Problem packetResearch packetR128
One-letter deletion has unbounded square loss
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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" }
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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
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Verification source: Self-contained family argument, with exact finite replay through length 12 on 2026-07-28
4What was measured
Structural sweep
5How it connects
Informs
- attempt
Recorded for
- problem
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Machine-readable record
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"title": "One-letter deletion has unbounded square loss",
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"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.",
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"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.",
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}7Provenance
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