Problem packetWorkR128
[#R128] One-letter deletion has unbounded square loss
1Summary
For w=u^2 with pairwise-distinct letters in u, every one-letter deletion reduces the square count from |u| to one.
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.
Ruled out evidence. Recorded scope: all doubled words u^2 whose half u consists of k pairwise-distinct letters, for every k at least 2.
2Outcome
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
3Overview
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.
4What was measured
- Method that fails
- induction by deleting one circular position while charging a bounded number of lost distinct squares
- Failure boundary
- the best deletion loss is k-1 on doubled permutation words of length 2k
- Reusable residue
- Deletion arguments need a nonlocal potential that credits the whole conjugacy class of a full-length square; raw square-count loss cannot be bounded by a constant.
Structural sweep
5How it connects
Informs
- attempt
Recorded for
- problem
6Agent packet
A compact handoff with the evidence boundary, replay manifest, and relation pointers.
View structured packet
{
"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": {
"schema": "theoremdb-reproduction-v1",
"readiness": "source_only",
"kind": "attempt",
"citation": {
"locator": "Self-contained family argument, with exact finite replay through length 12 on 2026-07-28"
},
"missing": [
"source",
"command",
"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": [],
"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
View source, identifiers, and projection details
- Project
- circular-distinct-squares-three-halves-research
- Locator
- Self-contained family argument, with exact finite replay through length 12 on 2026-07-28
- License
- CC0-1.0
- Public record
- R128
- Stable alias
- cds-attempt-deletion-induction
- Projection
- Reproduction fields are derived from the immutable record.
A route someone took, recorded so the next person can reuse it or avoid it.