Notation by work
Machine Understanding
All notation · Glossary · Source coverage
273 symbol spellings in 827 mathematical passages across 4 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 273 entries are available below.
\bar FNotation table · 5 passages
Search aliases: bar F bar
Notation table · Machine_Understanding_v1.0.0.tex, line 98
bar F_u, bar q · An update and a query decoder operating on the retained state.
Exact mathematical passage
\bar F_u
\bar F_uNotation table · 4 passages
Search aliases: bar F_u bar
Notation table · Machine_Understanding_v1.0.0.tex, line 98
bar F_u, bar q · An update and a query decoder operating on the retained state.
Exact mathematical passage
\bar F_u
\bar qNotation table · 5 passages
Search aliases: bar q bar
Notation table · Machine_Understanding_v1.0.0.tex, line 98
bar F_u, bar q · An update and a query decoder operating on the retained state.
Exact mathematical passage
\bar q
CNotation table · 20 passages
Search aliases: C
Notation table · Machine_Understanding_v1.0.0.tex, line 101
C_e,o(V) · Candidates retained after experiment e returns observation o.
Exact mathematical passage
C_{e,o}(V)
\CNotation table · 13 passages
Search aliases: C mathcal
Notation table · Machine_Understanding_v1.0.0.tex, line 95
midrule C · Declared family of states, operations, and future questions.
Exact mathematical passage
\C
C_{e,o}Notation table · 2 passages
Search aliases: C_e,o
Notation table · Machine_Understanding_v1.0.0.tex, line 101
C_e,o(V) · Candidates retained after experiment e returns observation o.
Exact mathematical passage
C_{e,o}(V)
eNotation table · 48 passages
Search aliases: e
Notation table · Machine_Understanding_v1.0.0.tex, line 101
C_e,o(V) · Candidates retained after experiment e returns observation o.
Exact mathematical passage
C_{e,o}(V)
FNotation table · 13 passages
Search aliases: F
Notation table · Machine_Understanding_v1.0.0.tex, line 96
F_u, F_w · One implemented operation, or operations in the order specified by a word w.
Exact mathematical passage
F_u
F_uNotation table · 8 passages
Search aliases: F_u
Notation table · Machine_Understanding_v1.0.0.tex, line 96
F_u, F_w · One implemented operation, or operations in the order specified by a word w.
Exact mathematical passage
F_u
F_wNotation table · 5 passages
Search aliases: F_w
Notation table · Machine_Understanding_v1.0.0.tex, line 96
F_u, F_w · One implemented operation, or operations in the order specified by a word w.
Exact mathematical passage
F_w
\MNotation table · 18 passages
Search aliases: M mathcal
Notation table · Machine_Understanding_v1.0.0.tex, line 100
V⊆ M · Current set of candidate models, not a probability distribution.
Exact mathematical passage
V\subseteq\M
oNotation table · 54 passages
Search aliases: o
Notation table · Machine_Understanding_v1.0.0.tex, line 101
C_e,o(V) · Candidates retained after experiment e returns observation o.
Exact mathematical passage
C_{e,o}(V)
PNotation table · 19 passages
Search aliases: P
Notation table · Machine_Understanding_v1.0.0.tex, line 103
d_TV(P,Q) · tfrac12∑_x|P(x)-Q(x)| on a finite outcome space.
Exact mathematical passage
\TV(P,Q)
QNotation table · 16 passages
Search aliases: Q
Notation table · Machine_Understanding_v1.0.0.tex, line 103
d_TV(P,Q) · tfrac12∑_x|P(x)-Q(x)| on a finite outcome space.
Exact mathematical passage
\TV(P,Q)
rNotation table · 47 passages
Search aliases: r
Notation table · Machine_Understanding_v1.0.0.tex, line 97
r: X→ Z · Retained representation. Its acquisition procedure is a separate obligation.
Exact mathematical passage
r:\X\to\Z
TNotation table · 37 passages
Search aliases: T
Notation table · Machine_Understanding_v1.0.0.tex, line 99
T, τ · Translation of representations and the corresponding operation names.
Exact mathematical passage
T
\TVNotation table · 7 passages
Search aliases: TV d_TV mathrm
Notation table · Machine_Understanding_v1.0.0.tex, line 103
d_TV(P,Q) · tfrac12∑_x|P(x)-Q(x)| on a finite outcome space.
Exact mathematical passage
\TV(P,Q)
uNotation table · 30 passages
Search aliases: u
Notation table · Machine_Understanding_v1.0.0.tex, line 96
F_u, F_w · One implemented operation, or operations in the order specified by a word w.
Exact mathematical passage
F_u
VNotation table · 35 passages
Search aliases: V
Notation table · Machine_Understanding_v1.0.0.tex, line 100
V⊆ M · Current set of candidate models, not a probability distribution.
Exact mathematical passage
V\subseteq\M
wNotation table · 15 passages
Search aliases: w
Notation table · Machine_Understanding_v1.0.0.tex, line 96
F_u, F_w · One implemented operation, or operations in the order specified by a word w.
Exact mathematical passage
F_w
\XNotation table · 15 passages
Search aliases: X mathcal
Notation table · Machine_Understanding_v1.0.0.tex, line 97
r: X→ Z · Retained representation. Its acquisition procedure is a separate obligation.
Exact mathematical passage
r:\X\to\Z
xNotation table · 52 passages
Search aliases: x
Notation table · Machine_Understanding_v1.0.0.tex, line 103
d_TV(P,Q) · tfrac12∑_x|P(x)-Q(x)| on a finite outcome space.
Exact mathematical passage
\tfrac12\sum_x|P(x)-Q(x)|
\ZNotation table · 10 passages
Search aliases: Z mathcal
Notation table · Machine_Understanding_v1.0.0.tex, line 97
r: X→ Z · Retained representation. Its acquisition procedure is a separate obligation.
Exact mathematical passage
r:\X\to\Z
|Notation table · 15 passages
Search aliases: |
Notation table · Machine_Understanding_v1.0.0.tex, line 103
d_TV(P,Q) · tfrac12∑_x|P(x)-Q(x)| on a finite outcome space.
Exact mathematical passage
\tfrac12\sum_x|P(x)-Q(x)|
\etaNotation table · 14 passages
Search aliases: η eta
Notation table · Machine_Understanding_v1.0.0.tex, line 102
ϕ, η · Translation of candidate models and translation of observations.
Exact mathematical passage
\eta
\tauNotation table · 8 passages
Search aliases: τ tau
Notation table · Machine_Understanding_v1.0.0.tex, line 99
T, τ · Translation of representations and the corresponding operation names.
Exact mathematical passage
\tau
\phiNotation table · 11 passages
Search aliases: ϕ phi
Notation table · Machine_Understanding_v1.0.0.tex, line 102
ϕ, η · Translation of candidate models and translation of observations.
Exact mathematical passage
\phi
\toNotation table · 27 passages
Search aliases: → to
Notation table · Machine_Understanding_v1.0.0.tex, line 97
r: X→ Z · Retained representation. Its acquisition procedure is a separate obligation.
Exact mathematical passage
r:\X\to\Z
\sumNotation table · 9 passages
Search aliases: ∑ sum
Notation table · Machine_Understanding_v1.0.0.tex, line 103
d_TV(P,Q) · tfrac12∑_x|P(x)-Q(x)| on a finite outcome space.
Exact mathematical passage
\tfrac12\sum_x|P(x)-Q(x)|
\sum_xNotation table · 1 passage
Search aliases: ∑_x sum
Notation table · Machine_Understanding_v1.0.0.tex, line 103
d_TV(P,Q) · tfrac12∑_x|P(x)-Q(x)| on a finite outcome space.
Exact mathematical passage
\tfrac12\sum_x|P(x)-Q(x)|
\subseteqNotation table · 9 passages
Search aliases: ⊆ subseteq
Notation table · Machine_Understanding_v1.0.0.tex, line 100
V⊆ M · Current set of candidate models, not a probability distribution.
Exact mathematical passage
V\subseteq\M
=Definition passage · 133 passages
Search aliases: =
Definition passage: Exact reusable representation · Machine_Understanding_v1.0.0.tex, line 223
[Exact reusable representation] An exact reusable representation for C is a surjection r: X→ Z, updates bar F_u: Z→ Z, and decoders bar q: Z→ Y such that, for all x,u,q, r(F_u(x))=bar F_u(r(x)), q(x)=bar q(r(x)). Induction immediately gives q(F_w(x))=bar q(bar F_w(r(x))) for every finite word. The same initial record supports many future questions. It is not allowed to inspect the selected future query and replace itself with that query's answer before the challenge begins.
Exact mathematical passage
\label{eq:representation-contract} r(F_u(x))=\bar F_u(r(x)),\qquad q(x)=\bar q(r(x)).
BDefinition passage · 27 passages
Search aliases: B
Definition passage: Recipient-relative explanatory adequacy · Machine_Understanding_v1.0.0.tex, line 441
[Recipient-relative explanatory adequacy] Fix a recipient procedure L, background information B, resource allowance, and continuation family. An artifact E is adequate for a target model m when L(B,E) yields a representation and operations that answer the required continuations for m within the declared error and resource limits. The artifact's inputs, the recipient's prior access, and any target-specific background are included in the contract. This is an operational property of an artifact and…
Exact mathematical passage
B
EDefinition passage · 23 passages
Search aliases: E
Definition passage: Recipient-relative explanatory adequacy · Machine_Understanding_v1.0.0.tex, line 441
[Recipient-relative explanatory adequacy] Fix a recipient procedure L, background information B, resource allowance, and continuation family. An artifact E is adequate for a target model m when L(B,E) yields a representation and operations that answer the required continuations for m within the declared error and resource limits. The artifact's inputs, the recipient's prior access, and any target-specific background are included in the contract. This is an operational property of an artifact and…
Exact mathematical passage
E
LDefinition passage · 33 passages
Search aliases: L
Definition passage: Recipient-relative explanatory adequacy · Machine_Understanding_v1.0.0.tex, line 441
[Recipient-relative explanatory adequacy] Fix a recipient procedure L, background information B, resource allowance, and continuation family. An artifact E is adequate for a target model m when L(B,E) yields a representation and operations that answer the required continuations for m within the declared error and resource limits. The artifact's inputs, the recipient's prior access, and any target-specific background are included in the contract. This is an operational property of an artifact and…
Exact mathematical passage
L
mDefinition passage · 60 passages
Search aliases: m
Definition passage: Recipient-relative explanatory adequacy · Machine_Understanding_v1.0.0.tex, line 441
[Recipient-relative explanatory adequacy] Fix a recipient procedure L, background information B, resource allowance, and continuation family. An artifact E is adequate for a target model m when L(B,E) yields a representation and operations that answer the required continuations for m within the declared error and resource limits. The artifact's inputs, the recipient's prior access, and any target-specific background are included in the contract. This is an operational property of an artifact and…
Exact mathematical passage
m
qDefinition passage · 22 passages
Search aliases: q
Definition passage: Exact reusable representation · Machine_Understanding_v1.0.0.tex, line 222
[Exact reusable representation] An exact reusable representation for C is a surjection r: X→ Z, updates bar F_u: Z→ Z, and decoders bar q: Z→ Y such that, for all x,u,q, r(F_u(x))=bar F_u(r(x)), q(x)=bar q(r(x)). Induction immediately gives q(F_w(x))=bar q(bar F_w(r(x))) for every finite word. The same initial record supports many future questions. It is not allowed to inspect the selected future query and replace itself with that query's answer before the challenge begins.
Exact mathematical passage
x,u,q
\YDefinition passage · 4 passages
Search aliases: Y mathcal
Definition passage: Exact reusable representation · Machine_Understanding_v1.0.0.tex, line 222
[Exact reusable representation] An exact reusable representation for C is a surjection r: X→ Z, updates bar F_u: Z→ Z, and decoders bar q: Z→ Y such that, for all x,u,q, r(F_u(x))=bar F_u(r(x)), q(x)=bar q(r(x)). Induction immediately gives q(F_w(x))=bar q(bar F_w(r(x))) for every finite word. The same initial record supports many future questions. It is not allowed to inspect the selected future query and replace itself with that query's answer before the challenge begins.
Exact mathematical passage
\bar q:\Z\to\Y
+Introduction passage · 41 passages
Search aliases: +
Introduction passage · Machine_Understanding_v1.0.0.tex, line 583
Let r:X→ Z be the target record. A support is terminal when r is constant on it. Let h be the number of experiments still available. Write V_h(V) for the least worst-case sum of experiment costs among policies that terminate with the target known within h steps. Infeasibility is denoted +∞. Then define V_h(V) · =0 · · if |r(V)|=1,
Exact mathematical passage
+\infty
AIntroduction passage · 10 passages
Search aliases: A
Introduction passage · Machine_Understanding_v1.0.0.tex, line 584
Let r:X→ Z be the target record. A support is terminal when r is constant on it. Let h be the number of experiments still available. Write V_h(V) for the least worst-case sum of experiment costs among policies that terminate with the target known within h steps. Infeasibility is denoted +∞. Then define V_h(V) · =0 · · if |r(V)|=1, V_0(V) · =+∞ · · otherwise, V_h(V) · =min_e:V⊆ A_e (c_e+max_o:V_e,o≠∅ V_h-1(V_e,o)) · · substackotherwise, h>0. An empty minimum is +∞. The terminal case permits stopp…
Exact mathematical passage
\label{eq:acquisition-bellman} \begin{aligned} \mathcal V_h(V)&=0 &&\text{if }|r(V)|=1,\\ \mathcal V_0(V)&=+\infty &&\text{otherwise},\\ \mathcal V_h(V)&=\min_{e:V\subseteq A_e} \left(c_e+\max_{o:V_{e,o}\ne\varnothing}\mathcal V_{h-1}(V_{e,o})\right) &&\substack{\text{otherwise,}\\h>0.} \end{aligned}
A_eIntroduction passage · 3 passages
Search aliases: A_e
Introduction passage · Machine_Understanding_v1.0.0.tex, line 584
Let r:X→ Z be the target record. A support is terminal when r is constant on it. Let h be the number of experiments still available. Write V_h(V) for the least worst-case sum of experiment costs among policies that terminate with the target known within h steps. Infeasibility is denoted +∞. Then define V_h(V) · =0 · · if |r(V)|=1, V_0(V) · =+∞ · · otherwise, V_h(V) · =min_e:V⊆ A_e (c_e+max_o:V_e,o≠∅ V_h-1(V_e,o)) · · substackotherwise, h>0. An empty minimum is +∞. The terminal case permits stopp…
Exact mathematical passage
\label{eq:acquisition-bellman} \begin{aligned} \mathcal V_h(V)&=0 &&\text{if }|r(V)|=1,\\ \mathcal V_0(V)&=+\infty &&\text{otherwise},\\ \mathcal V_h(V)&=\min_{e:V\subseteq A_e} \left(c_e+\max_{o:V_{e,o}\ne\varnothing}\mathcal V_{h-1}(V_{e,o})\right) &&\substack{\text{otherwise,}\\h>0.} \end{aligned}
cIntroduction passage · 66 passages
Search aliases: c
Introduction passage · Machine_Understanding_v1.0.0.tex, line 584
Let r:X→ Z be the target record. A support is terminal when r is constant on it. Let h be the number of experiments still available. Write V_h(V) for the least worst-case sum of experiment costs among policies that terminate with the target known within h steps. Infeasibility is denoted +∞. Then define V_h(V) · =0 · · if |r(V)|=1, V_0(V) · =+∞ · · otherwise, V_h(V) · =min_e:V⊆ A_e (c_e+max_o:V_e,o≠∅ V_h-1(V_e,o)) · · substackotherwise, h>0. An empty minimum is +∞. The terminal case permits stopp…
Exact mathematical passage
\label{eq:acquisition-bellman} \begin{aligned} \mathcal V_h(V)&=0 &&\text{if }|r(V)|=1,\\ \mathcal V_0(V)&=+\infty &&\text{otherwise},\\ \mathcal V_h(V)&=\min_{e:V\subseteq A_e} \left(c_e+\max_{o:V_{e,o}\ne\varnothing}\mathcal V_{h-1}(V_{e,o})\right) &&\substack{\text{otherwise,}\\h>0.} \end{aligned}
\C'Introduction passage · 6 passages
Search aliases: C' C mathcal
Introduction passage · Machine_Understanding_v1.0.0.tex, line 276
The cost of expanding the contract. Let C' have the same complete state space and contain every operation and query of C, while possibly adding more. Write ≡_ C' for its finer behavioral equivalence. Take any representation r adequate for C. For each z, let k_z be the number of ≡_ C' classes that meet the fibre r^-1(z).
Exact mathematical passage
\C'
c_eIntroduction passage · 6 passages
Search aliases: c_e
Introduction passage · Machine_Understanding_v1.0.0.tex, line 584
Let r:X→ Z be the target record. A support is terminal when r is constant on it. Let h be the number of experiments still available. Write V_h(V) for the least worst-case sum of experiment costs among policies that terminate with the target known within h steps. Infeasibility is denoted +∞. Then define V_h(V) · =0 · · if |r(V)|=1, V_0(V) · =+∞ · · otherwise, V_h(V) · =min_e:V⊆ A_e (c_e+max_o:V_e,o≠∅ V_h-1(V_e,o)) · · substackotherwise, h>0. An empty minimum is +∞. The terminal case permits stopp…
Exact mathematical passage
\label{eq:acquisition-bellman} \begin{aligned} \mathcal V_h(V)&=0 &&\text{if }|r(V)|=1,\\ \mathcal V_0(V)&=+\infty &&\text{otherwise},\\ \mathcal V_h(V)&=\min_{e:V\subseteq A_e} \left(c_e+\max_{o:V_{e,o}\ne\varnothing}\mathcal V_{h-1}(V_{e,o})\right) &&\substack{\text{otherwise,}\\h>0.} \end{aligned}
dIntroduction passage · 18 passages
Search aliases: d
Introduction passage · Machine_Understanding_v1.0.0.tex, line 929
For a policy tree T and initial state x, write s_T(x) for its terminal sender transcript and x_T(x) for its final physical state. The tree has finitely many reachable leaves. Specify recipient backgrounds β_i^T(x) and demands d_i^T(x) by a declared rule. A demand may concern x_T(x), the original x, or a stated pair of past and present values. These are different contracts.
Exact mathematical passage
d_i^T(x)
d_i^TIntroduction passage · 1 passage
Search aliases: d_i^T
Introduction passage · Machine_Understanding_v1.0.0.tex, line 929
For a policy tree T and initial state x, write s_T(x) for its terminal sender transcript and x_T(x) for its final physical state. The tree has finitely many reachable leaves. Specify recipient backgrounds β_i^T(x) and demands d_i^T(x) by a declared rule. A demand may concern x_T(x), the original x, or a stated pair of past and present values. These are different contracts.
Exact mathematical passage
d_i^T(x)
hIntroduction passage · 29 passages
Search aliases: h
Introduction passage · Machine_Understanding_v1.0.0.tex, line 583
Let r:X→ Z be the target record. A support is terminal when r is constant on it. Let h be the number of experiments still available. Write V_h(V) for the least worst-case sum of experiment costs among policies that terminate with the target known within h steps. Infeasibility is denoted +∞. Then define V_h(V) · =0 · · if |r(V)|=1,
Exact mathematical passage
h
iIntroduction passage · 43 passages
Search aliases: i
Introduction passage · Machine_Understanding_v1.0.0.tex, line 929
For a policy tree T and initial state x, write s_T(x) for its terminal sender transcript and x_T(x) for its final physical state. The tree has finitely many reachable leaves. Specify recipient backgrounds β_i^T(x) and demands d_i^T(x) by a declared rule. A demand may concern x_T(x), the original x, or a stated pair of past and present values. These are different contracts.
Exact mathematical passage
\beta_i^T(x)
kIntroduction passage · 9 passages
Search aliases: k
Introduction passage · Machine_Understanding_v1.0.0.tex, line 276
The cost of expanding the contract. Let C' have the same complete state space and contain every operation and query of C, while possibly adding more. Write ≡_ C' for its finer behavioral equivalence. Take any representation r adequate for C. For each z, let k_z be the number of ≡_ C' classes that meet the fibre r^-1(z).
Exact mathematical passage
k_z
k_zIntroduction passage · 5 passages
Search aliases: k_z
Introduction passage · Machine_Understanding_v1.0.0.tex, line 276
The cost of expanding the contract. Let C' have the same complete state space and contain every operation and query of C, while possibly adding more. Write ≡_ C' for its finer behavioral equivalence. Take any representation r adequate for C. For each z, let k_z be the number of ≡_ C' classes that meet the fibre r^-1(z).
Exact mathematical passage
k_z
L_mIntroduction passage · 7 passages
Search aliases: L_m
Introduction passage · Machine_Understanding_v1.0.0.tex, line 686
Let M∈ M be a fixed unknown model in a finite family, with prior p(m). An experiment produces O∈ O with likelihood L_m(o). A common reporting channel B(o,z) produces Z∈ Z. Each likelihood row and each channel row is a probability distribution. The complete joint model is Pr(M=m,O=o,Z=z)=p(m)L_m(o)B(o,z). In particular, the channel does not secretly inspect M. Define Q_m(z)=∑_o L_m(o)B(o,z), p^o(m)=fracp(m)L_m(o)∑_n p(n)L_n(o), p^z(m)=fracp(m)Q_m(z)∑_n p(n)Q_n(z). A posterior is defined only when…
Exact mathematical passage
L_m(o)
L_nIntroduction passage · 1 passage
Search aliases: L_n
Introduction passage · Machine_Understanding_v1.0.0.tex, line 691
prior p(m). An experiment produces O∈ O with likelihood L_m(o). A common reporting channel B(o,z) produces Z∈ Z. Each likelihood row and each channel row is a probability distribution. The complete joint model is Pr(M=m,O=o,Z=z)=p(m)L_m(o)B(o,z). In particular, the channel does not secretly inspect M. Define Q_m(z)=∑_o L_m(o)B(o,z), p^o(m)=fracp(m)L_m(o)∑_n p(n)L_n(o), p^z(m)=fracp(m)Q_m(z)∑_n p(n)Q_n(z). A posterior is defined only when its denominator is positive. An event assigned probability…
Exact mathematical passage
Q_m(z)=\sum_o L_m(o)B(o,z),\qquad p^o(m)=\frac{p(m)L_m(o)}{\sum_n p(n)L_n(o)},\qquad p^z(m)=\frac{p(m)Q_m(z)}{\sum_n p(n)Q_n(z)}.
