Reading notes and page index
Self-Improving Intelligence — Visual Guide v1.0.0
These notes accompany the diagrams. Full hypotheses and proofs remain in the source manuscript.
01. Self-Improving Intelligence: A Visual Guide
A standalone visual companion.
Interpretation. Four different objects improve across the finite chain.
Boundary. No claim of general intelligence or indefinite recursive acceleration.
Source: [S] First-edition scope, pp. ii-iii
02. The route through this guide
Read the worked world first. Then follow the discovery procedure outward, keeping the supplied machinery in view.
Interpretation. Every plate identifies its source section. Manuscript page numbers are the printed numbers, not the PDF viewer index.
Boundary. A solid arrow is a specified transition or dependency. Dashed paths are explicitly labeled as recurrence or a qualified extension. Counts retain their original units.
Source: [S] Scope guide; §§1-3 | manuscript pp. iii, 1-4
03. A better system is not always a better improver
Changing a library and changing the procedure that uses it answer different questions. A crossed comparison keeps them separate.
Interpretation. Holding the library fixed isolates a change in the improvement procedure. The combined descendant may improve for a different reason.
Boundary. Use comparable starting solvers, task laws, installed information, and budgets. A stronger base model or an extra library is another changed component.
Source: [S] §3; Counterexample 3.2 | manuscript pp. 3-4
04. One defect. Eight possible cells.
The procedure cannot see the defect index. It receives only the reports of the questions it is allowed to ask.
Interpretation. Retain an executable way to find a justified repair in a later episode. Do not install the next episode’s hidden answer.
Boundary. The binary sensors and repair meanings come from a declared finite model. The 88 query-table entries are supplied, not learned through executed calibration.
Source: [S] §4; §32 | manuscript pp. 4, 49-51
05. Two investigations, the same installed sensors
Equality-first testing and coordinate queries use the same declared language. Their paths have different depths, but their complete trees have the same size.
Interpretation. Equality-first: seven questions, then repair. Coordinate queries: three questions, then repair. Total costs are eight and four calls.
Boundary. Both full binary trees have eight repair leaves and seven query nodes: fifteen records. These are literal tree records, not bytes or minimal program sizes.
Source: [S] §4; Theorem 32.1 | manuscript pp. 4, 49-50
06. Follow one episode: the defect is at cell 5
This page follows the canonical coordinate policy. The procedure knows the reports, not the hidden index shown to the reader.
Interpretation. The three reports select a repair for this episode. The reusable artifact is the branching procedure that works for every allowed starting cell.
Boundary. The shipped compiler can use an equality query at the last two-case branch. It has the same cost. This plate deliberately shows the canonical all-coordinate tree.
Source: [S] §4; §32; Appendix A | manuscript pp. 4, 49-54
07. The path is not the stored program
A single successful run touches four records. Its complete reusable controller still contains all fifteen.
Interpretation. Execution follows one branch. Literal storage sums every branch. The highlighted path for reports 1, 0, 1 reaches repair 5.
Boundary. Immutable cases, binary observations, a different successful repair for each case, and one record per node. Shared code or state-reset operations change the model.
Source: [S] §6; Theorem 11.1; §28 | manuscript pp. 6-7, 14-15, 44
08. A residual is a question, not a diagnosis
The same mismatch can be consistent with several causes. A useful next experiment separates repair-relevant explanations.
Interpretation. Retain competing explanations. Seek an admitted observation whose possible outcomes change which correction is justified.
Boundary. A representation cannot recover a distinction unavailable in its complete transcript. With disjoint repairs and identical transcript laws, equal-prior success is at most one half.
Source: [S] §5; Proposition 5.1 | manuscript pp. 5-6
09. The best question need not identify the most
A repair may work for several hidden cases. The right stopping condition is a common successful action, not complete knowledge of the state.
Interpretation. The 2/2 question has more information about four equally likely cases, but one branch still mixes incompatible repairs.
Boundary. Both questions cost one unit and preserve the state. The 3/1 question supports a correct repair after one observation in every case.
Source: [S] Counterexample 6.3; §5.1 | manuscript pp. 6-8
10. One tree must satisfy both resource limits
The synthesis combines complete child policies. Worst-case execution takes a maximum. Stored controller size takes a sum.
Interpretation. This illustrative root has cost 1. Its children cost (2 calls, 3 records) and (4 calls, 7 records). The combined tree costs (5 calls, 11 records).
