fill=white] (10.6,2.7) rectangle (11.45,3.55); node[font=small,align=center,font=small] at (11.025,3.125) 0; path[dra…
Exact mathematical passage
I-e_1e_1^T
E_2Use in source · 1 passage
Search aliases: E_2
Use in source · Read passage · 02_supplement_05.tex, line 202
A direct inverse bound. If C=(I+E)^-1 and norm E_2≤η<1, then C=I-EC. Consequently (1-η)norm C_2≤ 1, normI+E_2≤ 1+η. The constructed normalization is invertible because S and A are. Combining the estimates gives the actual condition number bound κ_2(B):=norm B_2normB^-1_2 ≤frac1+η1-η. This norm argument does not require a separately assumed Hermitian perturbation. Nonnegative η follows from the norm bound. Nonemptiness is explicit because the condition-number and norm-of-identity convention uses …
Exact mathematical passage
\norm E_2\leq\eta<1
e_2Use in source · 4 passages
Search aliases: e_2
Use in source · pages.tex, line 217
vpage13Errors enter at different placesFACTOR PERTURBATION VERSUS RECOMPRESSION Let the exact scalar factors be (2,3,4) and start from p_0=1. If additive errors (1/10,1/5,1/20) enter after each partial multiplication, then widehat p_j=a_jwidehat p_j-1+e_j, widehat p_3-p_3=12e_1+4e_2+e_3=frac4120. vfigerror_chain headThe same local numbers, a different model If instead the factors themselves are perturbed, the computed product is (4+e_3)(3+e_2)(2+e_1). Its error is 402/125, not 41/20. The factor-…
Exact mathematical passage
\widehat p_j=a_j\widehat p_{j-1}+e_j,\qquad \widehat p_3-p_3=12e_1+4e_2+e_3=\frac{41}{20}.
e_3Use in source · 4 passages
Search aliases: e_3
Use in source · pages.tex, line 217
vpage13Errors enter at different placesFACTOR PERTURBATION VERSUS RECOMPRESSION Let the exact scalar factors be (2,3,4) and start from p_0=1. If additive errors (1/10,1/5,1/20) enter after each partial multiplication, then widehat p_j=a_jwidehat p_j-1+e_j, widehat p_3-p_3=12e_1+4e_2+e_3=frac4120. vfigerror_chain headThe same local numbers, a different model If instead the factors themselves are perturbed, the computed product is (4+e_3)(3+e_2)(2+e_1). Its error is 402/125, not 41/20. The factor-…
Exact mathematical passage
\widehat p_j=a_j\widehat p_{j-1}+e_j,\qquad \widehat p_3-p_3=12e_1+4e_2+e_3=\frac{41}{20}.
E_AUse in source · 3 passages
Search aliases: E_A
Use in source · Read passage · 02_supplement_01.tex, line 10
0 · b≠ a, ∑_aP_a=1. Write A_ba=P_bAP_a and let E_A={(b,a):A_ba≠0}. These are exact algebraic zeros, not values discarded by a floating-point threshold.
Exact mathematical passage
E_A=\{(b,a):A_{ba}\ne0\}
E_eUse in source · 7 passages
Search aliases: E_e
Use in source · pages.tex, line 333
For orthogonal coordinate blocks, the squared Frobenius norms add exactly. Operator norms instead require a suitable bound or further spectral information. The maximum omitted block norm is generally insufficient. headInput sensitivity is additional information The packet perturbation estimate is (widehat A-A)x ≤∑_e E_e P_α(e)x . It can be much tighter for an input that scarcely occupies an affected source packet. To obtain a uniform operator bound, the input dependence must be bounded uniformly…
Exact mathematical passage
\|(\widehat A-A)x\|\le\sum_e\|E_e\|\,\|P_{\alpha(e)}x\|.
E_jUse in source · 20 passages
Search aliases: E_j
Use in source · pages.tex, line 24
E_j · An error operator · Is it a factor perturbation or an additive error after a product?
Exact mathematical passage
E_j
e_jUse in source · 3 passages
Search aliases: e_j
Use in source · pages.tex, line 217
vpage13Errors enter at different placesFACTOR PERTURBATION VERSUS RECOMPRESSION Let the exact scalar factors be (2,3,4) and start from p_0=1. If additive errors (1/10,1/5,1/20) enter after each partial multiplication, then widehat p_j=a_jwidehat p_j-1+e_j, widehat p_3-p_3=12e_1+4e_2+e_3=frac4120. vfigerror_chain headThe same local numbers, a different model If instead the factors themselves are perturbed, the computed product is (4+e_3)(3+e_2)(2+e_1). Its error is 402/125, not 41/20. The factor-…
Exact mathematical passage
\widehat p_j=a_j\widehat p_{j-1}+e_j,\qquad \widehat p_3-p_3=12e_1+4e_2+e_3=\frac{41}{20}.
e_mUse in source · 2 passages
Search aliases: e_m
Use in source · Read passage · 02_supplement_05.tex, line 222
onditioning corollary. Define q(κ)=fracsqrtκ-1sqrtκ+1. Monotonicity of square root and positive denominators prove that q is monotone. Lean also proves κ_2(B)≥1, q(κ)≥0 for κ≥1, and q(κ)<1. Thus q(κ_2(B))≤ q (frac1+η1-η). For 1≤κ and e_0≥0, an explicit classical premise e_m≤2q(κ)^m e_0 and a certified upper bound κ≤barκ yield e_m≤2q(barκ)^m e_0.
Exact mathematical passage
e_m\leq2q(\kappa)^m e_0
\epsUse in source · 10 passages
Search aliases: eps ε varepsilon
Use in source · Read passage · 01_manuscript.tex, line 1281
[Packet perturbation certificate] Let A= ∑_e∈ EA_e, widehat A = ∑_e∈ E(A_e+E_e), where edge e starts at packet α(e), and assume A_e=A_eP_α(e), E_e=E_eP_α(e). Then norm(widehat A-A)x ≤ ∑_e∈ E normE_e normP_α(e)x. If normE_e ≤ ε_enormA_e, then norm(widehat A-A)x ≤ ∑_e∈ E ε_enormA_e normP_α(e)x.
Exact mathematical passage
\norm{E_e} \leq \eps_e\norm{A_e},
\eps_eUse in source · 10 passages
Search aliases: eps_e ε_e eps varepsilon
Use in source · Read passage · 01_manuscript.tex, line 1281
[Packet perturbation certificate] Let A= ∑_e∈ EA_e, widehat A = ∑_e∈ E(A_e+E_e), where edge e starts at packet α(e), and assume A_e=A_eP_α(e), E_e=E_eP_α(e). Then norm(widehat A-A)x ≤ ∑_e∈ E normE_e normP_α(e)x. If normE_e ≤ ε_enormA_e, then norm(widehat A-A)x ≤ ∑_e∈ E ε_enormA_e normP_α(e)x.
Exact mathematical passage
\norm{E_e} \leq \eps_e\norm{A_e},
\errUse in source · 14 passages
Search aliases: err operatorname
Use in source · pages.tex, line 387
ne. Keeping either a or b costs three with error two. Splitting a costs four with zero error, and splitting b costs five with zero error. vfigtree_frontier The five complete prunings have (cost,error) pairs (1,6), (7,4), (9,2), (8,2), (10,0). At budget two the unique cheapest pruning is {a_0,a_1,b}, costing eight. It is Pareto optimal, but never minimizes cost+λerr for any λ≥0. headSee the incompatible inequalities To beat the root, that pruning would require 8+2λ≤1+6λ, hence λ≥7/4. To beat the …
Exact mathematical passage
\cost+\lambda\err
\expUse in source · 5 passages
Search aliases: exp
Use in source · Read passage · 01_manuscript.tex, line 522
[Circulant spectral representation] Let n≥1 be an integer and c∈ C^n, with indices taken modulo n, and define C(c)_jk=c_j-k. Let F be the unitary discrete Fourier matrix F_jk = frac1sqrt n exp(-frac2π i jkn). Then C(c) = F^*diag(sqrt n Fc)F. The operator is determined by n scalars and can be applied through two Fourier transforms and one pointwise product.
Exact mathematical passage
F_{jk} = \frac{1}{\sqrt n} \exp\left(-\frac{2\pi\ii jk}{n}\right).
\mathrm FUse in source · 5 passages
Search aliases: F mathrm
Use in source · Read passage · 02_supplement_06.tex, line 54
ubsection:002textitModule SmallCoreNorms; arbitrary ambient competitors For rectangular isometries Q,V define L(G)=QGV^*, C(B)=Q^*BV, Q^*Q=I, V^*V=I. The following statements are proved for the actual matrix rank and actual matrix norms: C( L(G))=G, rank L(G)=rank G, rank C(B)≤rank B, norm L(G)_ν=normG_ν, norm C(B)_ν≤norm B_ν, ν∈{2, F}. For the spectral norm, the proof uses the Gram matrix identity (QG)^*(QG)=G^*G and the C^*-norm square identity, together with right-side adjoint symmetry. Compr…
Exact mathematical passage
\boxed{\norm{\mathcal L(G)}_\nu=\norm{G}_\nu, \qquad \norm{\mathcal C(B)}_\nu\leq\norm B_\nu, \quad \nu\in\{2,\mathrm F\}.}
fUse in source · 2 passages
Search aliases: f
Use in source · Read passage · 01_manuscript.tex, line 1346
[Continuous precision allocation] Let E be a finite nonempty edge set, let s_e>0 be sensitivities and c_e>0 be costs, and let τ>0. Consider min_ε_e>0 ∑_e c_elogfrac1ε_e subject to ∑_es_eε_e=τ. The unique minimizer is ε_e = fracτ c_es_e∑_fc_f.
Exact mathematical passage
\boxed{ \eps_e = \frac{\tau c_e}{s_e\sum_fc_f}. }
F^*Use in source · 10 passages
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Use in source · pages.tex, line 74
vpage5Four coefficients, four exact frequenciesCIRCULANT MULTIPLICATION Take c=(2,1,0,-1)^T and define C(c)_jk=c_j-kbmod4. The unitary Fourier matrix is F_jk= i^-jk/2. Its normalization is visible in the exact diagonalization: C(c)=F^*diag(2Fc)F, 2Fc=(2,2-2 i,2,2+2 i)^T. vfigcirculant headFollow an actual input For x=(1,2,-1,0)^T, dense multiplication gives (0,6,0,-2)^T. Transforming x, multiplying by the four displayed eigenvalues, and applying F^* gives exactly the same vector. Every entry bel…
Exact mathematical passage
C(c)=F^*\diag(2Fc)F,\qquad 2Fc=(2,2-2\mathrm i,2,2+2\mathrm i)^T.
F^2Use in source · 3 passages
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Use in source · pages.tex, line 328
e coherent errorDISCOVERY AND PERTURBATION CERTIFICATES Split QQ^4 into scalar packets and let A=I+E, E=frac1100(onesones^T-I). Dropping every off-diagonal block leaves I. Each discarded entry is only 1/100, but the total error is not bounded by that single-entry size. vfigblock_error headTwo norms answer different questions There are twelve omitted entries, so E _F^2=12/10^4. The all-one vector is an eigenvector with eigenvalue 3/100, while the three orthogonal directions have eigenvalue -1/100…
Exact mathematical passage
\|E\|_F^2=12/10^4
F_0Use in source · 2 passages
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Use in source · Read passage · 01_manuscript.tex, line 1431
[Exact discrete precision program] Let the total integer error budget be B. Define F_j(b) as the minimum cost for assigning precisions to the first j edges with total weight exactly b. Then F_0(0)=0, F_0(b)=+∞ for b≠0. Set F_j-1(u)=+∞ whenever u<0 or the state is unreachable. Then F_j(b) = min_k { F_j-1(b-w_j,k)+c_j,k }. The minimum of F_m(b) over 0≤ b≤ B is the exact minimum cost for the quantized model.
Exact mathematical passage
F_0(0)=0, \qquad F_0(b)=+\infty \quad\text{for }b\neq0.
F_jUse in source · 2 passages
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Use in source · Read passage · 01_manuscript.tex, line 1427
[Exact discrete precision program] Let the total integer error budget be B. Define F_j(b) as the minimum cost for assigning precisions to the first j edges with total weight exactly b. Then F_0(0)=0, F_0(b)=+∞ for b≠0. Set F_j-1(u)=+∞ whenever u<0 or the state is unreachable. Then F_j(b) = min_k { F_j-1(b-w_j,k)+c_j,k }. The minimum of F_m(b) over 0≤ b≤ B is the exact minimum cost for the quantized model.
Exact mathematical passage
F_j(b)
F_{j-1}Use in source · 2 passages
Search aliases: F_j-1
Use in source · Read passage · 01_manuscript.tex, line 1437
[Exact discrete precision program] Let the total integer error budget be B. Define F_j(b) as the minimum cost for assigning precisions to the first j edges with total weight exactly b. Then F_0(0)=0, F_0(b)=+∞ for b≠0. Set F_j-1(u)=+∞ whenever u<0 or the state is unreachable. Then F_j(b) = min_k { F_j-1(b-w_j,k)+c_j,k }. The minimum of F_m(b) over 0≤ b≤ B is the exact minimum cost for the quantized model.
Exact mathematical passage
F_{j-1}(u)=+\infty
F_{jk}Use in source · 5 passages
Search aliases: F_jk
Use in source · pages.tex, line 73
vpage5Four coefficients, four exact frequenciesCIRCULANT MULTIPLICATION Take c=(2,1,0,-1)^T and define C(c)_jk=c_j-kbmod4. The unitary Fourier matrix is F_jk= i^-jk/2. Its normalization is visible in the exact diagonalization: C(c)=F^*diag(2Fc)F, 2Fc=(2,2-2 i,2,2+2 i)^T. vfigcirculant headFollow an actual input For x=(1,2,-1,0)^T, dense multiplication gives (0,6,0,-2)^T. Transforming x, multiplying by the four displayed eigenvalues, and applying F^* gives exactly the same vector. Every entry bel…
Exact mathematical passage
F_{jk}=\mathrm i^{-jk}/2
F_mUse in source · 1 passage
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Use in source · Read passage · 01_manuscript.tex, line 1448
total integer error budget be B. Define F_j(b) as the minimum cost for assigning precisions to the first j edges with total weight exactly b. Then F_0(0)=0, F_0(b)=+∞ for b≠0. Set F_j-1(u)=+∞ whenever u<0 or the state is unreachable. Then F_j(b) = min_k { F_j-1(b-w_j,k)+c_j,k }. The minimum of F_m(b) over 0≤ b≤ B is the exact minimum cost for the quantized model.
Exact mathematical passage
F_m(b)
F_nUse in source · 2 passages
Search aliases: F_n
Use in source · Read passage · 02_supplement_01.tex, line 49
Define the active contribution sum recursively by F_0(b,a) · =P_a · b=a, 0 · b≠ a, F_n+1(b,a) · =∑_c: A_bc≠0A_bcF_n(c,a). The checked identity is P_bA^nP_a=F_n(b,a) (n≥0). Its proof is induction using the exact composition theorem, not an assumed support recurrence. When no active walk exists, the corresponding contribution sum is zero, and hence the actual block of A^n vanishes.
Exact mathematical passage
F_0(b,a)&=\begin{cases}P_a&b=a,\\0&b\ne a,\end{cases}\\ F_{n+1}(b,a)&=\sum_{c:\,A_{bc}\ne0}A_{bc}F_n(c,a).
F_{n+1}Use in source · 1 passage
Search aliases: F_n+1
Use in source · Read passage · 02_supplement_01.tex, line 49
Define the active contribution sum recursively by F_0(b,a) · =P_a · b=a, 0 · b≠ a, F_n+1(b,a) · =∑_c: A_bc≠0A_bcF_n(c,a). The checked identity is P_bA^nP_a=F_n(b,a) (n≥0). Its proof is induction using the exact composition theorem, not an assumed support recurrence. When no active walk exists, the corresponding contribution sum is zero, and hence the actual block of A^n vanishes.
Exact mathematical passage
F_0(b,a)&=\begin{cases}P_a&b=a,\\0&b\ne a,\end{cases}\\ F_{n+1}(b,a)&=\sum_{c:\,A_{bc}\ne0}A_{bc}F_n(c,a).
F_{\rm root}Use in source · 1 passage
Search aliases: F_rm root rm
Use in source · Read passage · 01_manuscript.tex, line 1663
Assume every keep error is a nonnegative integer w_v. Let B∈ N. For each node v, define F_v(b) as the minimum cost of a pruning of the subtree at v with total error exactly b. The keep option contributes F_v(w_v)≤κ_v. The split option is obtained by min plus convolution of the child tables, followed by addition of σ_v. The smallest value among F_rm root(b), 0≤ b≤ B, is the exact minimum cost among prunings whose quantized error is at most B.
Exact mathematical passage
F_{\rm root}(b), \qquad 0\leq b\leq B,
F_{\mathrm{root}}Use in source · 1 passage
Search aliases: F_root mathrm
Use in source · Read passage · 02_supplement_03.tex, line 181
The root scan min_0≤ b≤ BF_root(b) is therefore the exact budget-constrained optimum. Increasing the budget cannot increase a finite optimum. The theorem retains feasibility hypotheses where necessary rather than treating codenone as an ordinary real cost.
Exact mathematical passage
\boxed{\min_{0\leq b\leq B}F_{\mathrm{root}}(b)}
F_tUse in source · 1 passage
Search aliases: F_t
Use in source · Read passage · 02_supplement_03.tex, line 202
Entrywise implementation theorem. For every b≤ B, Lean proves lookup(storedTree(B,t),b)=F_t(b). There is a parallel theorem for stored discrete precision. The proof establishes first that an output entry at b depends only on input entries at most b, then proceeds through cached convolution and the tree recursion. Consequently the optimality, attainment, and infeasibility specification applies to the actual stored result.
Exact mathematical passage
\operatorname{lookup}(\operatorname{storedTree}(B,t),b)=F_t(b).
F_vUse in source · 6 passages
Search aliases: F_v
Use in source · pages.tex, line 27
F_v(b) · A budget table · Does b mean exact weight, or an upper bound?
Exact mathematical passage
F_v(b)
F_{v_i}Use in source · 1 passage
Search aliases: F_v_i
Use in source · Read passage · 02_supplement_03.tex, line 168
Keeping node v costs κ_v and incurs integer error w_v. Splitting costs overhead σ_v in addition to the selected child cuts. The exact-weight recurrence is F_v(b)=min{ l κ_v if b=w_v, [2pt] displaystyleσ_v+ min_∑_i b_i=b∑_iF_v_i(b_i) if v is internal. . Nonexistent alternatives are infeasible. The child sum is implemented by the proved finite min-plus convolution, not by assuming an optimal allocation.
Exact mathematical passage
F_v(b)=\min\left\{ \begin{array}{l} \kappa_v\quad\text{if }b=w_v,\\[2pt] \displaystyle\sigma_v+ \min_{\sum_i b_i=b}\sum_iF_{v_i}(b_i)\quad\text{if }v\text{ is internal}. \end{array}\right.
F_\lambdaUse in source · 10 passages
Search aliases: F_λ lambda
Use in source · Read passage · monograph_examples.tex, line 79
The dynamic recurrence gives the same Lagrangian choices without listing prunings. At a, it compares 2+4λ with 4+λ. At r, it compares 1+6λ with 1+F_λ(a). Its exactness follows from the finite tree and additive model in [source reference]. Overlapping decompositions or nonlocal execution costs need another argument.
Exact mathematical passage
1+F_\lambda(a)
\operatorname{fill}Use in source · 1 passage
Search aliases: fill operatorname
Use in source · Read passage · 02_supplement_02.tex, line 101
Correctness of the actual update. For any edited subtree T' and valid context C, Lean proves updateValue(C,F(T'))=F(fill(C,T')). The proof inducts through the context. At each frame, the updated child contributes the new value, and each other child contributes its unchanged cached optimum. The frame recurrence is therefore exactly the full-tree recurrence.
Exact mathematical passage
\operatorname{updateValue}\bigl(C,F(T')\bigr)=F\bigl(\operatorname{fill}(C,T')\bigr).
\operatorname{Fin}Use in source · 1 passage
Search aliases: Fin operatorname
Use in source · Read passage · 02_supplement_02.tex, line 161
Explicit storage and operation models. textitSource items 3.1, 3.2, 3.4, 3.6, 3.8--3.10, and 4.2 For a finite collection of rectangular blocks, the scalar addresses are actual elements of coprod_e∈ E(Fin(m_e)×Fin(n_e)). The corresponding operation addresses include a right-hand-side index and one of two scalar-operation tags. Lean proves their cardinalities rather than merely defining a formula named ``cost.''
Exact mathematical passage
\coprod_{e\in E}\bigl(\operatorname{Fin}(m_e)\times\operatorname{Fin}(n_e)\bigr).
\flop_{\rm block}Use in source · 1 passage
Search aliases: flop_rm block flop rm operatorname rm
Use in source · Read passage · 01_manuscript.tex, line 284
[Equal diagonal packets] Let p≥1 and b≥1 be integers and n=pb. Suppose n is divided into p equal diagonal packets of width b. Then flop_rm block=2qnb, M_rm block=nb. The dense-to-block storage reduction factor is the following; the arithmetic reduction factor has the same value when q≥1: fracnb.
Exact mathematical passage
\flop_{\rm block}=2qnb, \qquad M_{\rm block}=nb.
\flop_{\rm dense}Use in source · 1 passage
Search aliases: flop_rm dense flop rm operatorname rm
Use in source · Read passage · 01_manuscript.tex, line 251
[Dense and packet application counts] A dense n by n matrix applied to q right hand sides has modeled cost flop_rm dense=2n^2q. A packet representation with active edge set E has modeled cost flop_rm packet = 2q ∑_(β,α)∈ E n_β n_α. It stores M_rm packet = ∑_(β,α)∈ E n_β n_α scalar entries.
Exact mathematical passage
\boxed{ \flop_{\rm dense}=2n^2q. }
\flop_{\rm packet}Use in source · 1 passage
Search aliases: flop_rm packet flop rm operatorname rm
Use in source · Read passage · 01_manuscript.tex, line 257
[Dense and packet application counts] A dense n by n matrix applied to q right hand sides has modeled cost flop_rm dense=2n^2q. A packet representation with active edge set E has modeled cost flop_rm packet = 2q ∑_(β,α)∈ E n_β n_α. It stores M_rm packet = ∑_(β,α)∈ E n_β n_α scalar entries.
Exact mathematical passage
\boxed{ \flop_{\rm packet} = 2q \sum_{(\beta,\alpha)\in E} n_\beta n_\alpha. }
G^*Use in source · 2 passages
Search aliases: G^*
Use in source · Read passage · 02_supplement_06.tex, line 60
*V=I. The following statements are proved for the actual matrix rank and actual matrix norms: C( L(G))=G, rank L(G)=rank G, rank C(B)≤rank B, norm L(G)_ν=normG_ν, norm C(B)_ν≤norm B_ν, ν∈{2, F}. For the spectral norm, the proof uses the Gram matrix identity (QG)^*(QG)=G^*G and the C^*-norm square identity, together with right-side adjoint symmetry. Compression follows from submultiplicativity and the isometry norm bound, including empty spaces.
Exact mathematical passage
(QG)^*(QG)=G^*G
G_kUse in source · 2 passages
Search aliases: G_k
Use in source · Read passage · 02_supplement_06.tex, line 72
Suppose G_k is a best rank-at-most-k approximation to G in either norm. Take an arbitrary ambient matrix B with rank B≤ k. Compression is an admissible small-core competitor and norm L(G)- L(G_k)_ν · =normG-G_k_ν
Exact mathematical passage
G_k
hUse in source · 25 passages
Search aliases: h
Use in source · extended_pages.tex, line 65
vpage32Safe integers can still be conservativeQUANTIZATION AND THE SIZE OF THE BUDGET TABLE Let one integer unit represent h=1/10 of certified error. For an individual nonnegative error e, choosing w=lceil e/hrceil ensures e≤ hw. Thus integer feasibility ∑ w≤ B implies the real bound ∑ e≤ hB. This implication is checked before applying the exact option or tree recurrence. vfigquantization headA false certificate and a conservative rejection For errors (3/50,2/25), rounding down assigns weights (…
Exact mathematical passage
h=1/10
H_{2n}Use in source · 2 passages
Search aliases: H_2n
Use in source · pages.tex, line 126
vpage8See every butterfly valueWALSH RECURSION IN EIGHT COORDINATES The unnormalized Walsh matrix satisfies H_1=[1] and H_2n=matH_n · H_n H_n · -H_n, H_nH_n^T=nI. Apply the three add/subtract stages to (1,2,3,4,5,6,7,8)^T. The full intermediate ledger is displayed below. vfigwalsh headArithmetic can be counted exactly Each stage visits eight coordinates, producing one sum or difference per coordinate. Three stages therefore use exactly 24 additions or subtractions. No multiplication by a dense 8…
Exact mathematical passage
H_{2n}=\mat{H_n&H_n\\H_n&-H_n},\qquad H_nH_n^T=nI.
H_8Use in source · 3 passages
Search aliases: H_8
Use in source · pages.tex, line 134
The result (36,-4,-8,0,-16,0,0,0)^T still reconstructs the input: the inverse is H_8^T/8. Because the displayed transform is unnormalized, its squared output norm is eight times the squared input norm. The normalized transform H_8/sqrt8 would instead be orthogonal. headWhat the zeros mean The zero coefficients arise from exact cancellation among the input values. They do not imply that the transform's defining matrix is sparse. The butterfly structure, not the final coefficient support, explains…
Exact mathematical passage
H_8/\sqrt8
H_8^TUse in source · 3 passages
Search aliases: H_8^T
Use in source · pages.tex, line 134
The result (36,-4,-8,0,-16,0,0,0)^T still reconstructs the input: the inverse is H_8^T/8. Because the displayed transform is unnormalized, its squared output norm is eight times the squared input norm. The normalized transform H_8/sqrt8 would instead be orthogonal. headWhat the zeros mean The zero coefficients arise from exact cancellation among the input values. They do not imply that the transform's defining matrix is sparse. The butterfly structure, not the final coefficient support, explains…
Exact mathematical passage
H_8^T/8
H_nUse in source · 2 passages
Search aliases: H_n
Use in source · pages.tex, line 126
vpage8See every butterfly valueWALSH RECURSION IN EIGHT COORDINATES The unnormalized Walsh matrix satisfies H_1=[1] and H_2n=matH_n · H_n H_n · -H_n, H_nH_n^T=nI. Apply the three add/subtract stages to (1,2,3,4,5,6,7,8)^T. The full intermediate ledger is displayed below. vfigwalsh headArithmetic can be counted exactly Each stage visits eight coordinates, producing one sum or difference per coordinate. Three stages therefore use exactly 24 additions or subtractions. No multiplication by a dense 8…
Exact mathematical passage
H_{2n}=\mat{H_n&H_n\\H_n&-H_n},\qquad H_nH_n^T=nI.
H_N^TUse in source · 3 passages
Search aliases: H_N^T
Use in source · Read passage · 01_manuscript.tex, line 640
H_N · -H_N . Then H_NH_N^T=NI. The product H_Nx can be evaluated in exactly Nlog_2N additions or subtractions through the recursion.
Exact mathematical passage
H_NH_N^T=NI.
H_n^TUse in source · 2 passages
Search aliases: H_n^T
Use in source · pages.tex, line 126
vpage8See every butterfly valueWALSH RECURSION IN EIGHT COORDINATES The unnormalized Walsh matrix satisfies H_1=[1] and H_2n=matH_n · H_n H_n · -H_n, H_nH_n^T=nI. Apply the three add/subtract stages to (1,2,3,4,5,6,7,8)^T. The full intermediate ledger is displayed below. vfigwalsh headArithmetic can be counted exactly Each stage visits eight coordinates, producing one sum or difference per coordinate. Three stages therefore use exactly 24 additions or subtractions. No multiplication by a dense 8…
Exact mathematical passage
H_{2n}=\mat{H_n&H_n\\H_n&-H_n},\qquad H_nH_n^T=nI.
\mathrm iUse in source · 17 passages
Search aliases: i mathrm
Use in source · extended_pages.tex, line 3
vpage29A complex range needs an adjointTHE INNER PRODUCT IS PART OF THE CONTRACT The earlier compact example is real. Over C, an isometry means Q^*Q=I, with conjugation. Take the exact Gaussian-rational factors Q_L=mat3/5 · 0 4 i/5 · 0 0 · 1, Q_R=mat1 · 0 0 · i, G=diag(5,2). Their product is a genuinely complex rectangular operator: A=Q_LGQ_R^*=mat3 · 0
Exact mathematical passage
Q_L=\mat{3/5&0\\4\mathrm i/5&0\\0&1},\qquad Q_R=\mat{1&0\\0&\mathrm i},\qquad G=\diag(5,2).
\mathrm i^{-jk}Use in source · 3 passages
Search aliases: i^-jk mathrm
Use in source · pages.tex, line 73
vpage5Four coefficients, four exact frequenciesCIRCULANT MULTIPLICATION Take c=(2,1,0,-1)^T and define C(c)_jk=c_j-kbmod4. The unitary Fourier matrix is F_jk= i^-jk/2. Its normalization is visible in the exact diagonalization: C(c)=F^*diag(2Fc)F, 2Fc=(2,2-2 i,2,2+2 i)^T. vfigcirculant headFollow an actual input For x=(1,2,-1,0)^T, dense multiplication gives (0,6,0,-2)^T. Transforming x, multiplying by the four displayed eigenvalues, and applying F^* gives exactly the same vector. Every entry bel…
Exact mathematical passage
F_{jk}=\mathrm i^{-jk}/2
I_2Use in source · 7 passages
Search aliases: I_2
Use in source · pages.tex, line 173
-1 · 3. Hence D^-1u=u/4 and S=1+u^TD^-1u=frac52. The Woodbury identity reduces the entire inverse to A^-1=D^-1-D^-1uS^-1u^TD^-1 =D^-1-frac140uu^T. vfigwoodbury headTwo hypotheses, not one Invertibility of D alone does not suffice. For D=I_2, u=e_1, and v=-e_1, the update is diag(0,1). The small matrix S is zero, exactly where the correction formula would require an invalid division.
Exact mathematical passage
D=I_2
\iffUse in source · 1 passage
Search aliases: iff
Use in source · Read passage · 02_supplement_02.tex, line 193
Capacity and roofline. Positive physical capacities and the inequalities F≤ PT, B≤ WT give T≥max{F/P,B/W}. For positive bytes and elapsed time, the rate form is F/T≤min{P,WF/B}. The dense memory bound is the exact equivalence n^2s≤ Miff n≤sqrtM/s for nonnegative n,M and positive s.
Exact mathematical passage
n^2s\leq M\iff n\leq\sqrt{M/s}
\iiUse in source · 3 passages
Search aliases: ii i mathrm
Use in source · Read passage · 01_manuscript.tex, line 522
[Circulant spectral representation] Let n≥1 be an integer and c∈ C^n, with indices taken modulo n, and define C(c)_jk=c_j-k. Let F be the unitary discrete Fourier matrix F_jk = frac1sqrt n exp(-frac2π i jkn). Then C(c) = F^*diag(sqrt n Fc)F. The operator is determined by n scalars and can be applied through two Fourier transforms and one pointwise product.
Exact mathematical passage
F_{jk} = \frac{1}{\sqrt n} \exp\left(-\frac{2\pi\ii jk}{n}\right).
JUse in source · 16 passages
Search aliases: J
Use in source · extended_pages.tex, line 53
bottomrule vfig[.45]shared_family With reuse weights (r_1,r_2)=(1,3), the aggregated objectives are J_rm keep=4+2+3·7=27, J_rm split=2+8+3·2=16. The exact shared optimum is split. Operator 1 alone prefers keep on its per-use costs, while operator 2 prefers split. Choosing independently and trying to combine those decisions would not solve the declared one-pruning problem. headA hand-computable decision threshold Fix r_1=1 and vary r_2=r≥0. Keeping costs 6+7r and splitting costs 10+2r. Splitting …
Exact mathematical passage
J_{\rm keep}=4+2+3\cdot7=27,\qquad J_{\rm split}=2+8+3\cdot2=16.
J_{\rm keep}Use in source · 2 passages
Search aliases: J_rm keep rm
Use in source · extended_pages.tex, line 53
bottomrule vfig[.45]shared_family With reuse weights (r_1,r_2)=(1,3), the aggregated objectives are J_rm keep=4+2+3·7=27, J_rm split=2+8+3·2=16. The exact shared optimum is split. Operator 1 alone prefers keep on its per-use costs, while operator 2 prefers split. Choosing independently and trying to combine those decisions would not solve the declared one-pruning problem. headA hand-computable decision threshold Fix r_1=1 and vary r_2=r≥0. Keeping costs 6+7r and splitting costs 10+2r. Splitting …
Exact mathematical passage
J_{\rm keep}=4+2+3\cdot7=27,\qquad J_{\rm split}=2+8+3\cdot2=16.
J_{\rm split}Use in source · 2 passages
Search aliases: J_rm split rm
Use in source · extended_pages.tex, line 53
bottomrule vfig[.45]shared_family With reuse weights (r_1,r_2)=(1,3), the aggregated objectives are J_rm keep=4+2+3·7=27, J_rm split=2+8+3·2=16. The exact shared optimum is split. Operator 1 alone prefers keep on its per-use costs, while operator 2 prefers split. Choosing independently and trying to combine those decisions would not solve the declared one-pruning problem. headA hand-computable decision threshold Fix r_1=1 and vary r_2=r≥0. Keeping costs 6+7r and splitting costs 10+2r. Splitting …
Exact mathematical passage
J_{\rm keep}=4+2+3\cdot7=27,\qquad J_{\rm split}=2+8+3\cdot2=16.
J_\lambdaUse in source · 3 passages
Search aliases: J_λ lambda
Use in source · Read passage · 01_manuscript.tex, line 1529
[Exact Lagrangian packet compiler] Fix λ≥0 and define J_λ( L) = cost( L) + λerr( L). For every node v, set K_λ(v) = κ_v+λϵ_v and S_λ(v) = σ_v + ∑_w∈child(v)F_λ(w). For a terminal node, set F_λ(v)=K_λ(v). For an internal node, the recurrence F_λ(v) = min{K_λ(v),S_λ(v)} computes the exact minimum Lagrangian objective over every admissible pruning of the subtree rooted at v.
Exact mathematical passage
J_\lambda(\Lcal) = \cost(\Lcal) + \lambda\err(\Lcal).
\mathcal KUse in source · 6 passages
Search aliases: K mathcal
Use in source · pages.tex, line 283
vpage17Invariant does not mean independently addressableA STRICT KRYLOV INCLUSION Take A=I_2, split into the two coordinate packets, and start with v=(1,1)^T. Every packet is invariant. Nevertheless every polynomial applies the same scalar to both coordinates: p(A)v=p(1)(1,1)^T, dim K_m(A,v)=1 (m≥1). vfigkrylov The direct sum of the two local Krylov spans is all of QQ^2. It permits independent coefficients in the two packets. The global Krylov space does not. This is why the manuscript states an…
Exact mathematical passage
p(A)v=p(1)(1,1)^T,\qquad \dim\mathcal K_m(A,v)=1\quad(m\ge1).
K'Use in source · 2 passages
Search aliases: K'
Use in source · Read passage · 02_supplement_02.tex, line 69
Strict half-gap stability. At a solved internal node write K=κ_v+λϵ_v, S=σ_v+∑_wF_λ(w), g=|K-S|. If the updated scores satisfy |K'-K|<g/2, |S'-S|<g/2, Lean proves K'≤ S' Longleftrightarrow K≤ S. Applying this condition recursively at every internal node proves equality of the entire returned cuts before and after perturbation, not only equality of one numerical minimum.
