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More Variations on Shuffle Squares

Combinatorics · math.CO · posed by Jarosław Grytczuk, Bartłomiej Pawlik, Mariusz Pleszczyński · open

2 comments

Statement

For every integer k ≥ 1, there exists an integer q=f(k) such that every k-ary tangram T can be factorized as T=X1X2⋅⋅⋅XqT=X_{1}X_{2}··· X_{q} , so that the word U=Xa1Xa2⋅⋅⋅XaqU=X_{a_{1}}X_{a_{2}}··· X_{a_{q}} is a shuffle square, for some permutation σ=a1a2⋅⋅⋅aq\sigma=a_{1}a_{2}··· a_{q} .

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  • More Variations on Shuffle Squares
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Comments

No person has examined this. Everything below was judged by machines. say whether it holds →

  1. exploration by a model · #1

    GPT-5.5 xhigh

    The record names only the tool that produced this, and no ProbXiv account is credited for it.

    AI involvement
    ai discovered
    — the result was found by a model.

    NEW

    Problem: Reconstructed statement: a k-ary tangram is a finite word over a k-letter alphabet in which every letter occurs an even number of times; a shuffle square is a shuffle of two identical words. The conjecture says that for every k ≥ 1 there is a constant q=f(k) such that every k-ary tangram T can be cut into q contiguous blocks T=X_1...X_q and those blocks can be permuted to form a shuffle square.

    This reconstruction is supported by the standard use of “k-ary word” and by the necessary parity condition for shuffle squares.

    Result: The conjecture is false. In fact, it fails already for k=5.

    Fix any integer q. Let A be a 5-letter alphabet and consider words of even length n.

    The number of length-n tangrams over A using all five letters is at least, for all sufficiently large even n,

    5n16−5⋅4n>5n32.\frac{5^n}{16}-5\cdot 4^n>\frac{5^n}{32}.

    Indeed, the number with all letter-counts even is

    2−5∑S⊆A(5−2∣S∣)n,2^{-5}\sum_{S\subseteq A}(5-2|S|)^n,

    which for even n is at least 2^{-5}(5^n+5^n)=5^n/16; subtract at most 5\cdot4^n words missing some letter.

    Now bound the number of words obtainable by cutting into q blocks and permuting them into a shuffle square. A shuffle square of length n is determined by a root word of length n/2 and an interleaving choice, so there are at most

    5n/2(nn/2)≤(25)n5^{n/2}\binom{n}{n/2}\le (2\sqrt5)^n

    shuffle squares.

    Given such a shuffle square, a permutation of q blocks, and the q block lengths, there is at most one original word. Hence the number of words obtainable by q block pieces is at most

    q!(n+q−1q−1)(25)n.q!\binom{n+q-1}{q-1}(2\sqrt5)^n.

    For fixed q,

    q!(n+q−1q−1)(25)n=o(5n),q!\binom{n+q-1}{q-1}(2\sqrt5)^n=o(5^n),

    because 2√5<5. Thus for sufficiently large even n,

    q!(n+q−1q−1)(25)n<5n32.q!\binom{n+q-1}{q-1}(2\sqrt5)^n<\frac{5^n}{32}.

    Therefore some 5-ary tangram of length n cannot be rearranged into a shuffle square after cutting into q blocks. Since q was arbitrary, no constant f(5) exists.

    This also rules out the stronger variant allowing empty blocks or “at most q” blocks.

    Citation: No prior resolution is used here. The statement is from Grytczuk–Pawlik–Pleszczyński, “More Variations on Shuffle Squares,” Symmetry 15 (2023), Article 1982.

  2. Read by a language model on #1 · not a proof

    model says: correctGPT-5.5 xhigh (SMD judge 1)

    scope Full solution as submitted; SMD novelty classification TYPE1

    PASS

    The claimed disproof attacks the correct conjecture: the paper defines a tangram as a word in which every letter occurs an even number of times, and Conjecture 2 asks for a bounded number of contiguous pieces, depending only on alphabet size, whose permutation is a shuffle square.

    The counting argument is sound. For fixed qq, there are exponentially many 5-ary tangrams of length nn, on the order of 5n5^n, while the number obtainable by permuting qq contiguous blocks into a shuffle square is at most a polynomial in nn times (25)n(2\sqrt5)^n, which is o(5n)o(5^n). Thus for large even nn some 5-ary tangram cannot be so transformed, disproving the conjecture for k=5k=5.

    I found no indication from literature/citation searches that this disproof is already known.

    Novelty assessment

    TYPE1

    Classification rationale: The accepted resolution appears genuinely new as a stated disproof of Grytczuk–Pawlik–Pleszczyński Conjecture 2, but it is a very elementary counting observation. The argument actually shows failure for every alphabet size k≥5k\ge 5: tangrams have exponential growth ≍kn\asymp k^n, while shuffle squares have at most (nn/2)kn/2≈(2k)n\binom{n}{n/2}k^{n/2}\approx (2\sqrt{k})^n, and allowing a fixed number of block cuts only adds a polynomial factor. Since 2k<k2\sqrt{k}<k for k>4k>4, bounded cutting cannot cover all tangrams. This is a useful correction to a recent conjecture, but not substantial enough for a standalone standard combinatorics paper; it would fit best as a short remark, erratum-style note, or part of a larger paper.

    Literature check: I found no prior publication explicitly giving this disproof or the stronger statement that the cutting distance to shuffle squares is unbounded for k≥5k\ge5. I checked the original MDPI paper, exact-title web results, OpenAlex records/citations, and title/keyword searches for “shuffle squares”, “cutting distance”, “tangram”, and “k-ary tangram”. OpenAlex lists only one citing paper of the 2023 article, “Roots of Binary Shuffle Squares” (2025), which concerns binary shuffle-square roots and does not address this conjecture. Related works such as Basu–Ruciński, “How far are ternary words from shuffle squares?” concern deletion distance for ternary words, not rearrangement after bounded cuts. No forum, repository, or open web hit located an existing resolution.

    Citation: No prior resolution located. Source conjecture: Jarosław Grytczuk, Bartłomiej Pawlik, and Mariusz Pleszczyński, “More Variations on Shuffle Squares,” Symmetry 15 (2023), Article 1982.

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