MIntroduction passage · 20 passages
Search aliases: M
Introduction passage · Machine_Understanding_v1.0.0.tex, line 686
Let M∈ M be a fixed unknown model in a finite family, with prior p(m). An experiment produces O∈ O with likelihood L_m(o). A common reporting channel B(o,z) produces Z∈ Z. Each likelihood row and each channel row is a probability distribution. The complete joint model is Pr(M=m,O=o,Z=z)=p(m)L_m(o)B(o,z). In particular, the channel does not secretly inspect M. Define Q_m(z)=∑_o L_m(o)B(o,z), p^o(m)=fracp(m)L_m(o)∑_n p(n)L_n(o), p^z(m)=fracp(m)Q_m(z)∑_n p(n)Q_n(z). A posterior is defined only when…
Exact mathematical passage
M\in\M
\maxIntroduction passage · 15 passages
Search aliases: max
Introduction passage · Machine_Understanding_v1.0.0.tex, line 584
Let r:X→ Z be the target record. A support is terminal when r is constant on it. Let h be the number of experiments still available. Write V_h(V) for the least worst-case sum of experiment costs among policies that terminate with the target known within h steps. Infeasibility is denoted +∞. Then define V_h(V) · =0 · · if |r(V)|=1, V_0(V) · =+∞ · · otherwise, V_h(V) · =min_e:V⊆ A_e (c_e+max_o:V_e,o≠∅ V_h-1(V_e,o)) · · substackotherwise, h>0. An empty minimum is +∞. The terminal case permits stopp…
Exact mathematical passage
\label{eq:acquisition-bellman} \begin{aligned} \mathcal V_h(V)&=0 &&\text{if }|r(V)|=1,\\ \mathcal V_0(V)&=+\infty &&\text{otherwise},\\ \mathcal V_h(V)&=\min_{e:V\subseteq A_e} \left(c_e+\max_{o:V_{e,o}\ne\varnothing}\mathcal V_{h-1}(V_{e,o})\right) &&\substack{\text{otherwise,}\\h>0.} \end{aligned}
\max_{o:V_{e,o}\ne\varnothing}Introduction passage · 1 passage
Search aliases: max_o:V_e,o≠∅ max ne varnothing
Introduction passage · Machine_Understanding_v1.0.0.tex, line 584
Let r:X→ Z be the target record. A support is terminal when r is constant on it. Let h be the number of experiments still available. Write V_h(V) for the least worst-case sum of experiment costs among policies that terminate with the target known within h steps. Infeasibility is denoted +∞. Then define V_h(V) · =0 · · if |r(V)|=1, V_0(V) · =+∞ · · otherwise, V_h(V) · =min_e:V⊆ A_e (c_e+max_o:V_e,o≠∅ V_h-1(V_e,o)) · · substackotherwise, h>0. An empty minimum is +∞. The terminal case permits stopp…
Exact mathematical passage
\label{eq:acquisition-bellman} \begin{aligned} \mathcal V_h(V)&=0 &&\text{if }|r(V)|=1,\\ \mathcal V_0(V)&=+\infty &&\text{otherwise},\\ \mathcal V_h(V)&=\min_{e:V\subseteq A_e} \left(c_e+\max_{o:V_{e,o}\ne\varnothing}\mathcal V_{h-1}(V_{e,o})\right) &&\substack{\text{otherwise,}\\h>0.} \end{aligned}
\minIntroduction passage · 9 passages
Search aliases: min
Introduction passage · Machine_Understanding_v1.0.0.tex, line 584
Let r:X→ Z be the target record. A support is terminal when r is constant on it. Let h be the number of experiments still available. Write V_h(V) for the least worst-case sum of experiment costs among policies that terminate with the target known within h steps. Infeasibility is denoted +∞. Then define V_h(V) · =0 · · if |r(V)|=1, V_0(V) · =+∞ · · otherwise, V_h(V) · =min_e:V⊆ A_e (c_e+max_o:V_e,o≠∅ V_h-1(V_e,o)) · · substackotherwise, h>0. An empty minimum is +∞. The terminal case permits stopp…
Exact mathematical passage
\label{eq:acquisition-bellman} \begin{aligned} \mathcal V_h(V)&=0 &&\text{if }|r(V)|=1,\\ \mathcal V_0(V)&=+\infty &&\text{otherwise},\\ \mathcal V_h(V)&=\min_{e:V\subseteq A_e} \left(c_e+\max_{o:V_{e,o}\ne\varnothing}\mathcal V_{h-1}(V_{e,o})\right) &&\substack{\text{otherwise,}\\h>0.} \end{aligned}
\min_{e:V\subseteq A_e}Introduction passage · 1 passage
Search aliases: min_e:V⊆ A_e min subseteq
Introduction passage · Machine_Understanding_v1.0.0.tex, line 584
Let r:X→ Z be the target record. A support is terminal when r is constant on it. Let h be the number of experiments still available. Write V_h(V) for the least worst-case sum of experiment costs among policies that terminate with the target known within h steps. Infeasibility is denoted +∞. Then define V_h(V) · =0 · · if |r(V)|=1, V_0(V) · =+∞ · · otherwise, V_h(V) · =min_e:V⊆ A_e (c_e+max_o:V_e,o≠∅ V_h-1(V_e,o)) · · substackotherwise, h>0. An empty minimum is +∞. The terminal case permits stopp…
Exact mathematical passage
\label{eq:acquisition-bellman} \begin{aligned} \mathcal V_h(V)&=0 &&\text{if }|r(V)|=1,\\ \mathcal V_0(V)&=+\infty &&\text{otherwise},\\ \mathcal V_h(V)&=\min_{e:V\subseteq A_e} \left(c_e+\max_{o:V_{e,o}\ne\varnothing}\mathcal V_{h-1}(V_{e,o})\right) &&\substack{\text{otherwise,}\\h>0.} \end{aligned}
nIntroduction passage · 43 passages
Search aliases: n
Introduction passage · Machine_Understanding_v1.0.0.tex, line 691
prior p(m). An experiment produces O∈ O with likelihood L_m(o). A common reporting channel B(o,z) produces Z∈ Z. Each likelihood row and each channel row is a probability distribution. The complete joint model is Pr(M=m,O=o,Z=z)=p(m)L_m(o)B(o,z). In particular, the channel does not secretly inspect M. Define Q_m(z)=∑_o L_m(o)B(o,z), p^o(m)=fracp(m)L_m(o)∑_n p(n)L_n(o), p^z(m)=fracp(m)Q_m(z)∑_n p(n)Q_n(z). A posterior is defined only when its denominator is positive. An event assigned probability…
Exact mathematical passage
Q_m(z)=\sum_o L_m(o)B(o,z),\qquad p^o(m)=\frac{p(m)L_m(o)}{\sum_n p(n)L_n(o)},\qquad p^z(m)=\frac{p(m)Q_m(z)}{\sum_n p(n)Q_n(z)}.
OIntroduction passage · 9 passages
Search aliases: O
Introduction passage · Machine_Understanding_v1.0.0.tex, line 686
Let M∈ M be a fixed unknown model in a finite family, with prior p(m). An experiment produces O∈ O with likelihood L_m(o). A common reporting channel B(o,z) produces Z∈ Z. Each likelihood row and each channel row is a probability distribution. The complete joint model is Pr(M=m,O=o,Z=z)=p(m)L_m(o)B(o,z). In particular, the channel does not secretly inspect M. Define Q_m(z)=∑_o L_m(o)B(o,z), p^o(m)=fracp(m)L_m(o)∑_n p(n)L_n(o), p^z(m)=fracp(m)Q_m(z)∑_n p(n)Q_n(z). A posterior is defined only when…
Exact mathematical passage
O\in\mathcal O
\mathcal OIntroduction passage · 5 passages
Search aliases: O mathcal
Introduction passage · Machine_Understanding_v1.0.0.tex, line 686
Let M∈ M be a fixed unknown model in a finite family, with prior p(m). An experiment produces O∈ O with likelihood L_m(o). A common reporting channel B(o,z) produces Z∈ Z. Each likelihood row and each channel row is a probability distribution. The complete joint model is Pr(M=m,O=o,Z=z)=p(m)L_m(o)B(o,z). In particular, the channel does not secretly inspect M. Define Q_m(z)=∑_o L_m(o)B(o,z), p^o(m)=fracp(m)L_m(o)∑_n p(n)L_n(o), p^z(m)=fracp(m)Q_m(z)∑_n p(n)Q_n(z). A posterior is defined only when…
Exact mathematical passage
O\in\mathcal O
pIntroduction passage · 29 passages
Search aliases: p
Introduction passage · Machine_Understanding_v1.0.0.tex, line 686
Let M∈ M be a fixed unknown model in a finite family, with prior p(m). An experiment produces O∈ O with likelihood L_m(o). A common reporting channel B(o,z) produces Z∈ Z. Each likelihood row and each channel row is a probability distribution. The complete joint model is Pr(M=m,O=o,Z=z)=p(m)L_m(o)B(o,z). In particular, the channel does not secretly inspect M. Define Q_m(z)=∑_o L_m(o)B(o,z), p^o(m)=fracp(m)L_m(o)∑_n p(n)L_n(o), p^z(m)=fracp(m)Q_m(z)∑_n p(n)Q_n(z). A posterior is defined only when…
Exact mathematical passage
p(m)
p^oIntroduction passage · 8 passages
Search aliases: p^o
Introduction passage · Machine_Understanding_v1.0.0.tex, line 691
prior p(m). An experiment produces O∈ O with likelihood L_m(o). A common reporting channel B(o,z) produces Z∈ Z. Each likelihood row and each channel row is a probability distribution. The complete joint model is Pr(M=m,O=o,Z=z)=p(m)L_m(o)B(o,z). In particular, the channel does not secretly inspect M. Define Q_m(z)=∑_o L_m(o)B(o,z), p^o(m)=fracp(m)L_m(o)∑_n p(n)L_n(o), p^z(m)=fracp(m)Q_m(z)∑_n p(n)Q_n(z). A posterior is defined only when its denominator is positive. An event assigned probability…
Exact mathematical passage
Q_m(z)=\sum_o L_m(o)B(o,z),\qquad p^o(m)=\frac{p(m)L_m(o)}{\sum_n p(n)L_n(o)},\qquad p^z(m)=\frac{p(m)Q_m(z)}{\sum_n p(n)Q_n(z)}.
p^zIntroduction passage · 6 passages
Search aliases: p^z
Introduction passage · Machine_Understanding_v1.0.0.tex, line 691
prior p(m). An experiment produces O∈ O with likelihood L_m(o). A common reporting channel B(o,z) produces Z∈ Z. Each likelihood row and each channel row is a probability distribution. The complete joint model is Pr(M=m,O=o,Z=z)=p(m)L_m(o)B(o,z). In particular, the channel does not secretly inspect M. Define Q_m(z)=∑_o L_m(o)B(o,z), p^o(m)=fracp(m)L_m(o)∑_n p(n)L_n(o), p^z(m)=fracp(m)Q_m(z)∑_n p(n)Q_n(z). A posterior is defined only when its denominator is positive. An event assigned probability…
Exact mathematical passage
Q_m(z)=\sum_o L_m(o)B(o,z),\qquad p^o(m)=\frac{p(m)L_m(o)}{\sum_n p(n)L_n(o)},\qquad p^z(m)=\frac{p(m)Q_m(z)}{\sum_n p(n)Q_n(z)}.
\PrIntroduction passage · 4 passages
Search aliases: Pr
Introduction passage · Machine_Understanding_v1.0.0.tex, line 687
Let M∈ M be a fixed unknown model in a finite family, with prior p(m). An experiment produces O∈ O with likelihood L_m(o). A common reporting channel B(o,z) produces Z∈ Z. Each likelihood row and each channel row is a probability distribution. The complete joint model is Pr(M=m,O=o,Z=z)=p(m)L_m(o)B(o,z). In particular, the channel does not secretly inspect M. Define Q_m(z)=∑_o L_m(o)B(o,z), p^o(m)=fracp(m)L_m(o)∑_n p(n)L_n(o), p^z(m)=fracp(m)Q_m(z)∑_n p(n)Q_n(z). A posterior is defined only when…
Exact mathematical passage
\label{eq:bayes-joint} \Pr(M=m,O=o,Z=z)=p(m)L_m(o)B(o,z).
Q_mIntroduction passage · 3 passages
Search aliases: Q_m
Introduction passage · Machine_Understanding_v1.0.0.tex, line 691
prior p(m). An experiment produces O∈ O with likelihood L_m(o). A common reporting channel B(o,z) produces Z∈ Z. Each likelihood row and each channel row is a probability distribution. The complete joint model is Pr(M=m,O=o,Z=z)=p(m)L_m(o)B(o,z). In particular, the channel does not secretly inspect M. Define Q_m(z)=∑_o L_m(o)B(o,z), p^o(m)=fracp(m)L_m(o)∑_n p(n)L_n(o), p^z(m)=fracp(m)Q_m(z)∑_n p(n)Q_n(z). A posterior is defined only when its denominator is positive. An event assigned probability…
Exact mathematical passage
Q_m(z)=\sum_o L_m(o)B(o,z),\qquad p^o(m)=\frac{p(m)L_m(o)}{\sum_n p(n)L_n(o)},\qquad p^z(m)=\frac{p(m)Q_m(z)}{\sum_n p(n)Q_n(z)}.
Q_nIntroduction passage · 1 passage
Search aliases: Q_n
Introduction passage · Machine_Understanding_v1.0.0.tex, line 691
prior p(m). An experiment produces O∈ O with likelihood L_m(o). A common reporting channel B(o,z) produces Z∈ Z. Each likelihood row and each channel row is a probability distribution. The complete joint model is Pr(M=m,O=o,Z=z)=p(m)L_m(o)B(o,z). In particular, the channel does not secretly inspect M. Define Q_m(z)=∑_o L_m(o)B(o,z), p^o(m)=fracp(m)L_m(o)∑_n p(n)L_n(o), p^z(m)=fracp(m)Q_m(z)∑_n p(n)Q_n(z). A posterior is defined only when its denominator is positive. An event assigned probability…
Exact mathematical passage
Q_m(z)=\sum_o L_m(o)B(o,z),\qquad p^o(m)=\frac{p(m)L_m(o)}{\sum_n p(n)L_n(o)},\qquad p^z(m)=\frac{p(m)Q_m(z)}{\sum_n p(n)Q_n(z)}.
r^{-1}Introduction passage · 1 passage
Search aliases: r^-1
Introduction passage · Machine_Understanding_v1.0.0.tex, line 276
The cost of expanding the contract. Let C' have the same complete state space and contain every operation and query of C, while possibly adding more. Write ≡_ C' for its finer behavioral equivalence. Take any representation r adequate for C. For each z, let k_z be the number of ≡_ C' classes that meet the fibre r^-1(z).
Exact mathematical passage
r^{-1}(z)
sIntroduction passage · 61 passages
Search aliases: s
Introduction passage · Machine_Understanding_v1.0.0.tex, line 929
For a policy tree T and initial state x, write s_T(x) for its terminal sender transcript and x_T(x) for its final physical state. The tree has finitely many reachable leaves. Specify recipient backgrounds β_i^T(x) and demands d_i^T(x) by a declared rule. A demand may concern x_T(x), the original x, or a stated pair of past and present values. These are different contracts.
Exact mathematical passage
s_T(x)
s_TIntroduction passage · 2 passages
Search aliases: s_T
Introduction passage · Machine_Understanding_v1.0.0.tex, line 929
For a policy tree T and initial state x, write s_T(x) for its terminal sender transcript and x_T(x) for its final physical state. The tree has finitely many reachable leaves. Specify recipient backgrounds β_i^T(x) and demands d_i^T(x) by a declared rule. A demand may concern x_T(x), the original x, or a stated pair of past and present values. These are different contracts.
Exact mathematical passage
s_T(x)
\mathcal VIntroduction passage · 3 passages
Search aliases: V mathcal
Introduction passage · Machine_Understanding_v1.0.0.tex, line 583
Let r:X→ Z be the target record. A support is terminal when r is constant on it. Let h be the number of experiments still available. Write V_h(V) for the least worst-case sum of experiment costs among policies that terminate with the target known within h steps. Infeasibility is denoted +∞. Then define V_h(V) · =0 · · if |r(V)|=1,
Exact mathematical passage
\mathcal V_h(V)
\mathcal V_0Introduction passage · 1 passage
Search aliases: V_0 mathcal
Introduction passage · Machine_Understanding_v1.0.0.tex, line 584
Let r:X→ Z be the target record. A support is terminal when r is constant on it. Let h be the number of experiments still available. Write V_h(V) for the least worst-case sum of experiment costs among policies that terminate with the target known within h steps. Infeasibility is denoted +∞. Then define V_h(V) · =0 · · if |r(V)|=1, V_0(V) · =+∞ · · otherwise, V_h(V) · =min_e:V⊆ A_e (c_e+max_o:V_e,o≠∅ V_h-1(V_e,o)) · · substackotherwise, h>0. An empty minimum is +∞. The terminal case permits stopp…
Exact mathematical passage
\label{eq:acquisition-bellman} \begin{aligned} \mathcal V_h(V)&=0 &&\text{if }|r(V)|=1,\\ \mathcal V_0(V)&=+\infty &&\text{otherwise},\\ \mathcal V_h(V)&=\min_{e:V\subseteq A_e} \left(c_e+\max_{o:V_{e,o}\ne\varnothing}\mathcal V_{h-1}(V_{e,o})\right) &&\substack{\text{otherwise,}\\h>0.} \end{aligned}
V_{e,o}Introduction passage · 4 passages
Search aliases: V_e,o
Introduction passage · Machine_Understanding_v1.0.0.tex, line 584
Let r:X→ Z be the target record. A support is terminal when r is constant on it. Let h be the number of experiments still available. Write V_h(V) for the least worst-case sum of experiment costs among policies that terminate with the target known within h steps. Infeasibility is denoted +∞. Then define V_h(V) · =0 · · if |r(V)|=1, V_0(V) · =+∞ · · otherwise, V_h(V) · =min_e:V⊆ A_e (c_e+max_o:V_e,o≠∅ V_h-1(V_e,o)) · · substackotherwise, h>0. An empty minimum is +∞. The terminal case permits stopp…
Exact mathematical passage
\label{eq:acquisition-bellman} \begin{aligned} \mathcal V_h(V)&=0 &&\text{if }|r(V)|=1,\\ \mathcal V_0(V)&=+\infty &&\text{otherwise},\\ \mathcal V_h(V)&=\min_{e:V\subseteq A_e} \left(c_e+\max_{o:V_{e,o}\ne\varnothing}\mathcal V_{h-1}(V_{e,o})\right) &&\substack{\text{otherwise,}\\h>0.} \end{aligned}
\mathcal V_hIntroduction passage · 2 passages
Search aliases: V_h mathcal
Introduction passage · Machine_Understanding_v1.0.0.tex, line 583
Let r:X→ Z be the target record. A support is terminal when r is constant on it. Let h be the number of experiments still available. Write V_h(V) for the least worst-case sum of experiment costs among policies that terminate with the target known within h steps. Infeasibility is denoted +∞. Then define V_h(V) · =0 · · if |r(V)|=1,
Exact mathematical passage
\mathcal V_h(V)
\mathcal V_{h-1}Introduction passage · 2 passages
Search aliases: V_h-1 mathcal
Introduction passage · Machine_Understanding_v1.0.0.tex, line 584
Let r:X→ Z be the target record. A support is terminal when r is constant on it. Let h be the number of experiments still available. Write V_h(V) for the least worst-case sum of experiment costs among policies that terminate with the target known within h steps. Infeasibility is denoted +∞. Then define V_h(V) · =0 · · if |r(V)|=1, V_0(V) · =+∞ · · otherwise, V_h(V) · =min_e:V⊆ A_e (c_e+max_o:V_e,o≠∅ V_h-1(V_e,o)) · · substackotherwise, h>0. An empty minimum is +∞. The terminal case permits stopp…
Exact mathematical passage
\label{eq:acquisition-bellman} \begin{aligned} \mathcal V_h(V)&=0 &&\text{if }|r(V)|=1,\\ \mathcal V_0(V)&=+\infty &&\text{otherwise},\\ \mathcal V_h(V)&=\min_{e:V\subseteq A_e} \left(c_e+\max_{o:V_{e,o}\ne\varnothing}\mathcal V_{h-1}(V_{e,o})\right) &&\substack{\text{otherwise,}\\h>0.} \end{aligned}
XIntroduction passage · 14 passages
Search aliases: X
Introduction passage · Machine_Understanding_v1.0.0.tex, line 583
Let r:X→ Z be the target record. A support is terminal when r is constant on it. Let h be the number of experiments still available. Write V_h(V) for the least worst-case sum of experiment costs among policies that terminate with the target known within h steps. Infeasibility is denoted +∞. Then define V_h(V) · =0 · · if |r(V)|=1,
Exact mathematical passage
r:X\to Z
x_TIntroduction passage · 2 passages
Search aliases: x_T
Introduction passage · Machine_Understanding_v1.0.0.tex, line 929
For a policy tree T and initial state x, write s_T(x) for its terminal sender transcript and x_T(x) for its final physical state. The tree has finitely many reachable leaves. Specify recipient backgrounds β_i^T(x) and demands d_i^T(x) by a declared rule. A demand may concern x_T(x), the original x, or a stated pair of past and present values. These are different contracts.