Exact mathematical passage
|K'-K|<g/2,\qquad |S'-S|<g/2,
K^2Use in source · 8 passages
Search aliases: K^2
Use in source · extended_pages.tex, line 26
0 · 1 · -1 · 1. The two routes from coordinate zero to coordinate three cancel, so K^2=0. Consequently A^-1=I-K=mat1 · 0 · 0 · 0
Exact mathematical passage
K^2=0
K^hUse in source · 2 passages
Search aliases: K^h
Use in source · pages.tex, line 239
dicts possible support but not actual support. Replacing the inclusion supp(AB)⊆supp(A)∘supp(B) by equality would be wrong even in this four-dimensional model. headWhat an acyclic graph does guarantee If the longest strict directed path has h edges, then no term of K^h+1 can survive. Thus K^h+1=0. This is an upper bound on the nilpotence index, not an assertion that K^h≠0. Here the graph has paths of length two, while coefficient cancellation already gives K^2=0.
Exact mathematical passage
K^h\ne0
K^{h+1}Use in source · 6 passages
Search aliases: K^h+1
Use in source · pages.tex, line 239
, K^2=0. The matrix of path counts sees the two possibilities and forgets the signs. It therefore predicts possible support but not actual support. Replacing the inclusion supp(AB)⊆supp(A)∘supp(B) by equality would be wrong even in this four-dimensional model. headWhat an acyclic graph does guarantee If the longest strict directed path has h edges, then no term of K^h+1 can survive. Thus K^h+1=0. This is an upper bound on the nilpotence index, not an assertion that K^h≠0. Here the graph has path…
Exact mathematical passage
K^{h+1}
K_{10}Use in source · 2 passages
Search aliases: K_10
Use in source · pages.tex, line 233
vpage14Two routes can cancel exactlySUPPORT IS AN INCLUSION Let K have transfers 0→1, 0→2, 1→3, and 2→3, with coefficients 1,1,1,-1. There are two length-two routes from packet zero to packet three, but their contributions sum to zero. vfigcancellation (K^2)_30=K_31K_10+K_32K_20=1-1=0, K^2=0. The matrix of path counts sees the two possibilities and forgets the signs. It therefore predicts possible support but not actual support. Replacing the inclusion supp(AB)⊆supp(A)∘supp(B) by equality would …
Exact mathematical passage
(K^2)_{30}=K_{31}K_{10}+K_{32}K_{20}=1-1=0, \qquad K^2=0.
K_{20}Use in source · 2 passages
Search aliases: K_20
Use in source · pages.tex, line 233
vpage14Two routes can cancel exactlySUPPORT IS AN INCLUSION Let K have transfers 0→1, 0→2, 1→3, and 2→3, with coefficients 1,1,1,-1. There are two length-two routes from packet zero to packet three, but their contributions sum to zero. vfigcancellation (K^2)_30=K_31K_10+K_32K_20=1-1=0, K^2=0. The matrix of path counts sees the two possibilities and forgets the signs. It therefore predicts possible support but not actual support. Replacing the inclusion supp(AB)⊆supp(A)∘supp(B) by equality would …
Exact mathematical passage
(K^2)_{30}=K_{31}K_{10}+K_{32}K_{20}=1-1=0, \qquad K^2=0.
K_{31}Use in source · 2 passages
Search aliases: K_31
Use in source · pages.tex, line 233
vpage14Two routes can cancel exactlySUPPORT IS AN INCLUSION Let K have transfers 0→1, 0→2, 1→3, and 2→3, with coefficients 1,1,1,-1. There are two length-two routes from packet zero to packet three, but their contributions sum to zero. vfigcancellation (K^2)_30=K_31K_10+K_32K_20=1-1=0, K^2=0. The matrix of path counts sees the two possibilities and forgets the signs. It therefore predicts possible support but not actual support. Replacing the inclusion supp(AB)⊆supp(A)∘supp(B) by equality would …
Exact mathematical passage
(K^2)_{30}=K_{31}K_{10}+K_{32}K_{20}=1-1=0, \qquad K^2=0.
K_{32}Use in source · 2 passages
Search aliases: K_32
Use in source · pages.tex, line 233
vpage14Two routes can cancel exactlySUPPORT IS AN INCLUSION Let K have transfers 0→1, 0→2, 1→3, and 2→3, with coefficients 1,1,1,-1. There are two length-two routes from packet zero to packet three, but their contributions sum to zero. vfigcancellation (K^2)_30=K_31K_10+K_32K_20=1-1=0, K^2=0. The matrix of path counts sees the two possibilities and forgets the signs. It therefore predicts possible support but not actual support. Replacing the inclusion supp(AB)⊆supp(A)∘supp(B) by equality would …
Exact mathematical passage
(K^2)_{30}=K_{31}K_{10}+K_{32}K_{20}=1-1=0, \qquad K^2=0.
\mathcal K_{j+1}Use in source · 1 passage
Search aliases: K_j+1 mathcal
Use in source · Read passage · 02_supplement_03.tex, line 120
Define the comparison space L_j+1=⊕_i K_j+1(P_iDP_i,P_iv). Packet diagonality is first shown to imply commutation. The exact theorem then proves D^jv∈ L_j+1. Using that vector as a witness in the infimum distance gives dist(A^jv, L_j+1) ≤ jM^j-1normKnormv. This statement uses Lean's actual metric infimum distance to the subspace. It does not assume that the comparison vector lies there. The block-Jacobi condition bound and the preconditioned conjugate-gradient corollary are not completed by thes…
Exact mathematical passage
\mathcal L_{j+1}=\bigoplus_i\mathcal K_{j+1}(P_iDP_i,P_iv).
\mathcal K_mUse in source · 5 passages
Search aliases: K_m mathcal
Use in source · pages.tex, line 283
vpage17Invariant does not mean independently addressableA STRICT KRYLOV INCLUSION Take A=I_2, split into the two coordinate packets, and start with v=(1,1)^T. Every packet is invariant. Nevertheless every polynomial applies the same scalar to both coordinates: p(A)v=p(1)(1,1)^T, dim K_m(A,v)=1 (m≥1). vfigkrylov The direct sum of the two local Krylov spans is all of QQ^2. It permits independent coefficients in the two packets. The global Krylov space does not. This is why the manuscript states an…
Exact mathematical passage
p(A)v=p(1)(1,1)^T,\qquad \dim\mathcal K_m(A,v)=1\quad(m\ge1).
K_{\beta\alpha}Use in source · 1 passage
Search aliases: K_βα beta alpha
Use in source · Read passage · 01_manuscript.tex, line 1026
The equation in packet β is D_β x_β = b_β - ∑_α→βK_βαx_α. Every predecessor of β lies in an earlier level. Once those values are available, x_β is obtained from the local solve with D_β. Vertices in the same level have no directed edge between them, so their solves are independent.
Exact mathematical passage
D_\beta x_\beta = b_\beta - \sum_{\alpha\to\beta}K_{\beta\alpha}x_\alpha.
\Kcal_{j+1}Use in source · 1 passage
Search aliases: Kcal_j+1 K_j+1 Kcal mathcal
Use in source · Read passage · 01_manuscript.tex, line 1160
[Approximate packet Krylov bound] Write A=D+K, where D is packet diagonal. For every positive integer j, normA^j-D^j ≤ jM^j-1normK, where M=max{normA,normD}. Consequently, dist(A^jv,⊕_α K_j+1(D_α,v_α)) ≤ jM^j-1normKnormv.
Exact mathematical passage
\dist(A^jv,\bigoplus_\alpha\Kcal_{j+1}(D_\alpha,v_\alpha)) \leq jM^{j-1}\norm{K}\norm{v}.
\kerUse in source · 1 passage
Search aliases: ker
Use in source · Read passage · 02_supplement_06.tex, line 124
Frobenius optimality from kernel weights. textitModules KernelFrame and FrobeniusOptimality For an arbitrary competitor B with rank B≤ k, construct an orthonormal matrix Q spanning its actual kernel. Then BQ=0, Q^*Q=I, q=dimker B=d-rank B. Define the row weights w_i=∑_j=1^q|Q_ij|^2. The proof derives 0≤ w_i≤1, ∑_iw_i=q≥max(d-k,0). The upper bound uses contraction of Q^* on a coordinate vector. The weight sum follows from the trace of Q^*Q and rank-nullity.
Exact mathematical passage
BQ=0,\qquad Q^*Q=I,\qquad q=\dim\ker B=d-\rank B.
LUse in source · 64 passages
Search aliases: L
Use in source · extended_pages.tex, line 3
vpage29A complex range needs an adjointTHE INNER PRODUCT IS PART OF THE CONTRACT The earlier compact example is real. Over C, an isometry means Q^*Q=I, with conjugation. Take the exact Gaussian-rational factors Q_L=mat3/5 · 0 4 i/5 · 0 0 · 1, Q_R=mat1 · 0 0 · i, G=diag(5,2). Their product is a genuinely complex rectangular operator: A=Q_LGQ_R^*=mat3 · 0
Exact mathematical passage
Q_L=\mat{3/5&0\\4\mathrm i/5&0\\0&1},\qquad Q_R=\mat{1&0\\0&\mathrm i},\qquad G=\diag(5,2).
\mathcal LUse in source · 7 passages
Search aliases: L mathcal
Use in source · Read passage · 02_supplement_03.tex, line 120
Define the comparison space L_j+1=⊕_i K_j+1(P_iDP_i,P_iv). Packet diagonality is first shown to imply commutation. The exact theorem then proves D^jv∈ L_j+1. Using that vector as a witness in the infimum distance gives dist(A^jv, L_j+1) ≤ jM^j-1normKnormv. This statement uses Lean's actual metric infimum distance to the subspace. It does not assume that the comparison vector lies there. The block-Jacobi condition bound and the preconditioned conjugate-gradient corollary are not completed by thes…
Exact mathematical passage
\mathcal L_{j+1}=\bigoplus_i\mathcal K_{j+1}(P_iDP_i,P_iv).
lUse in source · 5 passages
Search aliases: l
Use in source · extended_pages.tex, line 53
bottomrule vfig[.45]shared_family With reuse weights (r_1,r_2)=(1,3), the aggregated objectives are J_rm keep=4+2+3·7=27, J_rm split=2+8+3·2=16. The exact shared optimum is split. Operator 1 alone prefers keep on its per-use costs, while operator 2 prefers split. Choosing independently and trying to combine those decisions would not solve the declared one-pruning problem. headA hand-computable decision threshold Fix r_1=1 and vary r_2=r≥0. Keeping costs 6+7r and splitting costs 10+2r. Splitting …
Exact mathematical passage
J_{\rm keep}=4+2+3\cdot7=27,\qquad J_{\rm split}=2+8+3\cdot2=16.
L_1Use in source · 3 passages
Search aliases: L_1
Use in source · Read passage · 02_supplement_02.tex, line 40
Frontiers and decision stability. textitSource items 9.6, 9.7, and 9.11 Let L_1,L_2 be actual optima at 0≤λ_1<λ_2. Their optimality inequalities are cost(L_1)+λ_1err(L_1) · ≤cost(L_2)+λ_1err(L_2),
Exact mathematical passage
L_1,L_2
L_2Use in source · 3 passages
Search aliases: L_2
Use in source · Read passage · 02_supplement_02.tex, line 40
Frontiers and decision stability. textitSource items 9.6, 9.7, and 9.11 Let L_1,L_2 be actual optima at 0≤λ_1<λ_2. Their optimality inequalities are cost(L_1)+λ_1err(L_1) · ≤cost(L_2)+λ_1err(L_2),
Exact mathematical passage
L_1,L_2
\mathcal L_{j+1}Use in source · 3 passages
Search aliases: L_j+1 mathcal
Use in source · Read passage · 02_supplement_03.tex, line 120
Define the comparison space L_j+1=⊕_i K_j+1(P_iDP_i,P_iv). Packet diagonality is first shown to imply commutation. The exact theorem then proves D^jv∈ L_j+1. Using that vector as a witness in the infimum distance gives dist(A^jv, L_j+1) ≤ jM^j-1normKnormv. This statement uses Lean's actual metric infimum distance to the subspace. It does not assume that the comparison vector lies there. The block-Jacobi condition bound and the preconditioned conjugate-gradient corollary are not completed by thes…
Exact mathematical passage
\mathcal L_{j+1}=\bigoplus_i\mathcal K_{j+1}(P_iDP_i,P_iv).
L_kUse in source · 5 passages
Search aliases: L_k
Use in source · Read passage · 01_manuscript.tex, line 825
R=Q_RR_R, where the columns of Q_L and Q_R are orthonormal. Form the small core G=R_LR_R^* and let G=WΣ Z^* be an ordered thin singular value decomposition, with nonnegative nonincreasing singular values. For an integer k≥0, retain min{k,d} columns, where d is the number of supplied thin singular directions; tail singular values beyond d are zero. Define L_k=Q_LW_kΣ_k^1/2, R_k=Q_RZ_kΣ_k^1/2. Then L_kR_k^* is a best rank at most k approximation to LR^*. Its spectral and Frobenius errors are the s…
Exact mathematical passage
L_k=Q_LW_k\Sigma_k^{1/2}, \qquad R_k=Q_RZ_k\Sigma_k^{1/2}.
L_mUse in source · 3 passages
Search aliases: L_m
Use in source · Read passage · 02_supplement_03.tex, line 13
Low-rank factors through a product chain. textitSource item 5.2 Let A_j=D_j+U_jV_j^*, U_j,V_j∈ C^n× r_j, with all D_j diagonal in one packet layout. Set P_m=A_m⋯ A_1, D^(m)=D_m⋯ D_1. The formalization constructs thin matrices L_m,R_m such that P_m=D^(m)+L_mR_m^*. Their column index type is a recursively constructed disjoint sum. Its cardinality is proved to equal ∑_j=1^m r_j.
Exact mathematical passage
L_m,R_m
L_{m-1}Use in source · 2 passages
Search aliases: L_m-1
Use in source · Read passage · 02_supplement_03.tex, line 19
Construction. Suppose P_m-1=D^(m-1)+L_m-1R_m-1^*. The new factors are L_m · =[D_mL_m-1+U_m(V_m^*L_m-1) U_m],
Exact mathematical passage
P_{m-1}=D^{(m-1)}+L_{m-1}R_{m-1}^*
\LcalUse in source · 11 passages
Search aliases: Lcal L mathcal
Use in source · Read passage · 01_manuscript.tex, line 1513
The cost and error of a pruning L are cost( L) = ∑_v∈ Lκ_v + ∑_u splitσ_u and err( L)=∑_v∈ Lϵ_v. The model can include arithmetic, storage, launch, transfer, and precision costs inside κ_v. It remains exact only when the declared totals are additive.
Exact mathematical passage
\Lcal
\Lcal_1Use in source · 6 passages
Search aliases: Lcal_1 L_1 Lcal mathcal
Use in source · Read passage · 01_manuscript.tex, line 1601
[Monotone Lagrangian frontier] Let 0≤λ_1<λ_2. Choose optimal prunings L_1 and L_2 for the two Lagrangian parameters. Then err( L_2) ≤ err( L_1) and cost( L_2) ≥ cost( L_1).
Exact mathematical passage
\Lcal_1
\Lcal_2Use in source · 6 passages
Search aliases: Lcal_2 L_2 Lcal mathcal
Use in source · Read passage · 01_manuscript.tex, line 1601
[Monotone Lagrangian frontier] Let 0≤λ_1<λ_2. Choose optimal prunings L_1 and L_2 for the two Lagrangian parameters. Then err( L_2) ≤ err( L_1) and cost( L_2) ≥ cost( L_1).
Exact mathematical passage
\Lcal_2
\lceilUse in source · 2 passages
Search aliases: lceil
Use in source · extended_pages.tex, line 66
vpage32Safe integers can still be conservativeQUANTIZATION AND THE SIZE OF THE BUDGET TABLE Let one integer unit represent h=1/10 of certified error. For an individual nonnegative error e, choosing w=lceil e/hrceil ensures e≤ hw. Thus integer feasibility ∑ w≤ B implies the real bound ∑ e≤ hB. This implication is checked before applying the exact option or tree recurrence. vfigquantization headA false certificate and a conservative rejection For errors (3/50,2/25), rounding down assigns weights (…
Exact mathematical passage
w=\lceil e/h\rceil\qquad\text{ensures}\qquad e\le hw.
\leadsto_{H}Use in source · 1 passage
Search aliases: leadsto_H leadsto
Use in source · Read passage · 02_supplement_05.tex, line 101
Hidden elimination. For the actual Schur complement S=A_RR-A_RHA_HH^-1A_HR, the hidden inverse is expanded through the hidden projectors. A nonzero retained correction block requires indices j,k with nonzero entrance and exit blocks and a hidden path jleadsto k. Therefore a nonzero block S_ba requires either a direct retained block, or a⟶ j leadsto_H k⟶ b. The entrance and exit matrices may be rectangular; retained and hidden coordinate sizes and packet systems are independently typed. The hidde…
Exact mathematical passage
a\longrightarrow j\ \leadsto_{H}\ k\longrightarrow b.
\lfloorUse in source · 4 passages
Search aliases: lfloor
Use in source · pages.tex, line 427
ONE LINEAR MODEL Suppose the baseline costs seven units per use. A candidate costs 30 units to set up and two per use. These are synthetic constants chosen to expose the exact threshold: T_b(R)=7R, T_c(R)=30+2R. vfigamortization At R=6, both cost 42. The candidate becomes strictly faster at R=7, where it costs 44 instead of 49. The minimum profitable integer reuse count is lfloorfrac307-2rfloor+1=7. Replacing the floor-plus-one rule with a ceiling would be wrong when the ratio is already an inte…
Exact mathematical passage
\left\lfloor\frac{30}{7-2}\right\rfloor+1=7.
\logUse in source · 18 passages
Search aliases: log
Use in source · pages.tex, line 342
vpage20Spend error where it is cheapestCONTINUOUS AND CLIPPED PRECISION Take sensitivities s=(1,2,4), cost coefficients c=(1,1,2), and budget τ=1. Minimizing ∑_ec_elog(1/ϵ_e) subject to ∑_es_eϵ_e=1 gives ϵ_e=fracc_e4s_e, ϵ=(1/4,1/8,1/8). vfigprecision headThe budget is weighted The three contributions to the error allowance are (1/4,1/4,1/2). Equal cost coefficients give equal weighted contributions, not equal tolerances. The third edge has twice the cost coefficient of the second, which offsets…
Exact mathematical passage
\sum_ec_e\log(1/\epsilon_e)
\log_2Use in source · 4 passages
Search aliases: log_2 log
Use in source · Read passage · 01_manuscript.tex, line 643
H_N · -H_N . Then H_NH_N^T=NI. The product H_Nx can be evaluated in exactly Nlog_2N additions or subtractions through the recursion.
Exact mathematical passage
N\log_2N
\LongleftrightarrowUse in source · 2 passages
Search aliases: Longleftrightarrow
Use in source · Read passage · 02_supplement_02.tex, line 73
Strict half-gap stability. At a solved internal node write K=κ_v+λϵ_v, S=σ_v+∑_wF_λ(w), g=|K-S|. If the updated scores satisfy |K'-K|<g/2, |S'-S|<g/2, Lean proves K'≤ S' Longleftrightarrow K≤ S. Applying this condition recursively at every internal node proves equality of the entire returned cuts before and after perturbation, not only equality of one numerical minimum.
Exact mathematical passage
K'\leq S'\quad\Longleftrightarrow\quad K\leq S.
\LongrightarrowUse in source · 6 passages
Search aliases: Longrightarrow
Use in source · Read passage · 02_supplement_01.tex, line 64
Nilpotence certificate. If no walk of length h+1 exists between any pair of packets, then A^h+1=0. Every block vanishes, and reconstruction forces the whole operator to vanish. A second theorem derives the required absence of walks from a supplied level map ℓ:I⟶{0,…,h}, A_bc≠0 Longrightarrow ℓ(c)<ℓ(b). It proves that a length-n walk increases the level by at least n. This certifies the acyclic case with a bounded height. This block does not implement or verify an algorithm that constructs a topo…
Exact mathematical passage
\ell:I\longrightarrow\{0,\ldots,h\},\qquad A_{bc}\ne0\ \Longrightarrow\ \ell(c)<\ell(b).
\operatorname{lookup}Use in source · 1 passage
Search aliases: lookup operatorname
Use in source · Read passage · 02_supplement_03.tex, line 202
Entrywise implementation theorem. For every b≤ B, Lean proves lookup(storedTree(B,t),b)=F_t(b). There is a parallel theorem for stored discrete precision. The proof establishes first that an output entry at b depends only on input entries at most b, then proceeds through cached convolution and the tree recursion. Consequently the optimality, attainment, and infeasibility specification applies to the actual stored result.
Exact mathematical passage
\operatorname{lookup}(\operatorname{storedTree}(B,t),b)=F_t(b).
M^*Use in source · 2 passages
Search aliases: M^*
Use in source · Read passage · 02_supplement_05.tex, line 171
Lean identifies the actual Frobenius norm with its trace expression, normM_rm F^2=Retr(M^*M), then expands the finite sum and removes each cross term. The result is normA-Π(A)_rm F^2 =∑_b≠ anormP_bAP_a_rm F^2. No Pythagorean equality for the target residual is assumed.
Exact mathematical passage
\norm{M}_{\rm F}^2=\operatorname{Re}\operatorname{tr}(M^*M),
M^{-1/2}Use in source · 1 passage
Search aliases: M^-1/2
Use in source · Read passage · 01_manuscript.tex, line 1211
[Preconditioned conjugate gradient convergence] Let A and M be Hermitian positive definite on a nonempty finite coordinate space, and set κ = cond_2(M^-1/2AM^-1/2). In exact arithmetic, the energy norm error of preconditioned conjugate gradient satisfies normx_m-x_*_A ≤ 2 ( fracsqrtκ-1sqrtκ+1 )^m normx_0-x_*_A.
Exact mathematical passage
\kappa = \cond_2(M^{-1/2}AM^{-1/2}).
m^2Use in source · 11 passages
Search aliases: m^2
Use in source · pages.tex, line 119
9+12 i · 6. Replacing A^T with A^* changes the signs of the imaginary terms. A real-only test would conceal that error, which is why the worked model uses an imaginary coefficient. headWhy modewise contraction matters For A of order m and B of order n, the explicit tensor matrix stores m^2n^2 entries. Its factors store m^2+n^2. Applying through BXA^T costs 2n^2m+2nm^2 modeled operations, rather than forming the full tensor matrix first.
Exact mathematical passage
m^2n^2
M^{j-1}Use in source · 7 passages
Search aliases: M^j-1
Use in source · pages.tex, line 292
With A=diag(1,2), the vectors v and Av=(1,2)^T are independent. Two polynomial samples now distinguish the packet labels. Alternatively, allowing the packet projectors as extra address operators separates the coordinates even when A=I_2. Both changes enlarge the addressing power, but in different ways. headWhen invariance is only approximate If A=D+K with D packet diagonal, then A^j-D^j ≤ jM^j-1 K , M=max{ A , D }. This controls departure from the decoupled iterate. It is not a claim that the di…
Exact mathematical passage
\|A^j-D^j\|\le jM^{j-1}\|K\|, \qquad M=\max\{\|A\|,\|D\|\}.
m_eUse in source · 24 passages
Search aliases: m_e
Use in source · pages.tex, line 352
Now impose upper tolerances m=(1/10,1/2,1/2). The first edge must stop at its tighter cap. With multiplier λ=10/3, the clipped solution is ϵ_e=min{m_e,c_e/(λ s_e)}, ϵ=(1/10,3/20,3/20). Its weighted contributions are (1/10,3/10,3/5) and still sum to one. The checker verifies the active cap, the multiplier equations, and the budget exactly. headFrom tolerance to a real data type The continuous objective is a model. Actual kernels offer discrete precisions and may use different accumulation types. …
Exact mathematical passage
\epsilon_e=\min\{m_e,c_e/(\lambda s_e)\},\qquad \epsilon=(1/10,3/20,3/20).
M_{ij}Use in source · 1 passage
Search aliases: M_ij
Use in source · Read passage · 02_supplement_06.tex, line 172
W^*W=I, Z^*Z=I, σ_1≥⋯≥σ_d≥0. These are precisely the fields of the formal SVD structure. It contains no rank-optimality or tail-error field. A derived theorem checks the Gram eigenvector equation G^*GZ=Zdiag(σ_1^2,…,σ_d^2). Let h=min(d,k) and retain exactly the first h column indices. The rectangular square-root mask has entries M_ij=sqrtσ_j, · i=j, 0, · i≠ j, 1≤ i≤ d, 1≤ j≤ h. Lean proves its index cardinality and MM^*=D_k. Define L_k=Q_LWM, R_k=Q_RZM. The product equals Q_LWD_kZ^*Q_R^* and has…
Exact mathematical passage
M_{ij}=\begin{cases}\sqrt{\sigma_j},&i=j,\\0,&i\ne j,\end{cases} \qquad 1\leq i\leq d,\quad 1\leq j\leq h.
M_{\rm block}Use in source · 1 passage
Search aliases: M_rm block rm
Use in source · Read passage · 01_manuscript.tex, line 284
[Equal diagonal packets] Let p≥1 and b≥1 be integers and n=pb. Suppose n is divided into p equal diagonal packets of width b. Then flop_rm block=2qnb, M_rm block=nb. The dense-to-block storage reduction factor is the following; the arithmetic reduction factor has the same value when q≥1: fracnb.
Exact mathematical passage
\flop_{\rm block}=2qnb, \qquad M_{\rm block}=nb.
M_{\rm packet}Use in source · 1 passage
Search aliases: M_rm packet rm
Use in source · Read passage · 01_manuscript.tex, line 267
[Dense and packet application counts] A dense n by n matrix applied to q right hand sides has modeled cost flop_rm dense=2n^2q. A packet representation with active edge set E has modeled cost flop_rm packet = 2q ∑_(β,α)∈ E n_β n_α. It stores M_rm packet = ∑_(β,α)∈ E n_β n_α scalar entries.
Exact mathematical passage
\boxed{ M_{\rm packet} = \sum_{(\beta,\alpha)\in E} n_\beta n_\alpha }
\matUse in source · 38 passages
Search aliases: mat
Use in source · extended_pages.tex, line 3
vpage29A complex range needs an adjointTHE INNER PRODUCT IS PART OF THE CONTRACT The earlier compact example is real. Over C, an isometry means Q^*Q=I, with conjugation. Take the exact Gaussian-rational factors Q_L=mat3/5 · 0 4 i/5 · 0 0 · 1, Q_R=mat1 · 0 0 · i, G=diag(5,2). Their product is a genuinely complex rectangular operator: A=Q_LGQ_R^*=mat3 · 0
Exact mathematical passage
Q_L=\mat{3/5&0\\4\mathrm i/5&0\\0&1},\qquad Q_R=\mat{1&0\\0&\mathrm i},\qquad G=\diag(5,2).
\max_{v\in V}Use in source · 1 passage
Search aliases: max_v∈ V max in
Use in source · Read passage · 02_supplement_05.tex, line 31
A valid schedule s:V→ N assigns positive stages and strictly increases on each edge. Induction proves the stronger pointwise statement ℓ(v)≤ s(v) for every v. Consequently the constructed schedule achieves the least possible makespan, d=max_v∈ Vℓ(v). For a nonempty graph, d equals the number of vertices on an attained longest path. Vertices of the same level have no dependency edge in either direction.
Exact mathematical passage
\boxed{d=\max_{v\in V}\ell(v).}
\minUse in source · 26 passages
Search aliases: min
Use in source · pages.tex, line 352
Now impose upper tolerances m=(1/10,1/2,1/2). The first edge must stop at its tighter cap. With multiplier λ=10/3, the clipped solution is ϵ_e=min{m_e,c_e/(λ s_e)}, ϵ=(1/10,3/20,3/20). Its weighted contributions are (1/10,3/10,3/5) and still sum to one. The checker verifies the active cap, the multiplier equations, and the budget exactly. headFrom tolerance to a real data type The continuous objective is a model. Actual kernels offer discrete precisions and may use different accumulation types. …
Exact mathematical passage
\epsilon_e=\min\{m_e,c_e/(\lambda s_e)\},\qquad \epsilon=(1/10,3/20,3/20).
\min_{(w,c)\text{ allowed},\;w\leq b}Use in source · 1 passage
Search aliases: min_(w,c) allowed, w≤ b min text leq
Use in source · Read passage · 02_supplement_03.tex, line 153
If D is the already-solved tail table, the next edge has exact-weight recurrence F(b)=min_(w,c) allowed, w≤ b{D(b-w)+c}. The final answer is min_0≤ b≤ BF(b). Lean proves that this is the exact minimum over legal assignments with total error at most B, or codenone exactly when no such assignment exists.
Exact mathematical passage
F(b)=\min_{(w,c)\text{ allowed},\;w\leq b}\{D(b-w)+c\}.
\min_{0\leq b\leq B}Use in source · 2 passages
Search aliases: min_0≤ b≤ B min leq leq
Use in source · Read passage · 02_supplement_03.tex, line 156
If D is the already-solved tail table, the next edge has exact-weight recurrence F(b)=min_(w,c) allowed, w≤ b{D(b-w)+c}. The final answer is min_0≤ b≤ BF(b). Lean proves that this is the exact minimum over legal assignments with total error at most B, or codenone exactly when no such assignment exists.
Exact mathematical passage
\min_{0\leq b\leq B}F(b)
\min_{0\leq k\leq b}Use in source · 1 passage
Search aliases: min_0≤ k≤ b min leq leq
Use in source · Read passage · 02_supplement_03.tex, line 145
For exact-weight tables F,G, min-plus convolution is (F⋆ G)(b)=min_0≤ k≤ b{F(k)+G(b-k)}, with an infeasible sum whenever either operand is infeasible. Its correctness theorem identifies every finite answer with an actual feasible pair of component weights and costs. Repeated convolution handles arbitrary finite lists of components.
Exact mathematical passage
(F\star G)(b)=\min_{0\leq k\leq b}\{F(k)+G(b-k)\},
\min_eUse in source · 1 passage
Search aliases: min_e min
Use in source · Read passage · 01_manuscript.tex, line 1409
upper bound. The weighted sum ϕ(λ) = ∑_es_e min{ m_e, fracc_eλ s_e } is continuous and nonincreasing. It decreases from the maximal feasible value to zero. For a strictly submaximal budget, once any coordinate is unsaturated, ϕ is strictly decreasing, which gives a unique multiplier. At maximal budget the only feasible vector is (m_e)_e and every 0<λ≤min_e c_e/(s_em_e) produces it. Thus the minimizing vector is unique at this endpoint while its multiplier need not be. Strict convexity gives uniq…
Exact mathematical passage
0<\lambda\le\min_e c_e/(s_em_e)
\min_{\eps_e>0}Use in source · 1 passage
Search aliases: min_eps_e>0 min_ε_e>0 min eps min varepsilon
Use in source · Read passage · 01_manuscript.tex, line 1337
[Continuous precision allocation] Let E be a finite nonempty edge set, let s_e>0 be sensitivities and c_e>0 be costs, and let τ>0. Consider min_ε_e>0 ∑_e c_elogfrac1ε_e subject to ∑_es_eε_e=τ. The unique minimizer is ε_e = fracτ c_es_e∑_fc_f.
Exact mathematical passage
\min_{\eps_e>0} \sum_e c_e\log\frac1{\eps_e}
\min_kUse in source · 1 passage
Search aliases: min_k min
Use in source · Read passage · 01_manuscript.tex, line 1438
ct discrete precision program] Let the total integer error budget be B. Define F_j(b) as the minimum cost for assigning precisions to the first j edges with total weight exactly b. Then F_0(0)=0, F_0(b)=+∞ for b≠0. Set F_j-1(u)=+∞ whenever u<0 or the state is unreachable. Then F_j(b) = min_k { F_j-1(b-w_j,k)+c_j,k }. The minimum of F_m(b) over 0≤ b≤ B is the exact minimum cost for the quantized model.
Exact mathematical passage
\boxed{ F_j(b) = \min_k \left\{ F_{j-1}(b-w_{j,k})+c_{j,k} \right\}. }
\min_{\varepsilon_e>0}Use in source · 1 passage
Search aliases: min_ε_e>0 min varepsilon
Use in source · Read passage · 02_supplement_02.tex, line 120
The continuous allocation theorem is complete. textitSource item 8.4, with an explicit positive budget Let E be a finite nonempty edge set. Assume s_e>0, c_e>0, τ>0, and define C=∑_ec_e. The optimization problem is min_ε_e>0∑_ec_elogfrac1ε_e subject to ∑_es_eε_e=τ. Lean constructs the displayed candidate ε_e^*=fracτ c_es_eC and proves strict positivity and the exact budget equality. It then proves global optimality and uniqueness for the actual logarithmic objective.
Exact mathematical passage
\min_{\varepsilon_e>0}\sum_ec_e\log\frac1{\varepsilon_e} \qquad\text{subject to}\qquad \sum_es_e\varepsilon_e=\tau.
\min_{\sum_i b_i=b}Use in source · 1 passage
Search aliases: min_∑_i b_i=b min sum
Use in source · Read passage · 02_supplement_03.tex, line 168
Keeping node v costs κ_v and incurs integer error w_v. Splitting costs overhead σ_v in addition to the selected child cuts. The exact-weight recurrence is F_v(b)=min{ l κ_v if b=w_v, [2pt] displaystyleσ_v+ min_∑_i b_i=b∑_iF_v_i(b_i) if v is internal. . Nonexistent alternatives are infeasible. The child sum is implemented by the proved finite min-plus convolution, not by assuming an optimal allocation.
Exact mathematical passage
F_v(b)=\min\left\{ \begin{array}{l} \kappa_v\quad\text{if }b=w_v,\\[2pt] \displaystyle\sigma_v+ \min_{\sum_i b_i=b}\sum_iF_{v_i}(b_i)\quad\text{if }v\text{ is internal}. \end{array}\right.
\mathbb NUse in source · 1 passage
Search aliases: N mathbb
Use in source · Read passage · 02_supplement_05.tex, line 26
A valid schedule s:V→ N assigns positive stages and strictly increases on each edge. Induction proves the stronger pointwise statement ℓ(v)≤ s(v) for every v. Consequently the constructed schedule achieves the least possible makespan, d=max_v∈ Vℓ(v). For a nonempty graph, d equals the number of vertices on an attained longest path. Vertices of the same level have no dependency edge in either direction.
Exact mathematical passage
s:V\to\mathbb N
\nUse in source · 2 passages
Search aliases: n
Use in source · complex_core.tex, line 7
[font=sffamilysmall] node[anchor=west] at (0,3.4) A^*A=diag(25,4); draw[->] (0,0)--(8,0) node[right] retained singular directions; draw[->] (0,0)--(0,2.5) node[above] squared error; draw[thick,black!65] (1,2.03)--(3.5,.28)--(6,0); foreach x/y/n/v in 1/2.03/0/29,3.5/.28/1/4,6/0/2/0 fill (x,y) circle (2pt);node[below] at (x,0) n;node[above] at (x,y) v; node[anchor=west] at (0,-.9) Q_L^*Q_L=I Q_L^TQ_L=diag(-7/25,1);
Exact mathematical passage
\n
n^{-1/2}Use in source · 1 passage
Search aliases: n^-1/2
Use in source · Read passage · 02_supplement_04.tex, line 22
Normalization and inversion. The proof identifies matrix application with the ordinary counting-measure discrete Fourier transform: Fx=n^-1/2DFT(x). It separately identifies F^*x with the appropriately scaled inverse transform. Mathlib's checked Fourier inversion theorem and the proved scalar normalization yield both identities F^*F=I, FF^*=I. No unitarity assumption is inserted.
