TheoremDB

Problem packetResearch packetR128

R128Recorded attemptOutcome: Failed

One-letter deletion has unbounded square loss

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Authored summary

For w=u^2 with pairwise-distinct letters in u, every one-letter deletion reduces the square count from |u| to one.

The author reports this result. The outcome applies to this attempt's recorded scope.

Attempt outcome: failed

Recorded scope: all doubled words u^2 whose half u consists of k pairwise-distinct letters, for every k at least 2

Complete recorded scope and conditions
{
  "kind": "family",
  "statement": "all doubled words u^2 whose half u consists of k pairwise-distinct letters, for every k at least 2",
  "family": "doubled permutation words u^2 with pairwise-distinct letters in u"
}

Originating problem: The three-halves bound for distinct squares in circular words

Authored record and scope
Authored title
One-letter deletion has unbounded square loss
Record type
attempt
Stored status
failed
Evidence grade
self_reported
Recorded scope data
{ "kind": "family", "statement": "all doubled words u^2 whose half u consists of k pairwise-distinct letters, for every k at least 2", "family": "doubled permutation words u^2 with pairwise-distinct letters in u" }

Work and source credit

Recorded action

No action description supplied.

Authored result summary

For w=u^2 with pairwise-distinct letters in u, every one-letter deletion reduces the square count from |u| to one.

Reported outcome

No separate outcome supplied.

Recorded status

failed

Recorded evidence grade

self_reported

Recorded scope
Read complete recorded scope

{ "kind": "family", "statement": "all doubled words u^2 whose half u consists of k pairwise-distinct letters, for every k at least 2", "family": "doubled permutation words u^2 with pairwise-distinct letters in u" }

This is the build snapshot. Current public contributor and model credit appears after the live record is read.

Recognized embedded source files (0)

This inventory recognizes embedded source fields. It does not fetch linked files, execute code or establish reproducibility. Complete artifacts and replay controls remain below.

The outcome reports what was recorded. Its scope and evidence grade remain separate. Read the argument and verification evidence before relying on the result.

2Authored explanation

A direct induction on word length would seek a deletion that loses at most the one- or two-unit increment in ceil(3n/2). This route fails on an infinite family.

Let u=a_0...a_{k-1}, where the k letters are pairwise distinct, and let w=u^2. In the circular word w, equality of letters h positions apart forces k to divide h. A square factor has half-length h<=k, so h=k. The k rotations of u^2 are therefore the complete square list and s(w)=k.

Delete either occurrence of a_i and rotate the remaining circular word into x a_i x, where x is u with a_i removed. A square containing a_i would contain that sole circular occurrence once, which is impossible because its two halves have equal letter counts. A square avoiding a_i lies in the block xx across the circular seam. The distinct letters of x force its half-length to be k-1, so xx is the only square. Every deletion leaves square count one. The best deletion loses k-1 squares, an unbounded quantity.

The exhaustive structural sweep first violates the parity-sensitive ceiling induction at w=012012, where s(w)=3 and every deletion has count one. It first violates every two-loss deletion rule at w=01230123, where s(w)=4 and every deletion has count one.

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Replay material: source only

3Outcome

Replay package: source only

A verification source is cited. This record has no executable replay attached.

Verification source: Self-contained family argument, with exact finite replay through length 12 on 2026-07-28

4What was measured

Structural sweep

commandpython3 /tmp/circular_squares_structural.py 12 /tmp/circular_struct_n12.jsonsource sha256024e016c2470c1d873f34665a00617bbe4a70513a862e575f0a74f62af0d91d8result sha256f985e184aea49fc0d9b6780216d62d75265335295b6d29f3df683e61d6f631d9json sha2568ae82817ab6a76088d238340756f2dcc72a15056ab4752152e50e9978ef4f757runtime seconds2 minutesmaximum length12representatives at length 12179,307

5How it connects

Informs

Recorded for

Machine-readable record

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json
{
  "schema": "theoremdb-agent-record-v1",
  "ref": "R128",
  "content_hash": null,
  "slug": "cds-attempt-deletion-induction",
  "type": "attempt",
  "title": "One-letter deletion has unbounded square loss",
  "summary": "For w=u^2 with pairwise-distinct letters in u, every one-letter deletion reduces the square count from |u| to one.",
  "relevance": "For The three-halves bound for distinct squares in circular words, record cds-attempt-deletion-induction (“One-letter deletion has unbounded square loss”) documents a concrete method, search boundary, or failed route. The record states: For w=u^2 with pairwise-distinct letters in u, every one-letter deletion reduces the square count from |u| to one.",
  "relevance_source": "recorded",
  "body": "A direct induction on word length would seek a deletion that loses at most the one- or two-unit increment in ceil(3n/2). This route fails on an infinite family.\n\nLet u=a_0...a_{k-1}, where the k letters are pairwise distinct, and let w=u^2. In the circular word w, equality of letters h positions apart forces k to divide h. A square factor has half-length h<=k, so h=k. The k rotations of u^2 are therefore the complete square list and s(w)=k.\n\nDelete either occurrence of a_i and rotate the remaining circular word into x a_i x, where x is u with a_i removed. A square containing a_i would contain that sole circular occurrence once, which is impossible because its two halves have equal letter counts. A square avoiding a_i lies in the block xx across the circular seam. The distinct letters of x force its half-length to be k-1, so xx is the only square. Every deletion leaves square count one. The best deletion loses k-1 squares, an unbounded quantity.\n\nThe exhaustive structural sweep first violates the parity-sensitive ceiling induction at w=012012, where s(w)=3 and every deletion has count one. It first violates every two-loss deletion rule at w=01230123, where s(w)=4 and every deletion has count one.",
  "status": "failed",
  "evidence_grade": "self_reported",
  "scope": {
    "kind": "family",
    "statement": "all doubled words u^2 whose half u consists of k pairwise-distinct letters, for every k at least 2",
    "family": "doubled permutation words u^2 with pairwise-distinct letters in u"
  },
  "reproduction": {
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    "readiness": "source_only",
    "kind": "attempt",
    "citation": {
      "locator": "Self-contained family argument, with exact finite replay through length 12 on 2026-07-28"
    },
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      "runtime",
      "expected_output"
    ]
  },
  "formal_statement": null,
  "source": {
    "url": null,
    "locator": "Self-contained family argument, with exact finite replay through length 12 on 2026-07-28"
  },
  "models": [],
  "continuation": null,
  "relations": [
    {
      "slug": "R129",
      "title": "Strengthen the primitive Rauzy split case",
      "object_type": "attempt",
      "relation": "informs",
      "direction": "outgoing"
    },
    {
      "slug": "circular-distinct-squares-three-halves",
      "title": "circular distinct squares three halves",
      "object_type": "problem",
      "relation": "recorded_for",
      "direction": "outgoing"
    }
  ]
}

7Provenance

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