Exact mathematical passage
x_T(x)
ZIntroduction passage · 11 passages
Search aliases: Z
Introduction passage · Machine_Understanding_v1.0.0.tex, line 583
Let r:X→ Z be the target record. A support is terminal when r is constant on it. Let h be the number of experiments still available. Write V_h(V) for the least worst-case sum of experiment costs among policies that terminate with the target known within h steps. Infeasibility is denoted +∞. Then define V_h(V) · =0 · · if |r(V)|=1,
Exact mathematical passage
r:X\to Z
\mathcal ZIntroduction passage · 1 passage
Search aliases: Z mathcal
Introduction passage · Machine_Understanding_v1.0.0.tex, line 686
Let M∈ M be a fixed unknown model in a finite family, with prior p(m). An experiment produces O∈ O with likelihood L_m(o). A common reporting channel B(o,z) produces Z∈ Z. Each likelihood row and each channel row is a probability distribution. The complete joint model is Pr(M=m,O=o,Z=z)=p(m)L_m(o)B(o,z). In particular, the channel does not secretly inspect M. Define Q_m(z)=∑_o L_m(o)B(o,z), p^o(m)=fracp(m)L_m(o)∑_n p(n)L_n(o), p^z(m)=fracp(m)Q_m(z)∑_n p(n)Q_n(z). A posterior is defined only when…
Exact mathematical passage
Z\in\mathcal Z
zIntroduction passage · 56 passages
Search aliases: z
Introduction passage · Machine_Understanding_v1.0.0.tex, line 276
The cost of expanding the contract. Let C' have the same complete state space and contain every operation and query of C, while possibly adding more. Write ≡_ C' for its finer behavioral equivalence. Take any representation r adequate for C. For each z, let k_z be the number of ≡_ C' classes that meet the fibre r^-1(z).
Exact mathematical passage
z
\betaIntroduction passage · 7 passages
Search aliases: β beta
Introduction passage · Machine_Understanding_v1.0.0.tex, line 929
For a policy tree T and initial state x, write s_T(x) for its terminal sender transcript and x_T(x) for its final physical state. The tree has finitely many reachable leaves. Specify recipient backgrounds β_i^T(x) and demands d_i^T(x) by a declared rule. A demand may concern x_T(x), the original x, or a stated pair of past and present values. These are different contracts.
Exact mathematical passage
\beta_i^T(x)
\beta_i^TIntroduction passage · 2 passages
Search aliases: β_i^T beta
Introduction passage · Machine_Understanding_v1.0.0.tex, line 929
For a policy tree T and initial state x, write s_T(x) for its terminal sender transcript and x_T(x) for its final physical state. The tree has finitely many reachable leaves. Specify recipient backgrounds β_i^T(x) and demands d_i^T(x) by a declared rule. A demand may concern x_T(x), the original x, or a stated pair of past and present values. These are different contracts.
Exact mathematical passage
\beta_i^T(x)
\varnothingIntroduction passage · 2 passages
Search aliases: ∅ varnothing
Introduction passage · Machine_Understanding_v1.0.0.tex, line 584
Let r:X→ Z be the target record. A support is terminal when r is constant on it. Let h be the number of experiments still available. Write V_h(V) for the least worst-case sum of experiment costs among policies that terminate with the target known within h steps. Infeasibility is denoted +∞. Then define V_h(V) · =0 · · if |r(V)|=1, V_0(V) · =+∞ · · otherwise, V_h(V) · =min_e:V⊆ A_e (c_e+max_o:V_e,o≠∅ V_h-1(V_e,o)) · · substackotherwise, h>0. An empty minimum is +∞. The terminal case permits stopp…
Exact mathematical passage
\label{eq:acquisition-bellman} \begin{aligned} \mathcal V_h(V)&=0 &&\text{if }|r(V)|=1,\\ \mathcal V_0(V)&=+\infty &&\text{otherwise},\\ \mathcal V_h(V)&=\min_{e:V\subseteq A_e} \left(c_e+\max_{o:V_{e,o}\ne\varnothing}\mathcal V_{h-1}(V_{e,o})\right) &&\substack{\text{otherwise,}\\h>0.} \end{aligned}
\inIntroduction passage · 35 passages
Search aliases: ∈ in
Introduction passage · Machine_Understanding_v1.0.0.tex, line 686
Let M∈ M be a fixed unknown model in a finite family, with prior p(m). An experiment produces O∈ O with likelihood L_m(o). A common reporting channel B(o,z) produces Z∈ Z. Each likelihood row and each channel row is a probability distribution. The complete joint model is Pr(M=m,O=o,Z=z)=p(m)L_m(o)B(o,z). In particular, the channel does not secretly inspect M. Define Q_m(z)=∑_o L_m(o)B(o,z), p^o(m)=fracp(m)L_m(o)∑_n p(n)L_n(o), p^z(m)=fracp(m)Q_m(z)∑_n p(n)Q_n(z). A posterior is defined only when…
Exact mathematical passage
M\in\M
\sum_nIntroduction passage · 1 passage
Search aliases: ∑_n sum
Introduction passage · Machine_Understanding_v1.0.0.tex, line 691
prior p(m). An experiment produces O∈ O with likelihood L_m(o). A common reporting channel B(o,z) produces Z∈ Z. Each likelihood row and each channel row is a probability distribution. The complete joint model is Pr(M=m,O=o,Z=z)=p(m)L_m(o)B(o,z). In particular, the channel does not secretly inspect M. Define Q_m(z)=∑_o L_m(o)B(o,z), p^o(m)=fracp(m)L_m(o)∑_n p(n)L_n(o), p^z(m)=fracp(m)Q_m(z)∑_n p(n)Q_n(z). A posterior is defined only when its denominator is positive. An event assigned probability…
Exact mathematical passage
Q_m(z)=\sum_o L_m(o)B(o,z),\qquad p^o(m)=\frac{p(m)L_m(o)}{\sum_n p(n)L_n(o)},\qquad p^z(m)=\frac{p(m)Q_m(z)}{\sum_n p(n)Q_n(z)}.
\sum_oIntroduction passage · 4 passages
Search aliases: ∑_o sum
Introduction passage · Machine_Understanding_v1.0.0.tex, line 691
prior p(m). An experiment produces O∈ O with likelihood L_m(o). A common reporting channel B(o,z) produces Z∈ Z. Each likelihood row and each channel row is a probability distribution. The complete joint model is Pr(M=m,O=o,Z=z)=p(m)L_m(o)B(o,z). In particular, the channel does not secretly inspect M. Define Q_m(z)=∑_o L_m(o)B(o,z), p^o(m)=fracp(m)L_m(o)∑_n p(n)L_n(o), p^z(m)=fracp(m)Q_m(z)∑_n p(n)Q_n(z). A posterior is defined only when its denominator is positive. An event assigned probability…
Exact mathematical passage
Q_m(z)=\sum_o L_m(o)B(o,z),\qquad p^o(m)=\frac{p(m)L_m(o)}{\sum_n p(n)L_n(o)},\qquad p^z(m)=\frac{p(m)Q_m(z)}{\sum_n p(n)Q_n(z)}.
\inftyIntroduction passage · 3 passages
Search aliases: ∞ infty
Introduction passage · Machine_Understanding_v1.0.0.tex, line 583
Let r:X→ Z be the target record. A support is terminal when r is constant on it. Let h be the number of experiments still available. Write V_h(V) for the least worst-case sum of experiment costs among policies that terminate with the target known within h steps. Infeasibility is denoted +∞. Then define V_h(V) · =0 · · if |r(V)|=1,
Exact mathematical passage
+\infty
\neIntroduction passage · 3 passages
Search aliases: ≠ ne
Introduction passage · Machine_Understanding_v1.0.0.tex, line 584
Let r:X→ Z be the target record. A support is terminal when r is constant on it. Let h be the number of experiments still available. Write V_h(V) for the least worst-case sum of experiment costs among policies that terminate with the target known within h steps. Infeasibility is denoted +∞. Then define V_h(V) · =0 · · if |r(V)|=1, V_0(V) · =+∞ · · otherwise, V_h(V) · =min_e:V⊆ A_e (c_e+max_o:V_e,o≠∅ V_h-1(V_e,o)) · · substackotherwise, h>0. An empty minimum is +∞. The terminal case permits stopp…
Exact mathematical passage
\label{eq:acquisition-bellman} \begin{aligned} \mathcal V_h(V)&=0 &&\text{if }|r(V)|=1,\\ \mathcal V_0(V)&=+\infty &&\text{otherwise},\\ \mathcal V_h(V)&=\min_{e:V\subseteq A_e} \left(c_e+\max_{o:V_{e,o}\ne\varnothing}\mathcal V_{h-1}(V_{e,o})\right) &&\substack{\text{otherwise,}\\h>0.} \end{aligned}
\equivIntroduction passage · 5 passages
Search aliases: ≡ equiv
Introduction passage · Machine_Understanding_v1.0.0.tex, line 276
The cost of expanding the contract. Let C' have the same complete state space and contain every operation and query of C, while possibly adding more. Write ≡_ C' for its finer behavioral equivalence. Take any representation r adequate for C. For each z, let k_z be the number of ≡_ C' classes that meet the fibre r^-1(z).
Exact mathematical passage
\equiv_{\C'}
\equiv_{\C'}Introduction passage · 4 passages
Search aliases: ≡_C' ≡_ C' equiv C equiv mathcal
Introduction passage · Machine_Understanding_v1.0.0.tex, line 276
The cost of expanding the contract. Let C' have the same complete state space and contain every operation and query of C, while possibly adding more. Write ≡_ C' for its finer behavioral equivalence. Take any representation r adequate for C. For each z, let k_z be the number of ≡_ C' classes that meet the fibre r^-1(z).
Exact mathematical passage
\equiv_{\C'}
*Use in source · 9 passages
Search aliases: *
Use in source · Machine_Understanding_v1.0.0.tex, line 248
Behavioral compression and the price of new questions. Define x≡_ C y Longleftrightarrow q(F_w(x))=q(F_w(y)) for all q∈ Q, w∈ U^*. This is equality of all declared future answer profiles. It is an equivalence relation. It describes behavioral sufficiency under the contract, not equality of hidden mechanisms or identity outside that contract.
Exact mathematical passage
\label{eq:equivalence} x\Eq y\quad\Longleftrightarrow\quad q(F_w(x))=q(F_w(y))\ \text{for all }q\in\Q,\ w\in\U^*.
\mathcal AUse in source · 12 passages
Search aliases: A mathcal
Use in source · Machine_Understanding_v1.0.0.tex, line 1018
Let X be a finite nonempty set of cases, α:X→ S the sender's access map, and A a finite nonempty artifact catalogue. Recipient i has a background map β_i:X→ B_i, a demanded output d_i:X→ D_i, and a fixed procedure L_i. Evaluate L_i(β_i(x),a) under a declared instruction and memory bound. A timeout, an invalid output, or a rejected artifact counts as failure. Since the inputs and execution bounds are finite, the resulting acceptance table is finite even when L_i is written in a general programmin…
Exact mathematical passage
\mathcal A
aUse in source · 81 passages
Search aliases: a
Use in source · Machine_Understanding_v1.0.0.tex, line 143
A small instrument with a hidden register. Consider a mathematical instrument with three registers x=(a,b,c)∈ F_5^3. Each register contains one of 0,1,2,3,4. All arithmetic in this example is modulo five. The display reports q(a,b,c)=a+b. There are 125 possible complete states, but only five possible readings. The registers and their operations are fully specified analytical objects. They are not measurements from an agent experiment.
Exact mathematical passage
x=(a,b,c)\in\Ffive^3.
a'Use in source · 1 passage
Search aliases: a'
Use in source · Machine_Understanding_v1.0.0.tex, line 205
An actual execution of J, followed by a new reading, supplies the missing distinction. Because J changes the instrument, the recovered state is the state after that operation: (a',b',c')=(s_J-b, b, s_J-s). The old state can then be reconstructed because this particular operation is invertible. Neither reversibility nor a free observation should be assumed for other instruments.
Exact mathematical passage
\label{eq:recover-after-J} (a',b',c')=(s_J-b,\ b,\ s_J-s).
a_*Use in source · 2 passages
Search aliases: a_*
Use in source · Machine_Understanding_v1.0.0.tex, line 828
. If the expected-loss gap between a_i and every other action under r_i is strictly greater than 2L_maxe_i, then a_i is also the unique optimum under p. The triangle inequality bounds d_TV(p,r_i) by e_i. For a function taking values in [0,L_max], its expectations under these two distributions differ by at most L_maxe_i. Let a_* minimize expected loss under p. Insert the two risks under r_i between the risks of a_i and a_* under p. The middle difference is nonpositive by optimality under r_i, and…
Exact mathematical passage
a_*
a_iUse in source · 6 passages
Search aliases: a_i
Use in source · Machine_Understanding_v1.0.0.tex, line 820
[A joint posterior and recipient decision certificate] Let p be the reference posterior, widehat p an approximate or transmitted posterior, and r_i recipient i's reconstructed posterior. Suppose d_TV(p,widehat p)≤ϵ, d_TV(widehat p,r_i)≤η_i. If a_i minimizes expected loss under r_i, then, with e_i=min{1,ϵ+η_i}, E_pℓ(M,a_i)-min_a∈ A E_pℓ(M,a) ≤min{L_max,2L_maxe_i}. If the expected-loss gap between a_i and every other action under r_i is strictly greater than 2L_maxe_i, then a_i is also the unique …
Exact mathematical passage
a_i
\mathcal A_sUse in source · 7 passages
Search aliases: A_s mathcal
Use in source · Machine_Understanding_v1.0.0.tex, line 1026
Define A(x)={a∈ A:every recipient returns d_i(x) within its declared limits}, and, for each reachable sender value s, define A_s=⋂_x:α(x)=s A(x). An artifact selected from A_s must work for all cases the sender still considers possible. Choosing a different acceptable artifact after secretly learning which hidden case is true is not allowed.
Exact mathematical passage
\label{eq:decoder-intersection} \mathcal A_s=\bigcap_{x:\alpha(x)=s}\mathcal A(x).
bUse in source · 103 passages
Search aliases: b
Use in source · Machine_Understanding_v1.0.0.tex, line 143
A small instrument with a hidden register. Consider a mathematical instrument with three registers x=(a,b,c)∈ F_5^3. Each register contains one of 0,1,2,3,4. All arithmetic in this example is modulo five. The display reports q(a,b,c)=a+b. There are 125 possible complete states, but only five possible readings. The registers and their operations are fully specified analytical objects. They are not measurements from an agent experiment.
Exact mathematical passage
x=(a,b,c)\in\Ffive^3.
b'Use in source · 1 passage
Search aliases: b'
Use in source · Machine_Understanding_v1.0.0.tex, line 205
An actual execution of J, followed by a new reading, supplies the missing distinction. Because J changes the instrument, the recovered state is the state after that operation: (a',b',c')=(s_J-b, b, s_J-s). The old state can then be reconstructed because this particular operation is invertible. Neither reversibility nor a free observation should be assumed for other instruments.
Exact mathematical passage
\label{eq:recover-after-J} (a',b',c')=(s_J-b,\ b,\ s_J-s).
b^hUse in source · 2 passages
Search aliases: b^h
Use in source · Machine_Understanding_v1.0.0.tex, line 539
rates any two distinct remaining observational classes. Repeatedly selecting an informative experiment preserves m_* and reaches a single observational class after at most |V_0|-1 informative observations. If there are N distinguishable target classes, every adaptive procedure identifying all of them using experiments with at most b≥2 outcomes and at most h observations must satisfy b^h≥ N. The true model predicts each exact observation, so filtering never removes it. An informative experiment s…
Exact mathematical passage
b^h\geq N
B_eUse in source · 3 passages
Search aliases: B_e
Use in source · Machine_Understanding_v1.0.0.tex, line 844
Let the unknown model M remain fixed. At time t, a known policy selects experiment e_t using only the prior reported history and recorded experiment names. Randomization is permitted when its conditional law is independent of M given that history. Given M=m and e_t=e, a fresh raw observation has law L_e,m, and its reported form is drawn through B_e. Conditional on the model and selected experiment, these draws do not depend on earlier hidden observations. Write Q_e,m=L_e,mB_e.
Exact mathematical passage
B_e
B_{e_t}Use in source · 1 passage
Search aliases: B_e_t
Use in source · Machine_Understanding_v1.0.0.tex, line 855
cy, the posterior conditioned on the complete raw and reported record equals the posterior conditioned only on the reported outcomes and experiment names. The claim also holds at a bounded stopping time determined by that reported history. For a fixed realized history, its likelihood under model m is the product of the selected-experiment probabilities and the terms L_e_t,m(o_t)B_e_t(o_t,z_t). The policy factors are independent of m once the reported history is fixed. Applying the assumed factor…
Exact mathematical passage
L_{e_t,m}(o_t)B_{e_t}(o_t,z_t)
B_iUse in source · 1 passage
Search aliases: B_i
Use in source · Machine_Understanding_v1.0.0.tex, line 1018
Let X be a finite nonempty set of cases, α:X→ S the sender's access map, and A a finite nonempty artifact catalogue. Recipient i has a background map β_i:X→ B_i, a demanded output d_i:X→ D_i, and a fixed procedure L_i. Evaluate L_i(β_i(x),a) under a declared instruction and memory bound. A timeout, an invalid output, or a rejected artifact counts as failure. Since the inputs and execution bounds are finite, the resulting acceptance table is finite even when L_i is written in a general programmin…
Exact mathematical passage
\beta_i:X\to B_i
B_\thetaUse in source · 2 passages
Search aliases: B_θ theta
Use in source · Machine_Understanding_v1.0.0.tex, line 449
[The information an exact explanation must carry] If a fixed-length b-bit artifact lets the recipient answer the declared questions exactly for every target in Θ, then 2^b≥ N. With target-dependent background B_θ, the same argument gives 2^b≥max_B N_B, where N_B counts distinct answer profiles among targets sharing that exact background value. Two different required answer profiles cannot produce the same background and message. The deterministic recipient would then make the same responses for …
Exact mathematical passage
B_\theta
\bar F_wUse in source · 1 passage
Search aliases: bar F_w bar
Use in source · Machine_Understanding_v1.0.0.tex, line 227
[Exact reusable representation] An exact reusable representation for C is a surjection r: X→ Z, updates bar F_u: Z→ Z, and decoders bar q: Z→ Y such that, for all x,u,q, r(F_u(x))=bar F_u(r(x)), q(x)=bar q(r(x)). Induction immediately gives q(F_w(x))=bar q(bar F_w(r(x))) for every finite word. The same initial record supports many future questions. It is not allowed to inspect the selected future query and replace itself with that query's answer before the challenge begins.
Exact mathematical passage
q(F_w(x))=\bar q(\bar F_w(r(x)))
\bar HUse in source · 2 passages
Search aliases: bar H bar
Use in source · Machine_Understanding_v1.0.0.tex, line 170
A sufficient retained state for the original operation family is z=(s,b)=(a+b,b). Its updates are bar H(s,b)=(s+b,b), bar W(s,b)=(s,s-b), and its displayed output is its first coordinate. These equations retain the information needed for every finite H/W continuation without retaining c.
Exact mathematical passage
\label{eq:display-coordinates} \bar H(s,b)=(s+b,b),\qquad \bar W(s,b)=(s,s-b),
\bar JUse in source · 1 passage
Search aliases: bar J bar
Use in source · Machine_Understanding_v1.0.0.tex, line 557
The current record supplies (s,b). Executing J and observing s_J determines c=s_J-s, while~[source reference] updates the affected first register. The corrected extended coordinates are (s_J,b,c). Their operations are bar H(s,b,c)=(s+b,b,c), bar W(s,b,c)=(s,s-b,c), bar J(s,b,c)=(s+c,b,c). The base method survives as the restriction of the enlarged method. Only the newly relevant distinction is acquired.
Exact mathematical passage
\bar H(s,b,c)=(s+b,b,c),\quad \bar W(s,b,c)=(s,s-b,c),\quad \bar J(s,b,c)=(s+c,b,c).
\bar KUse in source · 2 passages
Search aliases: bar K bar
Use in source · Machine_Understanding_v1.0.0.tex, line 344
[Uniform controlled stochastic reuse] A single reduced kernel bar K_u reproduces every projected one-step law from every complete state if and only if, whenever r(x)=r(y), ∑_v:r(v)=z'K_u(x,v)=∑_v:r(v)=z'K_u(y,v) for every u,z'. With matched initial projected laws, the reduced model preserves the law of every finite sequence of reduced states and queries under any policy depending only on the reduced history. A common projected row is necessary by considering point-mass initial states. If the row…
Exact mathematical passage
\bar K_u
\bar K_uUse in source · 2 passages
Search aliases: bar K_u bar
Use in source · Machine_Understanding_v1.0.0.tex, line 344
[Uniform controlled stochastic reuse] A single reduced kernel bar K_u reproduces every projected one-step law from every complete state if and only if, whenever r(x)=r(y), ∑_v:r(v)=z'K_u(x,v)=∑_v:r(v)=z'K_u(y,v) for every u,z'. With matched initial projected laws, the reduced model preserves the law of every finite sequence of reduced states and queries under any policy depending only on the reduced history. A common projected row is necessary by considering point-mass initial states. If the row…
Exact mathematical passage
\bar K_u
\bar WUse in source · 2 passages
Search aliases: bar W bar
Use in source · Machine_Understanding_v1.0.0.tex, line 170
A sufficient retained state for the original operation family is z=(s,b)=(a+b,b). Its updates are bar H(s,b)=(s+b,b), bar W(s,b)=(s,s-b), and its displayed output is its first coordinate. These equations retain the information needed for every finite H/W continuation without retaining c.