Exact mathematical passage
Fx=n^{-1/2}\operatorname{DFT}(x).
n^2Use in source · 24 passages
Search aliases: n^2
Use in source · pages.tex, line 55
vpage4Work removed is not time measuredARITHMETIC, STORAGE, AND CAPACITY For q right-hand sides, the paper models dense application by 2n^2q operations. A block-diagonal-plus-rank-r representation with equal block width b uses F=2qn(b+2r), M=n(b+2r) stored scalars. These formulas count the dense block products and two thin products. They follow the manuscript convention rather than adding every scalar accumulation or index operation. vfigwork_storage For the running example, n=6,b=2,r=1. The mod…
Exact mathematical passage
2n^2q
n_eUse in source · 4 passages
Search aliases: n_e
Use in source · Read passage · 02_supplement_02.tex, line 161
Explicit storage and operation models. textitSource items 3.1, 3.2, 3.4, 3.6, 3.8--3.10, and 4.2 For a finite collection of rectangular blocks, the scalar addresses are actual elements of coprod_e∈ E(Fin(m_e)×Fin(n_e)). The corresponding operation addresses include a right-hand-side index and one of two scalar-operation tags. Lean proves their cardinalities rather than merely defining a formula named ``cost.''
Exact mathematical passage
\coprod_{e\in E}\bigl(\operatorname{Fin}(m_e)\times\operatorname{Fin}(n_e)\bigr).
n_i^2Use in source · 2 passages
Search aliases: n_i^2
Use in source · Read passage · 02_supplement_02.tex, line 175
BDLR · ∑_i n_i^2+2nr · 2q(∑_i n_i^2+2nr)
Exact mathematical passage
\sum_i n_i^2+2nr
N_{p,w}Use in source · 5 passages
Search aliases: N_p,w
Use in source · Read passage · 01_manuscript.tex, line 465
[Equal block banded count] Let p,b≥1 be integers, n=pb, and let p=n/b equal packets have width b. Suppose a block is active precisely when absi-j≤ w, 0≤ w≤ p-1. The number of active blocks is N_p,w = p(2w+1)-w(w+1). For q right hand sides, the modeled application cost and scalar storage are 2qb^2N_p,w and b^2N_p,w.
Exact mathematical passage
\boxed{ N_{p,w} = p(2w+1)-w(w+1). }
n_\alpha^2Use in source · 4 passages
Search aliases: n_α^2 alpha
Use in source · Read passage · 01_manuscript.tex, line 321
[BDLR application and storage] For q right hand sides, AZ=DZ+U(V^*Z) has modeled cost 2q ( ∑_α n_α^2+2nr ). The representation stores ∑_α n_α^2+2nr scalars. For n=pb, with integers p,b≥1, the ideal dense storage reduction factor is the following; the arithmetic reduction factor has the same value when q≥1: fracnb+2r.
Exact mathematical passage
\boxed{ 2q \left( \sum_\alpha n_\alpha^2+2nr \right). }
n_\betaUse in source · 6 passages
Search aliases: n_β beta
Use in source · Read passage · 01_manuscript.tex, line 257
[Dense and packet application counts] A dense n by n matrix applied to q right hand sides has modeled cost flop_rm dense=2n^2q. A packet representation with active edge set E has modeled cost flop_rm packet = 2q ∑_(β,α)∈ E n_β n_α. It stores M_rm packet = ∑_(β,α)∈ E n_β n_α scalar entries.
Exact mathematical passage
\boxed{ \flop_{\rm packet} = 2q \sum_{(\beta,\alpha)\in E} n_\beta n_\alpha. }
\NNUse in source · 3 passages
Search aliases: NN N mathbb
Use in source · Read passage · 01_manuscript.tex, line 1658
[Quantized budget dynamic program] Assume every keep error is a nonnegative integer w_v. Let B∈ N. For each node v, define F_v(b) as the minimum cost of a pruning of the subtree at v with total error exactly b. The keep option contributes F_v(w_v)≤κ_v. The split option is obtained by min plus convolution of the child tables, followed by addition of σ_v. The smallest value among F_rm root(b), 0≤ b≤ B, is the exact minimum cost among prunings whose quantized error is at most B.
Exact mathematical passage
B\in\NN
\normUse in source · 48 passages
Search aliases: norm
Use in source · Read passage · 01_manuscript.tex, line 765
[Inverse norm bound] Under the hypotheses of [source reference], norm(D+UV^*)^-1 ≤ normD^-1 + normD^-1U normS^-1 normV^*D^-1.
Exact mathematical passage
\norm{(D+UV^*)^{-1}} \leq \norm{D^{-1}} + \norm{D^{-1}U} \norm{S^{-1}} \norm{V^*D^{-1}}.
\notUse in source · 2 passages
Search aliases: not
Use in source · Read passage · 02_supplement_05.tex, line 85
Fix the source packet a. In the regular representation, define the submodule W_a={x:P_bx=0 whenever anotleadsto b}. It contains P_a. Completeness of the projectors expands each component of Ax into packet transfers. If a transfer into an unreachable packet were nonzero from a reachable packet, it would extend a path, a contradiction. Thus AW_a⊆ W_a.
Exact mathematical passage
W_a=\{x:P_bx=0\ \text{whenever }a\not\leadsto b\}.
OUse in source · 4 passages
Search aliases: O
Use in source · Read passage · 01_manuscript.tex, line 443
[Materialization cost] Materializing D+UV^* as an explicit matrix requires storage for n^2 scalars and modeled arithmetic 2n^2r+O(n^2) when the low rank correction is formed by dense multiplication.
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2n^2r+O(n^2)
\mathcal OUse in source · 2 passages
Search aliases: O mathcal
Use in source · Read passage · 02_supplement_05.tex, line 140
Arithmetic events. For q right-hand sides, the declared event type adds a right-hand-side index and one of two operation roles, multiplication or accumulation: O={0,1}×{0,…,q-1}× S. Hence | O|=2qb^2N_p,w. These identities include zero right-hand-side count. The abstract storage formula also admits zero block width. To use the manuscript's substitution p=n/b, one must separately require b>0 and n=pb.
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\mathcal O=\{0,1\}\times\{0,\ldots,q-1\}\times\mathcal S.
oUse in source · 6 passages
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Use in source · Read passage · 01_manuscript.tex, line 284
[Equal diagonal packets] Let p≥1 and b≥1 be integers and n=pb. Suppose n is divided into p equal diagonal packets of width b. Then flop_rm block=2qnb, M_rm block=nb. The dense-to-block storage reduction factor is the following; the arithmetic reduction factor has the same value when q≥1: fracnb.
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\flop_{\rm block}=2qnb, \qquad M_{\rm block}=nb.
\onesUse in source · 6 passages
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Use in source · pages.tex, line 39
1 · 3 and put u=v=ones∈QQ^6. Then A=D+uv^T has diagonal entries four, within-block off-diagonal entries two, and every cross-block entry one. All 36 entries are nonzero. x=(1,-1,2,0,-2,3)^T, v^Tx=3, Ax=Dx+3u. vfigfactored_application headWhat was avoided? The global part does not require 36 independent coefficients. It depends on the input through one scalar, v^Tx, which is broadcast through u. The diagonal part is computed by three separate local matrix products. Nothing is approximated and no …
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u=v=\ones\in\QQ^6
\ones^TUse in source · 2 passages
Search aliases: ones^T ones
Use in source · pages.tex, line 322
vpage19Twelve small blocks, one coherent errorDISCOVERY AND PERTURBATION CERTIFICATES Split QQ^4 into scalar packets and let A=I+E, E=frac1100(onesones^T-I). Dropping every off-diagonal block leaves I. Each discarded entry is only 1/100, but the total error is not bounded by that single-entry size. vfigblock_error headTwo norms answer different questions There are twelve omitted entries, so E _F^2=12/10^4. The all-one vector is an eigenvector with eigenvalue 3/100, while the three orthogonal dir…
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A=I+E,\qquad E=\frac1{100}(\ones\ones^T-I).
pUse in source · 56 passages
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Use in source · extended_pages.tex, line 53
bottomrule vfig[.45]shared_family With reuse weights (r_1,r_2)=(1,3), the aggregated objectives are J_rm keep=4+2+3·7=27, J_rm split=2+8+3·2=16. The exact shared optimum is split. Operator 1 alone prefers keep on its per-use costs, while operator 2 prefers split. Choosing independently and trying to combine those decisions would not solve the declared one-pruning problem. headA hand-computable decision threshold Fix r_1=1 and vary r_2=r≥0. Keeping costs 6+7r and splitting costs 10+2r. Splitting …
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J_{\rm keep}=4+2+3\cdot7=27,\qquad J_{\rm split}=2+8+3\cdot2=16.
p^2Use in source · 1 passage
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Use in source · Read passage · 02_supplement_05.tex, line 128
s N_p,w=| B_p,w|=p(2w+1)-w(w+1). The proof uses finite prefix counting, the newly exposed boundary row and column when p increases, and the saturated initial band. An intermediate subtraction-free identity avoids relying on natural-number subtraction without the required inequality. The diagonal case w=0 gives exactly p blocks, and the saturated case w=p-1 gives p^2.
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p^2
P_0Use in source · 2 passages
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Use in source · Read passage · 01_manuscript.tex, line 901
[Certified repeated recompression] Let P_0=I, P_j=A_jP_j-1, be exact partial products. Suppose an implemented chain satisfies widehat P_0=I, widehat P_j = A_jwidehat P_j-1+E_j, where E_j is the additive error introduced after the jth multiplication, for example by factor recompression. Then normwidehat P_m-P_m ≤ ∑_j=1^m ( ∏_i=j+1^mnormA_i ) normE_j. If normE_j≤δ_j, the same formula with δ_j gives a certified final error.
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P_0=I, \qquad P_j=A_jP_{j-1},
p_0Use in source · 2 passages
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Use in source · pages.tex, line 216
vpage13Errors enter at different placesFACTOR PERTURBATION VERSUS RECOMPRESSION Let the exact scalar factors be (2,3,4) and start from p_0=1. If additive errors (1/10,1/5,1/20) enter after each partial multiplication, then widehat p_j=a_jwidehat p_j-1+e_j, widehat p_3-p_3=12e_1+4e_2+e_3=frac4120. vfigerror_chain headThe same local numbers, a different model If instead the factors themselves are perturbed, the computed product is (4+e_3)(3+e_2)(2+e_1). Its error is 402/125, not 41/20. The factor-…
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p_0=1
p_3Use in source · 2 passages
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Use in source · pages.tex, line 217
vpage13Errors enter at different placesFACTOR PERTURBATION VERSUS RECOMPRESSION Let the exact scalar factors be (2,3,4) and start from p_0=1. If additive errors (1/10,1/5,1/20) enter after each partial multiplication, then widehat p_j=a_jwidehat p_j-1+e_j, widehat p_3-p_3=12e_1+4e_2+e_3=frac4120. vfigerror_chain headThe same local numbers, a different model If instead the factors themselves are perturbed, the computed product is (4+e_3)(3+e_2)(2+e_1). Its error is 402/125, not 41/20. The factor-…
Exact mathematical passage
\widehat p_j=a_j\widehat p_{j-1}+e_j,\qquad \widehat p_3-p_3=12e_1+4e_2+e_3=\frac{41}{20}.
P_aUse in source · 19 passages
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Use in source · Read passage · 02_supplement_01.tex, line 5
Exact packet semantics. textitSource items 2.3, 2.5, and 2.6 Let R be a ring, I a finite index type, and (P_a)_a∈ I a family satisfying P_bP_a=P_a · b=a,
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(P_a)_{a\in I}
P_{a(e)}Use in source · 5 passages
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Use in source · Read passage · 02_supplement_01.tex, line 127
Packet perturbation certificates. textitSource item 8.1 and two uniform consequences Let X be a normed space over a nontrivially normed field. The formal objects are actual continuous linear maps X→ X, with their operator norm. Let a(e) identify the source packet of edge e, and suppose Δ_e=Δ_e P_a(e). Set A=∑_e A_e, widehat A=∑_e(A_e+Δ_e). The exact operator and vector identities are proved first: widehat A-A=∑_eΔ_e P_a(e), (widehat A-A)x=∑_eΔ_e(P_a(e)x). Applying the finite triangle inequality …
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\Delta_e=\Delta_e P_{a(e)}.
P_bUse in source · 11 passages
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Use in source · Read passage · 02_supplement_01.tex, line 6
Exact packet semantics. textitSource items 2.3, 2.5, and 2.6 Let R be a ring, I a finite index type, and (P_a)_a∈ I a family satisfying P_bP_a=P_a · b=a, 0 · b≠ a, ∑_aP_a=1. Write A_ba=P_bAP_a and let E_A={(b,a):A_ba≠0}. These are exact algebraic zeros, not values discarded by a floating-point threshold.
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P_bP_a=\begin{cases}P_a&b=a,\\0&b\ne a,\end{cases} \qquad \sum_aP_a=1.
P_cUse in source · 2 passages
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Use in source · Read passage · 02_supplement_05.tex, line 165
Hilbert--Schmidt orthogonality. For distinct ordered packet pairs (b,a)≠(d,c), adjoint multiplication and projector orthogonality give tr((P_bAP_a)^*P_dBP_c)=0. If b≠ d, the central projector product vanishes. Otherwise a≠ c, and cyclicity of trace brings P_cP_a together. This is proved for arbitrary input matrices A,B.
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\operatorname{tr}\bigl((P_bAP_a)^*P_dBP_c\bigr)=0.
P_dUse in source · 1 passage
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Use in source · Read passage · 02_supplement_05.tex, line 165
Hilbert--Schmidt orthogonality. For distinct ordered packet pairs (b,a)≠(d,c), adjoint multiplication and projector orthogonality give tr((P_bAP_a)^*P_dBP_c)=0. If b≠ d, the central projector product vanishes. Otherwise a≠ c, and cyclicity of trace brings P_cP_a together. This is proved for arbitrary input matrices A,B.
Exact mathematical passage
\operatorname{tr}\bigl((P_bAP_a)^*P_dBP_c\bigr)=0.
P_iUse in source · 9 passages
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Use in source · Read passage · 02_supplement_03.tex, line 96
Exact and approximate packet Krylov spaces. textitSource items 7.2 and 7.3 Write K_m(A,v)=span{v,Av,…,A^m-1v}. For a complete finite projector family P_i, suppose A commutes with every P_i. Let A_i=P_iAP_i. Lean proves the actual orbit identity A^jv=∑_i A_i^jP_iv and therefore K_m(A,v)⊆ ⊕_i K_m(A_i,P_iv). The case j=0 uses ∑_iP_i=I. The zeroth local power is the ambient identity, applied to P_iv, rather than an incorrectly substituted local projector power.
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P_i
P_jUse in source · 5 passages
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Use in source · Read passage · 01_manuscript.tex, line 901
[Certified repeated recompression] Let P_0=I, P_j=A_jP_j-1, be exact partial products. Suppose an implemented chain satisfies widehat P_0=I, widehat P_j = A_jwidehat P_j-1+E_j, where E_j is the additive error introduced after the jth multiplication, for example by factor recompression. Then normwidehat P_m-P_m ≤ ∑_j=1^m ( ∏_i=j+1^mnormA_i ) normE_j. If normE_j≤δ_j, the same formula with δ_j gives a certified final error.
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P_0=I, \qquad P_j=A_jP_{j-1},
P_{j+1}Use in source · 2 passages
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Use in source · Read passage · 02_supplement_03.tex, line 35
Actual noncommutative products. The error module defines the left-accumulated product recursively, with P_0=I and P_j+1=A_j+1P_j. For a perturbed sequence widehat A_j, it derives the discrepancy recursion widehat P_j+1-P_j+1 =widehat A_j+1(widehat P_j-P_j) +(widehat A_j+1-A_j+1)P_j. The identity follows from the actual products. It is not supplied as an assumption to a norm estimate, and it does not commute the operators. Lean uses zero-based indices internally, so index j in source code corresp…
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P_{j+1}=A_{j+1}P_j
P_{j,a(e)}Use in source · 2 passages
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Use in source · Read passage · 02_supplement_03.tex, line 75
Packetwise mixed precision. At each stage let A_j=∑_e A_j,e, widehat A_j=∑_e(A_j,e+Δ_j,e), Δ_j,e=Δ_j,eP_j,a(e). The previous packet certificate supplies the uniform operator bound normwidehat A_j-A_j≤ δ_j:=∑_enormΔ_j,enormP_j,a(e). The new theorem substitutes these δ_j into the actual product-chain estimate. The resulting vector error is at most the operator certificate times normx. A vector-specific packet bound is not silently reused as an operator norm.
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A_j=\sum_e A_{j,e},\qquad \widehat A_j=\sum_e(A_{j,e}+\Delta_{j,e}),\qquad \Delta_{j,e}=\Delta_{j,e}P_{j,a(e)}.
P_{j-1}Use in source · 2 passages
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Use in source · Read passage · 01_manuscript.tex, line 901
[Certified repeated recompression] Let P_0=I, P_j=A_jP_j-1, be exact partial products. Suppose an implemented chain satisfies widehat P_0=I, widehat P_j = A_jwidehat P_j-1+E_j, where E_j is the additive error introduced after the jth multiplication, for example by factor recompression. Then normwidehat P_m-P_m ≤ ∑_j=1^m ( ∏_i=j+1^mnormA_i ) normE_j. If normE_j≤δ_j, the same formula with δ_j gives a certified final error.
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P_0=I, \qquad P_j=A_jP_{j-1},
P_mUse in source · 5 passages
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Use in source · Read passage · 01_manuscript.tex, line 915
ntheorem[Certified repeated recompression] Let P_0=I, P_j=A_jP_j-1, be exact partial products. Suppose an implemented chain satisfies widehat P_0=I, widehat P_j = A_jwidehat P_j-1+E_j, where E_j is the additive error introduced after the jth multiplication, for example by factor recompression. Then normwidehat P_m-P_m ≤ ∑_j=1^m ( ∏_i=j+1^mnormA_i ) normE_j. If normE_j≤δ_j, the same formula with δ_j gives a certified final error.
Exact mathematical passage
\boxed{ \norm{\widehat P_m-P_m} \leq \sum_{j=1}^{m} \left( \prod_{i=j+1}^{m}\norm{A_i} \right) \norm{E_j}. }
P_{m-1}Use in source · 1 passage
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Use in source · Read passage · 02_supplement_03.tex, line 19
Construction. Suppose P_m-1=D^(m-1)+L_m-1R_m-1^*. The new factors are L_m · =[D_mL_m-1+U_m(V_m^*L_m-1) U_m],
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P_{m-1}=D^{(m-1)}+L_{m-1}R_{m-1}^*
P_{\alpha}Use in source · 1 passage
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Use in source · Read passage · 01_manuscript.tex, line 1330
Take the supremum over unit inputs in [source reference], using P_αx ≤ P_α x . Substitute the resulting uniform operator error bounds into [source reference]; the action estimate follows from submultiplicativity.
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\|P_{\alpha}x\|\le\|P_{\alpha}\|\|x\|
P_{\alpha(e)}Use in source · 6 passages
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Use in source · pages.tex, line 333
For orthogonal coordinate blocks, the squared Frobenius norms add exactly. Operator norms instead require a suitable bound or further spectral information. The maximum omitted block norm is generally insufficient. headInput sensitivity is additional information The packet perturbation estimate is (widehat A-A)x ≤∑_e E_e P_α(e)x . It can be much tighter for an input that scarcely occupies an affected source packet. To obtain a uniform operator bound, the input dependence must be bounded uniformly…
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\|(\widehat A-A)x\|\le\sum_e\|E_e\|\,\|P_{\alpha(e)}x\|.
P_{\alpha(j,e)}Use in source · 2 passages
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Use in source · Read passage · 01_manuscript.tex, line 1322
[Mixed precision product chain] If each fixed linear factor A_j is replaced by a fixed packetwise linear approximation widehat A_j, write E_j=widehat A_j-A_j=∑_eΔ_j,eP_α(j,e). A uniform certificate is E_j ≤δ_j:=∑_e Δ_j,e P_α(j,e) . For orthogonal packet projectors, δ_j≤∑_e Δ_j,e . The final operator error is bounded by [source reference] with these uniform δ_j. The corresponding vector error is this operator bound multiplied by the input norm.
Exact mathematical passage
E_j=\widehat A_j-A_j=\sum_e\Delta_{j,e}P_{\alpha(j,e)}
P_\gammaUse in source · 2 passages
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Use in source · Read passage · 01_manuscript.tex, line 182
Insert the packet identity between A and B: P_γ ABP_α = ∑_β P_γ AP_β BP_α. A summand can be nonzero only when both block addresses are active.
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P_\gamma ABP_\alpha = \sum_\beta P_\gamma AP_\beta BP_\alpha.
QUse in source · 49 passages
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Use in source · extended_pages.tex, line 2
vpage29A complex range needs an adjointTHE INNER PRODUCT IS PART OF THE CONTRACT The earlier compact example is real. Over C, an isometry means Q^*Q=I, with conjugation. Take the exact Gaussian-rational factors Q_L=mat3/5 · 0
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Q^*Q=I
\QUse in source · 2 passages
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Use in source · Read passage · 02_supplement_02.tex, line 33
Shared operator families. For reuse weights r_j, the local keep model can aggregate setup and per-operator costs: κ_v=κ_v^setup+∑_jr_jκ_v^(j), ϵ_v=∑_jr_jϵ_v^(j). Lean proves the aggregation identity at the cut level, including split overhead, before applying the compiler theorem. The proof is valid over an ordered field, including Q and R. The intended nonnegative model is a specialization.
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\Q
qUse in source · 61 passages
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Use in source · pages.tex, line 55
vpage4Work removed is not time measuredARITHMETIC, STORAGE, AND CAPACITY For q right-hand sides, the paper models dense application by 2n^2q operations. A block-diagonal-plus-rank-r representation with equal block width b uses F=2qn(b+2r), M=n(b+2r) stored scalars. These formulas count the dense block products and two thin products. They follow the manuscript convention rather than adding every scalar accumulation or index operation. vfigwork_storage For the running example, n=6,b=2,r=1. The mod…
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q
Q^*Use in source · 7 passages
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Use in source · extended_pages.tex, line 2
vpage29A complex range needs an adjointTHE INNER PRODUCT IS PART OF THE CONTRACT The earlier compact example is real. Over C, an isometry means Q^*Q=I, with conjugation. Take the exact Gaussian-rational factors Q_L=mat3/5 · 0
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Q^*Q=I
Q_cUse in source · 1 passage
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Use in source · Read passage · 02_supplement_01.tex, line 33
Adjoint and tensor products. With an involution and self-adjoint projectors, the formalization proves (A_ba)^*=(A^*)_ab. For actual finite matrices and actual Kronecker products, it proves (P_b⊗ Q_d)(A⊗ B)(P_a⊗ Q_c) =A_ba⊗ B_dc. The self-adjointness hypothesis is explicit in the adjoint theorem. The tensor theorem is not a statement about an uninterpreted tensor symbol.
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(P_b\otimes Q_d)(A\otimes B)(P_a\otimes Q_c) =A_{ba}\otimes B_{dc}.
Q_dUse in source · 1 passage
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Use in source · Read passage · 02_supplement_01.tex, line 33
Adjoint and tensor products. With an involution and self-adjoint projectors, the formalization proves (A_ba)^*=(A^*)_ab. For actual finite matrices and actual Kronecker products, it proves (P_b⊗ Q_d)(A⊗ B)(P_a⊗ Q_c) =A_ba⊗ B_dc. The self-adjointness hypothesis is explicit in the adjoint theorem. The tensor theorem is not a statement about an uninterpreted tensor symbol.
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(P_b\otimes Q_d)(A\otimes B)(P_a\otimes Q_c) =A_{ba}\otimes B_{dc}.
Q_{ij}Use in source · 1 passage
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Use in source · Read passage · 02_supplement_06.tex, line 128
Frobenius optimality from kernel weights. textitModules KernelFrame and FrobeniusOptimality For an arbitrary competitor B with rank B≤ k, construct an orthonormal matrix Q spanning its actual kernel. Then BQ=0, Q^*Q=I, q=dimker B=d-rank B. Define the row weights w_i=∑_j=1^q|Q_ij|^2. The proof derives 0≤ w_i≤1, ∑_iw_i=q≥max(d-k,0). The upper bound uses contraction of Q^* on a coordinate vector. The weight sum follows from the trace of Q^*Q and rank-nullity.
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w_i=\sum_{j=1}^{q}|Q_{ij}|^2.
Q_LUse in source · 29 passages
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Use in source · extended_pages.tex, line 3
vpage29A complex range needs an adjointTHE INNER PRODUCT IS PART OF THE CONTRACT The earlier compact example is real. Over C, an isometry means Q^*Q=I, with conjugation. Take the exact Gaussian-rational factors Q_L=mat3/5 · 0 4 i/5 · 0 0 · 1, Q_R=mat1 · 0 0 · i, G=diag(5,2). Their product is a genuinely complex rectangular operator: A=Q_LGQ_R^*=mat3 · 0
Exact mathematical passage
Q_L=\mat{3/5&0\\4\mathrm i/5&0\\0&1},\qquad Q_R=\mat{1&0\\0&\mathrm i},\qquad G=\diag(5,2).
Q_L^*Use in source · 5 passages
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Use in source · pages.tex, line 208
0 · 0 · 0. Keeping only its first singular direction gives a rank-one approximation with spectral and Frobenius error exactly two. Keeping none gives errors five and sqrt29; keeping both gives zero. vfigoptimality headWhy an outside competitor cannot do better For isometric outer factors, compression C(B)=Q_L^*BQ_R cannot increase rank or either matrix norm. Every ambient rank-k competitor therefore supplies an admissible core competitor. The lifted core residual has exactly the core residual no…
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C(B)=Q_L^*BQ_R
Q_L^TUse in source · 3 passages
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Use in source · extended_pages.tex, line 14
0 · -2 i, A^*A=diag(25,4). vfigcomplex_core headThe small calculation exposes a wrong convention The first column of Q_L has Hermitian squared length 9/25+16/25=1. Its bilinear square without conjugation is 9/25-16/25=-7/25. Thus Q_L^TQ_L is not the identity. The portable negative control replaces the adjoint by transpose and rejects the resulting isometry claim.
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Q_L^TQ_L
Q_RUse in source · 12 passages
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Use in source · extended_pages.tex, line 3
vpage29A complex range needs an adjointTHE INNER PRODUCT IS PART OF THE CONTRACT The earlier compact example is real. Over C, an isometry means Q^*Q=I, with conjugation. Take the exact Gaussian-rational factors Q_L=mat3/5 · 0 4 i/5 · 0 0 · 1, Q_R=mat1 · 0 0 · i, G=diag(5,2). Their product is a genuinely complex rectangular operator: A=Q_LGQ_R^*=mat3 · 0
Exact mathematical passage
Q_L=\mat{3/5&0\\4\mathrm i/5&0\\0&1},\qquad Q_R=\mat{1&0\\0&\mathrm i},\qquad G=\diag(5,2).
Q_R^*Use in source · 5 passages
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Use in source · extended_pages.tex, line 8
0 · i, G=diag(5,2). Their product is a genuinely complex rectangular operator: A=Q_LGQ_R^*=mat3 · 0 4 i · 0 0 · -2 i, A^*A=diag(25,4). vfigcomplex_core headThe small calculation exposes a wrong convention The first column of Q_L has Hermitian squared length 9/25+16/25=1. Its bilinear square without conjugation is 9/25-16/25=-7/25. Thus Q_L^TQ_L is not the identity. The portable negative control replaces the adjoint by transpose and rejects the resulting isometry claim.
Exact mathematical passage
A=Q_LGQ_R^*=\mat{3&0\\4\mathrm i&0\\0&-2\mathrm i}, \qquad A^*A=\diag(25,4).
Q_\gammaUse in source · 1 passage
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Use in source · Read passage · 01_manuscript.tex, line 210
Both identities follow by inserting the corresponding coordinate projectors and using (P_β AP_α)^*=P_α A^*P_β. The tensor identity follows from (P_β⊗ Q_δ) (A⊗ B) (P_α⊗ Q_γ) = A_βα⊗ B_δγ.
Exact mathematical passage
(P_\beta\otimes Q_\delta) (A\otimes B) (P_\alpha\otimes Q_\gamma) = A_{\beta\alpha}\otimes B_{\delta\gamma}.
Q_\deltaUse in source · 1 passage
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Use in source · Read passage · 01_manuscript.tex, line 210
Both identities follow by inserting the corresponding coordinate projectors and using (P_β AP_α)^*=P_α A^*P_β. The tensor identity follows from (P_β⊗ Q_δ) (A⊗ B) (P_α⊗ Q_γ) = A_βα⊗ B_δγ.
Exact mathematical passage
(P_\beta\otimes Q_\delta) (A\otimes B) (P_\alpha\otimes Q_\gamma) = A_{\beta\alpha}\otimes B_{\delta\gamma}.
\QQUse in source · 10 passages
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Use in source · extended_pages.tex, line 16
Keeping the first diagonal entry of G gives A_1 with its second column zero. The residual has squared Frobenius norm four and Gram matrix diag(0,4), hence spectral norm two. Both errors agree with the discarded core singular value. Every displayed arithmetic operation occurs in QQ( i). headWhat the finite check establishes It checks both isometries, the reconstructed ambient matrix, its Gram matrix, and the explicit rank-one residual. Global best-rank optimality still comes from the manuscript's…
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\QQ(\mathrm i)
\QQ^2Use in source · 2 passages
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Use in source · pages.tex, line 287
t does not mean independently addressableA STRICT KRYLOV INCLUSION Take A=I_2, split into the two coordinate packets, and start with v=(1,1)^T. Every packet is invariant. Nevertheless every polynomial applies the same scalar to both coordinates: p(A)v=p(1)(1,1)^T, dim K_m(A,v)=1 (m≥1). vfigkrylov The direct sum of the two local Krylov spans is all of QQ^2. It permits independent coefficients in the two packets. The global Krylov space does not. This is why the manuscript states an inclusion, not…
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\QQ^2
\QQ^4Use in source · 2 passages
Search aliases: QQ^4 QQ
Use in source · pages.tex, line 321
vpage19Twelve small blocks, one coherent errorDISCOVERY AND PERTURBATION CERTIFICATES Split QQ^4 into scalar packets and let A=I+E, E=frac1100(onesones^T-I). Dropping every off-diagonal block leaves I. Each discarded entry is only 1/100, but the total error is not bounded by that single-entry size. vfigblock_error headTwo norms answer different questions There are twelve omitted entries, so E _F^2=12/10^4. The all-one vector is an eigenvector with eigenvalue 3/100, while the three orthogonal dir…
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\QQ^4
\QQ^6Use in source · 2 passages
Search aliases: QQ^6 QQ
Use in source · pages.tex, line 39
1 · 3 and put u=v=ones∈QQ^6. Then A=D+uv^T has diagonal entries four, within-block off-diagonal entries two, and every cross-block entry one. All 36 entries are nonzero. x=(1,-1,2,0,-2,3)^T, v^Tx=3, Ax=Dx+3u. vfigfactored_application headWhat was avoided? The global part does not require 36 independent coefficients. It depends on the input through one scalar, v^Tx, which is broadcast through u. The diagonal part is computed by three separate local matrix products. Nothing is approximated and no …
Exact mathematical passage
u=v=\ones\in\QQ^6
RUse in source · 96 passages
Search aliases: R
Use in source · extended_pages.tex, line 3
vpage29A complex range needs an adjointTHE INNER PRODUCT IS PART OF THE CONTRACT The earlier compact example is real. Over C, an isometry means Q^*Q=I, with conjugation. Take the exact Gaussian-rational factors Q_L=mat3/5 · 0 4 i/5 · 0 0 · 1, Q_R=mat1 · 0 0 · i, G=diag(5,2). Their product is a genuinely complex rectangular operator: A=Q_LGQ_R^*=mat3 · 0
Exact mathematical passage
Q_L=\mat{3/5&0\\4\mathrm i/5&0\\0&1},\qquad Q_R=\mat{1&0\\0&\mathrm i},\qquad G=\diag(5,2).
\RUse in source · 1 passage
Search aliases: R mathbb
Use in source · Read passage · 02_supplement_02.tex, line 33
erator families. For reuse weights r_j, the local keep model can aggregate setup and per-operator costs: κ_v=κ_v^setup+∑_jr_jκ_v^(j), ϵ_v=∑_jr_jϵ_v^(j). Lean proves the aggregation identity at the cut level, including split overhead, before applying the compiler theorem. The proof is valid over an ordered field, including Q and R. The intended nonnegative model is a specialization.
Exact mathematical passage
\R
R^*Use in source · 5 passages
Search aliases: R^*
Use in source · Read passage · 01_manuscript.tex, line 834
* be an ordered thin singular value decomposition, with nonnegative nonincreasing singular values. For an integer k≥0, retain min{k,d} columns, where d is the number of supplied thin singular directions; tail singular values beyond d are zero. Define L_k=Q_LW_kΣ_k^1/2, R_k=Q_RZ_kΣ_k^1/2. Then L_kR_k^* is a best rank at most k approximation to LR^*. Its spectral and Frobenius errors are the singular value tails in [source reference]. The core dimensions equal the column ranks of the two input fac…
Exact mathematical passage
LR^*
R^TUse in source · 6 passages
Search aliases: R^T
Use in source · pages.tex, line 190
1 · 0 · -1. Set L=Q_LR_L and R=R_R. The two equal columns of L do not create a third range direction. In contrast, R has rank three. vfigcompact_core The actual core is G=R_LR_R^T=mat5 · 0 · 0 0 · 2 · 0, LR^T=Q_LG. Its dimensions are the two actual input ranks, two and three. This is not inferred by merely discarding a column with a small norm. The range factor Q_L has orthonormal columns and reconstructs L exactly. headThe preserved formal boundary The sixth release constructs orthonormal bases…
Exact mathematical passage
G=R_LR_R^T=\mat{5&0&0\\0&2&0},\qquad LR^T=Q_LG.
r_0Use in source · 1 passage
Search aliases: r_0
Use in source · Read passage · 02_supplement_04.tex, line 55
r_j-i, · i<j. The diagonal is supplied by c_0; the coordinate r_0 is unused. The constructed generator of length 2n is g=(c_0,…,c_n-1,0,r_n-1,…,r_1), and the padding map is the actual finite-index sum eliminator that appends n zeros.
Exact mathematical passage
r_0
r_AUse in source · 4 passages
Search aliases: r_A
Use in source · Read passage · 01_manuscript.tex, line 694
[Exact BDLR product] Let A=D_A+U_AV_A^*, B=D_B+U_BV_B^* use the same diagonal packet layout. Then AB=D_AD_B+U_CV_C^*, where U_C = D_AU_B+U_A(V_A^*U_B) · U_A and V_C = V_B · D_B^*V_A . The unrecompressed correction rank is at most r_A+r_B.
Exact mathematical passage
r_A+r_B
r_BUse in source · 4 passages
Search aliases: r_B
Use in source · Read passage · 01_manuscript.tex, line 694
[Exact BDLR product] Let A=D_A+U_AV_A^*, B=D_B+U_BV_B^* use the same diagonal packet layout. Then AB=D_AD_B+U_CV_C^*, where U_C = D_AU_B+U_A(V_A^*U_B) · U_A and V_C = V_B · D_B^*V_A . The unrecompressed correction rank is at most r_A+r_B.
Exact mathematical passage
r_A+r_B
r_eUse in source · 5 passages
Search aliases: r_e
Use in source · Read passage · 02_supplement_02.tex, line 132
Global comparison. For any other feasible ε, set r_e=ε_e/ε_e^*>0. The budget identities give ∑_ec_er_e=C. Using the standard analytic inequality log r≤ r-1 yields ∑_ec_elog r_e≤∑_ec_e(r_e-1)=0. Lean separately proves the exact objective-difference identity J(ε)-J(ε^*)=-∑_ec_elog r_e. Therefore J(ε)≥ J(ε^*).
Exact mathematical passage
r_e=\varepsilon_e/\varepsilon_e^*>0
r_jUse in source · 10 passages
Search aliases: r_j
Use in source · Read passage · 01_manuscript.tex, line 726
[Rank growth in a product chain] Let A_j=D_j+U_jV_j^*, j=1,…,m, share one diagonal packet layout. Before recompression, the product A_m⋯ A_1 has a BDLR representation whose correction rank is at most ∑_j=1^m r_j.
