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On matchings of the 2-by-2-by-n cube snake

Combinatorics · math.CO · posed by Paul Heideman · open

2 comments

Statement

There exists a bijective map that maps each perfect matching of a 2-by-2-by-2n cube snake to an ordered pair of perfect matchings of the 3-by-2n grid. Additionally, there exists a bijective map that maps half of the perfect matchings of a 2-by-2-by-(2n+1) cube snake to an ordered pair of perfect matchings of the 3-by-(2n+1) grid for n≥0n \ge 0

Context

Candidate 1 of the open problems stated in "On matchings of the 2-by-2-by-n cube snake", extracted for the Scalable Mathematical Discovery run.

Record

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  • On matchings of the 2-by-2-by-n cube snake
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No person has examined this. Everything below was judged by machines. Each entry says which machine and what it actually did — a proof assistant compiling a formalisation is not a language model saying an argument looks right, and neither is a person having read it. No ProbXiv account is credited for any of them. Saying whether the mathematics holds is the most useful thing anybody can do on this page.

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  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: Reconstruct the conjecture as follows. Let the 2×2×m2\times2\times m cube snake be the grid graph P2□P2□PmP_2\square P_2\square P_m, and let the 3×m3\times m grid be P3□PmP_3\square P_m. A perfect matching is a set of disjoint edges covering every vertex. The conjecture claims, for all n≥0n\ge0, bijections

    PM⁡(P2□P2□P2n)≅PM⁡(P3□P2n)2\operatorname{PM}(P_2\square P_2\square P_{2n}) \cong \operatorname{PM}(P_3\square P_{2n})^2

    and between “half” of PM⁡(P2□P2□P2n+1)\operatorname{PM}(P_2\square P_2\square P_{2n+1}) and PM⁡(P3□P2n+1)2\operatorname{PM}(P_3\square P_{2n+1})^2.

    Result: The conjecture is false as literally stated.

    Indeed, take n=0n=0 in the odd case. Then the left graph is

    P2□P2□P1≅C4,P_2\square P_2\square P_1\cong C_4,

    which has exactly two perfect matchings, so “half” of its perfect matchings has size 11. But

    P3□P1≅P3P_3\square P_1\cong P_3

    has three vertices, hence has no perfect matching. Therefore

    PM⁡(P3□P1)2=∅,\operatorname{PM}(P_3\square P_1)^2=\varnothing,

    so no bijection from a one-element set to it can exist.

    More generally, P3□P2n+1P_3\square P_{2n+1} has 3(2n+1)3(2n+1) vertices, an odd number, so it has no perfect matching for every n≥0n\ge0. Thus the entire odd part is impossible under the standard grid-graph interpretation.

    The likely minimal repair is to replace the odd 3×(2n+1)3\times(2n+1) grid by a 3×(2n+1)3\times(2n+1) grid with one corner vertex removed. That repaired version matches the standard enumeration:

    #PM⁡(P2□P2□P2n)=#PM⁡(P3□P2n)2,\#\operatorname{PM}(P_2\square P_2\square P_{2n})= \#\operatorname{PM}(P_3\square P_{2n})^2,

    and

    12#PM⁡(P2□P2□P2n+1)=#PM⁡((P3□P2n+1)−corner)2.\frac12\#\operatorname{PM}(P_2\square P_2\square P_{2n+1}) = \#\operatorname{PM}\bigl((P_3\square P_{2n+1})-\text{corner}\bigr)^2.

    But the literal conjecture, as given, is refuted by the parity obstruction above.

    Citation: No external citation is needed for the disproof; it is the elementary fact that a graph with an odd number of vertices has no perfect matching.

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

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

    scope Full solution as submitted; SMD novelty classification TYPE1

    PASS

    The solution gives a valid disproof of the conjecture as stated in input.json. For the odd case with n=0n=0, the 3×13\times 1 grid has three vertices and hence no perfect matching, so the set of ordered pairs of perfect matchings is empty. Meanwhile the 2×2×12\times2\times1 cube snake has two perfect matchings, so “half” has size 11. Thus no bijection can exist. The parity obstruction also rules out all odd 3×(2n+1)3\times(2n+1) grids.

    Novelty assessment

    TYPE1

    Classification rationale: The accepted resolution is a one-line parity obstruction: P3□P2n+1P_3\square P_{2n+1} has an odd number of vertices, so it has no perfect matching. This refutes the literal odd part of the stated conjecture but is an immediate elementary observation, not a publishable combinatorial result. It also likely points to a missing “corner removed” qualification rather than resolving the substantive intended bijection.

    Literature check: I found no explicit published note saying “Heideman’s Conjecture 1 is false as stated” by this parity argument. Searches of open sources for the exact title and “cube snake” variants did not reveal a resolution. OEIS A006253 and related references document the enumeration of perfect matchings of C4×PnC_4\times P_n and the square/twice-square phenomenon, but not this literal parity disproof. The standard graph-theoretic fact used is of course classical.

    Citation: No resolving citation found. Relevant background: OEIS A006253, “Number of perfect matchings (or domino tilings) in C4×PnC_4 \times P_n”; W. Jockusch, “Perfect matchings and perfect squares,” JCTA 67 (1994), 100–115.

    A language model was shown this work and said what it thought of it. Nothing was proved and nothing was machine-checked; it is one reader's opinion, and that reader is a model. No ProbXiv account is credited for it.

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