Exact mathematical passage
\label{eq:display-coordinates} \bar H(s,b)=(s+b,b),\qquad \bar W(s,b)=(s,s-b),
c'Use in source · 1 passage
Search aliases: c'
Use in source · Machine_Understanding_v1.0.0.tex, line 205
An actual execution of J, followed by a new reading, supplies the missing distinction. Because J changes the instrument, the recovered state is the state after that operation: (a',b',c')=(s_J-b, b, s_J-s). The old state can then be reconstructed because this particular operation is invertible. Neither reversibility nor a free observation should be assumed for other instruments.
Exact mathematical passage
\label{eq:recover-after-J} (a',b',c')=(s_J-b,\ b,\ s_J-s).
C'_{o'}Use in source · 1 passage
Search aliases: C'_o'
Use in source · Machine_Understanding_v1.0.0.tex, line 640
Write ϕ[V]={ϕ(m):m∈ V}. Let C_o(V)={m∈ V:f(m)=o} and C'_o'(W)={n∈ W:g(n)=o'}. Both are ordinary consistency filters.
Exact mathematical passage
C'_{o'}(W)=\{n\in W:g(n)=o'\}
C'_{\eta(o)}Use in source · 2 passages
Search aliases: C'_η(o) eta
Use in source · Machine_Understanding_v1.0.0.tex, line 644
[Correction transport and evidence reflection] Under~[source reference], for every V⊆ M and every attainable source outcome o∈ f( M), ϕ[C_o(V)]⊆ C'_η(o)(ϕ[V]). Equality holds for all such V and o if and only if η is injective on the attainable outcome set f( M). The claim includes candidate sets contradicted by the observation, so that preservation of an empty update is part of the contract. If n∈ϕ[C_o(V)], then n=ϕ(m) for some m∈ V with f(m)=o. Equation~[source reference] gives g(n)=η(o), provi…
Exact mathematical passage
\label{eq:correction-inclusion} \phi[C_o(V)]\subseteq C'_{\eta(o)}(\phi[V]).
C_oUse in source · 4 passages
Search aliases: C_o
Use in source · Machine_Understanding_v1.0.0.tex, line 640
Write ϕ[V]={ϕ(m):m∈ V}. Let C_o(V)={m∈ V:f(m)=o} and C'_o'(W)={n∈ W:g(n)=o'}. Both are ordinary consistency filters.
Exact mathematical passage
C_o(V)=\{m\in V:f(m)=o\}
DUse in source · 8 passages
Search aliases: D
Use in source · Machine_Understanding_v1.0.0.tex, line 480
One encoder E:S→ W produces a common artifact. A decoder D_i for each recipient must satisfy D_i(E(α(m)),β_i(m))=d_i(m) for every m,i. The encoder is not allowed to consult a recipient-specific hidden state. Recipients do not receive an uncharged transcript or a separate description of the selected case. Here decoders may be designed once for the declared finite contract and then remain fixed. If an existing interpreter is prescribed, the criterion below is only a necessary information condition…
Exact mathematical passage
D_i
D_iUse in source · 3 passages
Search aliases: D_i
Use in source · Machine_Understanding_v1.0.0.tex, line 480
One encoder E:S→ W produces a common artifact. A decoder D_i for each recipient must satisfy D_i(E(α(m)),β_i(m))=d_i(m) for every m,i. The encoder is not allowed to consult a recipient-specific hidden state. Recipients do not receive an uncharged transcript or a separate description of the selected case. Here decoders may be designed once for the declared finite contract and then remain fixed. If an existing interpreter is prescribed, the criterion below is only a necessary information condition…
Exact mathematical passage
D_i
d_iUse in source · 5 passages
Search aliases: d_i
Use in source · Machine_Understanding_v1.0.0.tex, line 478
An accessible sender and a family of recipients. Let M be a finite nonempty set of possible cases. The sender observes only α(m), where α: M→ S is a declared access map. Recipient i∈{1,…,k} receives background β_i(m) and must recover d_i(m). A demand can be a single answer, a reusable state, or the complete finite answer profile for a declared task. The maps and recipients are fixed before the case is selected.
Exact mathematical passage
d_i(m)
D_{\mathrm{KL}}Use in source · 2 passages
Search aliases: D_KL mathrm
Use in source · Machine_Understanding_v1.0.0.tex, line 741
[Expected posterior loss is conditional information] Under~[source reference], E_O,Z [D_KL(p^OVert p^Z)] =I(M;Omid Z)=I(M;O)-I(M;Z). The common value is nonnegative. It is zero exactly when the two posteriors agree on every positive-probability observation pair for that prior. For a fully supported prior, zero is equivalent to the factorization in Theorem~[source reference]. On the support of the joint law, p(mmid o,z)=p^o(m). Substituting this into the finite definition of conditional mutual in…
Exact mathematical passage
\label{eq:posterior-kl} \mathbb E_{O,Z}\!\left[D_{\mathrm{KL}}(p^O\Vert p^Z)\right] =I(M;O\mid Z)=I(M;O)-I(M;Z).
\mathbb EUse in source · 2 passages
Search aliases: E mathbb
Use in source · Machine_Understanding_v1.0.0.tex, line 741
[Expected posterior loss is conditional information] Under~[source reference], E_O,Z [D_KL(p^OVert p^Z)] =I(M;Omid Z)=I(M;O)-I(M;Z). The common value is nonnegative. It is zero exactly when the two posteriors agree on every positive-probability observation pair for that prior. For a fully supported prior, zero is equivalent to the factorization in Theorem~[source reference]. On the support of the joint law, p(mmid o,z)=p^o(m). Substituting this into the finite definition of conditional mutual in…
Exact mathematical passage
\label{eq:posterior-kl} \mathbb E_{O,Z}\!\left[D_{\mathrm{KL}}(p^O\Vert p^Z)\right] =I(M;O\mid Z)=I(M;O)-I(M;Z).
\mathcal EUse in source · 1 passage
Search aliases: E mathcal
Use in source · Machine_Understanding_v1.0.0.tex, line 530
A finite model of informative correction. Let M be a finite nonempty set of deterministic candidate models. Let E be a finite set of permitted experiments. Each experiment e has an outcome map f_e: M→ O_e. The current information is a candidate set V⊆ M. On observing o from e, update it by C_e,o(V)={m∈ V:f_e(m)=o}. This is consistency filtering, not Bayesian inference and not a general model of scientific reasoning. An empty set is meaningful: the observation contradicts every currently retained…
Exact mathematical passage
\mathcal E
e_0Use in source · 4 passages
Search aliases: e_0
Use in source · Machine_Understanding_v1.0.0.tex, line 407
[Finite-horizon transfer budget] For matched source and target operation sequences, start from z_0,w_0 with d(Tz_0,w_0)≤ e_0. After n operations, d(Tz_n,w_n) · ≤ L^n e_0+ϵ∑_j=0^n-1L^j,
Exact mathematical passage
d(Tz_0,w_0)\leq e_0
e_iUse in source · 6 passages
Search aliases: e_i
Use in source · Machine_Understanding_v1.0.0.tex, line 820
[A joint posterior and recipient decision certificate] Let p be the reference posterior, widehat p an approximate or transmitted posterior, and r_i recipient i's reconstructed posterior. Suppose d_TV(p,widehat p)≤ϵ, d_TV(widehat p,r_i)≤η_i. If a_i minimizes expected loss under r_i, then, with e_i=min{1,ϵ+η_i}, E_pℓ(M,a_i)-min_a∈ A E_pℓ(M,a) ≤min{L_max,2L_maxe_i}. If the expected-loss gap between a_i and every other action under r_i is strictly greater than 2L_maxe_i, then a_i is also the unique …
Exact mathematical passage
e_i=\min\{1,\epsilon+\eta_i\}
e_nUse in source · 3 passages
Search aliases: e_n
Use in source · Machine_Understanding_v1.0.0.tex, line 415
|q(z_n)-q'(w_n)| · ≤δ+M(L^n e_0+ϵ∑_j=0^n-1L^j). The empty sum is zero and L^0=1, including when L=0. At L=1 the state bound is e_0+nϵ. Writing e_n=d(Tz_n,w_n), insert g_τ(u)(Tz_n) and use the triangle inequality and the Lipschitz bound to obtain e_n+1≤ϵ+Le_n. Iterating gives the stated expression. Insert q'(Tz_n) between the two outputs and use the decoder mismatch and its Lipschitz constant. The bound concerns the matched operation sequence. It does not automatically cover two controllers that …
Exact mathematical passage
e_n=d(Tz_n,w_n)
e_{n+1}Use in source · 2 passages
Search aliases: e_n+1
Use in source · Machine_Understanding_v1.0.0.tex, line 415
|q(z_n)-q'(w_n)| · ≤δ+M(L^n e_0+ϵ∑_j=0^n-1L^j). The empty sum is zero and L^0=1, including when L=0. At L=1 the state bound is e_0+nϵ. Writing e_n=d(Tz_n,w_n), insert g_τ(u)(Tz_n) and use the triangle inequality and the Lipschitz bound to obtain e_n+1≤ϵ+Le_n. Iterating gives the stated expression. Insert q'(Tz_n) between the two outputs and use the decoder mismatch and its Lipschitz constant. The bound concerns the matched operation sequence. It does not automatically cover two controllers that …
Exact mathematical passage
e_{n+1}\leq\epsilon+Le_n
\mathbb E_{O,Z}Use in source · 1 passage
Search aliases: E_O,Z mathbb
Use in source · Machine_Understanding_v1.0.0.tex, line 741
[Expected posterior loss is conditional information] Under~[source reference], E_O,Z [D_KL(p^OVert p^Z)] =I(M;Omid Z)=I(M;O)-I(M;Z). The common value is nonnegative. It is zero exactly when the two posteriors agree on every positive-probability observation pair for that prior. For a fully supported prior, zero is equivalent to the factorization in Theorem~[source reference]. On the support of the joint law, p(mmid o,z)=p^o(m). Substituting this into the finite definition of conditional mutual in…
Exact mathematical passage
\label{eq:posterior-kl} \mathbb E_{O,Z}\!\left[D_{\mathrm{KL}}(p^O\Vert p^Z)\right] =I(M;O\mid Z)=I(M;O)-I(M;Z).
\mathbb E_pUse in source · 1 passage
Search aliases: E_p mathbb
Use in source · Machine_Understanding_v1.0.0.tex, line 821
int posterior and recipient decision certificate] Let p be the reference posterior, widehat p an approximate or transmitted posterior, and r_i recipient i's reconstructed posterior. Suppose d_TV(p,widehat p)≤ϵ, d_TV(widehat p,r_i)≤η_i. If a_i minimizes expected loss under r_i, then, with e_i=min{1,ϵ+η_i}, E_pℓ(M,a_i)-min_a∈ A E_pℓ(M,a) ≤min{L_max,2L_maxe_i}. If the expected-loss gap between a_i and every other action under r_i is strictly greater than 2L_maxe_i, then a_i is also the unique optim…
Exact mathematical passage
\label{eq:recipient-regret} \mathbb E_p\ell(M,a_i)-\min_{a\in A}\mathbb E_p\ell(M,a) \leq\min\{L_{\max},2L_{\max}e_i\}.
e_tUse in source · 5 passages
Search aliases: e_t
Use in source · Machine_Understanding_v1.0.0.tex, line 844
Let the unknown model M remain fixed. At time t, a known policy selects experiment e_t using only the prior reported history and recorded experiment names. Randomization is permitted when its conditional law is independent of M given that history. Given M=m and e_t=e, a fresh raw observation has law L_e,m, and its reported form is drawn through B_e. Conditional on the model and selected experiment, these draws do not depend on earlier hidden observations. Write Q_e,m=L_e,mB_e.
Exact mathematical passage
e_t
e_\thetaUse in source · 3 passages
Search aliases: e_θ theta
Use in source · Machine_Understanding_v1.0.0.tex, line 988
Task-dependent interpreter changes are information. Let Θ be a finite nonempty family of tasks and let each task have a required answer profile over a fixed query family. A fixed interpreter U receives background B_θ, an installed program string p_θ, an artifact e_θ, and a query. It must return the required answer. The interpreter is fixed across tasks. The strings may depend on the task, but their lengths are charged. Let N_B be the number of distinct required answer profiles among tasks having…
Exact mathematical passage
e_\theta
\EqUse in source · 8 passages
Search aliases: Eq ≡_ C equiv mathcal
Use in source · Machine_Understanding_v1.0.0.tex, line 248
Behavioral compression and the price of new questions. Define x≡_ C y Longleftrightarrow q(F_w(x))=q(F_w(y)) for all q∈ Q, w∈ U^*. This is equality of all declared future answer profiles. It is an equivalence relation. It describes behavioral sufficiency under the contract, not equality of hidden mechanisms or identity outside that contract.
Exact mathematical passage
\label{eq:equivalence} x\Eq y\quad\Longleftrightarrow\quad q(F_w(x))=q(F_w(y))\ \text{for all }q\in\Q,\ w\in\U^*.
\mathbb FUse in source · 1 passage
Search aliases: F mathbb
Use in source · Machine_Understanding_v1.0.0.tex, line 1110
A costed first-edition instrument certificate. The common realization can now be described without a designable recipient or an omniscient sender. The instrument is still (a,b,c)∈ F_5^3. Before the diagnostic, the sender knows only (s,b)=(a+b,b). Recipient A has the old register pair (a,b). Recipient B has the old display pair (s,b). Neither recipient has c. All three parties know the operation laws and the fixed diagnostic protocol before the particular hidden state is selected.
Exact mathematical passage
(a,b,c)\in\mathbb F_5^3
fUse in source · 20 passages
Search aliases: f
Use in source · Machine_Understanding_v1.0.0.tex, line 359
Let a source representation have state space Z, updates f_u, and query decoders q. Let a target have state space W, updates g_v, and decoders q'. A map τ translates operation names. Query pairs (q,q') and a representation map T:Z→ W are fixed in advance. The operation names alone do not establish that the target implements the intended source operation.
Exact mathematical passage
f_u
\mathbb F_5^3Use in source · 1 passage
Search aliases: F_5^3 mathbb
Use in source · Machine_Understanding_v1.0.0.tex, line 1110
A costed first-edition instrument certificate. The common realization can now be described without a designable recipient or an omniscient sender. The instrument is still (a,b,c)∈ F_5^3. Before the diagnostic, the sender knows only (s,b)=(a+b,b). Recipient A has the old register pair (a,b). Recipient B has the old display pair (s,b). Neither recipient has c. All three parties know the operation laws and the fixed diagnostic protocol before the particular hidden state is selected.
Exact mathematical passage
(a,b,c)\in\mathbb F_5^3
f_eUse in source · 2 passages
Search aliases: f_e
Use in source · Machine_Understanding_v1.0.0.tex, line 530
A finite model of informative correction. Let M be a finite nonempty set of deterministic candidate models. Let E be a finite set of permitted experiments. Each experiment e has an outcome map f_e: M→ O_e. The current information is a candidate set V⊆ M. On observing o from e, update it by C_e,o(V)={m∈ V:f_e(m)=o}. This is consistency filtering, not Bayesian inference and not a general model of scientific reasoning. An empty set is meaningful: the observation contradicts every currently retained…
Exact mathematical passage
f_e:\M\to\mathcal O_e
f_uUse in source · 5 passages
Search aliases: f_u
Use in source · Machine_Understanding_v1.0.0.tex, line 359
Let a source representation have state space Z, updates f_u, and query decoders q. Let a target have state space W, updates g_v, and decoders q'. A map τ translates operation names. Query pairs (q,q') and a representation map T:Z→ W are fixed in advance. The operation names alone do not establish that the target implements the intended source operation.
Exact mathematical passage
f_u
F_{u_1}Use in source · 1 passage
Search aliases: F_u_1
Use in source · Machine_Understanding_v1.0.0.tex, line 217
t for reusable understanding Let X and Y be finite nonempty sets. Let U be a finite operation alphabet, possibly empty, and let Q be a finite nonempty family of queries q: X→ Y. Each u∈ U has an implemented total transition F_u: X→ X. The deterministic mathematical contract is C=( X, U,(F_u)_u∈ U, Q, Y). For a word w=u_1⋯ u_n, define F_w=F_u_n∘⋯∘ F_u_1. The empty word acts as the identity. Thus q(F_w(x)) asks a question after a stated continuation.
Exact mathematical passage
F_w=F_{u_n}\circ\cdots\circ F_{u_1}
F_{u_n}Use in source · 1 passage
Search aliases: F_u_n
Use in source · Machine_Understanding_v1.0.0.tex, line 217
t for reusable understanding Let X and Y be finite nonempty sets. Let U be a finite operation alphabet, possibly empty, and let Q be a finite nonempty family of queries q: X→ Y. Each u∈ U has an implemented total transition F_u: X→ X. The deterministic mathematical contract is C=( X, U,(F_u)_u∈ U, Q, Y). For a word w=u_1⋯ u_n, define F_w=F_u_n∘⋯∘ F_u_1. The empty word acts as the identity. Thus q(F_w(x)) asks a question after a stated continuation.
Exact mathematical passage
F_w=F_{u_n}\circ\cdots\circ F_{u_1}
f_wUse in source · 2 passages
Search aliases: f_w
Use in source · Machine_Understanding_v1.0.0.tex, line 367
[Exact transport of reusable operations] Suppose, for every source operation and matched query, T f_u=g_τ(u)T, q'T=q. Then, for every source state and every finite source word w, T f_w=g_τ(w)T, q'(g_τ(w)(Tz))=q(f_w(z)). Two such transports compose, with their operation and query correspondences composed as well. No injectivity assumption on T is needed for this forward guarantee. The empty word is immediate. If the identity holds for w, then T f_u f_w=g_τ(u)T f_w =g_τ(u)g_τ(w)T. Induction proves…
Exact mathematical passage
T f_w=g_{\tau(w)}T,\qquad q'(g_{\tau(w)}(Tz))=q(f_w(z)).
\FfiveUse in source · 1 passage
Search aliases: Ffive F_5 mathbb
Use in source · Machine_Understanding_v1.0.0.tex, line 143
A small instrument with a hidden register. Consider a mathematical instrument with three registers x=(a,b,c)∈ F_5^3. Each register contains one of 0,1,2,3,4. All arithmetic in this example is modulo five. The display reports q(a,b,c)=a+b. There are 125 possible complete states, but only five possible readings. The registers and their operations are fully specified analytical objects. They are not measurements from an agent experiment.
Exact mathematical passage
x=(a,b,c)\in\Ffive^3.
\Ffive^3Use in source · 1 passage
Search aliases: Ffive^3 F_5^3 Ffive mathbb
Use in source · Machine_Understanding_v1.0.0.tex, line 143
A small instrument with a hidden register. Consider a mathematical instrument with three registers x=(a,b,c)∈ F_5^3. Each register contains one of 0,1,2,3,4. All arithmetic in this example is modulo five. The display reports q(a,b,c)=a+b. There are 125 possible complete states, but only five possible readings. The registers and their operations are fully specified analytical objects. They are not measurements from an agent experiment.
Exact mathematical passage
x=(a,b,c)\in\Ffive^3.
GUse in source · 6 passages
Search aliases: G
Use in source · Machine_Understanding_v1.0.0.tex, line 486
Build a graph G_α on α( M). Join s and t when some cases m,n and recipient i satisfy α(m)=s, α(n)=t, β_i(m)=β_i(n), d_i(m)≠ d_i(n). An edge can be a loop when s=t. It then records a distinction that neither the sender's access nor that recipient's background can resolve.
Exact mathematical passage
G_\alpha
gUse in source · 14 passages
Search aliases: g
Use in source · Machine_Understanding_v1.0.0.tex, line 359
Let a source representation have state space Z, updates f_u, and query decoders q. Let a target have state space W, updates g_v, and decoders q'. A map τ translates operation names. Query pairs (q,q') and a representation map T:Z→ W are fixed in advance. The operation names alone do not establish that the target implements the intended source operation.
Exact mathematical passage
g_v
g_aUse in source · 2 passages
Search aliases: g_a
Use in source · Machine_Understanding_v1.0.0.tex, line 1065
For each state x and artifact a accepted by all fixed recipient procedures, let γ(x,a)∈ N^r be the terminal charge. It includes transmission and the declared source and recipient execution work. Invalid or out-of-budget decoder executions have no terminal entry. At support V, a stop using a is allowed only if it works for every x∈ V, with worst-case vector g_a(V)=max_x∈ Vγ(x,a), where maxima and comparisons of vectors are componentwise. Different coordinates may attain their maxima on different …
Exact mathematical passage
g_a(V)=\max_{x\in V}\gamma(x,a),
G_TUse in source · 2 passages
Search aliases: G_T
Use in source · Machine_Understanding_v1.0.0.tex, line 931
The recipient sees only the common artifact and its specified background. If an experiment name, its timing, or another transcript is visible to the recipient, that information belongs in β_i^T. It must not be used as an uncharged auxiliary message. Let C(T) be the maximum total experimental cost across initial states. Build G_T on the reachable sender transcripts using the graph construction of Section~[source reference], with α=s_T.