Exact mathematical passage
\boxed{ \sum_{j=1}^m r_j. }
r_{j-i}Use in source · 1 passage
Search aliases: r_j-i
Use in source · Read passage · 02_supplement_04.tex, line 48
Toeplitz padding and column-major vectorization. textitSource items 4.4 and 4.5 Toeplitz embedding. For n>0, define T_ij= c_i-j, · j≤ i, r_j-i, · i<j. The diagonal is supplied by c_0; the coordinate r_0 is unused. The constructed generator of length 2n is g=(c_0,…,c_n-1,0,r_n-1,…,r_1), and the padding map is the actual finite-index sum eliminator that appends n zeros.
Exact mathematical passage
T_{ij}= \begin{cases} c_{i-j},&j\le i,\\ r_{j-i},&i<j. \end{cases}
R_kUse in source · 2 passages
Search aliases: R_k
Use in source · Read passage · 01_manuscript.tex, line 825
R=Q_RR_R, where the columns of Q_L and Q_R are orthonormal. Form the small core G=R_LR_R^* and let G=WΣ Z^* be an ordered thin singular value decomposition, with nonnegative nonincreasing singular values. For an integer k≥0, retain min{k,d} columns, where d is the number of supplied thin singular directions; tail singular values beyond d are zero. Define L_k=Q_LW_kΣ_k^1/2, R_k=Q_RZ_kΣ_k^1/2. Then L_kR_k^* is a best rank at most k approximation to LR^*. Its spectral and Frobenius errors are the s…
Exact mathematical passage
L_k=Q_LW_k\Sigma_k^{1/2}, \qquad R_k=Q_RZ_k\Sigma_k^{1/2}.
r_{k-j}Use in source · 3 passages
Search aliases: r_k-j
Use in source · pages.tex, line 94
2,-1)^T. The diagonal pattern is Toeplitz, not circulant. To use a circular convolution, embed it into dimension six with first column g=(2,3,4,0,5,-1)^T and pad the input with three zeros. vfigtoeplitz headRead the wrapped indices For output index j<3, an input index k≤ j uses the first-column coefficient c_j-k. An input index k>j wraps to the stored first-row coefficient r_k-j. The padded input positions contribute nothing. Keeping the first three outputs therefore gives Tx=(-5,8,8)^T. The rem…
Exact mathematical passage
r_{k-j}
R_k^*Use in source · 3 passages
Search aliases: R_k^*
Use in source · Read passage · 01_manuscript.tex, line 831
Form the small core G=R_LR_R^* and let G=WΣ Z^* be an ordered thin singular value decomposition, with nonnegative nonincreasing singular values. For an integer k≥0, retain min{k,d} columns, where d is the number of supplied thin singular directions; tail singular values beyond d are zero. Define L_k=Q_LW_kΣ_k^1/2, R_k=Q_RZ_kΣ_k^1/2. Then L_kR_k^* is a best rank at most k approximation to LR^*. Its spectral and Frobenius errors are the singular value tails in [source reference]. The core dimensio…
Exact mathematical passage
L_kR_k^*
R_LUse in source · 12 passages
Search aliases: R_L
Use in source · pages.tex, line 182
vpage11Stored width is not intrinsic rankACTUAL RANGE COORDINATES Choose the isometry Q_L=mat3/5 · 0 4/5 · 0 0 · 1 0 · 0, R_L=mat1 · 0 · 1 0 · 1 · 0, R_R=mat3 · 0 · 2 0 · 2 · 0 1 · 0 · -1. Set L=Q_LR_L and R=R_R. The two equal columns of L do not create a third range direction. In contrast, R has rank three. vfigcompact_core The actual core is G=R_LR_R^T=mat5 · 0 · 0
Exact mathematical passage
Q_L=\mat{3/5&0\\4/5&0\\0&1\\0&0},\quad R_L=\mat{1&0&1\\0&1&0},\quad R_R=\mat{3&0&2\\0&2&0\\1&0&-1}.
r_LUse in source · 3 passages
Search aliases: r_L
Use in source · Read passage · 02_supplement_06.tex, line 10
Compact factors from actual column spaces. textitModules CompactRangeFactors and SmallCoreTransfer Let L∈ C^m× t, R∈ C^n× t, A=LR^*. Use the ordinary Euclidean inner products on the coordinate spaces. Define S_L=range(L), r_L=dim S_L, S_R=range(R), r_R=dim S_R. The implementation constructs an orthonormal basis of each actual range subspace. Let Q_L and Q_R be the matrices of the corresponding isometric inclusions. Express the range-restricted input maps in these bases to obtain R_L,R_R.
Exact mathematical passage
S_L=\operatorname{range}(L),\quad r_L=\dim S_L, \qquad S_R=\operatorname{range}(R),\quad r_R=\dim S_R.
R_mUse in source · 2 passages
Search aliases: R_m
Use in source · Read passage · 02_supplement_03.tex, line 13
Low-rank factors through a product chain. textitSource item 5.2 Let A_j=D_j+U_jV_j^*, U_j,V_j∈ C^n× r_j, with all D_j diagonal in one packet layout. Set P_m=A_m⋯ A_1, D^(m)=D_m⋯ D_1. The formalization constructs thin matrices L_m,R_m such that P_m=D^(m)+L_mR_m^*. Their column index type is a recursively constructed disjoint sum. Its cardinality is proved to equal ∑_j=1^m r_j.
Exact mathematical passage
L_m,R_m
R_{m-1}Use in source · 1 passage
Search aliases: R_m-1
Use in source · Read passage · 02_supplement_03.tex, line 20
Construction. Suppose P_m-1=D^(m-1)+L_m-1R_m-1^*. The new factors are L_m · =[D_mL_m-1+U_m(V_m^*L_m-1) U_m], R_m · =[R_m-1 (D^(m-1))^*V_m]. Substitution into the already checked two-factor product identity proves exactness. This is not a rank-only existence argument. The recursively defined factors are themselves audited.
Exact mathematical passage
L_m&=\bigl[D_mL_{m-1}+U_m(V_m^*L_{m-1})\quad U_m\bigr],\\ R_m&=\bigl[R_{m-1}\quad (D^{(m-1)})^*V_m\bigr].
R_{m-1}^*Use in source · 1 passage
Search aliases: R_m-1^*
Use in source · Read passage · 02_supplement_03.tex, line 19
Construction. Suppose P_m-1=D^(m-1)+L_m-1R_m-1^*. The new factors are L_m · =[D_mL_m-1+U_m(V_m^*L_m-1) U_m],
Exact mathematical passage
P_{m-1}=D^{(m-1)}+L_{m-1}R_{m-1}^*
R_m^*Use in source · 1 passage
Search aliases: R_m^*
Use in source · Read passage · 02_supplement_03.tex, line 14
Low-rank factors through a product chain. textitSource item 5.2 Let A_j=D_j+U_jV_j^*, U_j,V_j∈ C^n× r_j, with all D_j diagonal in one packet layout. Set P_m=A_m⋯ A_1, D^(m)=D_m⋯ D_1. The formalization constructs thin matrices L_m,R_m such that P_m=D^(m)+L_mR_m^*. Their column index type is a recursively constructed disjoint sum. Its cardinality is proved to equal ∑_j=1^m r_j.
Exact mathematical passage
\boxed{P_m=D^{(m)}+L_mR_m^*.}
R_{\min}Use in source · 2 passages
Search aliases: R_min min
Use in source · Read passage · 01_manuscript.tex, line 1759
[Amortization threshold] Let s≥0 be a one time setup cost. Let t_b,t_c≥0 be the baseline and candidate times per use and let R∈ N be a reuse count. If t_b>t_c, the candidate is faster after R uses exactly when R> fracst_b-t_c. The smallest integer reuse count is R_min = lfloor fracst_b-t_c rfloor+1. If t_b≤ t_c, no positive reuse count amortizes a positive setup cost.
Exact mathematical passage
\boxed{ R_{\min} = \left\lfloor \frac{s}{t_b-t_c} \right\rfloor+1. }
r_{n-1}Use in source · 3 passages
Search aliases: r_n-1
Use in source · Read passage · 01_manuscript.tex, line 569
[Toeplitz circulant embedding] Let n≥1 be an integer and T∈ C^n× n be Toeplitz, with first column (c_0,c_1,…,c_n-1)^T and first row (c_0,r_1,…,r_n-1). Set g = (c_0,c_1,…,c_n-1,0,r_n-1,…,r_1)^T ∈ C^2n and append n zeros to x∈ C^n. The first n entries of the circulant product C(g)(x,0)^T equal Tx.
Exact mathematical passage
(c_0,r_1,\ldots,r_{n-1}).
R_RUse in source · 6 passages
Search aliases: R_R
Use in source · pages.tex, line 182
vpage11Stored width is not intrinsic rankACTUAL RANGE COORDINATES Choose the isometry Q_L=mat3/5 · 0 4/5 · 0 0 · 1 0 · 0, R_L=mat1 · 0 · 1 0 · 1 · 0, R_R=mat3 · 0 · 2 0 · 2 · 0 1 · 0 · -1. Set L=Q_LR_L and R=R_R. The two equal columns of L do not create a third range direction. In contrast, R has rank three. vfigcompact_core The actual core is G=R_LR_R^T=mat5 · 0 · 0
Exact mathematical passage
Q_L=\mat{3/5&0\\4/5&0\\0&1\\0&0},\quad R_L=\mat{1&0&1\\0&1&0},\quad R_R=\mat{3&0&2\\0&2&0\\1&0&-1}.
r_RUse in source · 3 passages
Search aliases: r_R
Use in source · Read passage · 02_supplement_06.tex, line 10
Compact factors from actual column spaces. textitModules CompactRangeFactors and SmallCoreTransfer Let L∈ C^m× t, R∈ C^n× t, A=LR^*. Use the ordinary Euclidean inner products on the coordinate spaces. Define S_L=range(L), r_L=dim S_L, S_R=range(R), r_R=dim S_R. The implementation constructs an orthonormal basis of each actual range subspace. Let Q_L and Q_R be the matrices of the corresponding isometric inclusions. Express the range-restricted input maps in these bases to obtain R_L,R_R.
Exact mathematical passage
S_L=\operatorname{range}(L),\quad r_L=\dim S_L, \qquad S_R=\operatorname{range}(R),\quad r_R=\dim S_R.
R_R^*Use in source · 3 passages
Search aliases: R_R^*
Use in source · Read passage · 01_manuscript.tex, line 817
[Small core recompression] Let L∈ C^m× r, R∈ C^n× r. Take compact QR factorizations L=Q_LR_L, R=Q_RR_R, where the columns of Q_L and Q_R are orthonormal. Form the small core G=R_LR_R^* and let G=WΣ Z^* be an ordered thin singular value decomposition, with nonnegative nonincreasing singular values. For an integer k≥0, retain min{k,d} columns, where d is the number of supplied thin singular directions; tail singular values beyond d are zero. Define L_k=Q_LW_kΣ_k^1/2, R_k=Q_RZ_kΣ_k^1/2. Then L_kR_k…
Exact mathematical passage
G=R_LR_R^*
R_R^TUse in source · 2 passages
Search aliases: R_R^T
Use in source · pages.tex, line 190
1 · 0 · -1. Set L=Q_LR_L and R=R_R. The two equal columns of L do not create a third range direction. In contrast, R has rank three. vfigcompact_core The actual core is G=R_LR_R^T=mat5 · 0 · 0 0 · 2 · 0, LR^T=Q_LG. Its dimensions are the two actual input ranks, two and three. This is not inferred by merely discarding a column with a small norm. The range factor Q_L has orthonormal columns and reconstructs L exactly. headThe preserved formal boundary The sixth release constructs orthonormal bases…
Exact mathematical passage
G=R_LR_R^T=\mat{5&0&0\\0&2&0},\qquad LR^T=Q_LG.
\operatorname{range}Use in source · 1 passage
Search aliases: range operatorname
Use in source · Read passage · 02_supplement_06.tex, line 10
Compact factors from actual column spaces. textitModules CompactRangeFactors and SmallCoreTransfer Let L∈ C^m× t, R∈ C^n× t, A=LR^*. Use the ordinary Euclidean inner products on the coordinate spaces. Define S_L=range(L), r_L=dim S_L, S_R=range(R), r_R=dim S_R. The implementation constructs an orthonormal basis of each actual range subspace. Let Q_L and Q_R be the matrices of the corresponding isometric inclusions. Express the range-restricted input maps in these bases to obtain R_L,R_R.
Exact mathematical passage
S_L=\operatorname{range}(L),\quad r_L=\dim S_L, \qquad S_R=\operatorname{range}(R),\quad r_R=\dim S_R.
\operatorname{rank}Use in source · 1 passage
Search aliases: rank operatorname
Use in source · Read passage · 02_supplement_03.tex, line 27
Applying the ordinary matrix-rank inequality gives rank(P_m-D^(m))≤∑_j=1^m r_j. The bound concerns the actual correction matrix, not just the number of allocated columns. The proof also shows that D^(m) stays packet diagonal. The empty chain has identity base, empty column type, and zero correction. Individual r_j=0 are permitted.
Exact mathematical passage
\boxed{\operatorname{rank}(P_m-D^{(m)})\leq\sum_{j=1}^m r_j.}
\rankUse in source · 9 passages
Search aliases: rank operatorname
Use in source · Read passage · 02_supplement_01.tex, line 95
V_C · =V_B · D_B^*V_A and proves AB=D_AD_B+U_CV_C^*. These are actual concatenated matrices with r_B+r_A columns. Expanding the product and grouping the terms with common right factors proves the identity. The separate rank theorem establishes rank(AB-D_AD_B)≤ r_A+r_B. It bounds actual matrix rank, not just the number of stored columns. Packet diagonality of D_AD_B follows from diagonality of both factors and the checked block-composition formula.
Exact mathematical passage
\rank(AB-D_AD_B)\le r_A+r_B.
\rceilUse in source · 2 passages
Search aliases: rceil
Use in source · extended_pages.tex, line 66
vpage32Safe integers can still be conservativeQUANTIZATION AND THE SIZE OF THE BUDGET TABLE Let one integer unit represent h=1/10 of certified error. For an individual nonnegative error e, choosing w=lceil e/hrceil ensures e≤ hw. Thus integer feasibility ∑ w≤ B implies the real bound ∑ e≤ hB. This implication is checked before applying the exact option or tree recurrence. vfigquantization headA false certificate and a conservative rejection For errors (3/50,2/25), rounding down assigns weights (…
Exact mathematical passage
w=\lceil e/h\rceil\qquad\text{ensures}\qquad e\le hw.
\operatorname{Re}Use in source · 1 passage
Search aliases: Re operatorname
Use in source · Read passage · 02_supplement_05.tex, line 171
Lean identifies the actual Frobenius norm with its trace expression, normM_rm F^2=Retr(M^*M), then expands the finite sum and removes each cross term. The result is normA-Π(A)_rm F^2 =∑_b≠ anormP_bAP_a_rm F^2. No Pythagorean equality for the target residual is assumed.
Exact mathematical passage
\norm{M}_{\rm F}^2=\operatorname{Re}\operatorname{tr}(M^*M),
\rfloorUse in source · 4 passages
Search aliases: rfloor
Use in source · pages.tex, line 427
ONE LINEAR MODEL Suppose the baseline costs seven units per use. A candidate costs 30 units to set up and two per use. These are synthetic constants chosen to expose the exact threshold: T_b(R)=7R, T_c(R)=30+2R. vfigamortization At R=6, both cost 42. The candidate becomes strictly faster at R=7, where it costs 44 instead of 49. The minimum profitable integer reuse count is lfloorfrac307-2rfloor+1=7. Replacing the floor-plus-one rule with a ceiling would be wrong when the ratio is already an inte…
Exact mathematical passage
\left\lfloor\frac{30}{7-2}\right\rfloor+1=7.
\rmUse in source · 15 passages
Search aliases: rm
Use in source · extended_pages.tex, line 53
bottomrule vfig[.45]shared_family With reuse weights (r_1,r_2)=(1,3), the aggregated objectives are J_rm keep=4+2+3·7=27, J_rm split=2+8+3·2=16. The exact shared optimum is split. Operator 1 alone prefers keep on its per-use costs, while operator 2 prefers split. Choosing independently and trying to combine those decisions would not solve the declared one-pruning problem. headA hand-computable decision threshold Fix r_1=1 and vary r_2=r≥0. Keeping costs 6+7r and splitting costs 10+2r. Splitting …
Exact mathematical passage
J_{\rm keep}=4+2+3\cdot7=27,\qquad J_{\rm split}=2+8+3\cdot2=16.
\mathrm{root}Use in source · 1 passage
Search aliases: root mathrm
Use in source · Read passage · 02_supplement_03.tex, line 181
The root scan min_0≤ b≤ BF_root(b) is therefore the exact budget-constrained optimum. Increasing the budget cannot increase a finite optimum. The theorem retains feasibility hypotheses where necessary rather than treating codenone as an ordinary real cost.
Exact mathematical passage
\boxed{\min_{0\leq b\leq B}F_{\mathrm{root}}(b)}
\RRUse in source · 2 passages
Search aliases: RR R mathbb
Use in source · Read passage · 01_manuscript.tex, line 1309
[Floating point matrix product model] For real A∈ R^m× p and B∈ R^p× n, consider ordinary dot-product matrix multiplication in uniform precision with unit roundoff u, no overflow or underflow, and pu<1. Its computed output widehat C satisfies the componentwise forward-error bound |widehat C-AB|≤γ_p|A||B|, γ_p=fracpu1-pu. Absolute values and the inequality are entrywise. Complex arithmetic, fused kernels, mixed accumulation, and tensor kernels require their own stated constants. This fact does no…
Exact mathematical passage
A\in\RR^{m\times p}
\RR^{m\times p}Use in source · 1 passage
Search aliases: RR^m× p R^m× p RR times mathbb times
Use in source · Read passage · 01_manuscript.tex, line 1309
[Floating point matrix product model] For real A∈ R^m× p and B∈ R^p× n, consider ordinary dot-product matrix multiplication in uniform precision with unit roundoff u, no overflow or underflow, and pu<1. Its computed output widehat C satisfies the componentwise forward-error bound |widehat C-AB|≤γ_p|A||B|, γ_p=fracpu1-pu. Absolute values and the inequality are entrywise. Complex arithmetic, fused kernels, mixed accumulation, and tensor kernels require their own stated constants. This fact does no…
Exact mathematical passage
A\in\RR^{m\times p}
\RR^{p\times n}Use in source · 1 passage
Search aliases: RR^p× n R^p× n RR times mathbb times
Use in source · Read passage · 01_manuscript.tex, line 1309
[Floating point matrix product model] For real A∈ R^m× p and B∈ R^p× n, consider ordinary dot-product matrix multiplication in uniform precision with unit roundoff u, no overflow or underflow, and pu<1. Its computed output widehat C satisfies the componentwise forward-error bound |widehat C-AB|≤γ_p|A||B|, γ_p=fracpu1-pu. Absolute values and the inequality are entrywise. Complex arithmetic, fused kernels, mixed accumulation, and tensor kernels require their own stated constants. This fact does no…
Exact mathematical passage
B\in\RR^{p\times n}
\mathrm{run}Use in source · 1 passage
Search aliases: run mathrm
Use in source · Read passage · 02_supplement_02.tex, line 203
Contracts, grouping, and amortization. textitSource items 2.9, 10.2, 10.4--10.6, and 11.3 Output conversion. A decoded run returns the actual pair (decode(compressed(x)), t_run(x)+t_decode(compressed(x))). Under the declared output equality, Lean proves that this returns the same explicit result and charges the conversion time. It also proves the downstream composition identity when compressed output is consumed directly. Omitting a positive decode cost strictly understates the modeled explicit-…
Exact mathematical passage
\bigl(\operatorname{decode}(\operatorname{compressed}(x)), \;t_{\mathrm{run}}(x)+t_{\mathrm{decode}}(\operatorname{compressed}(x))\bigr).
\mathcal SUse in source · 3 passages
Search aliases: S mathcal
Use in source · Read passage · 02_supplement_05.tex, line 131
Stored scalars. With equal block width b, the storage-address type is S= B_p,w×{0,…,b-1}^2. Its cardinality is the actual number of represented scalar entries: | S|=b^2N_p,w.
Exact mathematical passage
\mathcal S=\mathcal B_{p,w}\times\{0,\ldots,b-1\}^2.
S'Use in source · 2 passages
Search aliases: S'
Use in source · Read passage · 02_supplement_02.tex, line 69
Strict half-gap stability. At a solved internal node write K=κ_v+λϵ_v, S=σ_v+∑_wF_λ(w), g=|K-S|. If the updated scores satisfy |K'-K|<g/2, |S'-S|<g/2, Lean proves K'≤ S' Longleftrightarrow K≤ S. Applying this condition recursively at every internal node proves equality of the entire returned cuts before and after perturbation, not only equality of one numerical minimum.
Exact mathematical passage
|K'-K|<g/2,\qquad |S'-S|<g/2,
S^{-1}Use in source · 11 passages
Search aliases: S^-1
Use in source · pages.tex, line 167
-1 · 3. Hence D^-1u=u/4 and S=1+u^TD^-1u=frac52. The Woodbury identity reduces the entire inverse to A^-1=D^-1-D^-1uS^-1u^TD^-1 =D^-1-frac140uu^T. vfigwoodbury headTwo hypotheses, not one Invertibility of D alone does not suffice. For D=I_2, u=e_1, and v=-e_1, the update is diag(0,1). The small matrix S is zero, exactly where the correction formula would require an invalid division.
Exact mathematical passage
A^{-1}=D^{-1}-D^{-1}uS^{-1}u^TD^{-1} =D^{-1}-\frac1{40}uu^T.
S_{ba}Use in source · 1 passage
Search aliases: S_ba
Use in source · Read passage · 02_supplement_05.tex, line 100
Hidden elimination. For the actual Schur complement S=A_RR-A_RHA_HH^-1A_HR, the hidden inverse is expanded through the hidden projectors. A nonzero retained correction block requires indices j,k with nonzero entrance and exit blocks and a hidden path jleadsto k. Therefore a nonzero block S_ba requires either a direct retained block, or a⟶ j leadsto_H k⟶ b. The entrance and exit matrices may be rectangular; retained and hidden coordinate sizes and packet systems are independently typed. The hidde…
Exact mathematical passage
S_{ba}
s_eUse in source · 35 passages
Search aliases: s_e
Use in source · pages.tex, line 25
s_e,ϵ_e · Sensitivity and tolerance · Is the budget input-dependent or uniform over unit inputs?
Exact mathematical passage
s_e,\epsilon_e
S_LUse in source · 1 passage
Search aliases: S_L
Use in source · Read passage · 02_supplement_06.tex, line 10
Compact factors from actual column spaces. textitModules CompactRangeFactors and SmallCoreTransfer Let L∈ C^m× t, R∈ C^n× t, A=LR^*. Use the ordinary Euclidean inner products on the coordinate spaces. Define S_L=range(L), r_L=dim S_L, S_R=range(R), r_R=dim S_R. The implementation constructs an orthonormal basis of each actual range subspace. Let Q_L and Q_R be the matrices of the corresponding isometric inclusions. Express the range-restricted input maps in these bases to obtain R_L,R_R.
Exact mathematical passage
S_L=\operatorname{range}(L),\quad r_L=\dim S_L, \qquad S_R=\operatorname{range}(R),\quad r_R=\dim S_R.
S_RUse in source · 1 passage
Search aliases: S_R
Use in source · Read passage · 02_supplement_06.tex, line 10
Compact factors from actual column spaces. textitModules CompactRangeFactors and SmallCoreTransfer Let L∈ C^m× t, R∈ C^n× t, A=LR^*. Use the ordinary Euclidean inner products on the coordinate spaces. Define S_L=range(L), r_L=dim S_L, S_R=range(R), r_R=dim S_R. The implementation constructs an orthonormal basis of each actual range subspace. Let Q_L and Q_R be the matrices of the corresponding isometric inclusions. Express the range-restricted input maps in these bases to obtain R_L,R_R.
Exact mathematical passage
S_L=\operatorname{range}(L),\quad r_L=\dim S_L, \qquad S_R=\operatorname{range}(R),\quad r_R=\dim S_R.
\mathrm{setup}Use in source · 1 passage
Search aliases: setup mathrm
Use in source · Read passage · 02_supplement_02.tex, line 29
Shared operator families. For reuse weights r_j, the local keep model can aggregate setup and per-operator costs: κ_v=κ_v^setup+∑_jr_jκ_v^(j), ϵ_v=∑_jr_jϵ_v^(j). Lean proves the aggregation identity at the cut level, including split overhead, before applying the compiler theorem. The proof is valid over an ordered field, including Q and R. The intended nonnegative model is a specialization.
Exact mathematical passage
\kappa_v=\kappa_v^{\mathrm{setup}}+\sum_jr_j\kappa_v^{(j)}, \qquad \epsilon_v=\sum_jr_j\epsilon_v^{(j)}.
\operatorname{step}Use in source · 1 passage
Search aliases: step operatorname
Use in source · Read passage · 02_supplement_02.tex, line 86
Let v be the new optimum of the edited subtree, and let j be the edited child slot at one frame. The update performed at that frame is step(v)= min{K, σ+ ∑_i v, · i=j, widehat F_i, · i≠ j. }. This result is passed to the next ancestor. The update function does not recursively evaluate sibling subtrees. It reads their stored values widehat F_i.
Exact mathematical passage
\boxed{ \operatorname{step}(v)= \min\left\{K,\;\sigma+ \sum_i\begin{cases} v,&i=j,\\ \widehat F_i,&i\ne j. \end{cases}\right\}. }
\operatorname{storedTree}Use in source · 1 passage
Search aliases: storedTree operatorname
Use in source · Read passage · 02_supplement_03.tex, line 202
Entrywise implementation theorem. For every b≤ B, Lean proves lookup(storedTree(B,t),b)=F_t(b). There is a parallel theorem for stored discrete precision. The proof establishes first that an output entry at b depends only on input entries at most b, then proceeds through cached convolution and the tree recursion. Consequently the optimality, attainment, and infeasibility specification applies to the actual stored result.
Exact mathematical passage
\operatorname{lookup}(\operatorname{storedTree}(B,t),b)=F_t(b).
\mathrm{stream}Use in source · 1 passage
Search aliases: stream mathrm
Use in source · Read passage · 02_supplement_02.tex, line 184
Streaming. The explicit event type counts one block read, one source read, and a target read plus write for every edge. With scalar size s, its byte count is B_stream=s∑_e(m_en_e+n_eq+2m_eq). No cross-edge cache reuse is assumed. Sparse index metadata and temporary workspace remain separate.
Exact mathematical passage
B_{\mathrm{stream}}=s\sum_e(m_en_e+n_eq+2m_eq).
tUse in source · 49 passages
Search aliases: t
Use in source · extended_pages.tex, line 53
bottomrule vfig[.45]shared_family With reuse weights (r_1,r_2)=(1,3), the aggregated objectives are J_rm keep=4+2+3·7=27, J_rm split=2+8+3·2=16. The exact shared optimum is split. Operator 1 alone prefers keep on its per-use costs, while operator 2 prefers split. Choosing independently and trying to combine those decisions would not solve the declared one-pruning problem. headA hand-computable decision threshold Fix r_1=1 and vary r_2=r≥0. Keeping costs 6+7r and splitting costs 10+2r. Splitting …
Exact mathematical passage
J_{\rm keep}=4+2+3\cdot7=27,\qquad J_{\rm split}=2+8+3\cdot2=16.
T'Use in source · 2 passages
Search aliases: T'
Use in source · Read passage · 02_supplement_02.tex, line 100
Correctness of the actual update. For any edited subtree T' and valid context C, Lean proves updateValue(C,F(T'))=F(fill(C,T')). The proof inducts through the context. At each frame, the updated child contributes the new value, and each other child contributes its unchanged cached optimum. The frame recurrence is therefore exactly the full-tree recurrence.
Exact mathematical passage
T'
T_bUse in source · 2 passages
Search aliases: T_b
Use in source · pages.tex, line 422
vpage25The first win comes after the tieSETUP AND REUSE IN ONE LINEAR MODEL Suppose the baseline costs seven units per use. A candidate costs 30 units to set up and two per use. These are synthetic constants chosen to expose the exact threshold: T_b(R)=7R, T_c(R)=30+2R. vfigamortization At R=6, both cost 42. The candidate becomes strictly faster at R=7, where it costs 44 instead of 49. The minimum profitable integer reuse count is lfloorfrac307-2rfloor+1=7. Replacing the floor-plus-one rule with…
Exact mathematical passage
T_b(R)=7R,\qquad T_c(R)=30+2R.
t_bUse in source · 10 passages
Search aliases: t_b
Use in source · Read passage · 01_manuscript.tex, line 1753
[Amortization threshold] Let s≥0 be a one time setup cost. Let t_b,t_c≥0 be the baseline and candidate times per use and let R∈ N be a reuse count. If t_b>t_c, the candidate is faster after R uses exactly when R> fracst_b-t_c. The smallest integer reuse count is R_min = lfloor fracst_b-t_c rfloor+1. If t_b≤ t_c, no positive reuse count amortizes a positive setup cost.
Exact mathematical passage
t_b,t_c\ge0
T_cUse in source · 2 passages
Search aliases: T_c
Use in source · pages.tex, line 422
vpage25The first win comes after the tieSETUP AND REUSE IN ONE LINEAR MODEL Suppose the baseline costs seven units per use. A candidate costs 30 units to set up and two per use. These are synthetic constants chosen to expose the exact threshold: T_b(R)=7R, T_c(R)=30+2R. vfigamortization At R=6, both cost 42. The candidate becomes strictly faster at R=7, where it costs 44 instead of 49. The minimum profitable integer reuse count is lfloorfrac307-2rfloor+1=7. Replacing the floor-plus-one rule with…
Exact mathematical passage
T_b(R)=7R,\qquad T_c(R)=30+2R.
t_cUse in source · 13 passages
Search aliases: t_c
Use in source · Read passage · 01_manuscript.tex, line 1753
[Amortization threshold] Let s≥0 be a one time setup cost. Let t_b,t_c≥0 be the baseline and candidate times per use and let R∈ N be a reuse count. If t_b>t_c, the candidate is faster after R uses exactly when R> fracst_b-t_c. The smallest integer reuse count is R_min = lfloor fracst_b-t_c rfloor+1. If t_b≤ t_c, no positive reuse count amortizes a positive setup cost.
Exact mathematical passage
t_b,t_c\ge0
t_{\mathrm{decode}}Use in source · 1 passage
Search aliases: t_decode mathrm
Use in source · Read passage · 02_supplement_02.tex, line 203
Contracts, grouping, and amortization. textitSource items 2.9, 10.2, 10.4--10.6, and 11.3 Output conversion. A decoded run returns the actual pair (decode(compressed(x)), t_run(x)+t_decode(compressed(x))). Under the declared output equality, Lean proves that this returns the same explicit result and charges the conversion time. It also proves the downstream composition identity when compressed output is consumed directly. Omitting a positive decode cost strictly understates the modeled explicit-…
Exact mathematical passage
\bigl(\operatorname{decode}(\operatorname{compressed}(x)), \;t_{\mathrm{run}}(x)+t_{\mathrm{decode}}(\operatorname{compressed}(x))\bigr).
t_eUse in source · 3 passages
Search aliases: t_e
Use in source · Read passage · 01_manuscript.tex, line 1796
[Materialization threshold] Suppose a compressed representation costs t_c≥0 per downstream use, an explicit representation costs t_e≥0 per use, and one time materialization costs m≥0, with reuse count R∈ N. Materialization is profitable exactly when R> fracmt_c-t_e and t_c>t_e.
Exact mathematical passage
t_e\ge0
T_{ij}Use in source · 2 passages
Search aliases: T_ij
Use in source · Read passage · 02_supplement_04.tex, line 48
Toeplitz padding and column-major vectorization. textitSource items 4.4 and 4.5 Toeplitz embedding. For n>0, define T_ij= c_i-j, · j≤ i, r_j-i, · i<j. The diagonal is supplied by c_0; the coordinate r_0 is unused. The constructed generator of length 2n is g=(c_0,…,c_n-1,0,r_n-1,…,r_1), and the padding map is the actual finite-index sum eliminator that appends n zeros.
Exact mathematical passage
T_{ij}= \begin{cases} c_{i-j},&j\le i,\\ r_{j-i},&i<j. \end{cases}
t_oUse in source · 3 passages
Search aliases: t_o
Use in source · Read passage · 01_manuscript.tex, line 1784
[Graph replay threshold] If graph capture costs c≥0, ordinary execution costs t_o≥0, and replay costs 0≤ t_r<t_o, then capture is profitable when R> fracct_o-t_r.
Exact mathematical passage
t_o\ge0
t_rUse in source · 2 passages
Search aliases: t_r
Use in source · Read passage · 01_manuscript.tex, line 1784
[Graph replay threshold] If graph capture costs c≥0, ordinary execution costs t_o≥0, and replay costs 0≤ t_r<t_o, then capture is profitable when R> fracct_o-t_r.
Exact mathematical passage
0\le t_r<t_o
t_{\mathrm{run}}Use in source · 1 passage
Search aliases: t_run mathrm
Use in source · Read passage · 02_supplement_02.tex, line 203
Contracts, grouping, and amortization. textitSource items 2.9, 10.2, 10.4--10.6, and 11.3 Output conversion. A decoded run returns the actual pair (decode(compressed(x)), t_run(x)+t_decode(compressed(x))). Under the declared output equality, Lean proves that this returns the same explicit result and charges the conversion time. It also proves the downstream composition identity when compressed output is consumed directly. Omitting a positive decode cost strictly understates the modeled explicit-…
Exact mathematical passage
\bigl(\operatorname{decode}(\operatorname{compressed}(x)), \;t_{\mathrm{run}}(x)+t_{\mathrm{decode}}(\operatorname{compressed}(x))\bigr).
\operatorname{tr}Use in source · 2 passages
Search aliases: tr operatorname
Use in source · Read passage · 02_supplement_05.tex, line 165
Hilbert--Schmidt orthogonality. For distinct ordered packet pairs (b,a)≠(d,c), adjoint multiplication and projector orthogonality give tr((P_bAP_a)^*P_dBP_c)=0. If b≠ d, the central projector product vanishes. Otherwise a≠ c, and cyclicity of trace brings P_cP_a together. This is proved for arbitrary input matrices A,B.
Exact mathematical passage
\operatorname{tr}\bigl((P_bAP_a)^*P_dBP_c\bigr)=0.
u^TUse in source · 6 passages
Search aliases: u^T
Use in source · pages.tex, line 163
-1 · 3. Hence D^-1u=u/4 and S=1+u^TD^-1u=frac52. The Woodbury identity reduces the entire inverse to A^-1=D^-1-D^-1uS^-1u^TD^-1 =D^-1-frac140uu^T. vfigwoodbury headTwo hypotheses, not one Invertibility of D alone does not suffice. For D=I_2, u=e_1, and v=-e_1, the update is diag(0,1). The small matrix S is zero, exactly where the correction formula would require an invalid division.
Exact mathematical passage
S=1+u^TD^{-1}u=\frac52.
U_AUse in source · 7 passages
Search aliases: U_A
Use in source · Read passage · 01_manuscript.tex, line 669
[Exact BDLR product] Let A=D_A+U_AV_A^*, B=D_B+U_BV_B^* use the same diagonal packet layout. Then AB=D_AD_B+U_CV_C^*, where U_C = D_AU_B+U_A(V_A^*U_B) · U_A and V_C = V_B · D_B^*V_A . The unrecompressed correction rank is at most r_A+r_B.