Boundary. This is an illustration of the recurrence, not a new optimality result. A real child must be valid on the actual post-experiment support, and every possible branch needs a witness.
Source: [S] Theorem 6.2 | manuscript pp. 6-7
11. An unknown sensor needs a reference
A correct procedure for each known model need not give a common procedure when the model itself is unknown.
Interpretation. The report o = x ⊕ θ cannot distinguish (x,θ) = (0,0) from (1,1). They require different repairs. Repeating that same reading adds no information.
Boundary. A reference reports θ at cost 2. An ordinary query costs 1, then repair costs 1. Total work is 4, and the complete binary controller stores 7 records.
Source: [S] Theorem 7.1; Counterexample 7.2 | manuscript pp. 8-10
12. Transport the investigation, not only the repair
A translated answer can be correct while the observation needed to obtain it is unavailable in the target system.
Interpretation. Experiment permissions, report labels, state effects, successful-repair meanings, and costs. Correspondences are supplied in this finite theorem.
Boundary. Copying the two possible repairs into a world with no bit-revealing observation does not transport the policy that chooses between them.
Source: [S] Theorem 9.1; Counterexample 9.2 | manuscript pp. 12-13
13. Now improve the procedure that builds the tree
A checked candidate can end an exhaustive search only when it is valid and attains the relevant global lower bounds.
Interpretation. For n immutable cases with binary unit-cost queries and singleton repairs, a valid tree attaining both 1 + ceil(log₂ n) calls and 2n − 1 records closes the frontier.
Boundary. An invalid scope, missing candidate, or unattained bound sends control to the exact parent. That preliminary work still counts.
Source: [S] Theorem 11.1; §11.2 | manuscript pp. 14-15
14. What the shortcut actually reads
On the sixteen-case model, both compilers return an optimal controller with 5 execution calls and 31 stored records.
Interpretation. The recorded compilation count is 576,128 reads for the parent and 1,728 for the shortcut. Both include completed witness checking.
Boundary. Each also pays 320 common input-validation reads. Selection among all candidates costs 21,124 development reads. Table accesses are not total CPU instructions.
Source: [S] Proposition 12.1; §13.3 | manuscript pp. 16, 19
15. A fallback preserves meaning, not free speed
A preliminary shortcut can fail to prove an optimum and then spend extra work before invoking the same exact parent.
Interpretation. With four cells and only singleton queries, the shortcut falls back. Both return the same optimum. The wrapper uses 472 reads instead of 402.
Boundary. A successful witness, a proved finite-horizon impossibility, and an interrupted search are different results. A timeout cannot certify infeasibility.
Source: [S] Counterexample 13.1; §16.2 | manuscript pp. 18, 24-25
16. Sometimes the search state is too coarse
No longer formula can recover a feature the learner was never given. Here, the missing distinction is remaining branch capacity.
Interpretation. Both contexts have (a,b,m) = (0,3,4). With capacity 2 the root must fail. With capacity 4, a completion with branch sizes (1,3) is possible.
Boundary. Capacity is exposed from a supplied feature registry. This repairs an internal search representation, not a physical sensor model or the definition of depth.
Source: [S] Proposition 14.1; Counterexample 14.2 | manuscript pp. 20-21
17. Four proposals, three counterexamples
The learner composes an arithmetic predicate. Retained counterexamples prevent it from proposing the same rejected behavior again.
Interpretation. The final five-node expression is max(a,b) > C. The reference learner visits 190 of 3,960 candidates and makes four teacher calls.
Boundary. The teacher checks 198 developmental contexts. That certifies this finite agreement, not validity at every larger capacity. Installation needs a separate scope argument.
Source: [S] Theorem 15.1; §15.2; rule trace | manuscript pp. 21-23
18. Stop reading when no suffix can rescue the root
Eight cases must fit into two branches of capacity four. After five equal reports, the outcome is already decided.
Interpretation. Counts cannot decrease. Once one count exceeds C, no unread report can bring that branch back within capacity. The suffix cannot change rejection.
Boundary. Passing the capacity test does not prove that the remaining query library can separate either child. The compiler must still search and verify complete subtrees.
Source: [S] Proposition 16.1; Theorem 16.3 | manuscript pp. 23-25
19. A training-perfect rule can prune the answer
A dangerous shortcut is not made safe by a final answer checker: it may delete the only successful candidate before that checker sees it.
Interpretation. At (a,b,m,C) = (0,5,16,8), the bounded-cap formula rejects. But the unread cases can complete both branches to eight, so capacity remains feasible.