Exact mathematical passage
G_T
g_vUse in source · 1 passage
Search aliases: g_v
Use in source · Machine_Understanding_v1.0.0.tex, line 359
Let a source representation have state space Z, updates f_u, and query decoders q. Let a target have state space W, updates g_v, and decoders q'. A map τ translates operation names. Query pairs (q,q') and a representation map T:Z→ W are fixed in advance. The operation names alone do not establish that the target implements the intended source operation.
Exact mathematical passage
g_v
G_\alphaUse in source · 4 passages
Search aliases: G_α alpha
Use in source · Machine_Understanding_v1.0.0.tex, line 486
Build a graph G_α on α( M). Join s and t when some cases m,n and recipient i satisfy α(m)=s, α(n)=t, β_i(m)=β_i(n), d_i(m)≠ d_i(n). An edge can be a loop when s=t. It then records a distinction that neither the sender's access nor that recipient's background can resolve.
Exact mathematical passage
G_\alpha
g_{\tau(u)}Use in source · 4 passages
Search aliases: g_τ(u) tau
Use in source · Machine_Understanding_v1.0.0.tex, line 363
[Exact transport of reusable operations] Suppose, for every source operation and matched query, T f_u=g_τ(u)T, q'T=q. Then, for every source state and every finite source word w, T f_w=g_τ(w)T, q'(g_τ(w)(Tz))=q(f_w(z)). Two such transports compose, with their operation and query correspondences composed as well. No injectivity assumption on T is needed for this forward guarantee. The empty word is immediate. If the identity holds for w, then T f_u f_w=g_τ(u)T f_w =g_τ(u)g_τ(w)T. Induction proves…
Exact mathematical passage
\label{eq:transport-exact} T f_u=g_{\tau(u)}T,\qquad q'T=q.
g_{\tau(w)}Use in source · 2 passages
Search aliases: g_τ(w) tau
Use in source · Machine_Understanding_v1.0.0.tex, line 367
[Exact transport of reusable operations] Suppose, for every source operation and matched query, T f_u=g_τ(u)T, q'T=q. Then, for every source state and every finite source word w, T f_w=g_τ(w)T, q'(g_τ(w)(Tz))=q(f_w(z)). Two such transports compose, with their operation and query correspondences composed as well. No injectivity assumption on T is needed for this forward guarantee. The empty word is immediate. If the identity holds for w, then T f_u f_w=g_τ(u)T f_w =g_τ(u)g_τ(w)T. Induction proves…
Exact mathematical passage
T f_w=g_{\tau(w)}T,\qquad q'(g_{\tau(w)}(Tz))=q(f_w(z)).
HUse in source · 28 passages
Search aliases: H
Use in source · Machine_Understanding_v1.0.0.tex, line 153
Initially, two operations are available: H(a,b,c)=(a+b,b,c), W(a,b,c)=(b,a,c). The first adds the second register into the first. The second exchanges the first two registers. The third register is preserved by both and does not affect the display. The operation order is public. A prediction concerns what the display would show after a named finite sequence of these operations.
Exact mathematical passage
\label{eq:instrument-ops} H(a,b,c)=(a+b,b,c),\qquad W(a,b,c)=(b,a,c).
h_{\sigma(\tau(u))}Use in source · 1 passage
Search aliases: h_σ(τ(u)) sigma tau
Use in source · Machine_Understanding_v1.0.0.tex, line 378
ll. No injectivity assumption on T is needed for this forward guarantee. The empty word is immediate. If the identity holds for w, then T f_u f_w=g_τ(u)T f_w =g_τ(u)g_τ(w)T. Induction proves the word identity, and applying the matched decoder proves the answer identity. For a second transport S with operation map σ, substitution gives STf_u=h_σ(τ(u))ST and the corresponding decoder identity. This is the same condition for the composite. This is an intertwining argument. Related composition quest…
Exact mathematical passage
STf_u=h_{\sigma(\tau(u))}ST
IUse in source · 1 passage
Search aliases: I
Use in source · Machine_Understanding_v1.0.0.tex, line 741
[Expected posterior loss is conditional information] Under~[source reference], E_O,Z [D_KL(p^OVert p^Z)] =I(M;Omid Z)=I(M;O)-I(M;Z). The common value is nonnegative. It is zero exactly when the two posteriors agree on every positive-probability observation pair for that prior. For a fully supported prior, zero is equivalent to the factorization in Theorem~[source reference]. On the support of the joint law, p(mmid o,z)=p^o(m). Substituting this into the finite definition of conditional mutual in…
Exact mathematical passage
\label{eq:posterior-kl} \mathbb E_{O,Z}\!\left[D_{\mathrm{KL}}(p^O\Vert p^Z)\right] =I(M;O\mid Z)=I(M;O)-I(M;Z).
JUse in source · 48 passages
Search aliases: J
Use in source · Machine_Understanding_v1.0.0.tex, line 183
A new operation changes what must be known. Now admit J(a,b,c)=(a+c,b,c). The old representation remains correct for H and W. It is no longer sufficient for the enlarged contract. States with the same (a,b) but different c now produce different readings after J.
Exact mathematical passage
\label{eq:new-operation} J(a,b,c)=(a+c,b,c).
jUse in source · 1 passage
Search aliases: j
Use in source · Machine_Understanding_v1.0.0.tex, line 408
[Finite-horizon transfer budget] For matched source and target operation sequences, start from z_0,w_0 with d(Tz_0,w_0)≤ e_0. After n operations, d(Tz_n,w_n) · ≤ L^n e_0+ϵ∑_j=0^n-1L^j, |q(z_n)-q'(w_n)| · ≤δ+M(L^n e_0+ϵ∑_j=0^n-1L^j). The empty sum is zero and L^0=1, including when L=0. At L=1 the state bound is e_0+nϵ. Writing e_n=d(Tz_n,w_n), insert g_τ(u)(Tz_n) and use the triangle inequality and the Lipschitz bound to obtain e_n+1≤ϵ+Le_n. Iterating gives the stated expression. Insert q'(Tz_n) …
Exact mathematical passage
d(Tz_n,w_n)&\leq L^n e_0+\epsilon\sum_{j=0}^{n-1}L^j,\\ |q(z_n)-q'(w_n)|&\leq\delta+M\left(L^n e_0+\epsilon\sum_{j=0}^{n-1}L^j\right).
KUse in source · 8 passages
Search aliases: K
Use in source · Machine_Understanding_v1.0.0.tex, line 341
A bounded stochastic extension. The distinction between a current answer and a reusable state also survives stochastic dynamics. Let K_u(x,x') be finite transition kernels and let r: X→ Z be surjective. Assume the current query values are constant on fibres of r.
Exact mathematical passage
K_u(x,x')
K_uUse in source · 2 passages
Search aliases: K_u
Use in source · Machine_Understanding_v1.0.0.tex, line 341
A bounded stochastic extension. The distinction between a current answer and a reusable state also survives stochastic dynamics. Let K_u(x,x') be finite transition kernels and let r: X→ Z be surjective. Assume the current query values are constant on fibres of r.
Exact mathematical passage
K_u(x,x')
\mathrm{KL}Use in source · 2 passages
Search aliases: KL mathrm
Use in source · Machine_Understanding_v1.0.0.tex, line 741
[Expected posterior loss is conditional information] Under~[source reference], E_O,Z [D_KL(p^OVert p^Z)] =I(M;Omid Z)=I(M;O)-I(M;Z). The common value is nonnegative. It is zero exactly when the two posteriors agree on every positive-probability observation pair for that prior. For a fully supported prior, zero is equivalent to the factorization in Theorem~[source reference]. On the support of the joint law, p(mmid o,z)=p^o(m). Substituting this into the finite definition of conditional mutual in…
Exact mathematical passage
\label{eq:posterior-kl} \mathbb E_{O,Z}\!\left[D_{\mathrm{KL}}(p^O\Vert p^Z)\right] =I(M;O\mid Z)=I(M;O)-I(M;Z).
L^0Use in source · 1 passage
Search aliases: L^0
Use in source · Machine_Understanding_v1.0.0.tex, line 412
|q(z_n)-q'(w_n)| · ≤δ+M(L^n e_0+ϵ∑_j=0^n-1L^j). The empty sum is zero and L^0=1, including when L=0. At L=1 the state bound is e_0+nϵ. Writing e_n=d(Tz_n,w_n), insert g_τ(u)(Tz_n) and use the triangle inequality and the Lipschitz bound to obtain e_n+1≤ϵ+Le_n. Iterating gives the stated expression. Insert q'(Tz_n) between the two outputs and use the decoder mismatch and its Lipschitz constant. The bound concerns the matched operation sequence. It does not automatically cover two controllers that …
Exact mathematical passage
L^0=1
L^jUse in source · 1 passage
Search aliases: L^j
Use in source · Machine_Understanding_v1.0.0.tex, line 408
[Finite-horizon transfer budget] For matched source and target operation sequences, start from z_0,w_0 with d(Tz_0,w_0)≤ e_0. After n operations, d(Tz_n,w_n) · ≤ L^n e_0+ϵ∑_j=0^n-1L^j, |q(z_n)-q'(w_n)| · ≤δ+M(L^n e_0+ϵ∑_j=0^n-1L^j). The empty sum is zero and L^0=1, including when L=0. At L=1 the state bound is e_0+nϵ. Writing e_n=d(Tz_n,w_n), insert g_τ(u)(Tz_n) and use the triangle inequality and the Lipschitz bound to obtain e_n+1≤ϵ+Le_n. Iterating gives the stated expression. Insert q'(Tz_n) …
Exact mathematical passage
d(Tz_n,w_n)&\leq L^n e_0+\epsilon\sum_{j=0}^{n-1}L^j,\\ |q(z_n)-q'(w_n)|&\leq\delta+M\left(L^n e_0+\epsilon\sum_{j=0}^{n-1}L^j\right).
L^nUse in source · 1 passage
Search aliases: L^n
Use in source · Machine_Understanding_v1.0.0.tex, line 408
[Finite-horizon transfer budget] For matched source and target operation sequences, start from z_0,w_0 with d(Tz_0,w_0)≤ e_0. After n operations, d(Tz_n,w_n) · ≤ L^n e_0+ϵ∑_j=0^n-1L^j, |q(z_n)-q'(w_n)| · ≤δ+M(L^n e_0+ϵ∑_j=0^n-1L^j). The empty sum is zero and L^0=1, including when L=0. At L=1 the state bound is e_0+nϵ. Writing e_n=d(Tz_n,w_n), insert g_τ(u)(Tz_n) and use the triangle inequality and the Lipschitz bound to obtain e_n+1≤ϵ+Le_n. Iterating gives the stated expression. Insert q'(Tz_n) …
Exact mathematical passage
d(Tz_n,w_n)&\leq L^n e_0+\epsilon\sum_{j=0}^{n-1}L^j,\\ |q(z_n)-q'(w_n)|&\leq\delta+M\left(L^n e_0+\epsilon\sum_{j=0}^{n-1}L^j\right).
L_AUse in source · 1 passage
Search aliases: L_A
Use in source · Machine_Understanding_v1.0.0.tex, line 1115
The two frozen decoders are L_A((a,b),m)=(a+m,b,m), L_B((s,b),m)=(s+m,b,m). Recipient A obtains the post-J register description. Recipient B obtains the post-J display-centered description. Both compute one modular addition. Their installed update procedures then support all H/W/J continuations in their respective coordinates. These procedures are supplied, not learned by the diagnostic.
Exact mathematical passage
\label{eq:fixed-instrument-decoders} L_A((a,b),m)=(a+m,b,m),\qquad L_B((s,b),m)=(s+m,b,m).
L_BUse in source · 1 passage
Search aliases: L_B
Use in source · Machine_Understanding_v1.0.0.tex, line 1115
The two frozen decoders are L_A((a,b),m)=(a+m,b,m), L_B((s,b),m)=(s+m,b,m). Recipient A obtains the post-J register description. Recipient B obtains the post-J display-centered description. Both compute one modular addition. Their installed update procedures then support all H/W/J continuations in their respective coordinates. These procedures are supplied, not learned by the diagnostic.
Exact mathematical passage
\label{eq:fixed-instrument-decoders} L_A((a,b),m)=(a+m,b,m),\qquad L_B((s,b),m)=(s+m,b,m).
L_{e,m}Use in source · 3 passages
Search aliases: L_e,m
Use in source · Machine_Understanding_v1.0.0.tex, line 844
Let the unknown model M remain fixed. At time t, a known policy selects experiment e_t using only the prior reported history and recorded experiment names. Randomization is permitted when its conditional law is independent of M given that history. Given M=m and e_t=e, a fresh raw observation has law L_e,m, and its reported form is drawn through B_e. Conditional on the model and selected experiment, these draws do not depend on earlier hidden observations. Write Q_e,m=L_e,mB_e.
Exact mathematical passage
L_{e,m}
L_{e_t,m}Use in source · 1 passage
Search aliases: L_e_t,m
Use in source · Machine_Understanding_v1.0.0.tex, line 855
cy, the posterior conditioned on the complete raw and reported record equals the posterior conditioned only on the reported outcomes and experiment names. The claim also holds at a bounded stopping time determined by that reported history. For a fixed realized history, its likelihood under model m is the product of the selected-experiment probabilities and the terms L_e_t,m(o_t)B_e_t(o_t,z_t). The policy factors are independent of m once the reported history is fixed. Applying the assumed factor…
Exact mathematical passage
L_{e_t,m}(o_t)B_{e_t}(o_t,z_t)
L_iUse in source · 3 passages
Search aliases: L_i
Use in source · Machine_Understanding_v1.0.0.tex, line 1018
Let X be a finite nonempty set of cases, α:X→ S the sender's access map, and A a finite nonempty artifact catalogue. Recipient i has a background map β_i:X→ B_i, a demanded output d_i:X→ D_i, and a fixed procedure L_i. Evaluate L_i(β_i(x),a) under a declared instruction and memory bound. A timeout, an invalid output, or a rejected artifact counts as failure. Since the inputs and execution bounds are finite, the resulting acceptance table is finite even when L_i is written in a general programmin…
Exact mathematical passage
L_i
L_{\max}Use in source · 7 passages
Search aliases: L_max max
Use in source · Machine_Understanding_v1.0.0.tex, line 812
From approximate beliefs to a recipient's decisions. Not every posterior distinction matters equally to an action. To link correction with use, fix one finite nonempty action set A and losses ℓ(m,a)∈[0,L_max]. A receiver evaluates actions from the distribution its actual interpreter reconstructs. The target loss function and available actions must be the same in the comparison.
Exact mathematical passage
\ell(m,a)\in[0,L_{\max}]
\lceilUse in source · 2 passages
Search aliases: lceil
Use in source · Machine_Understanding_v1.0.0.tex, line 258
[The minimal exact reusable state] The quotient X/≡_ C carries an exact reusable representation for C. Every other exact reusable representation r: X→ Z refines this quotient: if r(x)=r(y), then x≡_ C y. There is a unique map h: Z→ X/≡_ C with h(r(x))=[x]. In particular, | Z|≥| X/≡_ C|. A fixed-length binary state needs at least lceillog_2| X/≡_ C|rceil bits. The quotient can be computed by finite partition refinement. If x≡_ C y, the empty word gives q(x)=q(y) for every query. Moreover, F_u(x)≡…
Exact mathematical passage
\lceil\log_2|\X/{\Eq}|\rceil
\logUse in source · 5 passages
Search aliases: log
Use in source · Machine_Understanding_v1.0.0.tex, line 258
[The minimal exact reusable state] The quotient X/≡_ C carries an exact reusable representation for C. Every other exact reusable representation r: X→ Z refines this quotient: if r(x)=r(y), then x≡_ C y. There is a unique map h: Z→ X/≡_ C with h(r(x))=[x]. In particular, | Z|≥| X/≡_ C|. A fixed-length binary state needs at least lceillog_2| X/≡_ C|rceil bits. The quotient can be computed by finite partition refinement. If x≡_ C y, the empty word gives q(x)=q(y) for every query. Moreover, F_u(x)≡…
Exact mathematical passage
\lceil\log_2|\X/{\Eq}|\rceil
\log_2Use in source · 2 passages
Search aliases: log_2 log
Use in source · Machine_Understanding_v1.0.0.tex, line 258
[The minimal exact reusable state] The quotient X/≡_ C carries an exact reusable representation for C. Every other exact reusable representation r: X→ Z refines this quotient: if r(x)=r(y), then x≡_ C y. There is a unique map h: Z→ X/≡_ C with h(r(x))=[x]. In particular, | Z|≥| X/≡_ C|. A fixed-length binary state needs at least lceillog_2| X/≡_ C|rceil bits. The quotient can be computed by finite partition refinement. If x≡_ C y, the empty word gives q(x)=q(y) for every query. Moreover, F_u(x)≡…
Exact mathematical passage
\lceil\log_2|\X/{\Eq}|\rceil
\LongleftrightarrowUse in source · 1 passage
Search aliases: Longleftrightarrow
Use in source · Machine_Understanding_v1.0.0.tex, line 248
Behavioral compression and the price of new questions. Define x≡_ C y Longleftrightarrow q(F_w(x))=q(F_w(y)) for all q∈ Q, w∈ U^*. This is equality of all declared future answer profiles. It is an equivalence relation. It describes behavioral sufficiency under the contract, not equality of hidden mechanisms or identity outside that contract.
Exact mathematical passage
\label{eq:equivalence} x\Eq y\quad\Longleftrightarrow\quad q(F_w(x))=q(F_w(y))\ \text{for all }q\in\Q,\ w\in\U^*.
m_*Use in source · 2 passages
Search aliases: m_*
Use in source · Machine_Understanding_v1.0.0.tex, line 539
[Finite correction under separating evidence] Suppose the true model m_* belongs to the initial set V_0, every observation is exact, and the allowed experiment family separates any two distinct remaining observational classes. Repeatedly selecting an informative experiment preserves m_* and reaches a single observational class after at most |V_0|-1 informative observations. If there are N distinguishable target classes, every adaptive procedure identifying all of them using experiments with at m…
Exact mathematical passage
m_*
m_0Use in source · 2 passages
Search aliases: m_0
Use in source · Machine_Understanding_v1.0.0.tex, line 661
[A translated observation preserves both mistaken candidates] Take M={m_0,m_1} with f(m_0)=0 and f(m_1)=1. Translate models bijectively to {n_0,n_1}, but translate both observations to a single symbol *. Let g(n_0)=g(n_1)=*. The observation square commutes. Starting from both candidates, source observation 1 leaves only m_1, whose image is {n_1}. Translating first and then filtering by * leaves both n_0 and n_1.
Exact mathematical passage
\M=\{m_0,m_1\}
m_1Use in source · 3 passages
Search aliases: m_1
Use in source · Machine_Understanding_v1.0.0.tex, line 661
[A translated observation preserves both mistaken candidates] Take M={m_0,m_1} with f(m_0)=0 and f(m_1)=1. Translate models bijectively to {n_0,n_1}, but translate both observations to a single symbol *. Let g(n_0)=g(n_1)=*. The observation square commutes. Starting from both candidates, source observation 1 leaves only m_1, whose image is {n_1}. Translating first and then filtering by * leaves both n_0 and n_1.
Exact mathematical passage
\M=\{m_0,m_1\}
\max_BUse in source · 1 passage
Search aliases: max_B max
Use in source · Machine_Understanding_v1.0.0.tex, line 992
[Charging a task-dependent interpreter] Suppose p_θ and e_θ have fixed lengths d and b bits, respectively, and the deterministic interpreter answers every required query exactly. Then 2^d+b≥max_B N_B. In particular, representing every Boolean function on n input bits under one fixed background requires d+b≥2^n. This counts task-dependent information, not the running time of the interpreter or the information needed for any one specially chosen function. Fix a background B. If two distinct requir…
Exact mathematical passage
2^{d+b}\geq\max_B N_B.
\max_{B}Use in source · 1 passage
Search aliases: max_B max
Use in source · Machine_Understanding_v1.0.0.tex, line 450
[The information an exact explanation must carry] If a fixed-length b-bit artifact lets the recipient answer the declared questions exactly for every target in Θ, then 2^b≥ N. With target-dependent background B_θ, the same argument gives 2^b≥max_B N_B, where N_B counts distinct answer profiles among targets sharing that exact background value. Two different required answer profiles cannot produce the same background and message. The deterministic recipient would then make the same responses for …
Exact mathematical passage
2^b\geq\max_{B}\ N_B,
\max_oUse in source · 1 passage
Search aliases: max_o max
Use in source · Machine_Understanding_v1.0.0.tex, line 1077
For a tree T, let v(T) be its componentwise worst-case charged online cost. Let P_h(V) be the attainable pairs (v(T),n(T)). The stop pairs are (g_a(V),1) for each artifact accepted throughout V. For h>0, each uniformly admissible experiment gives the additional pairs (c_e+max_o v_o, 1+∑_o n_o), (v_o,n_o)∈ P_h-1(V_e,o), with one child choice for every nonempty observation branch. An experiment contributes no pair when even one possible branch has no completion. At h=0 only stops are allowed. Sets…
Exact mathematical passage
\label{eq:resource-branch} \left(c_e+\max_o v_o,\ 1+\sum_o n_o\right), \quad (v_o,n_o)\in\mathcal P_{h-1}(V_{e,o}),
\max_{x\in V}Use in source · 1 passage
Search aliases: max_x∈ V max in
Use in source · Machine_Understanding_v1.0.0.tex, line 1065
For each state x and artifact a accepted by all fixed recipient procedures, let γ(x,a)∈ N^r be the terminal charge. It includes transmission and the declared source and recipient execution work. Invalid or out-of-budget decoder executions have no terminal entry. At support V, a stop using a is allowed only if it works for every x∈ V, with worst-case vector g_a(V)=max_x∈ Vγ(x,a), where maxima and comparisons of vectors are componentwise. Different coordinates may attain their maxima on different …
Exact mathematical passage
g_a(V)=\max_{x\in V}\gamma(x,a),
\max_zUse in source · 1 passage
Search aliases: max_z max
Use in source · Machine_Understanding_v1.0.0.tex, line 280
[Contract expansion and supplementary information] The relation ≡_ C' refines ≡_ C. Any supplementary code s: X→ S for which (r,s) supports exact reusable behavior under C' must satisfy | S|≥max_z k_z. If every ≡_ C' class lies within a single fibre of r, this bound is attained by an exact reusable pair (r,s). This condition holds when r is the old minimal behavioral quotient. A supplementary value computed solely from r(x) cannot repair a fibre meeting more than one expanded class. Agreement on…
Exact mathematical passage
|\mathcal S|\geq\max_z k_z.