Exact mathematical passage
A=D_A+U_AV_A^*, \qquad B=D_B+U_BV_B^*
u_aUse in source · 10 passages
Search aliases: u_a
Use in source · pages.tex, line 142
vpage9The cross term has somewhere to goCLOSED LOW-RANK PRODUCT ALGEBRA Let A=D_A+u_av_a^T be the running operator. Take D_B=diag(1,2,3,4,5,6), u_b=(1,0,1,0,1,0)^T, and v_b=(0,1,0,1,0,1)^T. Define B=D_B+u_bv_b^T. U_C=[D_Au_b+u_a(v_a^Tu_b) u_a], V_C=[v_b D_B^Tv_a]. vfigbdlr_product headWhy there are only two columns Expanding AB gives three correction terms, but two share the right factor v_b^T: D_Au_bv_b^T+u_a(v_a^Tu_b)v_b^T =(D_Au_b+u_a(v_a^Tu_b))v_b^T. The remaining term is u_av_a^TD_B. Thus A…
Exact mathematical passage
A=D_A+u_av_a^T
U_BUse in source · 6 passages
Search aliases: U_B
Use in source · Read passage · 01_manuscript.tex, line 669
[Exact BDLR product] Let A=D_A+U_AV_A^*, B=D_B+U_BV_B^* use the same diagonal packet layout. Then AB=D_AD_B+U_CV_C^*, where U_C = D_AU_B+U_A(V_A^*U_B) · U_A and V_C = V_B · D_B^*V_A . The unrecompressed correction rank is at most r_A+r_B.
Exact mathematical passage
A=D_A+U_AV_A^*, \qquad B=D_B+U_BV_B^*
u_bUse in source · 10 passages
Search aliases: u_b
Use in source · pages.tex, line 142
vpage9The cross term has somewhere to goCLOSED LOW-RANK PRODUCT ALGEBRA Let A=D_A+u_av_a^T be the running operator. Take D_B=diag(1,2,3,4,5,6), u_b=(1,0,1,0,1,0)^T, and v_b=(0,1,0,1,0,1)^T. Define B=D_B+u_bv_b^T. U_C=[D_Au_b+u_a(v_a^Tu_b) u_a], V_C=[v_b D_B^Tv_a]. vfigbdlr_product headWhy there are only two columns Expanding AB gives three correction terms, but two share the right factor v_b^T: D_Au_bv_b^T+u_a(v_a^Tu_b)v_b^T =(D_Au_b+u_a(v_a^Tu_b))v_b^T. The remaining term is u_av_a^TD_B. Thus A…
Exact mathematical passage
u_b=(1,0,1,0,1,0)^T
U_CUse in source · 15 passages
Search aliases: U_C
Use in source · pages.tex, line 143
vpage9The cross term has somewhere to goCLOSED LOW-RANK PRODUCT ALGEBRA Let A=D_A+u_av_a^T be the running operator. Take D_B=diag(1,2,3,4,5,6), u_b=(1,0,1,0,1,0)^T, and v_b=(0,1,0,1,0,1)^T. Define B=D_B+u_bv_b^T. U_C=[D_Au_b+u_a(v_a^Tu_b) u_a], V_C=[v_b D_B^Tv_a]. vfigbdlr_product headWhy there are only two columns Expanding AB gives three correction terms, but two share the right factor v_b^T: D_Au_bv_b^T+u_a(v_a^Tu_b)v_b^T =(D_Au_b+u_a(v_a^Tu_b))v_b^T. The remaining term is u_av_a^TD_B. Thus A…
Exact mathematical passage
U_C=[D_Au_b+u_a(v_a^Tu_b)\ \ u_a],\qquad V_C=[v_b\ \ D_B^Tv_a].
u_eUse in source · 7 passages
Search aliases: u_e
Use in source · Read passage · 02_supplement_04.tex, line 182
Constructive parameterization. Write t=1/λ and set u_e(t)=min{m_e,tc_e/s_e}, Φ(t)=∑_e s_eu_e(t). Lean proves that Φ is continuous and nondecreasing, with Φ(0)=0. At the finite positive scale T=∑_efracm_es_ec_e, every coordinate is saturated, so u(T)=m and Φ(T)=∑_es_em_e. The intermediate value theorem produces t>0 with Φ(t)=τ. The multiplier is not assumed to exist.
Exact mathematical passage
u_e(t)=\min\{m_e,tc_e/s_e\},\qquad \Phi(t)=\sum_e s_eu_e(t).
U_jUse in source · 2 passages
Search aliases: U_j
Use in source · Read passage · 01_manuscript.tex, line 716
[Rank growth in a product chain] Let A_j=D_j+U_jV_j^*, j=1,…,m, share one diagonal packet layout. Before recompression, the product A_m⋯ A_1 has a BDLR representation whose correction rank is at most ∑_j=1^m r_j.
Exact mathematical passage
A_j=D_j+U_jV_j^*, \qquad j=1,\ldots,m,
U_mUse in source · 1 passage
Search aliases: U_m
Use in source · Read passage · 02_supplement_03.tex, line 20
Construction. Suppose P_m-1=D^(m-1)+L_m-1R_m-1^*. The new factors are L_m · =[D_mL_m-1+U_m(V_m^*L_m-1) U_m], R_m · =[R_m-1 (D^(m-1))^*V_m]. Substitution into the already checked two-factor product identity proves exactness. This is not a rank-only existence argument. The recursively defined factors are themselves audited.
Exact mathematical passage
L_m&=\bigl[D_mL_{m-1}+U_m(V_m^*L_{m-1})\quad U_m\bigr],\\ R_m&=\bigl[R_{m-1}\quad (D^{(m-1)})^*V_m\bigr].
\operatorname{updateValue}Use in source · 1 passage
Search aliases: updateValue operatorname
Use in source · Read passage · 02_supplement_02.tex, line 101
Correctness of the actual update. For any edited subtree T' and valid context C, Lean proves updateValue(C,F(T'))=F(fill(C,T')). The proof inducts through the context. At each frame, the updated child contributes the new value, and each other child contributes its unchanged cached optimum. The frame recurrence is therefore exactly the full-tree recurrence.
Exact mathematical passage
\operatorname{updateValue}\bigl(C,F(T')\bigr)=F\bigl(\operatorname{fill}(C,T')\bigr).
\vUse in source · 2 passages
Search aliases: v
Use in source · complex_core.tex, line 7
ont=sffamilysmall] node[anchor=west] at (0,3.4) A^*A=diag(25,4); draw[->] (0,0)--(8,0) node[right] retained singular directions; draw[->] (0,0)--(0,2.5) node[above] squared error; draw[thick,black!65] (1,2.03)--(3.5,.28)--(6,0); foreach x/y/n/v in 1/2.03/0/29,3.5/.28/1/4,6/0/2/0 fill (x,y) circle (2pt);node[below] at (x,0) n;node[above] at (x,y) v; node[anchor=west] at (0,-.9) Q_L^*Q_L=I Q_L^TQ_L=diag(-7/25,1);
Exact mathematical passage
\v
v^{(k)}Use in source · 2 passages
Search aliases: v^(k)
Use in source · Read passage · 01_manuscript.tex, line 548
For each frequency k, let v^(k)_j = frac1sqrt n exp(frac2π i jkn). Then (C(c)v^(k))_j = ∑_r=0^n-1c_rv^(k)_j-r = ( ∑_r=0^n-1 c_rexp(-frac2π i rkn) )v^(k)_j. The eigenvalue is sqrt n(Fc)_k. The vectors v^(k) are the columns of F^*, which gives the formula.
Exact mathematical passage
(C(c)v^{(k)})_j = \sum_{r=0}^{n-1}c_rv^{(k)}_{j-r} = \left( \sum_{r=0}^{n-1} c_r\exp\left(-\frac{2\pi\ii rk}{n}\right) \right)v^{(k)}_j.
v^{(k)}_jUse in source · 2 passages
Search aliases: v^(k)_j
Use in source · Read passage · 01_manuscript.tex, line 541
For each frequency k, let v^(k)_j = frac1sqrt n exp(frac2π i jkn). Then (C(c)v^(k))_j = ∑_r=0^n-1c_rv^(k)_j-r = ( ∑_r=0^n-1 c_rexp(-frac2π i rkn) )v^(k)_j. The eigenvalue is sqrt n(Fc)_k. The vectors v^(k) are the columns of F^*, which gives the formula.
Exact mathematical passage
v^{(k)}_j = \frac{1}{\sqrt n} \exp\left(\frac{2\pi\ii jk}{n}\right).
v^{(k)}_{j-r}Use in source · 1 passage
Search aliases: v^(k)_j-r
Use in source · Read passage · 01_manuscript.tex, line 548
For each frequency k, let v^(k)_j = frac1sqrt n exp(frac2π i jkn). Then (C(c)v^(k))_j = ∑_r=0^n-1c_rv^(k)_j-r = ( ∑_r=0^n-1 c_rexp(-frac2π i rkn) )v^(k)_j. The eigenvalue is sqrt n(Fc)_k. The vectors v^(k) are the columns of F^*, which gives the formula.
Exact mathematical passage
(C(c)v^{(k)})_j = \sum_{r=0}^{n-1}c_rv^{(k)}_{j-r} = \left( \sum_{r=0}^{n-1} c_r\exp\left(-\frac{2\pi\ii rk}{n}\right) \right)v^{(k)}_j.
V_AUse in source · 3 passages
Search aliases: V_A
Use in source · Read passage · 01_manuscript.tex, line 687
[Exact BDLR product] Let A=D_A+U_AV_A^*, B=D_B+U_BV_B^* use the same diagonal packet layout. Then AB=D_AD_B+U_CV_C^*, where U_C = D_AU_B+U_A(V_A^*U_B) · U_A and V_C = V_B · D_B^*V_A . The unrecompressed correction rank is at most r_A+r_B.
Exact mathematical passage
V_C = \begin{bmatrix} V_B&D_B^*V_A \end{bmatrix}.
v_aUse in source · 2 passages
Search aliases: v_a
Use in source · pages.tex, line 143
vpage9The cross term has somewhere to goCLOSED LOW-RANK PRODUCT ALGEBRA Let A=D_A+u_av_a^T be the running operator. Take D_B=diag(1,2,3,4,5,6), u_b=(1,0,1,0,1,0)^T, and v_b=(0,1,0,1,0,1)^T. Define B=D_B+u_bv_b^T. U_C=[D_Au_b+u_a(v_a^Tu_b) u_a], V_C=[v_b D_B^Tv_a]. vfigbdlr_product headWhy there are only two columns Expanding AB gives three correction terms, but two share the right factor v_b^T: D_Au_bv_b^T+u_a(v_a^Tu_b)v_b^T =(D_Au_b+u_a(v_a^Tu_b))v_b^T. The remaining term is u_av_a^TD_B. Thus A…
Exact mathematical passage
U_C=[D_Au_b+u_a(v_a^Tu_b)\ \ u_a],\qquad V_C=[v_b\ \ D_B^Tv_a].
V_A^*Use in source · 5 passages
Search aliases: V_A^*
Use in source · Read passage · 01_manuscript.tex, line 669
[Exact BDLR product] Let A=D_A+U_AV_A^*, B=D_B+U_BV_B^* use the same diagonal packet layout. Then AB=D_AD_B+U_CV_C^*, where U_C = D_AU_B+U_A(V_A^*U_B) · U_A and V_C = V_B · D_B^*V_A . The unrecompressed correction rank is at most r_A+r_B.
Exact mathematical passage
A=D_A+U_AV_A^*, \qquad B=D_B+U_BV_B^*
V_A^TUse in source · 1 passage
Search aliases: V_A^T
Use in source · bdlr_product.tex, line 54
1.46) 5; path[draw=black!30,fill=black!8] (5.1,0.68) rectangle (5.62,1.2); node[font=small,align=center,font=small] at (5.36,0.94) 1; path[draw=black!30,fill=black!8] (5.62,0.68) rectangle (6.14,1.2); node[font=small,align=center,font=small] at (5.88,0.94) 6; node[font=small,align=center] at (10.3,3.45) AB=D_AD_B+U_CV_C^T; node[font=small,align=center] at (10.3,2.15) U_A(V_A^TU_B)V_B^T; node[font=small,align=center,font=small] at (10.3,1.35) must be retained; node[font=small,align=center,font=sc…
Exact mathematical passage
U_A(V_A^TU_B)V_B^T
v_a^TUse in source · 10 passages
Search aliases: v_a^T
Use in source · pages.tex, line 142
vpage9The cross term has somewhere to goCLOSED LOW-RANK PRODUCT ALGEBRA Let A=D_A+u_av_a^T be the running operator. Take D_B=diag(1,2,3,4,5,6), u_b=(1,0,1,0,1,0)^T, and v_b=(0,1,0,1,0,1)^T. Define B=D_B+u_bv_b^T. U_C=[D_Au_b+u_a(v_a^Tu_b) u_a], V_C=[v_b D_B^Tv_a]. vfigbdlr_product headWhy there are only two columns Expanding AB gives three correction terms, but two share the right factor v_b^T: D_Au_bv_b^T+u_a(v_a^Tu_b)v_b^T =(D_Au_b+u_a(v_a^Tu_b))v_b^T. The remaining term is u_av_a^TD_B. Thus A…
Exact mathematical passage
A=D_A+u_av_a^T
V_BUse in source · 2 passages
Search aliases: V_B
Use in source · Read passage · 01_manuscript.tex, line 687
[Exact BDLR product] Let A=D_A+U_AV_A^*, B=D_B+U_BV_B^* use the same diagonal packet layout. Then AB=D_AD_B+U_CV_C^*, where U_C = D_AU_B+U_A(V_A^*U_B) · U_A and V_C = V_B · D_B^*V_A . The unrecompressed correction rank is at most r_A+r_B.
Exact mathematical passage
V_C = \begin{bmatrix} V_B&D_B^*V_A \end{bmatrix}.
v_bUse in source · 4 passages
Search aliases: v_b
Use in source · pages.tex, line 142
vpage9The cross term has somewhere to goCLOSED LOW-RANK PRODUCT ALGEBRA Let A=D_A+u_av_a^T be the running operator. Take D_B=diag(1,2,3,4,5,6), u_b=(1,0,1,0,1,0)^T, and v_b=(0,1,0,1,0,1)^T. Define B=D_B+u_bv_b^T. U_C=[D_Au_b+u_a(v_a^Tu_b) u_a], V_C=[v_b D_B^Tv_a]. vfigbdlr_product headWhy there are only two columns Expanding AB gives three correction terms, but two share the right factor v_b^T: D_Au_bv_b^T+u_a(v_a^Tu_b)v_b^T =(D_Au_b+u_a(v_a^Tu_b))v_b^T. The remaining term is u_av_a^TD_B. Thus A…
Exact mathematical passage
v_b=(0,1,0,1,0,1)^T
V_B^*Use in source · 4 passages
Search aliases: V_B^*
Use in source · Read passage · 01_manuscript.tex, line 669
[Exact BDLR product] Let A=D_A+U_AV_A^*, B=D_B+U_BV_B^* use the same diagonal packet layout. Then AB=D_AD_B+U_CV_C^*, where U_C = D_AU_B+U_A(V_A^*U_B) · U_A and V_C = V_B · D_B^*V_A . The unrecompressed correction rank is at most r_A+r_B.
Exact mathematical passage
A=D_A+U_AV_A^*, \qquad B=D_B+U_BV_B^*
V_B^TUse in source · 1 passage
Search aliases: V_B^T
Use in source · bdlr_product.tex, line 54
1.46) 5; path[draw=black!30,fill=black!8] (5.1,0.68) rectangle (5.62,1.2); node[font=small,align=center,font=small] at (5.36,0.94) 1; path[draw=black!30,fill=black!8] (5.62,0.68) rectangle (6.14,1.2); node[font=small,align=center,font=small] at (5.88,0.94) 6; node[font=small,align=center] at (10.3,3.45) AB=D_AD_B+U_CV_C^T; node[font=small,align=center] at (10.3,2.15) U_A(V_A^TU_B)V_B^T; node[font=small,align=center,font=small] at (10.3,1.35) must be retained; node[font=small,align=center,font=sc…
Exact mathematical passage
U_A(V_A^TU_B)V_B^T
v_b^TUse in source · 8 passages
Search aliases: v_b^T
Use in source · pages.tex, line 142
vpage9The cross term has somewhere to goCLOSED LOW-RANK PRODUCT ALGEBRA Let A=D_A+u_av_a^T be the running operator. Take D_B=diag(1,2,3,4,5,6), u_b=(1,0,1,0,1,0)^T, and v_b=(0,1,0,1,0,1)^T. Define B=D_B+u_bv_b^T. U_C=[D_Au_b+u_a(v_a^Tu_b) u_a], V_C=[v_b D_B^Tv_a]. vfigbdlr_product headWhy there are only two columns Expanding AB gives three correction terms, but two share the right factor v_b^T: D_Au_bv_b^T+u_a(v_a^Tu_b)v_b^T =(D_Au_b+u_a(v_a^Tu_b))v_b^T. The remaining term is u_av_a^TD_B. Thus A…
Exact mathematical passage
B=D_B+u_bv_b^T
V_CUse in source · 7 passages
Search aliases: V_C
Use in source · pages.tex, line 143
vpage9The cross term has somewhere to goCLOSED LOW-RANK PRODUCT ALGEBRA Let A=D_A+u_av_a^T be the running operator. Take D_B=diag(1,2,3,4,5,6), u_b=(1,0,1,0,1,0)^T, and v_b=(0,1,0,1,0,1)^T. Define B=D_B+u_bv_b^T. U_C=[D_Au_b+u_a(v_a^Tu_b) u_a], V_C=[v_b D_B^Tv_a]. vfigbdlr_product headWhy there are only two columns Expanding AB gives three correction terms, but two share the right factor v_b^T: D_Au_bv_b^T+u_a(v_a^Tu_b)v_b^T =(D_Au_b+u_a(v_a^Tu_b))v_b^T. The remaining term is u_av_a^TD_B. Thus A…
Exact mathematical passage
U_C=[D_Au_b+u_a(v_a^Tu_b)\ \ u_a],\qquad V_C=[v_b\ \ D_B^Tv_a].
V_C^*Use in source · 2 passages
Search aliases: V_C^*
Use in source · Read passage · 01_manuscript.tex, line 675
[Exact BDLR product] Let A=D_A+U_AV_A^*, B=D_B+U_BV_B^* use the same diagonal packet layout. Then AB=D_AD_B+U_CV_C^*, where U_C = D_AU_B+U_A(V_A^*U_B) · U_A and V_C = V_B · D_B^*V_A . The unrecompressed correction rank is at most r_A+r_B.
Exact mathematical passage
AB=D_AD_B+U_CV_C^*,
V_C^TUse in source · 7 passages
Search aliases: V_C^T
Use in source · pages.tex, line 154
0,1,0,1,0)^T, and v_b=(0,1,0,1,0,1)^T. Define B=D_B+u_bv_b^T. U_C=[D_Au_b+u_a(v_a^Tu_b) u_a], V_C=[v_b D_B^Tv_a]. vfigbdlr_product headWhy there are only two columns Expanding AB gives three correction terms, but two share the right factor v_b^T: D_Au_bv_b^T+u_a(v_a^Tu_b)v_b^T =(D_Au_b+u_a(v_a^Tu_b))v_b^T. The remaining term is u_av_a^TD_B. Thus AB=D_AD_B+U_CV_C^T with at most two correction directions. Counting the three expanded summands as independent ranks would overestimate this representat…
Exact mathematical passage
AB=D_AD_B+U_CV_C^T
v_iUse in source · 5 passages
Search aliases: v_i
Use in source · Read passage · 02_supplement_03.tex, line 168
Keeping node v costs κ_v and incurs integer error w_v. Splitting costs overhead σ_v in addition to the selected child cuts. The exact-weight recurrence is F_v(b)=min{ l κ_v if b=w_v, [2pt] displaystyleσ_v+ min_∑_i b_i=b∑_iF_v_i(b_i) if v is internal. . Nonexistent alternatives are infeasible. The child sum is implemented by the proved finite min-plus convolution, not by assuming an optimal allocation.
Exact mathematical passage
F_v(b)=\min\left\{ \begin{array}{l} \kappa_v\quad\text{if }b=w_v,\\[2pt] \displaystyle\sigma_v+ \min_{\sum_i b_i=b}\sum_iF_{v_i}(b_i)\quad\text{if }v\text{ is internal}. \end{array}\right.
V_jUse in source · 1 passage
Search aliases: V_j
Use in source · Read passage · 02_supplement_03.tex, line 6
Low-rank factors through a product chain. textitSource item 5.2 Let A_j=D_j+U_jV_j^*, U_j,V_j∈ C^n× r_j, with all D_j diagonal in one packet layout. Set P_m=A_m⋯ A_1, D^(m)=D_m⋯ D_1. The formalization constructs thin matrices L_m,R_m such that P_m=D^(m)+L_mR_m^*. Their column index type is a recursively constructed disjoint sum. Its cardinality is proved to equal ∑_j=1^m r_j.
Exact mathematical passage
A_j=D_j+U_jV_j^*,\qquad U_j,V_j\in\mathbb C^{n\times r_j},
V_j^*Use in source · 2 passages
Search aliases: V_j^*
Use in source · Read passage · 01_manuscript.tex, line 716
[Rank growth in a product chain] Let A_j=D_j+U_jV_j^*, j=1,…,m, share one diagonal packet layout. Before recompression, the product A_m⋯ A_1 has a BDLR representation whose correction rank is at most ∑_j=1^m r_j.
Exact mathematical passage
A_j=D_j+U_jV_j^*, \qquad j=1,\ldots,m,
V_mUse in source · 1 passage
Search aliases: V_m
Use in source · Read passage · 02_supplement_03.tex, line 20
Construction. Suppose P_m-1=D^(m-1)+L_m-1R_m-1^*. The new factors are L_m · =[D_mL_m-1+U_m(V_m^*L_m-1) U_m], R_m · =[R_m-1 (D^(m-1))^*V_m]. Substitution into the already checked two-factor product identity proves exactness. This is not a rank-only existence argument. The recursively defined factors are themselves audited.
Exact mathematical passage
L_m&=\bigl[D_mL_{m-1}+U_m(V_m^*L_{m-1})\quad U_m\bigr],\\ R_m&=\bigl[R_{m-1}\quad (D^{(m-1)})^*V_m\bigr].
V_m^*Use in source · 1 passage
Search aliases: V_m^*
Use in source · Read passage · 02_supplement_03.tex, line 20
Construction. Suppose P_m-1=D^(m-1)+L_m-1R_m-1^*. The new factors are L_m · =[D_mL_m-1+U_m(V_m^*L_m-1) U_m], R_m · =[R_m-1 (D^(m-1))^*V_m]. Substitution into the already checked two-factor product identity proves exactness. This is not a rank-only existence argument. The recursively defined factors are themselves audited.
Exact mathematical passage
L_m&=\bigl[D_mL_{m-1}+U_m(V_m^*L_{m-1})\quad U_m\bigr],\\ R_m&=\bigl[R_{m-1}\quad (D^{(m-1)})^*V_m\bigr].
v_\alphaUse in source · 4 passages
Search aliases: v_α alpha
Use in source · Read passage · 01_manuscript.tex, line 1110
[Invariant packet Krylov decomposition] Assume [A,P_α]=0 for every packet. If v=∑_α v_α, v_α=P_α v, then A^jv = ∑_α A_α^jv_α for every j≥0, where A_α=P_α AP_α. Hence K_m(A,v) ⊆ ⊕_α K_m(A_α,v_α).
Exact mathematical passage
v=\sum_\alpha v_\alpha, \qquad v_\alpha=P_\alpha v,
\operatorname{val}Use in source · 1 passage
Search aliases: val operatorname
Use in source · Read passage · 02_supplement_04.tex, line 16
n:001textitSource item 4.3; module CirculantFourier Let n>0, with coordinates indexed by Z/n Z. Define the actual circulant by C(c)_jk=c_j-k. The formal Fourier matrix is F_jk=frac1sqrt n χ(-jk), χ(a)=exp (frac2π i an). The standard character is not left abstract: a checked theorem identifies these entries with frac1sqrt nexp (-frac2π i val(j)val(k)n). The value representatives enter only this explicit exponential formula. All modular algebra uses the residue classes themselves.
Exact mathematical passage
\frac{1}{\sqrt n}\exp\!\left(-\frac{2\pi i\,\operatorname{val}(j)\operatorname{val}(k)}n\right).
\operatorname{vec}Use in source · 7 passages
Search aliases: vec operatorname
Use in source · pages.tex, line 107
vpage7A transpose, not an adjointKRONECKER ACTION AND ARRAY ORDER With column-major vectorization, the tensor identity is (A⊗ B)vec(X)=vec(BXA^T). Here the transpose records index order, not a Hermitian inner product. Use the explicitly complex matrix A=mat1 · i
Exact mathematical passage
(A\otimes B)\operatorname{vec}(X)=\operatorname{vec}(BXA^T).
WUse in source · 27 passages
Search aliases: W
Use in source · pages.tex, line 417
Given operation capacity P and bandwidth W, T≥max(F/P,B/W) is a lower bound, not an achieved runtime. Launch overhead, imperfect overlap, and utilization can make the actual time much larger. takeawayGrouping can remove launches while leaving the same products and data movement. Arithmetic, traffic, and scheduling are related constraints, not one interchangeable cost.
Exact mathematical passage
W
W^*Use in source · 1 passage
Search aliases: W^*
Use in source · Read passage · 02_supplement_06.tex, line 163
Thin output factors and the final theorem. textitModules CompactTruncation and SmallCoreRecompression For the actual core G=R_LR_R^*, supply ordered thin SVD data G=Wdiag(σ)Z^*, W^*W=I, Z^*Z=I, σ_1≥⋯≥σ_d≥0. These are precisely the fields of the formal SVD structure. It contains no rank-optimality or tail-error field. A derived theorem checks the Gram eigenvector equation G^*GZ=Zdiag(σ_1^2,…,σ_d^2). Let h=min(d,k) and retain exactly the first h column indices. The rectangular square-root mask has…
Exact mathematical passage
G=W\diag(\sigma)Z^*,\qquad W^*W=I, \qquad Z^*Z=I,\qquad \sigma_1\geq\cdots\geq\sigma_d\geq0.
W_0Use in source · 1 passage
Search aliases: W_0
Use in source · Read passage · 02_supplement_04.tex, line 106
Operation events. The event type for k+1 modes is a disjoint union of d copies of the trailing event type and the events for the final fiber contraction. Its cardinality obeys W_0=0, W_k+1=dW_k+2d^k+2. Lean proves the closed form W_k=2kd^k+1. The factor-address type has cardinality k d^2. These statements include k=0 and d=0. With no factors, the action is the identity on the singleton coordinate type and incurs zero modeled arithmetic. With d=0 and at least one factor, the coordinate type is em…
Exact mathematical passage
W_0=0,\qquad W_{k+1}=dW_k+2d^{k+2}.
W_aUse in source · 6 passages
Search aliases: W_a
Use in source · Read passage · 02_supplement_05.tex, line 85
Fix the source packet a. In the regular representation, define the submodule W_a={x:P_bx=0 whenever anotleadsto b}. It contains P_a. Completeness of the projectors expands each component of Ax into packet transfers. If a transfer into an unreachable packet were nonzero from a reachable packet, it would extend a path, a contradiction. Thus AW_a⊆ W_a.
Exact mathematical passage
W_a=\{x:P_bx=0\ \text{whenever }a\not\leadsto b\}.
W_BUse in source · 3 passages
Search aliases: W_B
Use in source · Read passage · 02_supplement_03.tex, line 217
Let W_B(t) be the declared unit-cost count of table-cell constructions and tested min-plus candidates. The model charges the empty child-combination table as well as the output cells and merge candidates. Lean proves W_B(t)≤2(B+1)N+((B+1)+(B+1)^2)(N-1) and hence W_B(t)≤ N(3(B+1)+(B+1)^2). The root minimum scan adds a linear pass. Thus the declared tabulation model is pseudo-polynomial in N and the numeric budget B.
Exact mathematical passage
W_B(t)
w_iUse in source · 8 passages
Search aliases: w_i
Use in source · Read passage · 02_supplement_06.tex, line 128
Frobenius optimality from kernel weights. textitModules KernelFrame and FrobeniusOptimality For an arbitrary competitor B with rank B≤ k, construct an orthonormal matrix Q spanning its actual kernel. Then BQ=0, Q^*Q=I, q=dimker B=d-rank B. Define the row weights w_i=∑_j=1^q|Q_ij|^2. The proof derives 0≤ w_i≤1, ∑_iw_i=q≥max(d-k,0). The upper bound uses contraction of Q^* on a coordinate vector. The weight sum follows from the trace of Q^*Q and rank-nullity.
Exact mathematical passage
w_i=\sum_{j=1}^{q}|Q_{ij}|^2.
w_{j,k}Use in source · 2 passages
Search aliases: w_j,k
Use in source · Read passage · 01_manuscript.tex, line 1438
ct discrete precision program] Let the total integer error budget be B. Define F_j(b) as the minimum cost for assigning precisions to the first j edges with total weight exactly b. Then F_0(0)=0, F_0(b)=+∞ for b≠0. Set F_j-1(u)=+∞ whenever u<0 or the state is unreachable. Then F_j(b) = min_k { F_j-1(b-w_j,k)+c_j,k }. The minimum of F_m(b) over 0≤ b≤ B is the exact minimum cost for the quantized model.
Exact mathematical passage
\boxed{ F_j(b) = \min_k \left\{ F_{j-1}(b-w_{j,k})+c_{j,k} \right\}. }
W_kUse in source · 3 passages
Search aliases: W_k
Use in source · Read passage · 01_manuscript.tex, line 825
R=Q_RR_R, where the columns of Q_L and Q_R are orthonormal. Form the small core G=R_LR_R^* and let G=WΣ Z^* be an ordered thin singular value decomposition, with nonnegative nonincreasing singular values. For an integer k≥0, retain min{k,d} columns, where d is the number of supplied thin singular directions; tail singular values beyond d are zero. Define L_k=Q_LW_kΣ_k^1/2, R_k=Q_RZ_kΣ_k^1/2. Then L_kR_k^* is a best rank at most k approximation to LR^*. Its spectral and Frobenius errors are the s…
Exact mathematical passage
L_k=Q_LW_k\Sigma_k^{1/2}, \qquad R_k=Q_RZ_k\Sigma_k^{1/2}.
W_{k+1}Use in source · 1 passage
Search aliases: W_k+1
Use in source · Read passage · 02_supplement_04.tex, line 106
Operation events. The event type for k+1 modes is a disjoint union of d copies of the trailing event type and the events for the final fiber contraction. Its cardinality obeys W_0=0, W_k+1=dW_k+2d^k+2. Lean proves the closed form W_k=2kd^k+1. The factor-address type has cardinality k d^2. These statements include k=0 and d=0. With no factors, the action is the identity on the singleton coordinate type and incurs zero modeled arithmetic. With d=0 and at least one factor, the coordinate type is em…
Exact mathematical passage
W_0=0,\qquad W_{k+1}=dW_k+2d^{k+2}.
w_vUse in source · 4 passages
Search aliases: w_v
Use in source · Read passage · 01_manuscript.tex, line 1658
[Quantized budget dynamic program] Assume every keep error is a nonnegative integer w_v. Let B∈ N. For each node v, define F_v(b) as the minimum cost of a pruning of the subtree at v with total error exactly b. The keep option contributes F_v(w_v)≤κ_v. The split option is obtained by min plus convolution of the child tables, followed by addition of σ_v. The smallest value among F_rm root(b), 0≤ b≤ B, is the exact minimum cost among prunings whose quantized error is at most B.
Exact mathematical passage
w_v
\operatorname{Walsh}Use in source · 1 passage
Search aliases: Walsh operatorname
Use in source · Read passage · 02_supplement_04.tex, line 140
Actual butterfly. For the two halves x_0,x_1 of the input, the independently defined procedure returns Walsh_2N(x_0,x_1) =(Walsh_N(x_0+x_1), Walsh_N(x_0-x_1)). Lean proves that this vector equals H_2Nx. The argument expands the two first-coordinate cases and applies finite-sum distributivity. It does not assume that the recursive algorithm already represents the matrix.
Exact mathematical passage
\operatorname{Walsh}_{2N}(x_0,x_1) =\bigl(\operatorname{Walsh}_N(x_0+x_1), \operatorname{Walsh}_N(x_0-x_1)\bigr).
\operatorname{Walsh}_{2N}Use in source · 1 passage
Search aliases: Walsh_2N operatorname
Use in source · Read passage · 02_supplement_04.tex, line 140
Actual butterfly. For the two halves x_0,x_1 of the input, the independently defined procedure returns Walsh_2N(x_0,x_1) =(Walsh_N(x_0+x_1), Walsh_N(x_0-x_1)). Lean proves that this vector equals H_2Nx. The argument expands the two first-coordinate cases and applies finite-sum distributivity. It does not assume that the recursive algorithm already represents the matrix.
Exact mathematical passage
\operatorname{Walsh}_{2N}(x_0,x_1) =\bigl(\operatorname{Walsh}_N(x_0+x_1), \operatorname{Walsh}_N(x_0-x_1)\bigr).
\operatorname{Walsh}_NUse in source · 1 passage
Search aliases: Walsh_N operatorname
Use in source · Read passage · 02_supplement_04.tex, line 140
Actual butterfly. For the two halves x_0,x_1 of the input, the independently defined procedure returns Walsh_2N(x_0,x_1) =(Walsh_N(x_0+x_1), Walsh_N(x_0-x_1)). Lean proves that this vector equals H_2Nx. The argument expands the two first-coordinate cases and applies finite-sum distributivity. It does not assume that the recursive algorithm already represents the matrix.
Exact mathematical passage
\operatorname{Walsh}_{2N}(x_0,x_1) =\bigl(\operatorname{Walsh}_N(x_0+x_1), \operatorname{Walsh}_N(x_0-x_1)\bigr).
\widehat AUse in source · 23 passages
Search aliases: widehat A widehat
Use in source · pages.tex, line 333
For orthogonal coordinate blocks, the squared Frobenius norms add exactly. Operator norms instead require a suitable bound or further spectral information. The maximum omitted block norm is generally insufficient. headInput sensitivity is additional information The packet perturbation estimate is (widehat A-A)x ≤∑_e E_e P_α(e)x . It can be much tighter for an input that scarcely occupies an affected source packet. To obtain a uniform operator bound, the input dependence must be bounded uniformly…
Exact mathematical passage
\|(\widehat A-A)x\|\le\sum_e\|E_e\|\,\|P_{\alpha(e)}x\|.
\widehat A_1Use in source · 3 passages
Search aliases: widehat A_1 widehat
Use in source · Read passage · 01_manuscript.tex, line 864
[Product chain perturbation] Let A_1,…,A_m be exact factors and let widehat A_j=A_j+E_j. Then widehat A_m⋯widehat A_1 - A_m⋯ A_1 ≤ ∑_j=1^m ( ∏_i=j+1^mnormwidehat A_i ) normE_j ( ∏_i=1^j-1normA_i ). Empty products are one.
Exact mathematical passage
\boxed{ \left\| \widehat A_m\cdots\widehat A_1 - A_m\cdots A_1 \right\| \leq \sum_{j=1}^m \left( \prod_{i=j+1}^m\norm{\widehat A_i} \right) \norm{E_j} \left( \prod_{i=1}^{j-1}\norm{A_i} \right). }
\widehat A_iUse in source · 2 passages
Search aliases: widehat A_i widehat
Use in source · Read passage · 01_manuscript.tex, line 864
[Product chain perturbation] Let A_1,…,A_m be exact factors and let widehat A_j=A_j+E_j. Then widehat A_m⋯widehat A_1 - A_m⋯ A_1 ≤ ∑_j=1^m ( ∏_i=j+1^mnormwidehat A_i ) normE_j ( ∏_i=1^j-1normA_i ). Empty products are one.