Boundary. Examples help select a formula. A proof with applicable premises authorizes deleting possibilities. The supplied structural gate accepts only the supported rule form.
Source: [S] Counterexample 16.2 | manuscript pp. 23-24
20. Fewer model reads can still mean more work
The same ordered compiler returns the same literal controller. Streaming changes when impossible roots are abandoned.
Interpretation. At 64 cases, streaming avoids 8,287 model rows. The returned controller still costs 7 online calls and 127 stored records.
Boundary. Streaming also adds 14,209 comparisons in the conservative account. At equal prices for reads and comparisons, that isolated modification costs more.
Source: [S] §18; Theorem 16.3 | manuscript pp. 24-28
21. Count repair distinctions, not hidden states
If one repair fixes a pair of cases, a diagnostic policy need not identify which member of the pair occurred.
Interpretation. Put nonnegative weights on cases so every repair-success set has total at most 1. A successful leaf then carries at most 1 unit. Binary depth h supports at most 2ʰ leaves.
Boundary. Adding a universal repair makes the old total weight of 4 violate its success-set constraint. Recheck the weights before using the old pruning rule.
Source: [S] Propositions 17.1, 22.1; Theorem 22.2 | manuscript pp. 25, 33-35
22. A useful equation begins as a proposal
The next acquired object is a task-independent library of equations. Evaluations suggest candidates; derivations decide which can be used.
Interpretation. The finite proposer constructs 1,068 equations from equal probe vectors. The limited proof procedure admits 336 with replayable arithmetic derivations.
Boundary. The remaining 732 may be false, valid only on a narrower scope, or simply beyond this proof search. The primitive arithmetic axioms and checking machinery are supplied.
Source: [S] §§19-20 | manuscript pp. 29-30
23. Keep the earliest checked representative
Equivalent expressions can share a search representative without changing the original program submitted to the teacher.
Interpretation. The exact proposed program trees, retained counterexamples, final result, and exhaustion outcome. Removed later equivalents would already fail a retained counterexample.
Boundary. Pure total arithmetic, sound equations, fixed order, an extensional deterministic teacher, and no forgotten counterexamples. It does not preserve every possible cost-sensitive objective.
Source: [S] Theorems 21.1-21.2 | manuscript pp. 30-32
24. Less repetition. The same corrective evidence.
The quotient changes which redundant candidates are evaluated. It does not change the task teacher or smuggle in the target rule.
Interpretation. Both learners begin the new task with no task-specific counterexamples. They submit the same four programs and receive the same three counterexamples.
Boundary. Equation discovery, proof admission, quotient preparation, dispatch, and storage must be charged separately. The ten targets are finite tasks in one arithmetic language.
Source: [S] §21.1; Theorem 21.2 | manuscript pp. 30-32
25. One chain, not multiplied headline gains
The representation that speeds up rule acquisition feeds a separately certified compiler change. Each link keeps its own comparison.
Interpretation. If the quotient preserves rule acquisition and the acquired rule passes the unchanged compiler gate, the complete chain returns the same rule and controller.
Boundary. Let K be extra preparation. The composite saves work in a common additive unit only if (A − A′) + (D − D′) > K. Do not multiply ratios from different baselines or units.
Source: [S] Proposition 23.1; §25 | manuscript pp. 35-40
26. A certificate has a dependency address
When a premise changes, mark the affected claims and their descendants. Reuse unaffected witnesses only under a complete dependency declaration.
Interpretation. Arithmetic equations and a repair-weight witness belong to different branches. A change in one branch need not invalidate everything.
Boundary. This is not automatic discovery of every dependency in arbitrary code. An omitted premise can leave a registry green while its claim becomes false. Hashes do not repair missing semantics.
Source: [S] Theorem 24.1; §24.2 | manuscript pp. 37-38
27. A stronger repair can invalidate an old proof
The conclusion “this controller is valid” and the conclusion “this change saves work” do not automatically survive together.
Interpretation. For the 32-case paired model, adding a universal repair invalidates the old weight constraints. Rebuilt weights permit a one-step, one-record controller.
Boundary. The earlier assessment referred to another controller and compilation. The replacement needs a new comparison. The unaffected arithmetic equation library remains applicable.
Source: [S] Counterexample 22.3; §24.3 | manuscript pp. 35, 38-39
28. The same controller, opposite cost conclusions
Before claiming improvement, choose a common ledger and state the prices used to combine its event counts.