\midUse in source · 15 passages
Search aliases: mid
Use in source · Machine_Understanding_v1.0.0.tex, line 702
When a coarser report preserves every update. The reporting model implies M→ O→ Z. Therefore, on every positive-probability pair (o,z), the full posterior after observing both records equals p^o. The coarse posterior is a mixture of these fine posteriors: p^z=∑_oPr(O=omid Z=z) p^o. Averaging is not generally the same as retaining each possible update. The distinction is classical statistical sufficiency, whose factorization foundations include Halmos and Savage~[source reference]. We state and p…
Exact mathematical passage
\label{eq:posterior-mixture} p^z=\sum_o\Pr(O=o\mid Z=z)\,p^o.
\min_{a\in A}Use in source · 1 passage
Search aliases: min_a∈ A min in
Use in source · Machine_Understanding_v1.0.0.tex, line 821
int posterior and recipient decision certificate] Let p be the reference posterior, widehat p an approximate or transmitted posterior, and r_i recipient i's reconstructed posterior. Suppose d_TV(p,widehat p)≤ϵ, d_TV(widehat p,r_i)≤η_i. If a_i minimizes expected loss under r_i, then, with e_i=min{1,ϵ+η_i}, E_pℓ(M,a_i)-min_a∈ A E_pℓ(M,a) ≤min{L_max,2L_maxe_i}. If the expected-loss gap between a_i and every other action under r_i is strictly greater than 2L_maxe_i, then a_i is also the unique optim…
Exact mathematical passage
\label{eq:recipient-regret} \mathbb E_p\ell(M,a_i)-\min_{a\in A}\mathbb E_p\ell(M,a) \leq\min\{L_{\max},2L_{\max}e_i\}.
NUse in source · 16 passages
Search aliases: N
Use in source · Machine_Understanding_v1.0.0.tex, line 446
An exact communication lower bound. Let Θ be a finite family of target models. Relative to the required continuation questions, divide these models into N classes with distinct answer profiles. Use one fixed recipient background and one fixed deterministic recipient. The teacher can select a message but cannot change the recipient's background separately for each target without that change being counted as communication.
Exact mathematical passage
N
\mathbb NUse in source · 3 passages
Search aliases: N mathbb
Use in source · Machine_Understanding_v1.0.0.tex, line 1062
The explicit finite implementation model Use a finite augmented state space X. Include every origin variable, history flag, recipient background, and permission needed to make the declared transition and terminal tables valid. Each experiment e has the transition, observation, and admissibility semantics of Section~[source reference], together with a fixed nonnegative integer cost vector c_e∈ N^r. Its coordinates have declared units. The charge includes the specified source-side interface work f…
Exact mathematical passage
c_e\in\mathbb N^r
\mathcal NUse in source · 3 passages
Search aliases: N mathcal
Use in source · Machine_Understanding_v1.0.0.tex, line 634
Let M and N be finite model families, with model translation ϕ: M→ N. For one source experiment e, let f: M→ O give its outcomes. A corresponding target experiment has outcome map g: N→ O'. Let η: O→ O' translate evidence. Assume the basic observation square commutes: g(ϕ(m))=η(f(m)) for all m∈ M. This condition says that a model's predicted evidence translates correctly. It permits η to merge source outcomes.
Exact mathematical passage
\mathcal N
\mathbb N^rUse in source · 3 passages
Search aliases: N^r mathbb
Use in source · Machine_Understanding_v1.0.0.tex, line 1062
The explicit finite implementation model Use a finite augmented state space X. Include every origin variable, history flag, recipient background, and permission needed to make the declared transition and terminal tables valid. Each experiment e has the transition, observation, and admissibility semantics of Section~[source reference], together with a fixed nonnegative integer cost vector c_e∈ N^r. Its coordinates have declared units. The charge includes the specified source-side interface work f…
Exact mathematical passage
c_e\in\mathbb N^r
n_0Use in source · 3 passages
Search aliases: n_0
Use in source · Machine_Understanding_v1.0.0.tex, line 661
[A translated observation preserves both mistaken candidates] Take M={m_0,m_1} with f(m_0)=0 and f(m_1)=1. Translate models bijectively to {n_0,n_1}, but translate both observations to a single symbol *. Let g(n_0)=g(n_1)=*. The observation square commutes. Starting from both candidates, source observation 1 leaves only m_1, whose image is {n_1}. Translating first and then filtering by * leaves both n_0 and n_1.
Exact mathematical passage
\{n_0,n_1\}
n_1Use in source · 4 passages
Search aliases: n_1
Use in source · Machine_Understanding_v1.0.0.tex, line 661
[A translated observation preserves both mistaken candidates] Take M={m_0,m_1} with f(m_0)=0 and f(m_1)=1. Translate models bijectively to {n_0,n_1}, but translate both observations to a single symbol *. Let g(n_0)=g(n_1)=*. The observation square commutes. Starting from both candidates, source observation 1 leaves only m_1, whose image is {n_1}. Translating first and then filtering by * leaves both n_0 and n_1.
Exact mathematical passage
\{n_0,n_1\}
N_BUse in source · 4 passages
Search aliases: N_B
Use in source · Machine_Understanding_v1.0.0.tex, line 450
[The information an exact explanation must carry] If a fixed-length b-bit artifact lets the recipient answer the declared questions exactly for every target in Θ, then 2^b≥ N. With target-dependent background B_θ, the same argument gives 2^b≥max_B N_B, where N_B counts distinct answer profiles among targets sharing that exact background value. Two different required answer profiles cannot produce the same background and message. The deterministic recipient would then make the same responses for …
Exact mathematical passage
2^b\geq\max_{B}\ N_B,
n_oUse in source · 1 passage
Search aliases: n_o
Use in source · Machine_Understanding_v1.0.0.tex, line 1077
For a tree T, let v(T) be its componentwise worst-case charged online cost. Let P_h(V) be the attainable pairs (v(T),n(T)). The stop pairs are (g_a(V),1) for each artifact accepted throughout V. For h>0, each uniformly admissible experiment gives the additional pairs (c_e+max_o v_o, 1+∑_o n_o), (v_o,n_o)∈ P_h-1(V_e,o), with one child choice for every nonempty observation branch. An experiment contributes no pair when even one possible branch has no completion. At h=0 only stops are allowed. Sets…
Exact mathematical passage
\label{eq:resource-branch} \left(c_e+\max_o v_o,\ 1+\sum_o n_o\right), \quad (v_o,n_o)\in\mathcal P_{h-1}(V_{e,o}),
\mathcal O'Use in source · 2 passages
Search aliases: O' mathcal
Use in source · Machine_Understanding_v1.0.0.tex, line 634
Let M and N be finite model families, with model translation ϕ: M→ N. For one source experiment e, let f: M→ O give its outcomes. A corresponding target experiment has outcome map g: N→ O'. Let η: O→ O' translate evidence. Assume the basic observation square commutes: g(ϕ(m))=η(f(m)) for all m∈ M. This condition says that a model's predicted evidence translates correctly. It permits η to merge source outcomes.
Exact mathematical passage
g:\mathcal N\to\mathcal O'
o'Use in source · 1 passage
Search aliases: o'
Use in source · Machine_Understanding_v1.0.0.tex, line 640
Write ϕ[V]={ϕ(m):m∈ V}. Let C_o(V)={m∈ V:f(m)=o} and C'_o'(W)={n∈ W:g(n)=o'}. Both are ordinary consistency filters.
Exact mathematical passage
C'_{o'}(W)=\{n\in W:g(n)=o'\}
O_eUse in source · 1 passage
Search aliases: O_e
Use in source · Machine_Understanding_v1.0.0.tex, line 577
Experiments change both information and state. Let X be a finite nonempty state space. Each experiment e has an admissibility set A_e⊆ X, a transition T_e:X→ X, an observation o_e:X→ O_e, and a known cost c_e≥0. The observation is indexed by the state before the experiment, but can describe a reading taken afterward: for example o_e=q∘ T_e. At an information state V⊆ X, the experiment is permitted only when V⊆ A_e. After outcome o, the new support is V_e,o={T_e(x):x∈ V, o_e(x)=o}. Only nonempty …
Exact mathematical passage
o_e:X\to O_e
\mathcal O_eUse in source · 1 passage
Search aliases: O_e mathcal
Use in source · Machine_Understanding_v1.0.0.tex, line 530
A finite model of informative correction. Let M be a finite nonempty set of deterministic candidate models. Let E be a finite set of permitted experiments. Each experiment e has an outcome map f_e: M→ O_e. The current information is a candidate set V⊆ M. On observing o from e, update it by C_e,o(V)={m∈ V:f_e(m)=o}. This is consistency filtering, not Bayesian inference and not a general model of scientific reasoning. An empty set is meaningful: the observation contradicts every currently retained…
Exact mathematical passage
f_e:\M\to\mathcal O_e
o_eUse in source · 3 passages
Search aliases: o_e
Use in source · Machine_Understanding_v1.0.0.tex, line 577
Experiments change both information and state. Let X be a finite nonempty state space. Each experiment e has an admissibility set A_e⊆ X, a transition T_e:X→ X, an observation o_e:X→ O_e, and a known cost c_e≥0. The observation is indexed by the state before the experiment, but can describe a reading taken afterward: for example o_e=q∘ T_e. At an information state V⊆ X, the experiment is permitted only when V⊆ A_e. After outcome o, the new support is V_e,o={T_e(x):x∈ V, o_e(x)=o}. Only nonempty …
Exact mathematical passage
o_e:X\to O_e
o_tUse in source · 2 passages
Search aliases: o_t
Use in source · Machine_Understanding_v1.0.0.tex, line 855
cy, the posterior conditioned on the complete raw and reported record equals the posterior conditioned only on the reported outcomes and experiment names. The claim also holds at a bounded stopping time determined by that reported history. For a fixed realized history, its likelihood under model m is the product of the selected-experiment probabilities and the terms L_e_t,m(o_t)B_e_t(o_t,z_t). The policy factors are independent of m once the reported history is fixed. Applying the assumed factor…
Exact mathematical passage
L_{e_t,m}(o_t)B_{e_t}(o_t,z_t)
\mathcal PUse in source · 3 passages
Search aliases: P mathcal
Use in source · Machine_Understanding_v1.0.0.tex, line 1072
For a tree T, let v(T) be its componentwise worst-case charged online cost. Let P_h(V) be the attainable pairs (v(T),n(T)). The stop pairs are (g_a(V),1) for each artifact accepted throughout V. For h>0, each uniformly admissible experiment gives the additional pairs (c_e+max_o v_o, 1+∑_o n_o), (v_o,n_o)∈ P_h-1(V_e,o), with one child choice for every nonempty observation branch. An experiment contributes no pair when even one possible branch has no completion. At h=0 only stops are allowed. Sets…
Exact mathematical passage
\mathcal P_h(V)
p^{-1}Use in source · 1 passage
Search aliases: p^-1
Use in source · Machine_Understanding_v1.0.0.tex, line 329
The interpreter must not secretly do all the work. Take any bijection p: X→ Z of a finite state space. Define bar F_u=p∘ F_u∘ p^-1, bar q=q∘ p^-1. All exact representation equations hold. The labels in Z may nevertheless be arbitrary, and the translated updates may require a complete lookup table. Exact representability alone does not explain why the representation is easier to acquire, use, or communicate.
Exact mathematical passage
\bar F_u=p\circ F_u\circ p^{-1},\qquad \bar q=q\circ p^{-1}.
p^OUse in source · 1 passage
Search aliases: p^O
Use in source · Machine_Understanding_v1.0.0.tex, line 741
[Expected posterior loss is conditional information] Under~[source reference], E_O,Z [D_KL(p^OVert p^Z)] =I(M;Omid Z)=I(M;O)-I(M;Z). The common value is nonnegative. It is zero exactly when the two posteriors agree on every positive-probability observation pair for that prior. For a fully supported prior, zero is equivalent to the factorization in Theorem~[source reference]. On the support of the joint law, p(mmid o,z)=p^o(m). Substituting this into the finite definition of conditional mutual in…
Exact mathematical passage
\label{eq:posterior-kl} \mathbb E_{O,Z}\!\left[D_{\mathrm{KL}}(p^O\Vert p^Z)\right] =I(M;O\mid Z)=I(M;O)-I(M;Z).
p^ZUse in source · 1 passage
Search aliases: p^Z
Use in source · Machine_Understanding_v1.0.0.tex, line 741
[Expected posterior loss is conditional information] Under~[source reference], E_O,Z [D_KL(p^OVert p^Z)] =I(M;Omid Z)=I(M;O)-I(M;Z). The common value is nonnegative. It is zero exactly when the two posteriors agree on every positive-probability observation pair for that prior. For a fully supported prior, zero is equivalent to the factorization in Theorem~[source reference]. On the support of the joint law, p(mmid o,z)=p^o(m). Substituting this into the finite definition of conditional mutual in…
Exact mathematical passage
\label{eq:posterior-kl} \mathbb E_{O,Z}\!\left[D_{\mathrm{KL}}(p^O\Vert p^Z)\right] =I(M;O\mid Z)=I(M;O)-I(M;Z).
P_0Use in source · 4 passages
Search aliases: P_0
Use in source · Machine_Understanding_v1.0.0.tex, line 265
For computation, let P_0 group states with identical current query vectors. Given a partition P_k, refine each block by the block of F_u(x) for every u. Induction shows that P_k identifies exactly the states agreeing on words of length at most k. Every strict refinement increases the block count. At most | X|-1 strict refinements are possible. At a fixed point the partition is closed under all operations, hence under every finite word, and is exactly~[source reference].
Exact mathematical passage
P_0
P_1Use in source · 3 passages
Search aliases: P_1
Use in source · Machine_Understanding_v1.0.0.tex, line 311
[The identification limit of a transcript] Suppose two proposed implementations induce complete transcript laws P_0 and P_1 on a finite transcript alphabet under a specified permitted experiment, including any adaptive choices made by its evaluator. With equal prior probability, the greatest identification accuracy available from that transcript is frac12(1+d_TV(P_0,P_1)). In particular, identical transcript laws permit no accuracy above 1/2. Let d(o)∈[0,1] be the probability of guessing impleme…
Exact mathematical passage
P_1
\mathcal P_hUse in source · 2 passages
Search aliases: P_h mathcal
Use in source · Machine_Understanding_v1.0.0.tex, line 1072
For a tree T, let v(T) be its componentwise worst-case charged online cost. Let P_h(V) be the attainable pairs (v(T),n(T)). The stop pairs are (g_a(V),1) for each artifact accepted throughout V. For h>0, each uniformly admissible experiment gives the additional pairs (c_e+max_o v_o, 1+∑_o n_o), (v_o,n_o)∈ P_h-1(V_e,o), with one child choice for every nonempty observation branch. An experiment contributes no pair when even one possible branch has no completion. At h=0 only stops are allowed. Sets…
Exact mathematical passage
\mathcal P_h(V)
\mathcal P_{h-1}Use in source · 1 passage
Search aliases: P_h-1 mathcal
Use in source · Machine_Understanding_v1.0.0.tex, line 1077
For a tree T, let v(T) be its componentwise worst-case charged online cost. Let P_h(V) be the attainable pairs (v(T),n(T)). The stop pairs are (g_a(V),1) for each artifact accepted throughout V. For h>0, each uniformly admissible experiment gives the additional pairs (c_e+max_o v_o, 1+∑_o n_o), (v_o,n_o)∈ P_h-1(V_e,o), with one child choice for every nonempty observation branch. An experiment contributes no pair when even one possible branch has no completion. At h=0 only stops are allowed. Sets…
Exact mathematical passage
\label{eq:resource-branch} \left(c_e+\max_o v_o,\ 1+\sum_o n_o\right), \quad (v_o,n_o)\in\mathcal P_{h-1}(V_{e,o}),
P_kUse in source · 2 passages
Search aliases: P_k
Use in source · Machine_Understanding_v1.0.0.tex, line 265
For computation, let P_0 group states with identical current query vectors. Given a partition P_k, refine each block by the block of F_u(x) for every u. Induction shows that P_k identifies exactly the states agreeing on words of length at most k. Every strict refinement increases the block count. At most | X|-1 strict refinements are possible. At a fixed point the partition is closed under all operations, hence under every finite word, and is exactly~[source reference].
Exact mathematical passage
P_k
p_mUse in source · 1 passage
Search aliases: p_m
Use in source · Machine_Understanding_v1.0.0.tex, line 797
ion Write u_m=P(M=m,E), v_m=Q(M=m,E), and let p=u/a, q=v/b. Exchanging P and Q if necessary, assume a≤ b. The variation inside E is at least tfrac12∑_m|u_m-v_m|. The variation outside E is at least |a-b|/2, since its total masses differ by that amount. Hence δ≥ tfrac12∑_m|u_m-v_m|+tfrac12(b-a) =b-∑_mmin(u_m,v_m). Since a≤ b, min(ap_m,bq_m)≤ bmin(p_m,q_m). Substitution gives δ≥ b d_TV(p,q). Division by b=max(a,b) and the universal upper bound one prove the claim.
Exact mathematical passage
\min(ap_m,bq_m)\leq b\min(p_m,q_m)
p_\thetaUse in source · 3 passages
Search aliases: p_θ theta
Use in source · Machine_Understanding_v1.0.0.tex, line 988
Task-dependent interpreter changes are information. Let Θ be a finite nonempty family of tasks and let each task have a required answer profile over a fixed query family. A fixed interpreter U receives background B_θ, an installed program string p_θ, an artifact e_θ, and a query. It must return the required answer. The interpreter is fixed across tasks. The strings may depend on the task, but their lengths are charged. Let N_B be the number of distinct required answer profiles among tasks having…
Exact mathematical passage
p_\theta
\QUse in source · 3 passages
Search aliases: Q mathcal
Use in source · Machine_Understanding_v1.0.0.tex, line 213
A contract for reusable understanding. Let X and Y be finite nonempty sets. Let U be a finite operation alphabet, possibly empty, and let Q be a finite nonempty family of queries q: X→ Y. Each u∈ U has an implemented total transition F_u: X→ X. The deterministic mathematical contract is C=( X, U,(F_u)_u∈ U, Q, Y). For a word w=u_1⋯ u_n, define F_w=F_u_n∘⋯∘ F_u_1. The empty word acts as the identity. Thus q(F_w(x)) asks a question after a stated continuation.
Exact mathematical passage
\Q
q'Use in source · 7 passages
Search aliases: q'
Use in source · Machine_Understanding_v1.0.0.tex, line 359
Let a source representation have state space Z, updates f_u, and query decoders q. Let a target have state space W, updates g_v, and decoders q'. A map τ translates operation names. Query pairs (q,q') and a representation map T:Z→ W are fixed in advance. The operation names alone do not establish that the target implements the intended source operation.
Exact mathematical passage
q'
Q_{e,m}Use in source · 2 passages
Search aliases: Q_e,m
Use in source · Machine_Understanding_v1.0.0.tex, line 844
t e_t using only the prior reported history and recorded experiment names. Randomization is permitted when its conditional law is independent of M given that history. Given M=m and e_t=e, a fresh raw observation has law L_e,m, and its reported form is drawn through B_e. Conditional on the model and selected experiment, these draws do not depend on earlier hidden observations. Write Q_e,m=L_e,mB_e.
Exact mathematical passage
Q_{e,m}=L_{e,m}B_e
Q_{e_t,m}Use in source · 1 passage
Search aliases: Q_e_t,m
Use in source · Machine_Understanding_v1.0.0.tex, line 855
at a bounded stopping time determined by that reported history. For a fixed realized history, its likelihood under model m is the product of the selected-experiment probabilities and the terms L_e_t,m(o_t)B_e_t(o_t,z_t). The policy factors are independent of m once the reported history is fixed. Applying the assumed factorization gives a product of the terms Q_e_t,m(z_t) multiplied by the model-independent factors ρ_e_t(o_tmid z_t) and the same policy factors. Posterior normalization cancels tho…
Exact mathematical passage
Q_{e_t,m}(z_t)
q_mUse in source · 1 passage
Search aliases: q_m
Use in source · Machine_Understanding_v1.0.0.tex, line 797
ion Write u_m=P(M=m,E), v_m=Q(M=m,E), and let p=u/a, q=v/b. Exchanging P and Q if necessary, assume a≤ b. The variation inside E is at least tfrac12∑_m|u_m-v_m|. The variation outside E is at least |a-b|/2, since its total masses differ by that amount. Hence δ≥ tfrac12∑_m|u_m-v_m|+tfrac12(b-a) =b-∑_mmin(u_m,v_m). Since a≤ b, min(ap_m,bq_m)≤ bmin(p_m,q_m). Substitution gives δ≥ b d_TV(p,q). Division by b=max(a,b) and the universal upper bound one prove the claim.