Exact mathematical passage
\boxed{ \left\| \widehat A_m\cdots\widehat A_1 - A_m\cdots A_1 \right\| \leq \sum_{j=1}^m \left( \prod_{i=j+1}^m\norm{\widehat A_i} \right) \norm{E_j} \left( \prod_{i=1}^{j-1}\norm{A_i} \right). }
\widehat A_jUse in source · 7 passages
Search aliases: widehat A_j widehat
Use in source · Read passage · 01_manuscript.tex, line 860
[Product chain perturbation] Let A_1,…,A_m be exact factors and let widehat A_j=A_j+E_j. Then widehat A_m⋯widehat A_1 - A_m⋯ A_1 ≤ ∑_j=1^m ( ∏_i=j+1^mnormwidehat A_i ) normE_j ( ∏_i=1^j-1normA_i ). Empty products are one.
Exact mathematical passage
\widehat A_j=A_j+E_j.
\widehat A_{j+1}Use in source · 2 passages
Search aliases: widehat A_j+1 widehat
Use in source · Read passage · 01_manuscript.tex, line 887
Use the telescoping identity widehat A_m⋯widehat A_1 -A_m⋯ A_1 = ∑_j=1^m widehat A_m⋯widehat A_j+1 E_j A_j-1⋯ A_1. Take norms term by term.
Exact mathematical passage
\widehat A_m\cdots\widehat A_1 -A_m\cdots A_1 = \sum_{j=1}^m \widehat A_m\cdots\widehat A_{j+1} E_j A_{j-1}\cdots A_1.
\widehat A_mUse in source · 3 passages
Search aliases: widehat A_m widehat
Use in source · Read passage · 01_manuscript.tex, line 864
[Product chain perturbation] Let A_1,…,A_m be exact factors and let widehat A_j=A_j+E_j. Then widehat A_m⋯widehat A_1 - A_m⋯ A_1 ≤ ∑_j=1^m ( ∏_i=j+1^mnormwidehat A_i ) normE_j ( ∏_i=1^j-1normA_i ). Empty products are one.
Exact mathematical passage
\boxed{ \left\| \widehat A_m\cdots\widehat A_1 - A_m\cdots A_1 \right\| \leq \sum_{j=1}^m \left( \prod_{i=j+1}^m\norm{\widehat A_i} \right) \norm{E_j} \left( \prod_{i=1}^{j-1}\norm{A_i} \right). }
\widehat CUse in source · 2 passages
Search aliases: widehat C widehat
Use in source · Read passage · 01_manuscript.tex, line 1309
[Floating point matrix product model] For real A∈ R^m× p and B∈ R^p× n, consider ordinary dot-product matrix multiplication in uniform precision with unit roundoff u, no overflow or underflow, and pu<1. Its computed output widehat C satisfies the componentwise forward-error bound |widehat C-AB|≤γ_p|A||B|, γ_p=fracpu1-pu. Absolute values and the inequality are entrywise. Complex arithmetic, fused kernels, mixed accumulation, and tensor kernels require their own stated constants. This fact does no…
Exact mathematical passage
\widehat C
\widehat FUse in source · 2 passages
Search aliases: widehat F widehat
Use in source · Read passage · 02_supplement_02.tex, line 86
Let v be the new optimum of the edited subtree, and let j be the edited child slot at one frame. The update performed at that frame is step(v)= min{K, σ+ ∑_i v, · i=j, widehat F_i, · i≠ j. }. This result is passed to the next ancestor. The update function does not recursively evaluate sibling subtrees. It reads their stored values widehat F_i.
Exact mathematical passage
\boxed{ \operatorname{step}(v)= \min\left\{K,\;\sigma+ \sum_i\begin{cases} v,&i=j,\\ \widehat F_i,&i\ne j. \end{cases}\right\}. }
\widehat F_iUse in source · 2 passages
Search aliases: widehat F_i widehat
Use in source · Read passage · 02_supplement_02.tex, line 86
Let v be the new optimum of the edited subtree, and let j be the edited child slot at one frame. The update performed at that frame is step(v)= min{K, σ+ ∑_i v, · i=j, widehat F_i, · i≠ j. }. This result is passed to the next ancestor. The update function does not recursively evaluate sibling subtrees. It reads their stored values widehat F_i.
Exact mathematical passage
\boxed{ \operatorname{step}(v)= \min\left\{K,\;\sigma+ \sum_i\begin{cases} v,&i=j,\\ \widehat F_i,&i\ne j. \end{cases}\right\}. }
\widehat PUse in source · 4 passages
Search aliases: widehat P widehat
Use in source · Read passage · 01_manuscript.tex, line 907
[Certified repeated recompression] Let P_0=I, P_j=A_jP_j-1, be exact partial products. Suppose an implemented chain satisfies widehat P_0=I, widehat P_j = A_jwidehat P_j-1+E_j, where E_j is the additive error introduced after the jth multiplication, for example by factor recompression. Then normwidehat P_m-P_m ≤ ∑_j=1^m ( ∏_i=j+1^mnormA_i ) normE_j. If normE_j≤δ_j, the same formula with δ_j gives a certified final error.
Exact mathematical passage
\widehat P_0=I, \qquad \widehat P_j = A_j\widehat P_{j-1}+E_j,
\widehat pUse in source · 3 passages
Search aliases: widehat p widehat
Use in source · pages.tex, line 217
vpage13Errors enter at different placesFACTOR PERTURBATION VERSUS RECOMPRESSION Let the exact scalar factors be (2,3,4) and start from p_0=1. If additive errors (1/10,1/5,1/20) enter after each partial multiplication, then widehat p_j=a_jwidehat p_j-1+e_j, widehat p_3-p_3=12e_1+4e_2+e_3=frac4120. vfigerror_chain headThe same local numbers, a different model If instead the factors themselves are perturbed, the computed product is (4+e_3)(3+e_2)(2+e_1). Its error is 402/125, not 41/20. The factor-…
Exact mathematical passage
\widehat p_j=a_j\widehat p_{j-1}+e_j,\qquad \widehat p_3-p_3=12e_1+4e_2+e_3=\frac{41}{20}.
\widehat P_0Use in source · 1 passage
Search aliases: widehat P_0 widehat
Use in source · Read passage · 01_manuscript.tex, line 907
[Certified repeated recompression] Let P_0=I, P_j=A_jP_j-1, be exact partial products. Suppose an implemented chain satisfies widehat P_0=I, widehat P_j = A_jwidehat P_j-1+E_j, where E_j is the additive error introduced after the jth multiplication, for example by factor recompression. Then normwidehat P_m-P_m ≤ ∑_j=1^m ( ∏_i=j+1^mnormA_i ) normE_j. If normE_j≤δ_j, the same formula with δ_j gives a certified final error.
Exact mathematical passage
\widehat P_0=I, \qquad \widehat P_j = A_j\widehat P_{j-1}+E_j,
\widehat p_3Use in source · 2 passages
Search aliases: widehat p_3 widehat
Use in source · pages.tex, line 217
vpage13Errors enter at different placesFACTOR PERTURBATION VERSUS RECOMPRESSION Let the exact scalar factors be (2,3,4) and start from p_0=1. If additive errors (1/10,1/5,1/20) enter after each partial multiplication, then widehat p_j=a_jwidehat p_j-1+e_j, widehat p_3-p_3=12e_1+4e_2+e_3=frac4120. vfigerror_chain headThe same local numbers, a different model If instead the factors themselves are perturbed, the computed product is (4+e_3)(3+e_2)(2+e_1). Its error is 402/125, not 41/20. The factor-…
Exact mathematical passage
\widehat p_j=a_j\widehat p_{j-1}+e_j,\qquad \widehat p_3-p_3=12e_1+4e_2+e_3=\frac{41}{20}.
\widehat P_jUse in source · 3 passages
Search aliases: widehat P_j widehat
Use in source · Read passage · 01_manuscript.tex, line 907
[Certified repeated recompression] Let P_0=I, P_j=A_jP_j-1, be exact partial products. Suppose an implemented chain satisfies widehat P_0=I, widehat P_j = A_jwidehat P_j-1+E_j, where E_j is the additive error introduced after the jth multiplication, for example by factor recompression. Then normwidehat P_m-P_m ≤ ∑_j=1^m ( ∏_i=j+1^mnormA_i ) normE_j. If normE_j≤δ_j, the same formula with δ_j gives a certified final error.
Exact mathematical passage
\widehat P_0=I, \qquad \widehat P_j = A_j\widehat P_{j-1}+E_j,
\widehat p_jUse in source · 3 passages
Search aliases: widehat p_j widehat
Use in source · pages.tex, line 217
vpage13Errors enter at different placesFACTOR PERTURBATION VERSUS RECOMPRESSION Let the exact scalar factors be (2,3,4) and start from p_0=1. If additive errors (1/10,1/5,1/20) enter after each partial multiplication, then widehat p_j=a_jwidehat p_j-1+e_j, widehat p_3-p_3=12e_1+4e_2+e_3=frac4120. vfigerror_chain headThe same local numbers, a different model If instead the factors themselves are perturbed, the computed product is (4+e_3)(3+e_2)(2+e_1). Its error is 402/125, not 41/20. The factor-…
Exact mathematical passage
\widehat p_j=a_j\widehat p_{j-1}+e_j,\qquad \widehat p_3-p_3=12e_1+4e_2+e_3=\frac{41}{20}.
\widehat P_{j+1}Use in source · 1 passage
Search aliases: widehat P_j+1 widehat
Use in source · Read passage · 02_supplement_03.tex, line 36
Actual noncommutative products. The error module defines the left-accumulated product recursively, with P_0=I and P_j+1=A_j+1P_j. For a perturbed sequence widehat A_j, it derives the discrepancy recursion widehat P_j+1-P_j+1 =widehat A_j+1(widehat P_j-P_j) +(widehat A_j+1-A_j+1)P_j. The identity follows from the actual products. It is not supplied as an assumption to a norm estimate, and it does not commute the operators. Lean uses zero-based indices internally, so index j in source code corresp…
Exact mathematical passage
\widehat P_{j+1}-P_{j+1} =\widehat A_{j+1}(\widehat P_j-P_j) +(\widehat A_{j+1}-A_{j+1})P_j.
\widehat P_{j-1}Use in source · 1 passage
Search aliases: widehat P_j-1 widehat
Use in source · Read passage · 01_manuscript.tex, line 907
[Certified repeated recompression] Let P_0=I, P_j=A_jP_j-1, be exact partial products. Suppose an implemented chain satisfies widehat P_0=I, widehat P_j = A_jwidehat P_j-1+E_j, where E_j is the additive error introduced after the jth multiplication, for example by factor recompression. Then normwidehat P_m-P_m ≤ ∑_j=1^m ( ∏_i=j+1^mnormA_i ) normE_j. If normE_j≤δ_j, the same formula with δ_j gives a certified final error.
Exact mathematical passage
\widehat P_0=I, \qquad \widehat P_j = A_j\widehat P_{j-1}+E_j,
\widehat p_{j-1}Use in source · 3 passages
Search aliases: widehat p_j-1 widehat
Use in source · pages.tex, line 217
vpage13Errors enter at different placesFACTOR PERTURBATION VERSUS RECOMPRESSION Let the exact scalar factors be (2,3,4) and start from p_0=1. If additive errors (1/10,1/5,1/20) enter after each partial multiplication, then widehat p_j=a_jwidehat p_j-1+e_j, widehat p_3-p_3=12e_1+4e_2+e_3=frac4120. vfigerror_chain headThe same local numbers, a different model If instead the factors themselves are perturbed, the computed product is (4+e_3)(3+e_2)(2+e_1). Its error is 402/125, not 41/20. The factor-…
Exact mathematical passage
\widehat p_j=a_j\widehat p_{j-1}+e_j,\qquad \widehat p_3-p_3=12e_1+4e_2+e_3=\frac{41}{20}.
\widehat P_mUse in source · 1 passage
Search aliases: widehat P_m widehat
Use in source · Read passage · 01_manuscript.tex, line 915
ntheorem[Certified repeated recompression] Let P_0=I, P_j=A_jP_j-1, be exact partial products. Suppose an implemented chain satisfies widehat P_0=I, widehat P_j = A_jwidehat P_j-1+E_j, where E_j is the additive error introduced after the jth multiplication, for example by factor recompression. Then normwidehat P_m-P_m ≤ ∑_j=1^m ( ∏_i=j+1^mnormA_i ) normE_j. If normE_j≤δ_j, the same formula with δ_j gives a certified final error.
Exact mathematical passage
\boxed{ \norm{\widehat P_m-P_m} \leq \sum_{j=1}^{m} \left( \prod_{i=j+1}^{m}\norm{A_i} \right) \norm{E_j}. }
\widetilde PUse in source · 3 passages
Search aliases: widetilde P widetilde
Use in source · Read passage · 02_supplement_03.tex, line 58
Post-product recompression is different. Define the implemented iterates by widetilde P_0=I, widetilde P_j=A_jwidetilde P_j-1+E_j. Here E_j is a residual introduced after multiplication. Lean derives widetilde P_j-P_j=A_j(widetilde P_j-1-P_j-1)+E_j and proves normwidetilde P_m-P_m ≤∑_j=1^m(∏_i=j+1^mnormA_i)δ_j, normE_j≤δ_j. This theorem works in a normed ring without the extra norm-one condition. It includes the empty iteration. There is no exact-prefix factor in the local residual term, because…
Exact mathematical passage
\widetilde P_0=I,\qquad \widetilde P_j=A_j\widetilde P_{j-1}+E_j.
\widetilde P_0Use in source · 1 passage
Search aliases: widetilde P_0 widetilde
Use in source · Read passage · 02_supplement_03.tex, line 58
Post-product recompression is different. Define the implemented iterates by widetilde P_0=I, widetilde P_j=A_jwidetilde P_j-1+E_j. Here E_j is a residual introduced after multiplication. Lean derives widetilde P_j-P_j=A_j(widetilde P_j-1-P_j-1)+E_j and proves normwidetilde P_m-P_m ≤∑_j=1^m(∏_i=j+1^mnormA_i)δ_j, normE_j≤δ_j. This theorem works in a normed ring without the extra norm-one condition. It includes the empty iteration. There is no exact-prefix factor in the local residual term, because…
Exact mathematical passage
\widetilde P_0=I,\qquad \widetilde P_j=A_j\widetilde P_{j-1}+E_j.
\widetilde P_jUse in source · 2 passages
Search aliases: widetilde P_j widetilde
Use in source · Read passage · 02_supplement_03.tex, line 58
Post-product recompression is different. Define the implemented iterates by widetilde P_0=I, widetilde P_j=A_jwidetilde P_j-1+E_j. Here E_j is a residual introduced after multiplication. Lean derives widetilde P_j-P_j=A_j(widetilde P_j-1-P_j-1)+E_j and proves normwidetilde P_m-P_m ≤∑_j=1^m(∏_i=j+1^mnormA_i)δ_j, normE_j≤δ_j. This theorem works in a normed ring without the extra norm-one condition. It includes the empty iteration. There is no exact-prefix factor in the local residual term, because…
Exact mathematical passage
\widetilde P_0=I,\qquad \widetilde P_j=A_j\widetilde P_{j-1}+E_j.
\widetilde P_{j-1}Use in source · 2 passages
Search aliases: widetilde P_j-1 widetilde
Use in source · Read passage · 02_supplement_03.tex, line 58
Post-product recompression is different. Define the implemented iterates by widetilde P_0=I, widetilde P_j=A_jwidetilde P_j-1+E_j. Here E_j is a residual introduced after multiplication. Lean derives widetilde P_j-P_j=A_j(widetilde P_j-1-P_j-1)+E_j and proves normwidetilde P_m-P_m ≤∑_j=1^m(∏_i=j+1^mnormA_i)δ_j, normE_j≤δ_j. This theorem works in a normed ring without the extra norm-one condition. It includes the empty iteration. There is no exact-prefix factor in the local residual term, because…
Exact mathematical passage
\widetilde P_0=I,\qquad \widetilde P_j=A_j\widetilde P_{j-1}+E_j.
\widetilde P_mUse in source · 1 passage
Search aliases: widetilde P_m widetilde
Use in source · Read passage · 02_supplement_03.tex, line 67
Post-product recompression is different. Define the implemented iterates by widetilde P_0=I, widetilde P_j=A_jwidetilde P_j-1+E_j. Here E_j is a residual introduced after multiplication. Lean derives widetilde P_j-P_j=A_j(widetilde P_j-1-P_j-1)+E_j and proves normwidetilde P_m-P_m ≤∑_j=1^m(∏_i=j+1^mnormA_i)δ_j, normE_j≤δ_j. This theorem works in a normed ring without the extra norm-one condition. It includes the empty iteration. There is no exact-prefix factor in the local residual term, because…
Exact mathematical passage
\boxed{\norm{\widetilde P_m-P_m} \leq\sum_{j=1}^m\left(\prod_{i=j+1}^m\norm{A_i}\right)\delta_j, \quad \norm{E_j}\leq\delta_j.}
\xUse in source · 1 passage
Search aliases: x
Use in source · shared_family.tex, line 8
tex] draw[->] (0,0)--(7,0) node[right] r_2; draw[->] (0,0)--(0,4.5) node[above] modeled cost; draw[thick] (0,.6)--(4,3.4) node[right] 6+7r_2 : keep; draw[thick,dashed] (0,1)--(4,1.8) node[right] 10+2r_2 : split; draw[dotted] (.8,0)--(.8,1.16);fill (.8,1.16) circle (2pt); node[below] at (.8,0) 4/5; foreach x in 0,2,3,4draw (x,0)--(x,-.08);node[below] at (x,-.1) x; draw[dotted] (3,0)--(3,2.7);fill (3,2.7) circle (2pt);fill (3,1.6) circle (2pt); node[above left] at (3,2.7) 27;node[below right] at (…
Exact mathematical passage
\x
x^{(0)}Use in source · 1 passage
Search aliases: x^(0)
Use in source · Read passage · 02_supplement_05.tex, line 58
Parallel stages. A second, independent definition is an explicit staged evaluator. Set x^(0)=0 and x^(k+1)_v= D_v^-1 (b_v-displaystyle∑_u→ vT_vux^(k)_u), · ℓ(v)=k+1,
Exact mathematical passage
x^{(0)}=0
x^{(d)}Use in source · 1 passage
Search aliases: x^(d)
Use in source · Read passage · 02_supplement_05.tex, line 72
[5pt] x^(k)_v, · ℓ(v)≠ k+1. It reads the previous-stage vector, not partially updated coordinates. The invariant is ℓ(v)≤ k Longrightarrow x^(k)_v=x_v. Every predecessor of a level-k+1 vertex has level at most k, so the recurrence uses only coordinates already proved correct. At the canonical depth d this yields Ax^(d)=b. The local independence theorem separately proves that changing coordinates outside the strict earlier levels cannot change a vertex's local solve.
Exact mathematical passage
\boxed{Ax^{(d)}=b.}
x^{(k)}_uUse in source · 1 passage
Search aliases: x^(k)_u
Use in source · Read passage · 02_supplement_05.tex, line 59
Parallel stages. A second, independent definition is an explicit staged evaluator. Set x^(0)=0 and x^(k+1)_v= D_v^-1 (b_v-displaystyle∑_u→ vT_vux^(k)_u), · ℓ(v)=k+1, [5pt] x^(k)_v, · ℓ(v)≠ k+1. It reads the previous-stage vector, not partially updated coordinates. The invariant is ℓ(v)≤ k Longrightarrow x^(k)_v=x_v. Every predecessor of a level-k+1 vertex has level at most k, so the recurrence uses only coordinates already proved correct. At the canonical depth d this yields Ax^(d)=b. The local …
Exact mathematical passage
x^{(k+1)}_v= \begin{cases} D_v^{-1}\!\left(b_v-\displaystyle\sum_{u\to v}T_{vu}x^{(k)}_u\right), &\ell(v)=k+1,\\[5pt] x^{(k)}_v,&\ell(v)\ne k+1. \end{cases}
x^{(k)}_vUse in source · 2 passages
Search aliases: x^(k)_v
Use in source · Read passage · 02_supplement_05.tex, line 59
Parallel stages. A second, independent definition is an explicit staged evaluator. Set x^(0)=0 and x^(k+1)_v= D_v^-1 (b_v-displaystyle∑_u→ vT_vux^(k)_u), · ℓ(v)=k+1, [5pt] x^(k)_v, · ℓ(v)≠ k+1. It reads the previous-stage vector, not partially updated coordinates. The invariant is ℓ(v)≤ k Longrightarrow x^(k)_v=x_v. Every predecessor of a level-k+1 vertex has level at most k, so the recurrence uses only coordinates already proved correct. At the canonical depth d this yields Ax^(d)=b. The local …
Exact mathematical passage
x^{(k+1)}_v= \begin{cases} D_v^{-1}\!\left(b_v-\displaystyle\sum_{u\to v}T_{vu}x^{(k)}_u\right), &\ell(v)=k+1,\\[5pt] x^{(k)}_v,&\ell(v)\ne k+1. \end{cases}
x^{(k+1)}_vUse in source · 1 passage
Search aliases: x^(k+1)_v
Use in source · Read passage · 02_supplement_05.tex, line 59
Parallel stages. A second, independent definition is an explicit staged evaluator. Set x^(0)=0 and x^(k+1)_v= D_v^-1 (b_v-displaystyle∑_u→ vT_vux^(k)_u), · ℓ(v)=k+1, [5pt] x^(k)_v, · ℓ(v)≠ k+1. It reads the previous-stage vector, not partially updated coordinates. The invariant is ℓ(v)≤ k Longrightarrow x^(k)_v=x_v. Every predecessor of a level-k+1 vertex has level at most k, so the recurrence uses only coordinates already proved correct. At the canonical depth d this yields Ax^(d)=b. The local …
Exact mathematical passage
x^{(k+1)}_v= \begin{cases} D_v^{-1}\!\left(b_v-\displaystyle\sum_{u\to v}T_{vu}x^{(k)}_u\right), &\ell(v)=k+1,\\[5pt] x^{(k)}_v,&\ell(v)\ne k+1. \end{cases}
x^*Use in source · 1 passage
Search aliases: x^*
Use in source · Read passage · 01_manuscript.tex, line 1203
For every unit vector x, absx^*Ex≤η. The Rayleigh quotient of I+E lies in [1-η,1+η]. The quotient of the extreme eigenvalue bounds gives the result.
Exact mathematical passage
\abs{x^*Ex}\leq\eta.
x^TUse in source · 3 passages
Search aliases: x^T
Use in source · pages.tex, line 308
1/2 · 1. Its exact eigenvalues are 1/2 and 3/2, giving condition number three. vfigpreconditioning headThe unit circles of the energy norms The left ellipse is x^TAx=1. Writing x=By gives the right ellipse y^T(BAB)y=1. The drawing samples these explicitly defined curves; the norm and eigenvalue equalities are checked symbolically.
Exact mathematical passage
x^TAx=1
x_*Use in source · 1 passage
Search aliases: x_*
Use in source · Read passage · 01_manuscript.tex, line 1217
[Preconditioned conjugate gradient convergence] Let A and M be Hermitian positive definite on a nonempty finite coordinate space, and set κ = cond_2(M^-1/2AM^-1/2). In exact arithmetic, the energy norm error of preconditioned conjugate gradient satisfies normx_m-x_*_A ≤ 2 ( fracsqrtκ-1sqrtκ+1 )^m normx_0-x_*_A.
Exact mathematical passage
\norm{x_m-x_*}_A \leq 2 \left( \frac{\sqrt\kappa-1}{\sqrt\kappa+1} \right)^m \norm{x_0-x_*}_A.
x_0Use in source · 5 passages
Search aliases: x_0
Use in source · Read passage · 01_manuscript.tex, line 647
The orthogonality identity follows by induction from block multiplication. One recursive stage forms x_0+x_1 and x_0-x_1, which uses N additions or subtractions, and then applies two transforms of size N/2. Thus T(N)=2T(N/2)+N, T(1)=0, whose solution is T(N)=Nlog_2N.
Exact mathematical passage
x_0+x_1
x_1Use in source · 4 passages
Search aliases: x_1
Use in source · Read passage · 01_manuscript.tex, line 647
The orthogonality identity follows by induction from block multiplication. One recursive stage forms x_0+x_1 and x_0-x_1, which uses N additions or subtractions, and then applies two transforms of size N/2. Thus T(N)=2T(N/2)+N, T(1)=0, whose solution is T(N)=Nlog_2N.
Exact mathematical passage
x_0+x_1
x_2Use in source · 2 passages
Search aliases: x_2
Use in source · pages.tex, line 249
, put two on the diagonal and -1 on each declared incoming edge. The edges are 0→2, 1→2, 1→3, 2→4, 3→4, 3→5, 4→6, and 5→6. Solve (2I+K)x=ones. vfigtriangular headRead the stages as dependencies Packets zero and one are ready first. Packets two and three use those values. Packets four and five then become available, followed by packet six. For instance, x_4=frac1+x_2+x_32=frac118, x_6=frac1+x_4+x_52=frac138. All entries in the displayed solution are checked by multiplying the original triangular …
Exact mathematical passage
x_4=\frac{1+x_2+x_3}{2}=\frac{11}{8},\qquad x_6=\frac{1+x_4+x_5}{2}=\frac{13}{8}.
x_3Use in source · 2 passages
Search aliases: x_3
Use in source · pages.tex, line 249
, put two on the diagonal and -1 on each declared incoming edge. The edges are 0→2, 1→2, 1→3, 2→4, 3→4, 3→5, 4→6, and 5→6. Solve (2I+K)x=ones. vfigtriangular headRead the stages as dependencies Packets zero and one are ready first. Packets two and three use those values. Packets four and five then become available, followed by packet six. For instance, x_4=frac1+x_2+x_32=frac118, x_6=frac1+x_4+x_52=frac138. All entries in the displayed solution are checked by multiplying the original triangular …
Exact mathematical passage
x_4=\frac{1+x_2+x_3}{2}=\frac{11}{8},\qquad x_6=\frac{1+x_4+x_5}{2}=\frac{13}{8}.
x_4Use in source · 2 passages
Search aliases: x_4
Use in source · pages.tex, line 249
, put two on the diagonal and -1 on each declared incoming edge. The edges are 0→2, 1→2, 1→3, 2→4, 3→4, 3→5, 4→6, and 5→6. Solve (2I+K)x=ones. vfigtriangular headRead the stages as dependencies Packets zero and one are ready first. Packets two and three use those values. Packets four and five then become available, followed by packet six. For instance, x_4=frac1+x_2+x_32=frac118, x_6=frac1+x_4+x_52=frac138. All entries in the displayed solution are checked by multiplying the original triangular …
Exact mathematical passage
x_4=\frac{1+x_2+x_3}{2}=\frac{11}{8},\qquad x_6=\frac{1+x_4+x_5}{2}=\frac{13}{8}.
x_5Use in source · 2 passages
Search aliases: x_5
Use in source · pages.tex, line 249
, put two on the diagonal and -1 on each declared incoming edge. The edges are 0→2, 1→2, 1→3, 2→4, 3→4, 3→5, 4→6, and 5→6. Solve (2I+K)x=ones. vfigtriangular headRead the stages as dependencies Packets zero and one are ready first. Packets two and three use those values. Packets four and five then become available, followed by packet six. For instance, x_4=frac1+x_2+x_32=frac118, x_6=frac1+x_4+x_52=frac138. All entries in the displayed solution are checked by multiplying the original triangular …
Exact mathematical passage
x_4=\frac{1+x_2+x_3}{2}=\frac{11}{8},\qquad x_6=\frac{1+x_4+x_5}{2}=\frac{13}{8}.
x_6Use in source · 2 passages
Search aliases: x_6
Use in source · pages.tex, line 249
, put two on the diagonal and -1 on each declared incoming edge. The edges are 0→2, 1→2, 1→3, 2→4, 3→4, 3→5, 4→6, and 5→6. Solve (2I+K)x=ones. vfigtriangular headRead the stages as dependencies Packets zero and one are ready first. Packets two and three use those values. Packets four and five then become available, followed by packet six. For instance, x_4=frac1+x_2+x_32=frac118, x_6=frac1+x_4+x_52=frac138. All entries in the displayed solution are checked by multiplying the original triangular …
Exact mathematical passage
x_4=\frac{1+x_2+x_3}{2}=\frac{11}{8},\qquad x_6=\frac{1+x_4+x_5}{2}=\frac{13}{8}.
X_{ij}Use in source · 1 passage
Search aliases: X_ij
Use in source · Read passage · 02_supplement_04.tex, line 77
Kronecker application. For A∈ C^m× m, B∈ C^n× n and X∈ C^n× m, Lean proves (A⊗ B)vec(X)=vec(BXA^T). The formal vectorization is explicitly column-major: its pair coordinate (j,i) stores X_ij. The right factor is the ordinary transpose, not the conjugate transpose. This distinction is essential over C.
Exact mathematical passage
X_{ij}
X_{k\ell}Use in source · 1 passage
Search aliases: X_kℓ ell
Use in source · Read passage · 01_manuscript.tex, line 606
The (i,j) entry of BXA^T is ∑_k=1^n∑_ℓ=1^m B_ikX_kℓA_jℓ. This is the coordinate formula for (A⊗ B)vec(X). Multiplying first by B and then by A^T gives the stated count.
Exact mathematical passage
\sum_{k=1}^{n}\sum_{\ell=1}^{m} B_{ik}X_{k\ell}A_{j\ell}.
x_mUse in source · 1 passage
Search aliases: x_m
Use in source · Read passage · 01_manuscript.tex, line 1217
[Preconditioned conjugate gradient convergence] Let A and M be Hermitian positive definite on a nonempty finite coordinate space, and set κ = cond_2(M^-1/2AM^-1/2). In exact arithmetic, the energy norm error of preconditioned conjugate gradient satisfies normx_m-x_*_A ≤ 2 ( fracsqrtκ-1sqrtκ+1 )^m normx_0-x_*_A.
Exact mathematical passage
\norm{x_m-x_*}_A \leq 2 \left( \frac{\sqrt\kappa-1}{\sqrt\kappa+1} \right)^m \norm{x_0-x_*}_A.
X_RUse in source · 5 passages
Search aliases: X_R
Use in source · Read passage · 01_manuscript.tex, line 1064
Let R be the set of packets reachable from α, including α, and set X_R=⊕_γ∈ RX_γ. No active edge leaves R, so AX_R⊆ X_R. The restriction of A to the finite dimensional space X_R is injective. It is therefore onto X_R. Hence A^-1X_R⊆ X_R. Since β∉ R, the stated block of A^-1 vanishes.
Exact mathematical passage
X_R=\bigoplus_{\gamma\in R}X_\gamma.
x_\alphaUse in source · 1 passage
Search aliases: x_α alpha
Use in source · Read passage · 01_manuscript.tex, line 1026
The equation in packet β is D_β x_β = b_β - ∑_α→βK_βαx_α. Every predecessor of β lies in an earlier level. Once those values are available, x_β is obtained from the local solve with D_β. Vertices in the same level have no directed edge between them, so their solves are independent.
Exact mathematical passage
D_\beta x_\beta = b_\beta - \sum_{\alpha\to\beta}K_{\beta\alpha}x_\alpha.
x_\betaUse in source · 2 passages
Search aliases: x_β beta
Use in source · Read passage · 01_manuscript.tex, line 1026
The equation in packet β is D_β x_β = b_β - ∑_α→βK_βαx_α. Every predecessor of β lies in an earlier level. Once those values are available, x_β is obtained from the local solve with D_β. Vertices in the same level have no directed edge between them, so their solves are independent.
Exact mathematical passage
D_\beta x_\beta = b_\beta - \sum_{\alpha\to\beta}K_{\beta\alpha}x_\alpha.
X_\gammaUse in source · 1 passage
Search aliases: X_γ gamma
Use in source · Read passage · 01_manuscript.tex, line 1064
Let R be the set of packets reachable from α, including α, and set X_R=⊕_γ∈ RX_γ. No active edge leaves R, so AX_R⊆ X_R. The restriction of A to the finite dimensional space X_R is injective. It is therefore onto X_R. Hence A^-1X_R⊆ X_R. Since β∉ R, the stated block of A^-1 vanishes.
Exact mathematical passage
X_R=\bigoplus_{\gamma\in R}X_\gamma.
YUse in source · 3 passages
Search aliases: Y
Use in source · Read passage · 01_manuscript.tex, line 196
[Adjoint and tensor semantics] The adjoint representation has blocks (A^*)_αβ=A_βα^*. If A and B are packet resolved on X and Y, then (A⊗ B)_(β,δ),(α,γ) = A_βα⊗ B_δγ.
Exact mathematical passage
Y
y^{2}Use in source · 1 passage
Search aliases: y^2
Use in source · Read passage · 02_supplement_06.tex, line 101
Monotonicity of the diagonal entries gives normDy^2 =∑_i=1^k+1σ_i^2|y_i|^2 ≥σ_k+1^2norm y^2. Because (D-B)y=Dy, the operator-norm inequality implies σ_k+1norm y ≤norm(D-B)y ≤normD-B_2norm y. Cancellation of the strictly positive norm of y proves σ_k+1≤normD-B_2. The norm of the actual diagonal residual D-D_k equals σ_k+1, so D_k attains the lower bound. If k≥ d, the residual is zero and optimality follows from norm nonnegativity.
Exact mathematical passage
\norm{Dy}^{2} =\sum_{i=1}^{k+1}\sigma_i^{2}|y_i|^{2} \geq\sigma_{k+1}^{2}\norm y^{2}.
y^TUse in source · 3 passages
Search aliases: y^T
Use in source · pages.tex, line 308
1/2 · 1. Its exact eigenvalues are 1/2 and 3/2, giving condition number three. vfigpreconditioning headThe unit circles of the energy norms The left ellipse is x^TAx=1. Writing x=By gives the right ellipse y^T(BAB)y=1. The drawing samples these explicitly defined curves; the norm and eigenvalue equalities are checked symbolically.
Exact mathematical passage
y^T(BAB)y=1
y_iUse in source · 1 passage
Search aliases: y_i
Use in source · Read passage · 02_supplement_06.tex, line 101
Monotonicity of the diagonal entries gives normDy^2 =∑_i=1^k+1σ_i^2|y_i|^2 ≥σ_k+1^2norm y^2. Because (D-B)y=Dy, the operator-norm inequality implies σ_k+1norm y ≤norm(D-B)y ≤normD-B_2norm y. Cancellation of the strictly positive norm of y proves σ_k+1≤normD-B_2. The norm of the actual diagonal residual D-D_k equals σ_k+1, so D_k attains the lower bound. If k≥ d, the residual is zero and optimality follows from norm nonnegativity.
Exact mathematical passage
\norm{Dy}^{2} =\sum_{i=1}^{k+1}\sigma_i^{2}|y_i|^{2} \geq\sigma_{k+1}^{2}\norm y^{2}.
ZUse in source · 15 passages
Search aliases: Z
Use in source · Read passage · 01_manuscript.tex, line 114
[Exact evaluation] Let A∈End(X). Set B_βα=π_β Aι_α and let E contain exactly the nonzero blocks. Then A= ∑_(β,α)∈ E P_β AP_α. For every matrix of right hand sides Z, AZ = ∑_(β,α)∈ E ι_β B_βαπ_α Z.
Exact mathematical passage
Z
\mathbb ZUse in source · 1 passage
Search aliases: Z mathbb
Use in source · Read passage · 02_supplement_04.tex, line 5
Circulant Fourier diagonalization. textitSource item 4.3; module CirculantFourier Let n>0, with coordinates indexed by Z/n Z. Define the actual circulant by C(c)_jk=c_j-k. The formal Fourier matrix is F_jk=frac1sqrt n χ(-jk), χ(a)=exp (frac2π i an). The standard character is not left abstract: a checked theorem identifies these entries with frac1sqrt nexp (-frac2π i val(j)val(k)n). The value representatives enter only this explicit exponential formula. All modular algebra uses the residue classe…
Exact mathematical passage
\mathbb Z/n\mathbb Z
zUse in source · 13 passages
Search aliases: z
Use in source · Read passage · 01_manuscript.tex, line 497
[Sparse plus low rank application] Let A=S+UV^*, where S has z stored scalar nonzeros and U,V∈ C^n× r. In the model that charges two operations per stored sparse entry and right hand side, applying A to q right hand sides costs 2q(z+2nr) modeled operations and stores z+2nr scalar values, apart from sparse index metadata.