Interpretation. These are old-minus-new search events on the same 32-case weighted problem. Scope and witness checks are equal and cancel only in this isolated difference.
Boundary. Weight preparation, equation acquisition, installed programs, storage, and revision management do not become free. Event counts alone do not determine processor time or energy.
Source: [S] Proposition 25.1; Counterexample 25.2 | manuscript pp. 39-40
29. Do not turn selection evidence into a fresh audit
What the improver has already inspected belongs to its developmental history, even when the inner solver never saw it.
Interpretation. A proof supports a conditional statement. A checker validates a finite witness. A fresh performance audit supports a comparison on a declared task law.
Boundary. Repeatedly reading a “holdout” exposes it to the outer loop. A correct proof does not establish its physical premises. Finite algorithm counters are not independent neural trials.
Source: [S] §26; Proposition 26.1 | manuscript pp. 41-42
30. An audit can stop early without betting in hindsight
Choose the evidence update before seeing its paired score. Stopping after observing evidence is a different operation.
Interpretation. Freeze the comparison, choose the next bet from past information, observe the next paired score, then update the evidence process.
Boundary. With four fair ± scores, betting only after seeing a positive doubles the evidence on positives. The nominal threshold 4 is crossed on 11 of 16 paths, not at most 1/4.
Source: [S] Theorem 27.1; Counterexample 27.2 | manuscript pp. 42-43
31. Safe repetition is not guaranteed progress
An authorized attempt must be selected before its outcome is known. One policy must meet the continuation requirement on every allowed branch.
Interpretation. Given fully observed lineage states, a finite attempt catalogue, and separately justified progress marks, the recurrence decides whether one policy guarantees k gains within h attempts.
Boundary. If every attempt can reject and return to the same safe state, any finite horizon admits an all-rejection path. Safety can hold without a bounded sure gain.
Source: [S] Theorem 29.1; Counterexample 29.2 | manuscript pp. 45-46
32. Positive gains need not accelerate
Recursion, continued improvement, and increasing productivity per unit resource are separate claims.
Interpretation. The gains 1/2, 1/4, 1/8, … are all positive. Their accumulated total remains below 1 after every finite number of steps.
Boundary. Faster improvement requires a claim about gain divided by complete discovery-and-acceptance cost. More headroom or a successful feedback loop does not establish it.
Source: [S] Proposition 30.1; §30 | manuscript pp. 46-47
33. The learned semantic bridge is still a research target
A commuting square specifies meaningful reuse. It does not discover the representation, learn the operations, or demonstrate their use by a neural system.
Interpretation. If the domain is invariant, every admitted operation commutes with q, and q preserves the required output, the relation extends to every finite operation sequence.
Boundary. Learning encoders, operations, experiment selectors, and proposers is a separate problem. The complete learned Neuralese–OSL bridge is not implemented in this edition.
Source: [S] §31; Proposition 31.1; §34 | manuscript pp. 48-49, 52-53
34. Locate the recursive achievement exactly
The finite chain acquires objects at several levels. It does not erase the supplied foundations of the next outer level.
Interpretation. A generated equation library changes the search representation used to acquire later discovery rules, while preserving the specified evidence and final programs.
Boundary. General intelligence, learned physical semantics, autonomous invention of the primitive proof system, and indefinite improvement of the equation learner remain open.
Source: [S] First-edition scope; §§23, 34 | manuscript pp. iii, 35-36, 52-53
35. A small vocabulary for reading the proofs
Use these distinctions to avoid turning several different achievements into one word such as “intelligence” or “success.”
Interpretation. The repair changes a target. The compiler discovers a policy. The learner can change a compiler rule. The equation library can change the learner’s candidate representation.
Boundary. A witness is checked under a model. A proof has hypotheses. A resource claim has a ledger. A generalization claim needs evidence beyond semantic relabeling.
Source: [S] Definitions 3.1, 6.1; §§5, 24-25 | manuscript pp. 3, 5-7, 37-40
36. Return to the source at the right level
This companion prioritizes mechanisms and counterexamples. The first-edition manuscript is the authority for complete hypotheses and proofs.
Interpretation. All substantive claims are restatements or worked illustrations of [S]. Source statistics are attributed to the supplied first-edition records, not presented as new learning runs.
Boundary. The guide’s examples and numerical transcriptions were checked separately. The full original 336,867-assertion suite was not rerun to make this visual companion.
Source: [S] Komorebi P., Self-Improving Intelligence, v1.0.0, 8 October 2026.