Exact mathematical passage
\min(ap_m,bq_m)\leq b\min(p_m,q_m)
r_iUse in source · 7 passages
Search aliases: r_i
Use in source · Machine_Understanding_v1.0.0.tex, line 815
[A joint posterior and recipient decision certificate] Let p be the reference posterior, widehat p an approximate or transmitted posterior, and r_i recipient i's reconstructed posterior. Suppose d_TV(p,widehat p)≤ϵ, d_TV(widehat p,r_i)≤η_i. If a_i minimizes expected loss under r_i, then, with e_i=min{1,ϵ+η_i}, E_pℓ(M,a_i)-min_a∈ A E_pℓ(M,a) ≤min{L_max,2L_maxe_i}. If the expected-loss gap between a_i and every other action under r_i is strictly greater than 2L_maxe_i, then a_i is also the unique …
Exact mathematical passage
r_i
\rceilUse in source · 2 passages
Search aliases: rceil
Use in source · Machine_Understanding_v1.0.0.tex, line 258
[The minimal exact reusable state] The quotient X/≡_ C carries an exact reusable representation for C. Every other exact reusable representation r: X→ Z refines this quotient: if r(x)=r(y), then x≡_ C y. There is a unique map h: Z→ X/≡_ C with h(r(x))=[x]. In particular, | Z|≥| X/≡_ C|. A fixed-length binary state needs at least lceillog_2| X/≡_ C|rceil bits. The quotient can be computed by finite partition refinement. If x≡_ C y, the empty word gives q(x)=q(y) for every query. Moreover, F_u(x)≡…
Exact mathematical passage
\lceil\log_2|\X/{\Eq}|\rceil
SUse in source · 10 passages
Search aliases: S
Use in source · Machine_Understanding_v1.0.0.tex, line 378
r operation and query correspondences composed as well. No injectivity assumption on T is needed for this forward guarantee. The empty word is immediate. If the identity holds for w, then T f_u f_w=g_τ(u)T f_w =g_τ(u)g_τ(w)T. Induction proves the word identity, and applying the matched decoder proves the answer identity. For a second transport S with operation map σ, substitution gives STf_u=h_σ(τ(u))ST and the corresponding decoder identity. This is the same condition for the composite. This is…
Exact mathematical passage
S
\mathcal SUse in source · 2 passages
Search aliases: S mathcal
Use in source · Machine_Understanding_v1.0.0.tex, line 279
[Contract expansion and supplementary information] The relation ≡_ C' refines ≡_ C. Any supplementary code s: X→ S for which (r,s) supports exact reusable behavior under C' must satisfy | S|≥max_z k_z. If every ≡_ C' class lies within a single fibre of r, this bound is attained by an exact reusable pair (r,s). This condition holds when r is the old minimal behavioral quotient. A supplementary value computed solely from r(x) cannot repair a fibre meeting more than one expanded class. Agreement on…
Exact mathematical passage
s:\X\to\mathcal S
s_HUse in source · 2 passages
Search aliases: s_H
Use in source · Machine_Understanding_v1.0.0.tex, line 175
The representation is also minimal in its number of states. The present reading s and the reading s_H after a hypothetical H determine b=s_H-s, a=2s-s_H. Two different pairs (a,b) therefore differ on at least one of these questions. Every exact reusable representation for this contract needs at least 25 distinguishable values. Section~[source reference] gives the general argument.
Exact mathematical passage
s_H
s_JUse in source · 9 passages
Search aliases: s_J
Use in source · Machine_Understanding_v1.0.0.tex, line 188
The additional reading s_J yields c=s_J-s. Together with the earlier distinctions for a and b, the enlarged question family distinguishes all 125 complete states. The useful state counts are therefore five for the current display, 25 for reusable H/W prediction, and 125 after J is admitted.
Exact mathematical passage
s_J
tUse in source · 13 passages
Search aliases: t
Use in source · Machine_Understanding_v1.0.0.tex, line 486
Build a graph G_α on α( M). Join s and t when some cases m,n and recipient i satisfy α(m)=s, α(n)=t, β_i(m)=β_i(n), d_i(m)≠ d_i(n). An edge can be a loop when s=t. It then records a distinction that neither the sender's access nor that recipient's background can resolve.
Exact mathematical passage
t
T_eUse in source · 3 passages
Search aliases: T_e
Use in source · Machine_Understanding_v1.0.0.tex, line 577
Experiments change both information and state. Let X be a finite nonempty state space. Each experiment e has an admissibility set A_e⊆ X, a transition T_e:X→ X, an observation o_e:X→ O_e, and a known cost c_e≥0. The observation is indexed by the state before the experiment, but can describe a reading taken afterward: for example o_e=q∘ T_e. At an information state V⊆ X, the experiment is permitted only when V⊆ A_e. After outcome o, the new support is V_e,o={T_e(x):x∈ V, o_e(x)=o}. Only nonempty …
Exact mathematical passage
T_e:X\to X
UUse in source · 1 passage
Search aliases: U
Use in source · Machine_Understanding_v1.0.0.tex, line 988
Task-dependent interpreter changes are information. Let Θ be a finite nonempty family of tasks and let each task have a required answer profile over a fixed query family. A fixed interpreter U receives background B_θ, an installed program string p_θ, an artifact e_θ, and a query. It must return the required answer. The interpreter is fixed across tasks. The strings may depend on the task, but their lengths are charged. Let N_B be the number of distinct required answer profiles among tasks having…
Exact mathematical passage
U
\UUse in source · 4 passages
Search aliases: U mathcal
Use in source · Machine_Understanding_v1.0.0.tex, line 213
A contract for reusable understanding. Let X and Y be finite nonempty sets. Let U be a finite operation alphabet, possibly empty, and let Q be a finite nonempty family of queries q: X→ Y. Each u∈ U has an implemented total transition F_u: X→ X. The deterministic mathematical contract is C=( X, U,(F_u)_u∈ U, Q, Y). For a word w=u_1⋯ u_n, define F_w=F_u_n∘⋯∘ F_u_1. The empty word acts as the identity. Thus q(F_w(x)) asks a question after a stated continuation.
Exact mathematical passage
\U
\U^*Use in source · 1 passage
Search aliases: U^* U mathcal
Use in source · Machine_Understanding_v1.0.0.tex, line 248
Behavioral compression and the price of new questions. Define x≡_ C y Longleftrightarrow q(F_w(x))=q(F_w(y)) for all q∈ Q, w∈ U^*. This is equality of all declared future answer profiles. It is an equivalence relation. It describes behavioral sufficiency under the contract, not equality of hidden mechanisms or identity outside that contract.
Exact mathematical passage
\label{eq:equivalence} x\Eq y\quad\Longleftrightarrow\quad q(F_w(x))=q(F_w(y))\ \text{for all }q\in\Q,\ w\in\U^*.
u_1Use in source · 2 passages
Search aliases: u_1
Use in source · Machine_Understanding_v1.0.0.tex, line 217
A contract for reusable understanding. Let X and Y be finite nonempty sets. Let U be a finite operation alphabet, possibly empty, and let Q be a finite nonempty family of queries q: X→ Y. Each u∈ U has an implemented total transition F_u: X→ X. The deterministic mathematical contract is C=( X, U,(F_u)_u∈ U, Q, Y). For a word w=u_1⋯ u_n, define F_w=F_u_n∘⋯∘ F_u_1. The empty word acts as the identity. Thus q(F_w(x)) asks a question after a stated continuation.
Exact mathematical passage
w=u_1\cdots u_n
u_mUse in source · 3 passages
Search aliases: u_m
Use in source · Machine_Understanding_v1.0.0.tex, line 792
ith d_TV(P,Q)≤δ. Let E be an observation event with a=P(E)>0 and b=Q(E)>0. Then d_TV(P(Mmid E),Q(Mmid E)) ≤min{1,fracδmax(a,b)}. In particular, a justified lower bound P(E)≥γ>0 supplies the weaker usable bound min{1,δ/γ}. Write u_m=P(M=m,E), v_m=Q(M=m,E), and let p=u/a, q=v/b. Exchanging P and Q if necessary, assume a≤ b. The variation inside E is at least tfrac12∑_m|u_m-v_m|. The variation outside E is at least |a-b|/2, since its total masses differ by that amount. Hence δ≥ tfrac12∑_m|u_m-v_m|+…
Exact mathematical passage
u_m=P(M=m,E)
u_nUse in source · 2 passages
Search aliases: u_n
Use in source · Machine_Understanding_v1.0.0.tex, line 217
A contract for reusable understanding. Let X and Y be finite nonempty sets. Let U be a finite operation alphabet, possibly empty, and let Q be a finite nonempty family of queries q: X→ Y. Each u∈ U has an implemented total transition F_u: X→ X. The deterministic mathematical contract is C=( X, U,(F_u)_u∈ U, Q, Y). For a word w=u_1⋯ u_n, define F_w=F_u_n∘⋯∘ F_u_1. The empty word acts as the identity. Thus q(F_w(x)) asks a question after a stated continuation.
Exact mathematical passage
w=u_1\cdots u_n
vUse in source · 11 passages
Search aliases: v
Use in source · Machine_Understanding_v1.0.0.tex, line 345
[Uniform controlled stochastic reuse] A single reduced kernel bar K_u reproduces every projected one-step law from every complete state if and only if, whenever r(x)=r(y), ∑_v:r(v)=z'K_u(x,v)=∑_v:r(v)=z'K_u(y,v) for every u,z'. With matched initial projected laws, the reduced model preserves the law of every finite sequence of reduced states and queries under any policy depending only on the reduced history. A common projected row is necessary by considering point-mass initial states. If the row…
Exact mathematical passage
\sum_{v:r(v)=z'}K_u(x,v)=\sum_{v:r(v)=z'}K_u(y,v) \quad\text{for every }u,z'.
V_0Use in source · 3 passages
Search aliases: V_0
Use in source · Machine_Understanding_v1.0.0.tex, line 539
[Finite correction under separating evidence] Suppose the true model m_* belongs to the initial set V_0, every observation is exact, and the allowed experiment family separates any two distinct remaining observational classes. Repeatedly selecting an informative experiment preserves m_* and reaches a single observational class after at most |V_0|-1 informative observations. If there are N distinguishable target classes, every adaptive procedure identifying all of them using experiments with at m…
Exact mathematical passage
V_0
v_mUse in source · 3 passages
Search aliases: v_m
Use in source · Machine_Understanding_v1.0.0.tex, line 792
qδ. Let E be an observation event with a=P(E)>0 and b=Q(E)>0. Then d_TV(P(Mmid E),Q(Mmid E)) ≤min{1,fracδmax(a,b)}. In particular, a justified lower bound P(E)≥γ>0 supplies the weaker usable bound min{1,δ/γ}. Write u_m=P(M=m,E), v_m=Q(M=m,E), and let p=u/a, q=v/b. Exchanging P and Q if necessary, assume a≤ b. The variation inside E is at least tfrac12∑_m|u_m-v_m|. The variation outside E is at least |a-b|/2, since its total masses differ by that amount. Hence δ≥ tfrac12∑_m|u_m-v_m|+tfrac12(b-a) …
Exact mathematical passage
v_m=Q(M=m,E)
v_oUse in source · 1 passage
Search aliases: v_o
Use in source · Machine_Understanding_v1.0.0.tex, line 1077
For a tree T, let v(T) be its componentwise worst-case charged online cost. Let P_h(V) be the attainable pairs (v(T),n(T)). The stop pairs are (g_a(V),1) for each artifact accepted throughout V. For h>0, each uniformly admissible experiment gives the additional pairs (c_e+max_o v_o, 1+∑_o n_o), (v_o,n_o)∈ P_h-1(V_e,o), with one child choice for every nonempty observation branch. An experiment contributes no pair when even one possible branch has no completion. At h=0 only stops are allowed. Sets…
Exact mathematical passage
\label{eq:resource-branch} \left(c_e+\max_o v_o,\ 1+\sum_o n_o\right), \quad (v_o,n_o)\in\mathcal P_{h-1}(V_{e,o}),
\VertUse in source · 2 passages
Search aliases: Vert
Use in source · Machine_Understanding_v1.0.0.tex, line 741
[Expected posterior loss is conditional information] Under~[source reference], E_O,Z [D_KL(p^OVert p^Z)] =I(M;Omid Z)=I(M;O)-I(M;Z). The common value is nonnegative. It is zero exactly when the two posteriors agree on every positive-probability observation pair for that prior. For a fully supported prior, zero is equivalent to the factorization in Theorem~[source reference]. On the support of the joint law, p(mmid o,z)=p^o(m). Substituting this into the finite definition of conditional mutual in…
Exact mathematical passage
\label{eq:posterior-kl} \mathbb E_{O,Z}\!\left[D_{\mathrm{KL}}(p^O\Vert p^Z)\right] =I(M;O\mid Z)=I(M;O)-I(M;Z).
WUse in source · 24 passages
Search aliases: W
Use in source · Machine_Understanding_v1.0.0.tex, line 153
Initially, two operations are available: H(a,b,c)=(a+b,b,c), W(a,b,c)=(b,a,c). The first adds the second register into the first. The second exchanges the first two registers. The third register is preserved by both and does not affect the display. The operation order is public. A prediction concerns what the display would show after a named finite sequence of these operations.
Exact mathematical passage
\label{eq:instrument-ops} H(a,b,c)=(a+b,b,c),\qquad W(a,b,c)=(b,a,c).
w_0Use in source · 2 passages
Search aliases: w_0
Use in source · Machine_Understanding_v1.0.0.tex, line 407
[Finite-horizon transfer budget] For matched source and target operation sequences, start from z_0,w_0 with d(Tz_0,w_0)≤ e_0. After n operations, d(Tz_n,w_n) · ≤ L^n e_0+ϵ∑_j=0^n-1L^j,
Exact mathematical passage
z_0,w_0
w_nUse in source · 2 passages
Search aliases: w_n
Use in source · Machine_Understanding_v1.0.0.tex, line 408
[Finite-horizon transfer budget] For matched source and target operation sequences, start from z_0,w_0 with d(Tz_0,w_0)≤ e_0. After n operations, d(Tz_n,w_n) · ≤ L^n e_0+ϵ∑_j=0^n-1L^j, |q(z_n)-q'(w_n)| · ≤δ+M(L^n e_0+ϵ∑_j=0^n-1L^j). The empty sum is zero and L^0=1, including when L=0. At L=1 the state bound is e_0+nϵ. Writing e_n=d(Tz_n,w_n), insert g_τ(u)(Tz_n) and use the triangle inequality and the Lipschitz bound to obtain e_n+1≤ϵ+Le_n. Iterating gives the stated expression. Insert q'(Tz_n) …
Exact mathematical passage
d(Tz_n,w_n)&\leq L^n e_0+\epsilon\sum_{j=0}^{n-1}L^j,\\ |q(z_n)-q'(w_n)|&\leq\delta+M\left(L^n e_0+\epsilon\sum_{j=0}^{n-1}L^j\right).
\widehat pUse in source · 3 passages
Search aliases: widehat p widehat
Use in source · Machine_Understanding_v1.0.0.tex, line 815
[A joint posterior and recipient decision certificate] Let p be the reference posterior, widehat p an approximate or transmitted posterior, and r_i recipient i's reconstructed posterior. Suppose d_TV(p,widehat p)≤ϵ, d_TV(widehat p,r_i)≤η_i. If a_i minimizes expected loss under r_i, then, with e_i=min{1,ϵ+η_i}, E_pℓ(M,a_i)-min_a∈ A E_pℓ(M,a) ≤min{L_max,2L_maxe_i}. If the expected-loss gap between a_i and every other action under r_i is strictly greater than 2L_maxe_i, then a_i is also the unique …
Exact mathematical passage
\widehat p
\widetilde WUse in source · 1 passage
Search aliases: widetilde W widetilde
Use in source · Machine_Understanding_v1.0.0.tex, line 395
[Correct current answers with a wrong transported operation] Retain the correct map T(a,b)=(a+b,b) and the correct display decoder. Now incorrectly define the target swap by widetilde W(s,b)=(s,b). The current display and the display immediately after a swap are still correct. Starting from (a,b)=(1,3), however, the source sequence W then H ends with display zero. The incorrect target sequence ends with display two. Matching a restricted collection of outputs did not preserve the method. The err…
Exact mathematical passage
\widetilde W(s,b)=(s,b)
x'Use in source · 1 passage
Search aliases: x'
Use in source · Machine_Understanding_v1.0.0.tex, line 341
A bounded stochastic extension. The distinction between a current answer and a reusable state also survives stochastic dynamics. Let K_u(x,x') be finite transition kernels and let r: X→ Z be surjective. Assume the current query values are constant on fibres of r.
Exact mathematical passage
K_u(x,x')
X_0Use in source · 2 passages
Search aliases: X_0
Use in source · Machine_Understanding_v1.0.0.tex, line 927
The graph belongs to the acquired transcript. Fix a finite set X_0 of possible initial states, a finite deterministic experiment catalogue as in Section~[source reference], and a finite horizon h. A deterministic experimental policy is an observation-history tree. Its choices cannot depend on the hidden initial state. Only uniformly admissible choices are permitted at each reachable node.
Exact mathematical passage
X_0
x_0Use in source · 4 passages
Search aliases: x_0
Use in source · Machine_Understanding_v1.0.0.tex, line 621
To require recovery of a past distinction, replace the state by (x_0,x_t), let every transition preserve x_0, and make the terminal target include the required part of x_0. The same recurrence then treats historical reconstruction honestly. In the reset example, the two origins remain indistinguishable at every horizon. If both current knowledge and preserved history matter, use their joint target rather than claim that either implies the other.
Exact mathematical passage
(x_0,x_t)
x_1Use in source · 1 passage
Search aliases: x_1
Use in source · Machine_Understanding_v1.0.0.tex, line 1048
[Information suffices but the installed decoder needs a larger codebook] Take three cases x_0,x_1,x_2 fully visible to the sender. One recipient already has background i in case x_i and always needs output zero. The sender-access graph has no edges, so a designable decoder could ignore a single common message and return zero. Now fix an installed decoder on artifacts A,B,C. It returns zero on A,B at background 0, on B,C at background 1, and on A,C at background 2. It fails on the remaining pairs…
Exact mathematical passage
x_0,x_1,x_2
x_2Use in source · 1 passage
Search aliases: x_2
Use in source · Machine_Understanding_v1.0.0.tex, line 1048
[Information suffices but the installed decoder needs a larger codebook] Take three cases x_0,x_1,x_2 fully visible to the sender. One recipient already has background i in case x_i and always needs output zero. The sender-access graph has no edges, so a designable decoder could ignore a single common message and return zero. Now fix an installed decoder on artifacts A,B,C. It returns zero on A,B at background 0, on B,C at background 1, and on A,C at background 2. It fails on the remaining pairs…
Exact mathematical passage
x_0,x_1,x_2
x_iUse in source · 1 passage
Search aliases: x_i
Use in source · Machine_Understanding_v1.0.0.tex, line 1048
[Information suffices but the installed decoder needs a larger codebook] Take three cases x_0,x_1,x_2 fully visible to the sender. One recipient already has background i in case x_i and always needs output zero. The sender-access graph has no edges, so a designable decoder could ignore a single common message and return zero. Now fix an installed decoder on artifacts A,B,C. It returns zero on A,B at background 0, on B,C at background 1, and on A,C at background 2. It fails on the remaining pairs…
Exact mathematical passage
x_i
x_tUse in source · 1 passage
Search aliases: x_t
Use in source · Machine_Understanding_v1.0.0.tex, line 621
To require recovery of a past distinction, replace the state by (x_0,x_t), let every transition preserve x_0, and make the terminal target include the required part of x_0. The same recurrence then treats historical reconstruction honestly. In the reset example, the two origins remain indistinguishable at every horizon. If both current knowledge and preserved history matter, use their joint target rather than claim that either implies the other.
Exact mathematical passage
(x_0,x_t)
YUse in source · 5 passages
Search aliases: Y
Use in source · Machine_Understanding_v1.0.0.tex, line 756
The noisy instrument and a harmless discarded flag. Return to the instrument immediately after a known J intervention. The earlier (s,b) is known, while c is initially uniform on five values. A read-only sensor reports the shifted value Y=reported display-s modulo five. It returns the correct c with probability 3/4 and each wrong value with probability 1/16. The physical state remains fixed during repeated read-only observations.
Exact mathematical passage
Y=\text{reported display}-s
yUse in source · 10 passages
Search aliases: y
Use in source · Machine_Understanding_v1.0.0.tex, line 160
Knowing the current display is insufficient. A record containing only s=a+b answers the current question exactly. It cannot, in general, predict the display after H. The two complete states x=(0,0,0), y=(1,4,0) both display zero. After H, their displays are zero and four. Any predictor receiving only the common old display must return the same answer for both, so it cannot be exact in both cases.