Exact mathematical passage
z
Z^*Use in source · 3 passages
Search aliases: Z^*
Use in source · Read passage · 01_manuscript.tex, line 821
[Small core recompression] Let L∈ C^m× r, R∈ C^n× r. Take compact QR factorizations L=Q_LR_L, R=Q_RR_R, where the columns of Q_L and Q_R are orthonormal. Form the small core G=R_LR_R^* and let G=WΣ Z^* be an ordered thin singular value decomposition, with nonnegative nonincreasing singular values. For an integer k≥0, retain min{k,d} columns, where d is the number of supplied thin singular directions; tail singular values beyond d are zero. Define L_k=Q_LW_kΣ_k^1/2, R_k=Q_RZ_kΣ_k^1/2. Then L_kR_k…
Exact mathematical passage
G=W\Sigma Z^*
z^jUse in source · 1 passage
Search aliases: z^j
Use in source · Read passage · 01_manuscript.tex, line 978
[Polynomial support] If p(z)=∑_j=0^dc_jz^j, then supp(p(A)) is contained in the union of the support relations of A^j for 0≤ j≤ d.
Exact mathematical passage
p(z)=\sum_{j=0}^dc_jz^j,
Z_kUse in source · 1 passage
Search aliases: Z_k
Use in source · Read passage · 01_manuscript.tex, line 825
R=Q_RR_R, where the columns of Q_L and Q_R are orthonormal. Form the small core G=R_LR_R^* and let G=WΣ Z^* be an ordered thin singular value decomposition, with nonnegative nonincreasing singular values. For an integer k≥0, retain min{k,d} columns, where d is the number of supplied thin singular directions; tail singular values beyond d are zero. Define L_k=Q_LW_kΣ_k^1/2, R_k=Q_RZ_kΣ_k^1/2. Then L_kR_k^* is a best rank at most k approximation to LR^*. Its spectral and Frobenius errors are the s…
Exact mathematical passage
L_k=Q_LW_k\Sigma_k^{1/2}, \qquad R_k=Q_RZ_k\Sigma_k^{1/2}.
|Use in source · 10 passages
Search aliases: |
Use in source · Read passage · 01_manuscript.tex, line 1310
[Floating point matrix product model] For real A∈ R^m× p and B∈ R^p× n, consider ordinary dot-product matrix multiplication in uniform precision with unit roundoff u, no overflow or underflow, and pu<1. Its computed output widehat C satisfies the componentwise forward-error bound |widehat C-AB|≤γ_p|A||B|, γ_p=fracpu1-pu. Absolute values and the inequality are entrywise. Complex arithmetic, fused kernels, mixed accumulation, and tensor kernels require their own stated constants. This fact does no…
Exact mathematical passage
|\widehat C-AB|\le\gamma_p|A||B|, \qquad\gamma_p=\frac{pu}{1-pu}.
|^2Use in source · 1 passage
Search aliases: |^2
Use in source · Read passage · 02_supplement_06.tex, line 128
Frobenius optimality from kernel weights. textitModules KernelFrame and FrobeniusOptimality For an arbitrary competitor B with rank B≤ k, construct an orthonormal matrix Q spanning its actual kernel. Then BQ=0, Q^*Q=I, q=dimker B=d-rank B. Define the row weights w_i=∑_j=1^q|Q_ij|^2. The proof derives 0≤ w_i≤1, ∑_iw_i=q≥max(d-k,0). The upper bound uses contraction of Q^* on a coordinate vector. The weight sum follows from the trace of Q^*Q and rank-nullity.
Exact mathematical passage
w_i=\sum_{j=1}^{q}|Q_{ij}|^2.
|^{2}Use in source · 1 passage
Search aliases: |^2
Use in source · Read passage · 02_supplement_06.tex, line 101
Monotonicity of the diagonal entries gives normDy^2 =∑_i=1^k+1σ_i^2|y_i|^2 ≥σ_k+1^2norm y^2. Because (D-B)y=Dy, the operator-norm inequality implies σ_k+1norm y ≤norm(D-B)y ≤normD-B_2norm y. Cancellation of the strictly positive norm of y proves σ_k+1≤normD-B_2. The norm of the actual diagonal residual D-D_k equals σ_k+1, so D_k attains the lower bound. If k≥ d, the residual is zero and optimality follows from norm nonnegativity.
Exact mathematical passage
\norm{Dy}^{2} =\sum_{i=1}^{k+1}\sigma_i^{2}|y_i|^{2} \geq\sigma_{k+1}^{2}\norm y^{2}.
\alpha_0Use in source · 2 passages
Search aliases: α_0 alpha
Use in source · Read passage · 01_manuscript.tex, line 956
[Path support of powers] For every positive integer m, P_γ A^mP_α is the sum over directed walks (vertices may repeat) α=α_0 ⟶ α_1 ⟶⋯⟶ α_m=γ in G_A, with contribution A_α_mα_m-1 ⋯ A_α_1α_0. If no such path exists, the block vanishes.
Exact mathematical passage
\alpha=\alpha_0 \longrightarrow \alpha_1 \longrightarrow\cdots\longrightarrow \alpha_m=\gamma
\alpha_1Use in source · 2 passages
Search aliases: α_1 alpha
Use in source · Read passage · 01_manuscript.tex, line 956
[Path support of powers] For every positive integer m, P_γ A^mP_α is the sum over directed walks (vertices may repeat) α=α_0 ⟶ α_1 ⟶⋯⟶ α_m=γ in G_A, with contribution A_α_mα_m-1 ⋯ A_α_1α_0. If no such path exists, the block vanishes.
Exact mathematical passage
\alpha=\alpha_0 \longrightarrow \alpha_1 \longrightarrow\cdots\longrightarrow \alpha_m=\gamma
\alpha_mUse in source · 2 passages
Search aliases: α_m alpha
Use in source · Read passage · 01_manuscript.tex, line 956
[Path support of powers] For every positive integer m, P_γ A^mP_α is the sum over directed walks (vertices may repeat) α=α_0 ⟶ α_1 ⟶⋯⟶ α_m=γ in G_A, with contribution A_α_mα_m-1 ⋯ A_α_1α_0. If no such path exists, the block vanishes.
Exact mathematical passage
\alpha=\alpha_0 \longrightarrow \alpha_1 \longrightarrow\cdots\longrightarrow \alpha_m=\gamma
\alpha_{m-1}Use in source · 1 passage
Search aliases: α_m-1 alpha
Use in source · Read passage · 01_manuscript.tex, line 964
[Path support of powers] For every positive integer m, P_γ A^mP_α is the sum over directed walks (vertices may repeat) α=α_0 ⟶ α_1 ⟶⋯⟶ α_m=γ in G_A, with contribution A_α_mα_m-1 ⋯ A_α_1α_0. If no such path exists, the block vanishes.
Exact mathematical passage
A_{\alpha_m\alpha_{m-1}} \cdots A_{\alpha_1\alpha_0}.
\gamma_pUse in source · 1 passage
Search aliases: γ_p gamma
Use in source · Read passage · 01_manuscript.tex, line 1310
[Floating point matrix product model] For real A∈ R^m× p and B∈ R^p× n, consider ordinary dot-product matrix multiplication in uniform precision with unit roundoff u, no overflow or underflow, and pu<1. Its computed output widehat C satisfies the componentwise forward-error bound |widehat C-AB|≤γ_p|A||B|, γ_p=fracpu1-pu. Absolute values and the inequality are entrywise. Complex arithmetic, fused kernels, mixed accumulation, and tensor kernels require their own stated constants. This fact does no…
Exact mathematical passage
|\widehat C-AB|\le\gamma_p|A||B|, \qquad\gamma_p=\frac{pu}{1-pu}.
\DeltaUse in source · 13 passages
Search aliases: Δ Delta
Use in source · Read passage · 01_manuscript.tex, line 1322
[Mixed precision product chain] If each fixed linear factor A_j is replaced by a fixed packetwise linear approximation widehat A_j, write E_j=widehat A_j-A_j=∑_eΔ_j,eP_α(j,e). A uniform certificate is E_j ≤δ_j:=∑_e Δ_j,e P_α(j,e) . For orthogonal packet projectors, δ_j≤∑_e Δ_j,e . The final operator error is bounded by [source reference] with these uniform δ_j. The corresponding vector error is this operator bound multiplied by the input norm.
Exact mathematical passage
E_j=\widehat A_j-A_j=\sum_e\Delta_{j,e}P_{\alpha(j,e)}
\deltaUse in source · 12 passages
Search aliases: δ delta
Use in source · Read passage · 01_manuscript.tex, line 10
This article begins with a finite direct sum X=⊕_α∈ IX_α and a complete family of coordinate projectors P_α P_β=δ_αβP_α, ∑_α∈ IP_α=I. An operator has packet blocks A_βα=P_β AP_α. The active addresses form a directed graph. Block dimensions, block types, approximation tolerances, and output requirements turn this graph into a computational intermediate representation.
Exact mathematical passage
P_\alpha P_\beta=\delta_{\alpha\beta}P_\alpha, \qquad \sum_{\alpha\in I}P_\alpha=I.
\Delta_eUse in source · 8 passages
Search aliases: Δ_e Delta
Use in source · Read passage · 02_supplement_01.tex, line 127
Packet perturbation certificates. textitSource item 8.1 and two uniform consequences Let X be a normed space over a nontrivially normed field. The formal objects are actual continuous linear maps X→ X, with their operator norm. Let a(e) identify the source packet of edge e, and suppose Δ_e=Δ_e P_a(e). Set A=∑_e A_e, widehat A=∑_e(A_e+Δ_e). The exact operator and vector identities are proved first: widehat A-A=∑_eΔ_e P_a(e), (widehat A-A)x=∑_eΔ_e(P_a(e)x). Applying the finite triangle inequality …
Exact mathematical passage
\Delta_e=\Delta_e P_{a(e)}.
\delta_jUse in source · 9 passages
Search aliases: δ_j delta
Use in source · Read passage · 01_manuscript.tex, line 926
ad P_j=A_jP_j-1, be exact partial products. Suppose an implemented chain satisfies widehat P_0=I, widehat P_j = A_jwidehat P_j-1+E_j, where E_j is the additive error introduced after the jth multiplication, for example by factor recompression. Then normwidehat P_m-P_m ≤ ∑_j=1^m ( ∏_i=j+1^mnormA_i ) normE_j. If normE_j≤δ_j, the same formula with δ_j gives a certified final error.
Exact mathematical passage
\norm{E_j}\leq\delta_j
\Delta_{j,e}Use in source · 5 passages
Search aliases: Δ_j,e Delta
Use in source · Read passage · 01_manuscript.tex, line 1322
[Mixed precision product chain] If each fixed linear factor A_j is replaced by a fixed packetwise linear approximation widehat A_j, write E_j=widehat A_j-A_j=∑_eΔ_j,eP_α(j,e). A uniform certificate is E_j ≤δ_j:=∑_e Δ_j,e P_α(j,e) . For orthogonal packet projectors, δ_j≤∑_e Δ_j,e . The final operator error is bounded by [source reference] with these uniform δ_j. The corresponding vector error is this operator bound multiplied by the input norm.
Exact mathematical passage
E_j=\widehat A_j-A_j=\sum_e\Delta_{j,e}P_{\alpha(j,e)}
\delta_{\alpha\beta}Use in source · 1 passage
Search aliases: δ_αβ delta alpha beta
Use in source · Read passage · 01_manuscript.tex, line 10
This article begins with a finite direct sum X=⊕_α∈ IX_α and a complete family of coordinate projectors P_α P_β=δ_αβP_α, ∑_α∈ IP_α=I. An operator has packet blocks A_βα=P_β AP_α. The active addresses form a directed graph. Block dimensions, block types, approximation tolerances, and output requirements turn this graph into a computational intermediate representation.
Exact mathematical passage
P_\alpha P_\beta=\delta_{\alpha\beta}P_\alpha, \qquad \sum_{\alpha\in I}P_\alpha=I.
\varepsilonUse in source · 22 passages
Search aliases: ε varepsilon
Use in source · Read passage · 02_supplement_01.tex, line 146
_a(e). Set A=∑_e A_e, widehat A=∑_e(A_e+Δ_e). The exact operator and vector identities are proved first: widehat A-A=∑_eΔ_e P_a(e), (widehat A-A)x=∑_eΔ_e(P_a(e)x). Applying the finite triangle inequality and the defining operator-norm estimate gives norm(widehat A-A)x ≤∑_enormΔ_e normP_a(e)x. If normΔ_e≤ε_enormA_e, then the formal relative certificate is norm(widehat A-A)x ≤∑_eε_enormA_e normP_a(e)x. The source additionally assumes that each exact block is source supported. The proof of this bou…
Exact mathematical passage
\norm{\Delta_e}\le\varepsilon_e\norm{A_e}
\varepsilon^*Use in source · 3 passages
Search aliases: ε^* varepsilon
Use in source · Read passage · 02_supplement_02.tex, line 141
Global comparison. For any other feasible ε, set r_e=ε_e/ε_e^*>0. The budget identities give ∑_ec_er_e=C. Using the standard analytic inequality log r≤ r-1 yields ∑_ec_elog r_e≤∑_ec_e(r_e-1)=0. Lean separately proves the exact objective-difference identity J(ε)-J(ε^*)=-∑_ec_elog r_e. Therefore J(ε)≥ J(ε^*).
Exact mathematical passage
J(\varepsilon)-J(\varepsilon^*)=-\sum_ec_e\log r_e.
\epsilon_eUse in source · 11 passages
Search aliases: ϵ_e epsilon
Use in source · pages.tex, line 25
s_e,ϵ_e · Sensitivity and tolerance · Is the budget input-dependent or uniform over unit inputs?
Exact mathematical passage
s_e,\epsilon_e
\varepsilon_eUse in source · 13 passages
Search aliases: ε_e varepsilon
Use in source · Read passage · 02_supplement_01.tex, line 146
_a(e). Set A=∑_e A_e, widehat A=∑_e(A_e+Δ_e). The exact operator and vector identities are proved first: widehat A-A=∑_eΔ_e P_a(e), (widehat A-A)x=∑_eΔ_e(P_a(e)x). Applying the finite triangle inequality and the defining operator-norm estimate gives norm(widehat A-A)x ≤∑_enormΔ_e normP_a(e)x. If normΔ_e≤ε_enormA_e, then the formal relative certificate is norm(widehat A-A)x ≤∑_eε_enormA_e normP_a(e)x. The source additionally assumes that each exact block is source supported. The proof of this bou…
Exact mathematical passage
\norm{\Delta_e}\le\varepsilon_e\norm{A_e}
\varepsilon_e^*Use in source · 2 passages
Search aliases: ε_e^* varepsilon
Use in source · Read passage · 02_supplement_02.tex, line 126
lete textitSource item 8.4, with an explicit positive budget Let E be a finite nonempty edge set. Assume s_e>0, c_e>0, τ>0, and define C=∑_ec_e. The optimization problem is min_ε_e>0∑_ec_elogfrac1ε_e subject to ∑_es_eε_e=τ. Lean constructs the displayed candidate ε_e^*=fracτ c_es_eC and proves strict positivity and the exact budget equality. It then proves global optimality and uniqueness for the actual logarithmic objective.
Exact mathematical passage
\boxed{\varepsilon_e^*=\frac{\tau c_e}{s_eC}}
\epsilon_v^{(j)}Use in source · 2 passages
Search aliases: ϵ_v^(j) epsilon
Use in source · Read passage · 01_manuscript.tex, line 1588
[Shared compilation for an operator family] Let operators A^(1),…,A^(m) use the same candidate tree. Give operator j a nonnegative reuse weight r_j. Replace the local keep cost and error by κ_v = κ_v^rm setup + ∑_j=1^m r_jκ_v^(j) and ϵ_v = ∑_j=1^m r_jϵ_v^(j). Then [source reference] gives the exact optimal shared pruning for the additive weighted objective.
Exact mathematical passage
\epsilon_v = \sum_{j=1}^m r_j\epsilon_v^{(j)}.
\etaUse in source · 12 passages
Search aliases: η eta
Use in source · pages.tex, line 311
Here E _2=1/2. The manuscript's general inequality κ_2(D^-1/2AD^-1/2) ≤frac1+η1-η, E _2≤η<1, is attained. The raw off-diagonal entry is one; its meaningful size is one-half only after normalization by the diagonal energy scales. headKeep the convergence input separate The condition bound can be inserted into a conjugate-gradient convergence estimate under that method's positive-definite assumptions. The CG estimate is one of the separately tracked classical inputs. The algebraic normalization th…
Exact mathematical passage
\kappa_2(D^{-1/2}AD^{-1/2}) \le\frac{1+\eta}{1-\eta},\qquad \|E\|_2\le\eta<1,
\kappa_2Use in source · 6 passages
Search aliases: κ_2 kappa
Use in source · pages.tex, line 311
Here E _2=1/2. The manuscript's general inequality κ_2(D^-1/2AD^-1/2) ≤frac1+η1-η, E _2≤η<1, is attained. The raw off-diagonal entry is one; its meaningful size is one-half only after normalization by the diagonal energy scales. headKeep the convergence input separate The condition bound can be inserted into a conjugate-gradient convergence estimate under that method's positive-definite assumptions. The CG estimate is one of the separately tracked classical inputs. The algebraic normalization th…
Exact mathematical passage
\kappa_2(D^{-1/2}AD^{-1/2}) \le\frac{1+\eta}{1-\eta},\qquad \|E\|_2\le\eta<1,
\kappa_v^{(j)}Use in source · 2 passages
Search aliases: κ_v^(j) kappa
Use in source · Read passage · 01_manuscript.tex, line 1580
[Shared compilation for an operator family] Let operators A^(1),…,A^(m) use the same candidate tree. Give operator j a nonnegative reuse weight r_j. Replace the local keep cost and error by κ_v = κ_v^rm setup + ∑_j=1^m r_jκ_v^(j) and ϵ_v = ∑_j=1^m r_jϵ_v^(j). Then [source reference] gives the exact optimal shared pruning for the additive weighted objective.
Exact mathematical passage
\kappa_v = \kappa_v^{\rm setup} + \sum_{j=1}^m r_j\kappa_v^{(j)}
\kappa_v^{\rm setup}Use in source · 1 passage
Search aliases: κ_v^rm setup kappa rm
Use in source · Read passage · 01_manuscript.tex, line 1580
[Shared compilation for an operator family] Let operators A^(1),…,A^(m) use the same candidate tree. Give operator j a nonnegative reuse weight r_j. Replace the local keep cost and error by κ_v = κ_v^rm setup + ∑_j=1^m r_jκ_v^(j) and ϵ_v = ∑_j=1^m r_jϵ_v^(j). Then [source reference] gives the exact optimal shared pruning for the additive weighted objective.
Exact mathematical passage
\kappa_v = \kappa_v^{\rm setup} + \sum_{j=1}^m r_j\kappa_v^{(j)}
\kappa_v^{\mathrm{setup}}Use in source · 1 passage
Search aliases: κ_v^setup kappa mathrm
Use in source · Read passage · 02_supplement_02.tex, line 29
Shared operator families. For reuse weights r_j, the local keep model can aggregate setup and per-operator costs: κ_v=κ_v^setup+∑_jr_jκ_v^(j), ϵ_v=∑_jr_jϵ_v^(j). Lean proves the aggregation identity at the cut level, including split overhead, before applying the compiler theorem. The proof is valid over an ordered field, including Q and R. The intended nonnegative model is a specialization.
Exact mathematical passage
\kappa_v=\kappa_v^{\mathrm{setup}}+\sum_jr_j\kappa_v^{(j)}, \qquad \epsilon_v=\sum_jr_j\epsilon_v^{(j)}.
\lambda_1Use in source · 5 passages
Search aliases: λ_1 lambda
Use in source · Read passage · 01_manuscript.tex, line 1601
[Monotone Lagrangian frontier] Let 0≤λ_1<λ_2. Choose optimal prunings L_1 and L_2 for the two Lagrangian parameters. Then err( L_2) ≤ err( L_1) and cost( L_2) ≥ cost( L_1).
Exact mathematical passage
0\leq\lambda_1<\lambda_2
\lambda_2Use in source · 5 passages
Search aliases: λ_2 lambda
Use in source · Read passage · 01_manuscript.tex, line 1601
[Monotone Lagrangian frontier] Let 0≤λ_1<λ_2. Choose optimal prunings L_1 and L_2 for the two Lagrangian parameters. Then err( L_2) ≤ err( L_1) and cost( L_2) ≥ cost( L_1).
Exact mathematical passage
0\leq\lambda_1<\lambda_2
\muUse in source · 4 passages
Search aliases: μ mu
Use in source · Read passage · 02_supplement_06.tex, line 141
For any nonnegative decreasing sequence a_i, weights with these properties satisfy ∑_i a_iw_i≥∑_i>ka_i. This finite weighted inequality is itself proved. For k<d, use μ=a_k+1 and compare the summands to tail weights v_i=0 for i≤ k and v_i=1 otherwise: a_iv_i+μ(w_i-v_i)≤ a_iw_i. For a retained coordinate the inequality uses a_i≥μ and w_i≥0. For a tail coordinate it uses a_i≤μ and w_i≤1. Summation and ∑ w_i≥∑ v_i give the result.
Exact mathematical passage
\mu=a_{k+1}
\nuUse in source · 3 passages
Search aliases: ν nu
Use in source · Read passage · 02_supplement_06.tex, line 54
ubsection:002textitModule SmallCoreNorms; arbitrary ambient competitors For rectangular isometries Q,V define L(G)=QGV^*, C(B)=Q^*BV, Q^*Q=I, V^*V=I. The following statements are proved for the actual matrix rank and actual matrix norms: C( L(G))=G, rank L(G)=rank G, rank C(B)≤rank B, norm L(G)_ν=normG_ν, norm C(B)_ν≤norm B_ν, ν∈{2, F}. For the spectral norm, the proof uses the Gram matrix identity (QG)^*(QG)=G^*G and the C^*-norm square identity, together with right-side adjoint symmetry. Compr…
Exact mathematical passage
\boxed{\norm{\mathcal L(G)}_\nu=\norm{G}_\nu, \qquad \norm{\mathcal C(B)}_\nu\leq\norm B_\nu, \quad \nu\in\{2,\mathrm F\}.}
\PiUse in source · 6 passages
Search aliases: Π Pi
Use in source · Read passage · 01_manuscript.tex, line 1469
[Exact block discovery certificate] For an orthogonal packet system (P_α), define the block diagonal compression Π_ P(A) = ∑_α P_α AP_α. Then normA-Π_ P(A)_rm F^2 = ∑_α≠β normP_β AP_α_rm F^2. In operator norm, normA-Π_ P(A)_2 ≤ ∑_α≠β normP_β AP_α_2. Thus any proposed packet discovery has an exact finite residual certificate after its projectors are known.
Exact mathematical passage
\Pi_{\Pcal}(A) = \sum_\alpha P_\alpha AP_\alpha.
\Pi_{\Pcal}Use in source · 3 passages
Search aliases: Π_Pcal Π_ P Pi Pcal Pi mathcal
Use in source · Read passage · 01_manuscript.tex, line 1469
[Exact block discovery certificate] For an orthogonal packet system (P_α), define the block diagonal compression Π_ P(A) = ∑_α P_α AP_α. Then normA-Π_ P(A)_rm F^2 = ∑_α≠β normP_β AP_α_rm F^2. In operator norm, normA-Π_ P(A)_2 ≤ ∑_α≠β normP_β AP_α_2. Thus any proposed packet discovery has an exact finite residual certificate after its projectors are known.
Exact mathematical passage
\Pi_{\Pcal}(A) = \sum_\alpha P_\alpha AP_\alpha.
\pi_\betaUse in source · 1 passage
Search aliases: π_β pi beta
Use in source · Read passage · 01_manuscript.tex, line 103
[Exact evaluation] Let A∈End(X). Set B_βα=π_β Aι_α and let E contain exactly the nonzero blocks. Then A= ∑_(β,α)∈ E P_β AP_α. For every matrix of right hand sides Z, AZ = ∑_(β,α)∈ E ι_β B_βαπ_α Z.
Exact mathematical passage
B_{\beta\alpha}=\pi_\beta A\iota_\alpha
\SigmaUse in source · 2 passages
Search aliases: Σ Sigma
Use in source · Read passage · 01_manuscript.tex, line 821
[Small core recompression] Let L∈ C^m× r, R∈ C^n× r. Take compact QR factorizations L=Q_LR_L, R=Q_RR_R, where the columns of Q_L and Q_R are orthonormal. Form the small core G=R_LR_R^* and let G=WΣ Z^* be an ordered thin singular value decomposition, with nonnegative nonincreasing singular values. For an integer k≥0, retain min{k,d} columns, where d is the number of supplied thin singular directions; tail singular values beyond d are zero. Define L_k=Q_LW_kΣ_k^1/2, R_k=Q_RZ_kΣ_k^1/2. Then L_kR_k…
Exact mathematical passage
G=W\Sigma Z^*
\sigma_1Use in source · 4 passages
Search aliases: σ_1 sigma
Use in source · Read passage · 01_manuscript.tex, line 784
[Eckart Young Mirsky theorem] If M has singular values σ_1≥σ_2≥⋯, pad this finite sequence with zeros. For each integer k≥0, its truncated singular value decomposition M_k is a best rank-at-most-k approximation. It satisfies normM-M_k_2=σ_k+1 and normM-M_k_rm F = ( ∑_j>kσ_j^2 )^1/2.
Exact mathematical passage
\sigma_1\geq\sigma_2\geq\cdots,
\sigma_1^2Use in source · 1 passage
Search aliases: σ_1^2 sigma
Use in source · Read passage · 02_supplement_06.tex, line 168
05textitModules CompactTruncation and SmallCoreRecompression For the actual core G=R_LR_R^*, supply ordered thin SVD data G=Wdiag(σ)Z^*, W^*W=I, Z^*Z=I, σ_1≥⋯≥σ_d≥0. These are precisely the fields of the formal SVD structure. It contains no rank-optimality or tail-error field. A derived theorem checks the Gram eigenvector equation G^*GZ=Zdiag(σ_1^2,…,σ_d^2). Let h=min(d,k) and retain exactly the first h column indices. The rectangular square-root mask has entries M_ij=sqrtσ_j, · i=j,
Exact mathematical passage
G^*GZ=Z\diag(\sigma_1^2,\ldots,\sigma_d^2).
\sigma_2Use in source · 1 passage
Search aliases: σ_2 sigma
Use in source · Read passage · 01_manuscript.tex, line 784
[Eckart Young Mirsky theorem] If M has singular values σ_1≥σ_2≥⋯, pad this finite sequence with zeros. For each integer k≥0, its truncated singular value decomposition M_k is a best rank-at-most-k approximation. It satisfies normM-M_k_2=σ_k+1 and normM-M_k_rm F = ( ∑_j>kσ_j^2 )^1/2.
Exact mathematical passage
\sigma_1\geq\sigma_2\geq\cdots,
\sigma_dUse in source · 3 passages
Search aliases: σ_d sigma
Use in source · Read passage · 02_supplement_06.tex, line 87
Spectral optimality from a kernel witness. textitModules DiagonalTruncation and SpectralOptimality Let σ_1≥⋯≥σ_d≥0 and D=diag(σ_1,…,σ_d), D_k=diag(σ_1,…,σ_min(k,d),0,…,0). Lean uses indices beginning at zero; the formulas on this page use the conventional one-based notation. The rank of D_k is proved to be at most k, even when retained singular values are zero.
Exact mathematical passage
\sigma_1\geq\cdots\geq\sigma_d\geq0
\sigma_d^2Use in source · 1 passage
Search aliases: σ_d^2 sigma
Use in source · Read passage · 02_supplement_06.tex, line 168
05textitModules CompactTruncation and SmallCoreRecompression For the actual core G=R_LR_R^*, supply ordered thin SVD data G=Wdiag(σ)Z^*, W^*W=I, Z^*Z=I, σ_1≥⋯≥σ_d≥0. These are precisely the fields of the formal SVD structure. It contains no rank-optimality or tail-error field. A derived theorem checks the Gram eigenvector equation G^*GZ=Zdiag(σ_1^2,…,σ_d^2). Let h=min(d,k) and retain exactly the first h column indices. The rectangular square-root mask has entries M_ij=sqrtσ_j, · i=j,
Exact mathematical passage
G^*GZ=Z\diag(\sigma_1^2,\ldots,\sigma_d^2).
\sigma_i^2Use in source · 3 passages
Search aliases: σ_i^2 sigma
Use in source · Read passage · 02_supplement_06.tex, line 147
Set a_i=σ_i^2. Right compression cannot increase the Frobenius norm, and BQ=0, hence normD-B_ F^2 · ≥norm(D-B)Q_ F^2
Exact mathematical passage
a_i=\sigma_i^2
\sigma_i^{2}Use in source · 1 passage
Search aliases: σ_i^2 sigma
Use in source · Read passage · 02_supplement_06.tex, line 101
Monotonicity of the diagonal entries gives normDy^2 =∑_i=1^k+1σ_i^2|y_i|^2 ≥σ_k+1^2norm y^2. Because (D-B)y=Dy, the operator-norm inequality implies σ_k+1norm y ≤norm(D-B)y ≤normD-B_2norm y. Cancellation of the strictly positive norm of y proves σ_k+1≤normD-B_2. The norm of the actual diagonal residual D-D_k equals σ_k+1, so D_k attains the lower bound. If k≥ d, the residual is zero and optimality follows from norm nonnegativity.
Exact mathematical passage
\norm{Dy}^{2} =\sum_{i=1}^{k+1}\sigma_i^{2}|y_i|^{2} \geq\sigma_{k+1}^{2}\norm y^{2}.
\sigma_jUse in source · 1 passage
Search aliases: σ_j sigma
Use in source · Read passage · 02_supplement_06.tex, line 172
W^*W=I, Z^*Z=I, σ_1≥⋯≥σ_d≥0. These are precisely the fields of the formal SVD structure. It contains no rank-optimality or tail-error field. A derived theorem checks the Gram eigenvector equation G^*GZ=Zdiag(σ_1^2,…,σ_d^2). Let h=min(d,k) and retain exactly the first h column indices. The rectangular square-root mask has entries M_ij=sqrtσ_j, · i=j, 0, · i≠ j, 1≤ i≤ d, 1≤ j≤ h. Lean proves its index cardinality and MM^*=D_k. Define L_k=Q_LWM, R_k=Q_RZM. The product equals Q_LWD_kZ^*Q_R^* and has…
Exact mathematical passage
M_{ij}=\begin{cases}\sqrt{\sigma_j},&i=j,\\0,&i\ne j,\end{cases} \qquad 1\leq i\leq d,\quad 1\leq j\leq h.
\sigma_j^2Use in source · 1 passage
Search aliases: σ_j^2 sigma
Use in source · Read passage · 01_manuscript.tex, line 792
[Eckart Young Mirsky theorem] If M has singular values σ_1≥σ_2≥⋯, pad this finite sequence with zeros. For each integer k≥0, its truncated singular value decomposition M_k is a best rank-at-most-k approximation. It satisfies normM-M_k_2=σ_k+1 and normM-M_k_rm F = ( ∑_j>kσ_j^2 )^1/2.
Exact mathematical passage
\norm{M-M_k}_{\rm F} = \left( \sum_{j>k}\sigma_j^2 \right)^{1/2}.
\sigma_{k+1}Use in source · 7 passages
Search aliases: σ_k+1 sigma
Use in source · pages.tex, line 210
For a d-dimensional diagonal core and 0≤ k<d, the spectral lower bound has a concrete witness. A rank-at-most-k competitor has a nonzero kernel vector in the first k+1 coordinates. On that vector the diagonal matrix acts with norm at least σ_k+1 times the vector norm. The competitor cannot cancel that direction.
Exact mathematical passage
\sigma_{k+1}
\sigma_{k+1}^{2}Use in source · 1 passage
Search aliases: σ_k+1^2 sigma
Use in source · Read passage · 02_supplement_06.tex, line 101
Monotonicity of the diagonal entries gives normDy^2 =∑_i=1^k+1σ_i^2|y_i|^2 ≥σ_k+1^2norm y^2. Because (D-B)y=Dy, the operator-norm inequality implies σ_k+1norm y ≤norm(D-B)y ≤normD-B_2norm y. Cancellation of the strictly positive norm of y proves σ_k+1≤normD-B_2. The norm of the actual diagonal residual D-D_k equals σ_k+1, so D_k attains the lower bound. If k≥ d, the residual is zero and optimality follows from norm nonnegativity.
Exact mathematical passage
\norm{Dy}^{2} =\sum_{i=1}^{k+1}\sigma_i^{2}|y_i|^{2} \geq\sigma_{k+1}^{2}\norm y^{2}.
\Sigma_k^{1/2}Use in source · 1 passage
Search aliases: Σ_k^1/2 Sigma
Use in source · Read passage · 01_manuscript.tex, line 825
R=Q_RR_R, where the columns of Q_L and Q_R are orthonormal. Form the small core G=R_LR_R^* and let G=WΣ Z^* be an ordered thin singular value decomposition, with nonnegative nonincreasing singular values. For an integer k≥0, retain min{k,d} columns, where d is the number of supplied thin singular directions; tail singular values beyond d are zero. Define L_k=Q_LW_kΣ_k^1/2, R_k=Q_RZ_kΣ_k^1/2. Then L_kR_k^* is a best rank at most k approximation to LR^*. Its spectral and Frobenius errors are the s…
Exact mathematical passage
L_k=Q_LW_k\Sigma_k^{1/2}, \qquad R_k=Q_RZ_k\Sigma_k^{1/2}.
\sigma_{\min(k,d)}Use in source · 1 passage
Search aliases: σ_min(k,d) sigma min
Use in source · Read passage · 02_supplement_06.tex, line 88
Spectral optimality from a kernel witness. textitModules DiagonalTruncation and SpectralOptimality Let σ_1≥⋯≥σ_d≥0 and D=diag(σ_1,…,σ_d), D_k=diag(σ_1,…,σ_min(k,d),0,…,0). Lean uses indices beginning at zero; the formulas on this page use the conventional one-based notation. The rank of D_k is proved to be at most k, even when retained singular values are zero.
Exact mathematical passage
D=\diag(\sigma_1,\ldots,\sigma_d),\qquad D_k=\diag(\sigma_1,\ldots,\sigma_{\min(k,d)},0,\ldots,0).
\sigma_uUse in source · 1 passage
Search aliases: σ_u sigma
Use in source · Read passage · 01_manuscript.tex, line 1514
The cost and error of a pruning L are cost( L) = ∑_v∈ Lκ_v + ∑_u splitσ_u and err( L)=∑_v∈ Lϵ_v. The model can include arithmetic, storage, launch, transfer, and precision costs inside κ_v. It remains exact only when the declared totals are additive.
Exact mathematical passage
\cost(\Lcal) = \sum_{v\in\Lcal}\kappa_v + \sum_{u\text{ split}}\sigma_u
\tauUse in source · 19 passages
Search aliases: τ tau
Use in source · pages.tex, line 342
vpage20Spend error where it is cheapestCONTINUOUS AND CLIPPED PRECISION Take sensitivities s=(1,2,4), cost coefficients c=(1,1,2), and budget τ=1. Minimizing ∑_ec_elog(1/ϵ_e) subject to ∑_es_eϵ_e=1 gives ϵ_e=fracc_e4s_e, ϵ=(1/4,1/8,1/8). vfigprecision headThe budget is weighted The three contributions to the error allowance are (1/4,1/4,1/2). Equal cost coefficients give equal weighted contributions, not equal tolerances. The third edge has twice the cost coefficient of the second, which offsets…
Exact mathematical passage
\tau=1
\PhiUse in source · 7 passages
Search aliases: Φ Phi
Use in source · Read passage · 02_supplement_04.tex, line 182
Constructive parameterization. Write t=1/λ and set u_e(t)=min{m_e,tc_e/s_e}, Φ(t)=∑_e s_eu_e(t). Lean proves that Φ is continuous and nondecreasing, with Φ(0)=0. At the finite positive scale T=∑_efracm_es_ec_e, every coordinate is saturated, so u(T)=m and Φ(T)=∑_es_em_e. The intermediate value theorem produces t>0 with Φ(t)=τ. The multiplier is not assumed to exist.
Exact mathematical passage
u_e(t)=\min\{m_e,tc_e/s_e\},\qquad \Phi(t)=\sum_e s_eu_e(t).
\phiUse in source · 2 passages
Search aliases: ϕ phi
Use in source · Read passage · 01_manuscript.tex, line 1400
The Karush Kuhn Tucker conditions give the unconstrained form on every nonactive coordinate and equality with m_e on every active upper bound. The weighted sum ϕ(λ) = ∑_es_e min{ m_e, fracc_eλ s_e } is continuous and nonincreasing. It decreases from the maximal feasible value to zero. For a strictly submaximal budget, once any coordinate is unsaturated, ϕ is strictly decreasing, which gives a unique multiplier. At maximal budget the only feasible vector is (m_e)_e and every 0<λ≤min_e c_e/(s_em_e…
Exact mathematical passage
\phi(\lambda) = \sum_es_e \min\left\{ m_e, \frac{c_e}{\lambda s_e} \right\}
\chiUse in source · 1 passage
Search aliases: χ chi
Use in source · Read passage · 02_supplement_04.tex, line 10
Circulant Fourier diagonalization. textitSource item 4.3; module CirculantFourier Let n>0, with coordinates indexed by Z/n Z. Define the actual circulant by C(c)_jk=c_j-k. The formal Fourier matrix is F_jk=frac1sqrt n χ(-jk), χ(a)=exp (frac2π i an). The standard character is not left abstract: a checked theorem identifies these entries with frac1sqrt nexp (-frac2π i val(j)val(k)n). The value representatives enter only this explicit exponential formula. All modular algebra uses the residue classe…
Exact mathematical passage
F_{jk}=\frac{1}{\sqrt n}\,\chi(-jk), \qquad \chi(a)=\exp\!\left(\frac{2\pi i a}{n}\right).
\existsUse in source · 1 passage
Search aliases: ∃ exists
Use in source · Read passage · 02_supplement_04.tex, line 193
Strictness below saturation. If t<u and Φ(t) is below the maximum, at least one coordinate is unsaturated at t. That coordinate strictly increases; every other coordinate is nondecreasing. Hence Φ(t)<Φ(u). This proves uniqueness of the scale and, by reciprocation, the exact manuscript statement ∃!λ>0: ∑_e s_emin {m_e,fracc_eλ s_e}=τ when 0<τ<∑_es_em_e.
Exact mathematical passage
\boxed{\exists!\lambda>0:\quad \sum_e s_e\min\!\left\{m_e,\frac{c_e}{\lambda s_e}\right\}=\tau}
\notinUse in source · 1 passage
Search aliases: ∉ notin
Use in source · Read passage · 01_manuscript.tex, line 1067
Let R be the set of packets reachable from α, including α, and set X_R=⊕_γ∈ RX_γ. No active edge leaves R, so AX_R⊆ X_R. The restriction of A to the finite dimensional space X_R is injective. It is therefore onto X_R. Hence A^-1X_R⊆ X_R. Since β∉ R, the stated block of A^-1 vanishes.
Exact mathematical passage
\beta\notin R
\prodUse in source · 4 passages
Search aliases: ∏ prod
Use in source · Read passage · 01_manuscript.tex, line 864
[Product chain perturbation] Let A_1,…,A_m be exact factors and let widehat A_j=A_j+E_j. Then widehat A_m⋯widehat A_1 - A_m⋯ A_1 ≤ ∑_j=1^m ( ∏_i=j+1^mnormwidehat A_i ) normE_j ( ∏_i=1^j-1normA_i ). Empty products are one.
Exact mathematical passage
\boxed{ \left\| \widehat A_m\cdots\widehat A_1 - A_m\cdots A_1 \right\| \leq \sum_{j=1}^m \left( \prod_{i=j+1}^m\norm{\widehat A_i} \right) \norm{E_j} \left( \prod_{i=1}^{j-1}\norm{A_i} \right). }
\prod_{i=1}^{j-1}Use in source · 2 passages
Search aliases: ∏_i=1^j-1 prod
Use in source · Read passage · 01_manuscript.tex, line 864
[Product chain perturbation] Let A_1,…,A_m be exact factors and let widehat A_j=A_j+E_j. Then widehat A_m⋯widehat A_1 - A_m⋯ A_1 ≤ ∑_j=1^m ( ∏_i=j+1^mnormwidehat A_i ) normE_j ( ∏_i=1^j-1normA_i ). Empty products are one.
Exact mathematical passage
\boxed{ \left\| \widehat A_m\cdots\widehat A_1 - A_m\cdots A_1 \right\| \leq \sum_{j=1}^m \left( \prod_{i=j+1}^m\norm{\widehat A_i} \right) \norm{E_j} \left( \prod_{i=1}^{j-1}\norm{A_i} \right). }
\prod_{i=j+1}^mUse in source · 3 passages
Search aliases: ∏_i=j+1^m prod
Use in source · Read passage · 01_manuscript.tex, line 864
[Product chain perturbation] Let A_1,…,A_m be exact factors and let widehat A_j=A_j+E_j. Then widehat A_m⋯widehat A_1 - A_m⋯ A_1 ≤ ∑_j=1^m ( ∏_i=j+1^mnormwidehat A_i ) normE_j ( ∏_i=1^j-1normA_i ). Empty products are one.
Exact mathematical passage
\boxed{ \left\| \widehat A_m\cdots\widehat A_1 - A_m\cdots A_1 \right\| \leq \sum_{j=1}^m \left( \prod_{i=j+1}^m\norm{\widehat A_i} \right) \norm{E_j} \left( \prod_{i=1}^{j-1}\norm{A_i} \right). }
\prod_{i=j+1}^{m}Use in source · 1 passage
Search aliases: ∏_i=j+1^m prod
Use in source · Read passage · 01_manuscript.tex, line 915
ntheorem[Certified repeated recompression] Let P_0=I, P_j=A_jP_j-1, be exact partial products. Suppose an implemented chain satisfies widehat P_0=I, widehat P_j = A_jwidehat P_j-1+E_j, where E_j is the additive error introduced after the jth multiplication, for example by factor recompression. Then normwidehat P_m-P_m ≤ ∑_j=1^m ( ∏_i=j+1^mnormA_i ) normE_j. If normE_j≤δ_j, the same formula with δ_j gives a certified final error.
Exact mathematical passage
\boxed{ \norm{\widehat P_m-P_m} \leq \sum_{j=1}^{m} \left( \prod_{i=j+1}^{m}\norm{A_i} \right) \norm{E_j}. }
\sum_{(b,a)\in E_A}Use in source · 2 passages
Search aliases: ∑_(b,a)∈ E_A sum in
Use in source · Read passage · 02_supplement_01.tex, line 13
Reconstruction and action. The checked finite sum is A=∑_(b,a)∈ E_A A_ba. The proof inserts the complete projector sum on both sides and removes only zero summands. Applying this identity to an arbitrary module gives exact action. A separate finite-matrix theorem establishes AZ=∑_(b,a)∈ E_AA_baZ for every rectangular batch of right-hand sides Z.
Exact mathematical passage
\boxed{A=\sum_{(b,a)\in E_A} A_{ba}.}
\sum_aUse in source · 2 passages
Search aliases: ∑_a sum
Use in source · Read passage · 02_supplement_01.tex, line 6
Exact packet semantics. textitSource items 2.3, 2.5, and 2.6 Let R be a ring, I a finite index type, and (P_a)_a∈ I a family satisfying P_bP_a=P_a · b=a, 0 · b≠ a, ∑_aP_a=1. Write A_ba=P_bAP_a and let E_A={(b,a):A_ba≠0}. These are exact algebraic zeros, not values discarded by a floating-point threshold.
Exact mathematical passage
P_bP_a=\begin{cases}P_a&b=a,\\0&b\ne a,\end{cases} \qquad \sum_aP_a=1.
\sum_{b=0}^BUse in source · 1 passage
Search aliases: ∑_b=0^B sum
Use in source · Read passage · 02_supplement_03.tex, line 212
Counted candidates. One full convolution table examines indices (b,k), 0≤ b≤ B,quad0≤ k≤ b. These form a concrete finite dependent-sum type. Its cardinality is proved to be ∑_b=0^B(b+1)≤(B+1)^2. For a nonempty finite tree, the number of child merges plus one equals the node count N. This identity is proved for arbitrary finite branching.
Exact mathematical passage
\sum_{b=0}^B(b+1)\leq(B+1)^2.
\sum_{b\ne a}Use in source · 3 passages
Search aliases: ∑_b≠ a sum ne
Use in source · Read passage · 02_supplement_05.tex, line 160
esidual certificates for supplied packets textitSource item 9.2; module DiscoveryCertificate Let (P_a) be a complete orthogonal projector system on a finite complex coordinate space. The projectors need not be coordinate masks. For an actual matrix A, define Π(A)=∑_aP_aAP_a, E=A-Π(A). Reconstruction gives the exact off-diagonal residual E=∑_b≠ aP_bAP_a.
Exact mathematical passage
E=\sum_{b\ne a}P_bAP_a.
\sum_cUse in source · 1 passage
Search aliases: ∑_c sum
Use in source · Read passage · 02_supplement_01.tex, line 23
Composition. Inserting the projector identity between factors yields (AB)_ba=∑_c A_bcB_ca. If the left side is nonzero, at least one intermediate c has both A_bc≠0 and B_ca≠0. Consequently supp(AB)⊆supp(A)∘supp(B). No equality of supports is asserted.
Exact mathematical passage
\boxed{(AB)_{ba}=\sum_c A_{bc}B_{ca}.}
\sum_{c:\,A_{bc}\ne0}Use in source · 1 passage
Search aliases: ∑_c: A_bc≠0 sum ne
Use in source · Read passage · 02_supplement_01.tex, line 49
Define the active contribution sum recursively by F_0(b,a) · =P_a · b=a, 0 · b≠ a, F_n+1(b,a) · =∑_c: A_bc≠0A_bcF_n(c,a). The checked identity is P_bA^nP_a=F_n(b,a) (n≥0). Its proof is induction using the exact composition theorem, not an assumed support recurrence. When no active walk exists, the corresponding contribution sum is zero, and hence the actual block of A^n vanishes.
Exact mathematical passage
F_0(b,a)&=\begin{cases}P_a&b=a,\\0&b\ne a,\end{cases}\\ F_{n+1}(b,a)&=\sum_{c:\,A_{bc}\ne0}A_{bc}F_n(c,a).
\sum_eUse in source · 44 passages
Search aliases: ∑_e sum
Use in source · pages.tex, line 333
For orthogonal coordinate blocks, the squared Frobenius norms add exactly. Operator norms instead require a suitable bound or further spectral information. The maximum omitted block norm is generally insufficient. headInput sensitivity is additional information The packet perturbation estimate is (widehat A-A)x ≤∑_e E_e P_α(e)x . It can be much tighter for an input that scarcely occupies an affected source packet. To obtain a uniform operator bound, the input dependence must be bounded uniformly…
Exact mathematical passage
\|(\widehat A-A)x\|\le\sum_e\|E_e\|\,\|P_{\alpha(e)}x\|.
\sum_{e\in E}Use in source · 3 passages
Search aliases: ∑_e∈ E sum in
Use in source · Read passage · 01_manuscript.tex, line 1256
[Packet perturbation certificate] Let A= ∑_e∈ EA_e, widehat A = ∑_e∈ E(A_e+E_e), where edge e starts at packet α(e), and assume A_e=A_eP_α(e), E_e=E_eP_α(e). Then norm(widehat A-A)x ≤ ∑_e∈ E normE_e normP_α(e)x. If normE_e ≤ ε_enormA_e, then norm(widehat A-A)x ≤ ∑_e∈ E ε_enormA_e normP_α(e)x.
Exact mathematical passage
A= \sum_{e\in E}A_e, \qquad \widehat A = \sum_{e\in E}(A_e+E_e),
\sum_fUse in source · 2 passages
Search aliases: ∑_f sum
Use in source · Read passage · 01_manuscript.tex, line 1346
[Continuous precision allocation] Let E be a finite nonempty edge set, let s_e>0 be sensitivities and c_e>0 be costs, and let τ>0. Consider min_ε_e>0 ∑_e c_elogfrac1ε_e subject to ∑_es_eε_e=τ. The unique minimizer is ε_e = fracτ c_es_e∑_fc_f.
Exact mathematical passage
\boxed{ \eps_e = \frac{\tau c_e}{s_e\sum_fc_f}. }
\sum_iUse in source · 9 passages
Search aliases: ∑_i sum
Use in source · Read passage · 02_supplement_02.tex, line 86
Let v be the new optimum of the edited subtree, and let j be the edited child slot at one frame. The update performed at that frame is step(v)= min{K, σ+ ∑_i v, · i=j, widehat F_i, · i≠ j. }. This result is passed to the next ancestor. The update function does not recursively evaluate sibling subtrees. It reads their stored values widehat F_i.
Exact mathematical passage
\boxed{ \operatorname{step}(v)= \min\left\{K,\;\sigma+ \sum_i\begin{cases} v,&i=j,\\ \widehat F_i,&i\ne j. \end{cases}\right\}. }
\sum_{i=1}^{k+1}Use in source · 1 passage
Search aliases: ∑_i=1^k+1 sum
Use in source · Read passage · 02_supplement_06.tex, line 101
Monotonicity of the diagonal entries gives normDy^2 =∑_i=1^k+1σ_i^2|y_i|^2 ≥σ_k+1^2norm y^2. Because (D-B)y=Dy, the operator-norm inequality implies σ_k+1norm y ≤norm(D-B)y ≤normD-B_2norm y. Cancellation of the strictly positive norm of y proves σ_k+1≤normD-B_2. The norm of the actual diagonal residual D-D_k equals σ_k+1, so D_k attains the lower bound. If k≥ d, the residual is zero and optimality follows from norm nonnegativity.
Exact mathematical passage
\norm{Dy}^{2} =\sum_{i=1}^{k+1}\sigma_i^{2}|y_i|^{2} \geq\sigma_{k+1}^{2}\norm y^{2}.
\sum_{i>k}Use in source · 3 passages
Search aliases: ∑_i>k sum
Use in source · Read passage · 02_supplement_06.tex, line 138
For any nonnegative decreasing sequence a_i, weights with these properties satisfy ∑_i a_iw_i≥∑_i>ka_i. This finite weighted inequality is itself proved. For k<d, use μ=a_k+1 and compare the summands to tail weights v_i=0 for i≤ k and v_i=1 otherwise: a_iv_i+μ(w_i-v_i)≤ a_iw_i. For a retained coordinate the inequality uses a_i≥μ and w_i≥0. For a tail coordinate it uses a_i≤μ and w_i≤1. Summation and ∑ w_i≥∑ v_i give the result.
Exact mathematical passage
\sum_i a_iw_i\geq\sum_{i>k}a_i.
\sum_jUse in source · 1 passage
Search aliases: ∑_j sum
Use in source · Read passage · 02_supplement_02.tex, line 29
Shared operator families. For reuse weights r_j, the local keep model can aggregate setup and per-operator costs: κ_v=κ_v^setup+∑_jr_jκ_v^(j), ϵ_v=∑_jr_jϵ_v^(j). Lean proves the aggregation identity at the cut level, including split overhead, before applying the compiler theorem. The proof is valid over an ordered field, including Q and R. The intended nonnegative model is a specialization.
Exact mathematical passage
\kappa_v=\kappa_v^{\mathrm{setup}}+\sum_jr_j\kappa_v^{(j)}, \qquad \epsilon_v=\sum_jr_j\epsilon_v^{(j)}.
\sum_{j=0}^dUse in source · 1 passage
Search aliases: ∑_j=0^d sum
Use in source · Read passage · 01_manuscript.tex, line 978
[Polynomial support] If p(z)=∑_j=0^dc_jz^j, then supp(p(A)) is contained in the union of the support relations of A^j for 0≤ j≤ d.
Exact mathematical passage
p(z)=\sum_{j=0}^dc_jz^j,
\sum_{j=0}^{d}Use in source · 1 passage
Search aliases: ∑_j=0^d sum
Use in source · Read passage · 02_supplement_01.tex, line 71
Polynomial support. For the actual finite sum p(A)=∑_j=0^dc_jA^j, the checked support lies in the union of the support relations of the powers. The formalization treats the j=0 contribution explicitly, so the constant term is not accidentally omitted.
Exact mathematical passage
p(A)=\sum_{j=0}^{d}c_jA^j,
\sum_{j=1}^mUse in source · 9 passages
Search aliases: ∑_j=1^m sum
Use in source · Read passage · 01_manuscript.tex, line 726
[Rank growth in a product chain] Let A_j=D_j+U_jV_j^*, j=1,…,m, share one diagonal packet layout. Before recompression, the product A_m⋯ A_1 has a BDLR representation whose correction rank is at most ∑_j=1^m r_j.
Exact mathematical passage
\boxed{ \sum_{j=1}^m r_j. }
\sum_{j=1}^{m}Use in source · 2 passages
Search aliases: ∑_j=1^m sum
Use in source · Read passage · 01_manuscript.tex, line 915
ntheorem[Certified repeated recompression] Let P_0=I, P_j=A_jP_j-1, be exact partial products. Suppose an implemented chain satisfies widehat P_0=I, widehat P_j = A_jwidehat P_j-1+E_j, where E_j is the additive error introduced after the jth multiplication, for example by factor recompression. Then normwidehat P_m-P_m ≤ ∑_j=1^m ( ∏_i=j+1^mnormA_i ) normE_j. If normE_j≤δ_j, the same formula with δ_j gives a certified final error.
Exact mathematical passage
\boxed{ \norm{\widehat P_m-P_m} \leq \sum_{j=1}^{m} \left( \prod_{i=j+1}^{m}\norm{A_i} \right) \norm{E_j}. }
\sum_{j=1}^{q}Use in source · 1 passage
Search aliases: ∑_j=1^q sum
Use in source · Read passage · 02_supplement_06.tex, line 128
Frobenius optimality from kernel weights. textitModules KernelFrame and FrobeniusOptimality For an arbitrary competitor B with rank B≤ k, construct an orthonormal matrix Q spanning its actual kernel. Then BQ=0, Q^*Q=I, q=dimker B=d-rank B. Define the row weights w_i=∑_j=1^q|Q_ij|^2. The proof derives 0≤ w_i≤1, ∑_iw_i=q≥max(d-k,0). The upper bound uses contraction of Q^* on a coordinate vector. The weight sum follows from the trace of Q^*Q and rank-nullity.
Exact mathematical passage
w_i=\sum_{j=1}^{q}|Q_{ij}|^2.
\sum_{j>k}Use in source · 1 passage
Search aliases: ∑_j>k sum
Use in source · Read passage · 01_manuscript.tex, line 792
[Eckart Young Mirsky theorem] If M has singular values σ_1≥σ_2≥⋯, pad this finite sequence with zeros. For each integer k≥0, its truncated singular value decomposition M_k is a best rank-at-most-k approximation. It satisfies normM-M_k_2=σ_k+1 and normM-M_k_rm F = ( ∑_j>kσ_j^2 )^1/2.
Exact mathematical passage
\norm{M-M_k}_{\rm F} = \left( \sum_{j>k}\sigma_j^2 \right)^{1/2}.
\sum_{k=0}^{j-1}Use in source · 1 passage
Search aliases: ∑_k=0^j-1 sum
Use in source · Read passage · 01_manuscript.tex, line 1169
Use the telescoping identity A^j-D^j = ∑_k=0^j-1 A^j-1-kKD^k. Each term has norm at most M^j-1normK. The vector D^jv lies in the direct sum of local Krylov spaces.
Exact mathematical passage
A^j-D^j = \sum_{k=0}^{j-1} A^{j-1-k}KD^k.
\sum_{k=1}^{n}Use in source · 1 passage
Search aliases: ∑_k=1^n sum
Use in source · Read passage · 01_manuscript.tex, line 606
The (i,j) entry of BXA^T is ∑_k=1^n∑_ℓ=1^m B_ikX_kℓA_jℓ. This is the coordinate formula for (A⊗ B)vec(X). Multiplying first by B and then by A^T gives the stated count.
Exact mathematical passage
\sum_{k=1}^{n}\sum_{\ell=1}^{m} B_{ik}X_{k\ell}A_{j\ell}.
\sum_{k=1}^{w}Use in source · 1 passage
Search aliases: ∑_k=1^w sum
Use in source · Read passage · 01_manuscript.tex, line 484
The diagonal contributes p blocks. At offset k, where 1≤ k≤ w, the upper and lower block diagonals contribute 2(p-k) blocks. Therefore p+2∑_k=1^w(p-k) = p(2w+1)-w(w+1). The cost and storage follow from [source reference].
Exact mathematical passage
p+2\sum_{k=1}^{w}(p-k) = p(2w+1)-w(w+1).
\sum_{r=0}^{n-1}Use in source · 1 passage
Search aliases: ∑_r=0^n-1 sum
Use in source · Read passage · 01_manuscript.tex, line 548
For each frequency k, let v^(k)_j = frac1sqrt n exp(frac2π i jkn). Then (C(c)v^(k))_j = ∑_r=0^n-1c_rv^(k)_j-r = ( ∑_r=0^n-1 c_rexp(-frac2π i rkn) )v^(k)_j. The eigenvalue is sqrt n(Fc)_k. The vectors v^(k) are the columns of F^*, which gives the formula.
Exact mathematical passage
(C(c)v^{(k)})_j = \sum_{r=0}^{n-1}c_rv^{(k)}_{j-r} = \left( \sum_{r=0}^{n-1} c_r\exp\left(-\frac{2\pi\ii rk}{n}\right) \right)v^{(k)}_j.
\sum_{u\text{ split}}Use in source · 1 passage
Search aliases: ∑_u split sum text
Use in source · Read passage · 01_manuscript.tex, line 1514
The cost and error of a pruning L are cost( L) = ∑_v∈ Lκ_v + ∑_u splitσ_u and err( L)=∑_v∈ Lϵ_v. The model can include arithmetic, storage, launch, transfer, and precision costs inside κ_v. It remains exact only when the declared totals are additive.
Exact mathematical passage
\cost(\Lcal) = \sum_{v\in\Lcal}\kappa_v + \sum_{u\text{ split}}\sigma_u
\sum_{v\in\Lcal}Use in source · 2 passages
Search aliases: ∑_v∈Lcal ∑_v∈ L sum in Lcal sum in mathcal
Use in source · Read passage · 01_manuscript.tex, line 1514
The cost and error of a pruning L are cost( L) = ∑_v∈ Lκ_v + ∑_u splitσ_u and err( L)=∑_v∈ Lϵ_v. The model can include arithmetic, storage, launch, transfer, and precision costs inside κ_v. It remains exact only when the declared totals are additive.
Exact mathematical passage
\cost(\Lcal) = \sum_{v\in\Lcal}\kappa_v + \sum_{u\text{ split}}\sigma_u
\sum_wUse in source · 1 passage
Search aliases: ∑_w sum
Use in source · Read passage · 02_supplement_02.tex, line 64
Strict half-gap stability. At a solved internal node write K=κ_v+λϵ_v, S=σ_v+∑_wF_λ(w), g=|K-S|. If the updated scores satisfy |K'-K|<g/2, |S'-S|<g/2, Lean proves K'≤ S' Longleftrightarrow K≤ S. Applying this condition recursively at every internal node proves equality of the entire returned cuts before and after perturbation, not only equality of one numerical minimum.
Exact mathematical passage
K=\kappa_v+\lambda\epsilon_v,\qquad S=\sigma_v+\sum_wF_\lambda(w),\qquad g=|K-S|.
\sum_{w\in\mathrm{child}(v)}Use in source · 1 passage
Search aliases: ∑_w∈child(v) sum in mathrm
Use in source · Read passage · 02_supplement_02.tex, line 20
Recurrence. For terminal nodes, F_λ(v)=κ_v+λϵ_v. For internal nodes, F_λ(v)=min{κ_v+λϵ_v, σ_v+∑_w∈child(v)F_λ(w)}. The checked result has two parts. First, F_λ(v)≤ J_λ(L) for every legal cut of the subtree. Second, the constructed function codecompile returns a legal cut with objective exactly F_λ(v). Thus the minimum is attained, not merely bounded below.
Exact mathematical passage
\boxed{F_\lambda(v)=\min\left\{\kappa_v+\lambda\epsilon_v, \;\sigma_v+\sum_{w\in\mathrm{child}(v)}F_\lambda(w)\right\}.}
\sum_{w\in\operatorname{child}(v)}Use in source · 1 passage
Search aliases: ∑_w∈child(v) sum in operatorname
Use in source · Read passage · 01_manuscript.tex, line 1543
[Exact Lagrangian packet compiler] Fix λ≥0 and define J_λ( L) = cost( L) + λerr( L). For every node v, set K_λ(v) = κ_v+λϵ_v and S_λ(v) = σ_v + ∑_w∈child(v)F_λ(w). For a terminal node, set F_λ(v)=K_λ(v). For an internal node, the recurrence F_λ(v) = min{K_λ(v),S_λ(v)} computes the exact minimum Lagrangian objective over every admissible pruning of the subtree rooted at v.
Exact mathematical passage
S_\lambda(v) = \sigma_v + \sum_{w\in\operatorname{child}(v)}F_\lambda(w).
\sum_\alphaUse in source · 8 passages
Search aliases: ∑_α sum alpha
Use in source · Read passage · 01_manuscript.tex, line 125
Insert the identity on both sides of A: A = (∑_β P_β) A (∑_α P_α) = ∑_α,βP_β AP_α. The zero blocks can be removed. Applying the identity to Z gives the second formula.
Exact mathematical passage
A = \left(\sum_\beta P_\beta\right) A \left(\sum_\alpha P_\alpha\right) = \sum_{\alpha,\beta}P_\beta AP_\alpha.
\sum_{\alpha,\beta}Use in source · 1 passage
Search aliases: ∑_α,β sum alpha beta
Use in source · Read passage · 01_manuscript.tex, line 125
Insert the identity on both sides of A: A = (∑_β P_β) A (∑_α P_α) = ∑_α,βP_β AP_α. The zero blocks can be removed. Applying the identity to Z gives the second formula.
Exact mathematical passage
A = \left(\sum_\beta P_\beta\right) A \left(\sum_\alpha P_\alpha\right) = \sum_{\alpha,\beta}P_\beta AP_\alpha.
\sum_{\alpha\to\beta}Use in source · 1 passage
Search aliases: ∑_α→β sum alpha to beta
Use in source · Read passage · 01_manuscript.tex, line 1026
The equation in packet β is D_β x_β = b_β - ∑_α→βK_βαx_α. Every predecessor of β lies in an earlier level. Once those values are available, x_β is obtained from the local solve with D_β. Vertices in the same level have no directed edge between them, so their solves are independent.
Exact mathematical passage
D_\beta x_\beta = b_\beta - \sum_{\alpha\to\beta}K_{\beta\alpha}x_\alpha.
\sum_{\alpha\neq\beta}Use in source · 2 passages
Search aliases: ∑_α≠β sum alpha neq beta
Use in source · Read passage · 01_manuscript.tex, line 1475
[Exact block discovery certificate] For an orthogonal packet system (P_α), define the block diagonal compression Π_ P(A) = ∑_α P_α AP_α. Then normA-Π_ P(A)_rm F^2 = ∑_α≠β normP_β AP_α_rm F^2. In operator norm, normA-Π_ P(A)_2 ≤ ∑_α≠β normP_β AP_α_2. Thus any proposed packet discovery has an exact finite residual certificate after its projectors are known.
Exact mathematical passage
\boxed{ \norm{A-\Pi_{\Pcal}(A)}_{\rm F}^2 = \sum_{\alpha\neq\beta} \norm{P_\beta AP_\alpha}_{\rm F}^2. }
\sum_\betaUse in source · 2 passages
Search aliases: ∑_β sum beta
Use in source · Read passage · 01_manuscript.tex, line 125
Insert the identity on both sides of A: A = (∑_β P_β) A (∑_α P_α) = ∑_α,βP_β AP_α. The zero blocks can be removed. Applying the identity to Z gives the second formula.
Exact mathematical passage
A = \left(\sum_\beta P_\beta\right) A \left(\sum_\alpha P_\alpha\right) = \sum_{\alpha,\beta}P_\beta AP_\alpha.
\sum_{\ell=1}^{m}Use in source · 1 passage
Search aliases: ∑_ℓ=1^m sum ell
Use in source · Read passage · 01_manuscript.tex, line 606
The (i,j) entry of BXA^T is ∑_k=1^n∑_ℓ=1^m B_ikX_kℓA_jℓ. This is the coordinate formula for (A⊗ B)vec(X). Multiplying first by B and then by A^T gives the stated count.
Exact mathematical passage
\sum_{k=1}^{n}\sum_{\ell=1}^{m} B_{ik}X_{k\ell}A_{j\ell}.
\inftyUse in source · 20 passages
Search aliases: ∞ infty
Use in source · pages.tex, line 375
The checker compares the recurrence with complete enumeration for every budget from zero through twelve. Infeasible states remain +∞; they are not zero-cost states. headQuantization must be conservative Suppose real errors are 3/50 and 2/25, and one integer unit represents 1/10. Rounding down gives two zero weights, even though the true sum is 7/50>1/10. Rounding up gives a certified upper envelope instead. The budget method is exact for its integers; it cannot repair unsafe integers supplied by…
Exact mathematical passage
+\infty
\neUse in source · 22 passages
Search aliases: ≠ ne
Use in source · pages.tex, line 239
dicts possible support but not actual support. Replacing the inclusion supp(AB)⊆supp(A)∘supp(B) by equality would be wrong even in this four-dimensional model. headWhat an acyclic graph does guarantee If the longest strict directed path has h edges, then no term of K^h+1 can survive. Thus K^h+1=0. This is an upper bound on the nilpotence index, not an assertion that K^h≠0. Here the graph has paths of length two, while coefficient cancellation already gives K^2=0.
Exact mathematical passage
K^h\ne0
\leUse in source · 58 passages
Search aliases: ≤ le
Use in source · extended_pages.tex, line 66
vpage32Safe integers can still be conservativeQUANTIZATION AND THE SIZE OF THE BUDGET TABLE Let one integer unit represent h=1/10 of certified error. For an individual nonnegative error e, choosing w=lceil e/hrceil ensures e≤ hw. Thus integer feasibility ∑ w≤ B implies the real bound ∑ e≤ hB. This implication is checked before applying the exact option or tree recurrence. vfigquantization headA false certificate and a conservative rejection For errors (3/50,2/25), rounding down assigns weights (…
Exact mathematical passage
w=\lceil e/h\rceil\qquad\text{ensures}\qquad e\le hw.
\leqUse in source · 92 passages
Search aliases: ≤ leq
Use in source · Read passage · 01_manuscript.tex, line 428
[Capacity threshold] Let a device have available memory M≥0 bytes and scalar size s>0, and let n≥0. The matrix representation alone can fit only when n≤sqrtM/s. The equal packet BDLR representation alone fits when n(b+2r)s≤ M.
Exact mathematical passage
n\leq\sqrt{M/s}.
\bigoplus_iUse in source · 2 passages
Search aliases: ⊕_i bigoplus
Use in source · Read passage · 02_supplement_03.tex, line 101
Exact and approximate packet Krylov spaces. textitSource items 7.2 and 7.3 Write K_m(A,v)=span{v,Av,…,A^m-1v}. For a complete finite projector family P_i, suppose A commutes with every P_i. Let A_i=P_iAP_i. Lean proves the actual orbit identity A^jv=∑_i A_i^jP_iv and therefore K_m(A,v)⊆ ⊕_i K_m(A_i,P_iv). The case j=0 uses ∑_iP_i=I. The zeroth local power is the ambient identity, applied to P_iv, rather than an incorrectly substituted local projector power.
Exact mathematical passage
\boxed{\mathcal K_m(A,v)\subseteq \bigoplus_i\mathcal K_m(A_i,P_iv).}
\bigoplus_\alphaUse in source · 2 passages
Search aliases: ⊕_α bigoplus alpha
Use in source · Read passage · 01_manuscript.tex, line 1126
[Invariant packet Krylov decomposition] Assume [A,P_α]=0 for every packet. If v=∑_α v_α, v_α=P_α v, then A^jv = ∑_α A_α^jv_α for every j≥0, where A_α=P_α AP_α. Hence K_m(A,v) ⊆ ⊕_α K_m(A_α,v_α).
Exact mathematical passage
\Kcal_m(A,v) \subseteq \bigoplus_\alpha \Kcal_m(A_\alpha,v_\alpha).
\bigoplus_{\gamma\in R}Use in source · 1 passage
Search aliases: ⊕_γ∈ R bigoplus gamma in
Use in source · Read passage · 01_manuscript.tex, line 1064
Let R be the set of packets reachable from α, including α, and set X_R=⊕_γ∈ RX_γ. No active edge leaves R, so AX_R⊆ X_R. The restriction of A to the finite dimensional space X_R is injective. It is therefore onto X_R. Hence A^-1X_R⊆ X_R. Since β∉ R, the stated block of A^-1 vanishes.
Exact mathematical passage
X_R=\bigoplus_{\gamma\in R}X_\gamma.
\starUse in source · 1 passage
Search aliases: ⋆ star
Use in source · Read passage · 02_supplement_03.tex, line 145
For exact-weight tables F,G, min-plus convolution is (F⋆ G)(b)=min_0≤ k≤ b{F(k)+G(b-k)}, with an infeasible sum whenever either operand is infeasible. Its correctness theorem identifies every finite answer with an actual feasible pair of component weights and costs. Repeated convolution handles arbitrary finite lists of components.
Exact mathematical passage
(F\star G)(b)=\min_{0\leq k\leq b}\{F(k)+G(b-k)\},
\cdotsUse in source · 11 passages
Search aliases: ⋯ cdots
Use in source · Read passage · 01_manuscript.tex, line 722
[Rank growth in a product chain] Let A_j=D_j+U_jV_j^*, j=1,…,m, share one diagonal packet layout. Before recompression, the product A_m⋯ A_1 has a BDLR representation whose correction rank is at most ∑_j=1^m r_j.
Exact mathematical passage
A_m\cdots A_1
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PDF reading editions (28)
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- Paper_VIII_Traditional_Final.pdf
- Paper_VIII_Visual_Guide_Final.pdf
- Paper_VIII_Lean_Guide_Final.pdf
- 01_exact_bdlr_action.pdf
- 02_exact_compiler_frontier.pdf
- 03_exact_precision_budget.pdf
- 01_packet_compilation_architecture.pdf
- 02_packet_ir_graph.pdf
- 03_ideal_arithmetic_reduction.pdf
- 04_roofline_schematic.pdf
- 05_bdlr_structure.pdf
- 06_recompression_spectrum.pdf
- 07_path_support.pdf
- 08_topological_schedule.pdf
- 09_packet_preconditioner.pdf
- 10_precision_allocation.pdf
- 11_tree_compiler.pdf
- 12_budget_compiler_frontier.pdf
- 22_benchmark_contracts.pdf
- 13_dense_gpu_throughput.pdf
- 14_accuracy_frontier.pdf
- 15_block_diagonal_speedup.pdf
- 16_bdlr_timings.pdf
- 17_output_contract_timings.pdf
- 18_discovery_amortization.pdf
- 19_transfer_pipeline.pdf
- 20_negative_controls.pdf
- 21_best_gpu_speedups.pdf