Exact mathematical passage
x=(0,0,0),\qquad y=(1,4,0)
z'Use in source · 2 passages
Search aliases: z'
Use in source · Machine_Understanding_v1.0.0.tex, line 345
[Uniform controlled stochastic reuse] A single reduced kernel bar K_u reproduces every projected one-step law from every complete state if and only if, whenever r(x)=r(y), ∑_v:r(v)=z'K_u(x,v)=∑_v:r(v)=z'K_u(y,v) for every u,z'. With matched initial projected laws, the reduced model preserves the law of every finite sequence of reduced states and queries under any policy depending only on the reduced history. A common projected row is necessary by considering point-mass initial states. If the row…
Exact mathematical passage
\sum_{v:r(v)=z'}K_u(x,v)=\sum_{v:r(v)=z'}K_u(y,v) \quad\text{for every }u,z'.
z_0Use in source · 2 passages
Search aliases: z_0
Use in source · Machine_Understanding_v1.0.0.tex, line 407
[Finite-horizon transfer budget] For matched source and target operation sequences, start from z_0,w_0 with d(Tz_0,w_0)≤ e_0. After n operations, d(Tz_n,w_n) · ≤ L^n e_0+ϵ∑_j=0^n-1L^j,
Exact mathematical passage
z_0,w_0
z_nUse in source · 4 passages
Search aliases: z_n
Use in source · Machine_Understanding_v1.0.0.tex, line 408
[Finite-horizon transfer budget] For matched source and target operation sequences, start from z_0,w_0 with d(Tz_0,w_0)≤ e_0. After n operations, d(Tz_n,w_n) · ≤ L^n e_0+ϵ∑_j=0^n-1L^j, |q(z_n)-q'(w_n)| · ≤δ+M(L^n e_0+ϵ∑_j=0^n-1L^j). The empty sum is zero and L^0=1, including when L=0. At L=1 the state bound is e_0+nϵ. Writing e_n=d(Tz_n,w_n), insert g_τ(u)(Tz_n) and use the triangle inequality and the Lipschitz bound to obtain e_n+1≤ϵ+Le_n. Iterating gives the stated expression. Insert q'(Tz_n) …
Exact mathematical passage
d(Tz_n,w_n)&\leq L^n e_0+\epsilon\sum_{j=0}^{n-1}L^j,\\ |q(z_n)-q'(w_n)|&\leq\delta+M\left(L^n e_0+\epsilon\sum_{j=0}^{n-1}L^j\right).
z_tUse in source · 3 passages
Search aliases: z_t
Use in source · Machine_Understanding_v1.0.0.tex, line 855
cy, the posterior conditioned on the complete raw and reported record equals the posterior conditioned only on the reported outcomes and experiment names. The claim also holds at a bounded stopping time determined by that reported history. For a fixed realized history, its likelihood under model m is the product of the selected-experiment probabilities and the terms L_e_t,m(o_t)B_e_t(o_t,z_t). The policy factors are independent of m once the reported history is fixed. Applying the assumed factor…
Exact mathematical passage
L_{e_t,m}(o_t)B_{e_t}(o_t,z_t)
\cdotUse in source · 2 passages
Search aliases: · cdot
Use in source · Machine_Understanding_v1.0.0.tex, line 727
id Z=z). If (o,z) has positive probability, posterior equality and Bayes' formula give p(m)L_m(o)B(o,z) =Pr(O=o,Z=z)p^o(m) =ρ(omid z)Pr(Z=z)p^z(m) =p(m)Q_m(z)ρ(omid z). Cancel the positive p(m). If the pair has probability zero, full prior support makes L_m(o)B(o,z)=0 for every model, and ρ(omid z)=0. If z is impossible under every model, choose any probability row ρ(·mid z) since both sides are zero. This proves the third condition.
Exact mathematical passage
\rho(\cdot\mid z)
\alphaUse in source · 13 passages
Search aliases: α alpha
Use in source · Machine_Understanding_v1.0.0.tex, line 478
An accessible sender and a family of recipients. Let M be a finite nonempty set of possible cases. The sender observes only α(m), where α: M→ S is a declared access map. Recipient i∈{1,…,k} receives background β_i(m) and must recover d_i(m). A demand can be a single answer, a reusable state, or the complete finite answer profile for a declared task. The maps and recipients are fixed before the case is selected.
Exact mathematical passage
\alpha(m)
\beta_iUse in source · 5 passages
Search aliases: β_i beta
Use in source · Machine_Understanding_v1.0.0.tex, line 478
An accessible sender and a family of recipients. Let M be a finite nonempty set of possible cases. The sender observes only α(m), where α: M→ S is a declared access map. Recipient i∈{1,…,k} receives background β_i(m) and must recover d_i(m). A demand can be a single answer, a reusable state, or the complete finite answer profile for a declared task. The maps and recipients are fixed before the case is selected.
Exact mathematical passage
\beta_i(m)
\gammaUse in source · 6 passages
Search aliases: γ gamma
Use in source · Machine_Understanding_v1.0.0.tex, line 789
evidence denominator] Let P and Q be finite joint laws of a model and an observation, with d_TV(P,Q)≤δ. Let E be an observation event with a=P(E)>0 and b=Q(E)>0. Then d_TV(P(Mmid E),Q(Mmid E)) ≤min{1,fracδmax(a,b)}. In particular, a justified lower bound P(E)≥γ>0 supplies the weaker usable bound min{1,δ/γ}. Write u_m=P(M=m,E), v_m=Q(M=m,E), and let p=u/a, q=v/b. Exchanging P and Q if necessary, assume a≤ b. The variation inside E is at least tfrac12∑_m|u_m-v_m|. The variation outside E is at lea…
Exact mathematical passage
P(E)\geq\gamma>0
\deltaUse in source · 10 passages
Search aliases: δ delta
Use in source · Machine_Understanding_v1.0.0.tex, line 404
te transfer accumulates error Exact equalities are not always available. For this result let the target state space carry a metric d. Suppose each target update is L-Lipschitz for a common L≥0, and suppose d(Tf_u(z),g_τ(u)(Tz))≤ϵ uniformly on every state and operation that the contract permits. Let a target query be M-Lipschitz with real-valued output and satisfy |q'(Tz)-q(z)|≤δ.
Exact mathematical passage
|q'(Tz)-q(z)|\leq\delta
\epsilonUse in source · 11 passages
Search aliases: ϵ epsilon
Use in source · Machine_Understanding_v1.0.0.tex, line 401
Approximate transfer accumulates error. Exact equalities are not always available. For this result let the target state space carry a metric d. Suppose each target update is L-Lipschitz for a common L≥0, and suppose d(Tf_u(z),g_τ(u)(Tz))≤ϵ uniformly on every state and operation that the contract permits. Let a target query be M-Lipschitz with real-valued output and satisfy |q'(Tz)-q(z)|≤δ.
Exact mathematical passage
d(Tf_u(z),g_{\tau(u)}(Tz))\leq\epsilon
\eta_iUse in source · 2 passages
Search aliases: η_i eta
Use in source · Machine_Understanding_v1.0.0.tex, line 816
[A joint posterior and recipient decision certificate] Let p be the reference posterior, widehat p an approximate or transmitted posterior, and r_i recipient i's reconstructed posterior. Suppose d_TV(p,widehat p)≤ϵ, d_TV(widehat p,r_i)≤η_i. If a_i minimizes expected loss under r_i, then, with e_i=min{1,ϵ+η_i}, E_pℓ(M,a_i)-min_a∈ A E_pℓ(M,a) ≤min{L_max,2L_maxe_i}. If the expected-loss gap between a_i and every other action under r_i is strictly greater than 2L_maxe_i, then a_i is also the unique …
Exact mathematical passage
\TV(p,\widehat p)\leq\epsilon,\qquad \TV(\widehat p,r_i)\leq\eta_i.
\ThetaUse in source · 3 passages
Search aliases: Θ Theta
Use in source · Machine_Understanding_v1.0.0.tex, line 446
An exact communication lower bound. Let Θ be a finite family of target models. Relative to the required continuation questions, divide these models into N classes with distinct answer profiles. Use one fixed recipient background and one fixed deterministic recipient. The teacher can select a message but cannot change the recipient's background separately for each target without that change being counted as communication.
Exact mathematical passage
\Theta
\thetaUse in source · 7 passages
Search aliases: θ theta
Use in source · Machine_Understanding_v1.0.0.tex, line 449
[The information an exact explanation must carry] If a fixed-length b-bit artifact lets the recipient answer the declared questions exactly for every target in Θ, then 2^b≥ N. With target-dependent background B_θ, the same argument gives 2^b≥max_B N_B, where N_B counts distinct answer profiles among targets sharing that exact background value. Two different required answer profiles cannot produce the same background and message. The deterministic recipient would then make the same responses for …
Exact mathematical passage
B_\theta
\rhoUse in source · 10 passages
Search aliases: ρ rho
Use in source · Machine_Understanding_v1.0.0.tex, line 712
act Bayesian correction through a reporting channel] For fixed finite likelihoods L and reporting channel B, the following are equivalent: [label=(roman*)] item For every prior p, p^o=p^z on every positive-probability pair (o,z). item This equality holds for one prior assigning positive mass to every model. item There exists a row-stochastic kernel ρ(omid z) such that, for all m,o,z, L_m(o)B(o,z)=Q_m(z)ρ(omid z). Thus a report preserves every posterior update exactly when the discarded observati…
Exact mathematical passage
\rho(o\mid z)
\rho_eUse in source · 2 passages
Search aliases: ρ_e rho
Use in source · Machine_Understanding_v1.0.0.tex, line 847
[Sequential preservation of noisy correction under adaptive experiments] Suppose each permitted experiment admits a kernel ρ_e with L_e,m(o)B_e(o,z)=Q_e,m(z)ρ_e(omid z) for all m,o,z. For every finite positive-probability history under the declared policy, the posterior conditioned on the complete raw and reported record equals the posterior conditioned only on the reported outcomes and experiment names. The claim also holds at a bounded stopping time determined by that reported history. For a f…
Exact mathematical passage
\rho_e
\rho_{e_t}Use in source · 1 passage
Search aliases: ρ_e_t rho
Use in source · Machine_Understanding_v1.0.0.tex, line 855
y. For a fixed realized history, its likelihood under model m is the product of the selected-experiment probabilities and the terms L_e_t,m(o_t)B_e_t(o_t,z_t). The policy factors are independent of m once the reported history is fixed. Applying the assumed factorization gives a product of the terms Q_e_t,m(z_t) multiplied by the model-independent factors ρ_e_t(o_tmid z_t) and the same policy factors. Posterior normalization cancels those factors. The remaining product is exactly the likelihood f…
Exact mathematical passage
\rho_{e_t}(o_t\mid z_t)
\sigmaUse in source · 2 passages
Search aliases: σ sigma
Use in source · Machine_Understanding_v1.0.0.tex, line 378
orrespondences composed as well. No injectivity assumption on T is needed for this forward guarantee. The empty word is immediate. If the identity holds for w, then T f_u f_w=g_τ(u)T f_w =g_τ(u)g_τ(w)T. Induction proves the word identity, and applying the matched decoder proves the answer identity. For a second transport S with operation map σ, substitution gives STf_u=h_σ(τ(u))ST and the corresponding decoder identity. This is the same condition for the composite. This is an intertwining argume…
Exact mathematical passage
\sigma
\chiUse in source · 3 passages
Search aliases: χ chi
Use in source · Machine_Understanding_v1.0.0.tex, line 494
[A common explanatory artifact under sender and recipient constraints] An exact artifact satisfying~[source reference] with at most K≥1 symbols exists if and only if G_α has no loop and admits a proper coloring with at most K colors. In the loop-free case, the minimum alphabet size is χ(G_α), and the minimum fixed binary length with designable decoders is lceillog_2χ(G_α)rceil. A loop makes the exact contract infeasible for every alphabet size. An encoder assigns a symbol to every accessible sen…
Exact mathematical passage
\chi(G_\alpha)
\ldotsUse in source · 3 passages
Search aliases: … ldots
Use in source · Machine_Understanding_v1.0.0.tex, line 286
ntary value computed solely from r(x) cannot repair a fibre meeting more than one expanded class. Agreement on the larger continuation family implies agreement on the smaller one. Within a fixed fibre of r, an adequate pair (r,s) must separate the k_z new classes, giving the lower bound. For attainment under the additional condition, assign distinct labels 1,…,k_z to the expanded classes within each old fibre and let s(x) be its class label. The pair (r(x),s(x)) then identifies exactly the expan…
Exact mathematical passage
1,\ldots,k_z
\ellUse in source · 2 passages
Search aliases: ℓ ell
Use in source · Machine_Understanding_v1.0.0.tex, line 812
From approximate beliefs to a recipient's decisions. Not every posterior distinction matters equally to an action. To link correction with use, fix one finite nonempty action set A and losses ℓ(m,a)∈[0,L_max]. A receiver evaluates actions from the distribution its actual interpreter reconstructs. The target loss function and available actions must be the same in the comparison.
Exact mathematical passage
\ell(m,a)\in[0,L_{\max}]
\mapstoUse in source · 2 passages
Search aliases: ↦ mapsto
Use in source · Machine_Understanding_v1.0.0.tex, line 261
(r(x))=[x]. In particular, | Z|≥| X/≡_ C|. A fixed-length binary state needs at least lceillog_2| X/≡_ C|rceil bits. The quotient can be computed by finite partition refinement. If x≡_ C y, the empty word gives q(x)=q(y) for every query. Moreover, F_u(x)≡_ C F_u(y) because any continuation after u is a continuation in the original contract. Thus [x]↦[F_u(x)] and [x]↦ q(x) are well-defined updates and decoders.
Exact mathematical passage
[x]\mapsto[F_u(x)]
\sum_{j=0}^{n-1}Use in source · 1 passage
Search aliases: ∑_j=0^n-1 sum
Use in source · Machine_Understanding_v1.0.0.tex, line 408
[Finite-horizon transfer budget] For matched source and target operation sequences, start from z_0,w_0 with d(Tz_0,w_0)≤ e_0. After n operations, d(Tz_n,w_n) · ≤ L^n e_0+ϵ∑_j=0^n-1L^j, |q(z_n)-q'(w_n)| · ≤δ+M(L^n e_0+ϵ∑_j=0^n-1L^j). The empty sum is zero and L^0=1, including when L=0. At L=1 the state bound is e_0+nϵ. Writing e_n=d(Tz_n,w_n), insert g_τ(u)(Tz_n) and use the triangle inequality and the Lipschitz bound to obtain e_n+1≤ϵ+Le_n. Iterating gives the stated expression. Insert q'(Tz_n) …
Exact mathematical passage
d(Tz_n,w_n)&\leq L^n e_0+\epsilon\sum_{j=0}^{n-1}L^j,\\ |q(z_n)-q'(w_n)|&\leq\delta+M\left(L^n e_0+\epsilon\sum_{j=0}^{n-1}L^j\right).
\sum_mUse in source · 2 passages
Search aliases: ∑_m sum
Use in source · Machine_Understanding_v1.0.0.tex, line 792
(Mmid E),Q(Mmid E)) ≤min{1,fracδmax(a,b)}. In particular, a justified lower bound P(E)≥γ>0 supplies the weaker usable bound min{1,δ/γ}. Write u_m=P(M=m,E), v_m=Q(M=m,E), and let p=u/a, q=v/b. Exchanging P and Q if necessary, assume a≤ b. The variation inside E is at least tfrac12∑_m|u_m-v_m|. The variation outside E is at least |a-b|/2, since its total masses differ by that amount. Hence δ≥ tfrac12∑_m|u_m-v_m|+tfrac12(b-a) =b-∑_mmin(u_m,v_m). Since a≤ b, min(ap_m,bq_m)≤ bmin(p_m,q_m). Substituti…
Exact mathematical passage
\tfrac12\sum_m|u_m-v_m|
\sum_{v:r(v)=z'}Use in source · 1 passage
Search aliases: ∑_v:r(v)=z' sum
Use in source · Machine_Understanding_v1.0.0.tex, line 345
[Uniform controlled stochastic reuse] A single reduced kernel bar K_u reproduces every projected one-step law from every complete state if and only if, whenever r(x)=r(y), ∑_v:r(v)=z'K_u(x,v)=∑_v:r(v)=z'K_u(y,v) for every u,z'. With matched initial projected laws, the reduced model preserves the law of every finite sequence of reduced states and queries under any policy depending only on the reduced history. A common projected row is necessary by considering point-mass initial states. If the row…
Exact mathematical passage
\sum_{v:r(v)=z'}K_u(x,v)=\sum_{v:r(v)=z'}K_u(y,v) \quad\text{for every }u,z'.
\circUse in source · 3 passages
Search aliases: ∘ circ
Use in source · Machine_Understanding_v1.0.0.tex, line 217
t for reusable understanding Let X and Y be finite nonempty sets. Let U be a finite operation alphabet, possibly empty, and let Q be a finite nonempty family of queries q: X→ Y. Each u∈ U has an implemented total transition F_u: X→ X. The deterministic mathematical contract is C=( X, U,(F_u)_u∈ U, Q, Y). For a word w=u_1⋯ u_n, define F_w=F_u_n∘⋯∘ F_u_1. The empty word acts as the identity. Thus q(F_w(x)) asks a question after a stated continuation.
Exact mathematical passage
F_w=F_{u_n}\circ\cdots\circ F_{u_1}
\capUse in source · 2 passages
Search aliases: ∩ cap
Use in source · Machine_Understanding_v1.0.0.tex, line 1033
[The fixed-recipient covering criterion] Under this finite contract, an exact encoder E:S→ A exists on the reachable sender values if and only if every A_s is nonempty. Its smallest possible image size is min{|C|:C⊆ A, C∩ A_s≠∅ for every reachable s}, when the feasible family is nonempty. Thus a codebook of at most K artifacts is possible exactly when a set of at most K artifacts meets all these requirements. Every valid encoder induces a proper coloring of the sender-access graph, so its image …
Exact mathematical passage
\label{eq:decoder-cover} \min\{|C|:C\subseteq\mathcal A,\ C\cap\mathcal A_s\ne\varnothing \text{ for every reachable }s\},
\equiv_{\C}Use in source · 1 passage
Search aliases: ≡_C ≡_ C equiv C equiv mathcal
Use in source · Machine_Understanding_v1.0.0.tex, line 279
[Contract expansion and supplementary information] The relation ≡_ C' refines ≡_ C. Any supplementary code s: X→ S for which (r,s) supports exact reusable behavior under C' must satisfy | S|≥max_z k_z. If every ≡_ C' class lies within a single fibre of r, this bound is attained by an exact reusable pair (r,s). This condition holds when r is the old minimal behavioral quotient. A supplementary value computed solely from r(x) cannot repair a fibre meeting more than one expanded class. Agreement on…
Exact mathematical passage
\equiv_{\C}
\leqUse in source · 17 passages
Search aliases: ≤ leq
Use in source · Machine_Understanding_v1.0.0.tex, line 401
Approximate transfer accumulates error. Exact equalities are not always available. For this result let the target state space carry a metric d. Suppose each target update is L-Lipschitz for a common L≥0, and suppose d(Tf_u(z),g_τ(u)(Tz))≤ϵ uniformly on every state and operation that the contract permits. Let a target query be M-Lipschitz with real-valued output and satisfy |q'(Tz)-q(z)|≤δ.
Exact mathematical passage
d(Tf_u(z),g_{\tau(u)}(Tz))\leq\epsilon
\geqUse in source · 18 passages
Search aliases: ≥ geq
Use in source · Machine_Understanding_v1.0.0.tex, line 255
[The minimal exact reusable state] The quotient X/≡_ C carries an exact reusable representation for C. Every other exact reusable representation r: X→ Z refines this quotient: if r(x)=r(y), then x≡_ C y. There is a unique map h: Z→ X/≡_ C with h(r(x))=[x]. In particular, | Z|≥| X/≡_ C|. A fixed-length binary state needs at least lceillog_2| X/≡_ C|rceil bits. The quotient can be computed by finite partition refinement. If x≡_ C y, the empty word gives q(x)=q(y) for every query. Moreover, F_u(x)≡…
Exact mathematical passage
|\Z|\geq|\X/{\Eq}|.
\oplusUse in source · 2 passages
Search aliases: ⊕ oplus
Use in source · Machine_Understanding_v1.0.0.tex, line 505
Complementary knowledge can reduce communication. For two bits (a,b), let the first recipient know a and need b, while the second knows b and needs a. A sender with access only to a⊕ b can broadcast that one bit. Both recipients recover their missing bit. The sender need not reconstruct the complete pair. An inference from successful explanation to possession of the recipient's full state would therefore be invalid.
Exact mathematical passage
a\oplus b
\bigcapUse in source · 1 passage
Search aliases: ⋂ bigcap
Use in source · Machine_Understanding_v1.0.0.tex, line 1026
Define A(x)={a∈ A:every recipient returns d_i(x) within its declared limits}, and, for each reachable sender value s, define A_s=⋂_x:α(x)=s A(x). An artifact selected from A_s must work for all cases the sender still considers possible. Choosing a different acceptable artifact after secretly learning which hidden case is true is not allowed.
Exact mathematical passage
\label{eq:decoder-intersection} \mathcal A_s=\bigcap_{x:\alpha(x)=s}\mathcal A(x).
\bigcap_{x:\alpha(x)=s}Use in source · 1 passage
Search aliases: ⋂_x:α(x)=s bigcap alpha
Use in source · Machine_Understanding_v1.0.0.tex, line 1026
Define A(x)={a∈ A:every recipient returns d_i(x) within its declared limits}, and, for each reachable sender value s, define A_s=⋂_x:α(x)=s A(x). An artifact selected from A_s must work for all cases the sender still considers possible. Choosing a different acceptable artifact after secretly learning which hidden case is true is not allowed.
Exact mathematical passage
\label{eq:decoder-intersection} \mathcal A_s=\bigcap_{x:\alpha(x)=s}\mathcal A(x).
\cdotsUse in source · 2 passages
Search aliases: ⋯ cdots
Use in source · Machine_Understanding_v1.0.0.tex, line 217
A contract for reusable understanding. Let X and Y be finite nonempty sets. Let U be a finite operation alphabet, possibly empty, and let Q be a finite nonempty family of queries q: X→ Y. Each u∈ U has an implemented total transition F_u: X→ X. The deterministic mathematical contract is C=( X, U,(F_u)_u∈ U, Q, Y). For a word w=u_1⋯ u_n, define F_w=F_u_n∘⋯∘ F_u_1. The empty word acts as the identity. Thus q(F_w(x)) asks a question after a stated continuation.
Exact mathematical passage
w=u_1\cdots u_n
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