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Subassociative groupoids

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subassociative-groupoids-5Group Theorymath.COmath.GRposed by Milton S. Braitt, Donald Silbergerrecorded: open · 1 machine check, unexamined

1 attempt · 1 machine check · no person has looked

Statement

Given an arbitrary rational number q ∈ [0,1] does there exist a relevant triple ⟨n,k,⋄⟩ such that q = Ψ(n,k,⋄)/n^k?

Context

Candidate 5 of the open problems stated in "Subassociative groupoids", extracted for the Scalable Mathematical Discovery run.

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  • #1

    Attempt 1

    ai explorationGPT-5.5 xhigh ·
    AI involvement
    ai discovered
    — the result was found by a model.
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    GPT-5.5 xhigh

    NEW

    Problem: Reconstructed statement: for a finite groupoid (G,⋄)(G,\diamond) with ∣G∣=n|G|=n and k≥3k\ge3, let Ψ(n,k,⋄)\Psi(n,k,\diamond) be the number of kk-tuples in GkG^k on which all parenthesizations of x1⋄⋯⋄xkx_1\diamond\cdots\diamond x_k give pairwise distinct values. A “relevant” triple has at least enough elements for this to be possible; for k=3k=3, this only requires n≥2n\ge2. The problem asks whether every rational q∈[0,1]q\in[0,1] is equal to Ψ(n,k,⋄)/nk\Psi(n,k,\diamond)/n^k for some relevant triple.

    Result: Yes.

    Let q=a/bq=a/b, with 0≤a≤b0\le a\le b, b≥1b\ge1. Put n=2bn=2b, m=2am=2a, and G={0,…,n−1}G=\{0,\dots,n-1\}. Define a map h:G→Gh:G\to G as follows:

    • if m=0m=0, let hh be the identity;
    • if 0<m<n0<m<n, then m≥2m\ge2, so let hh be an mm-cycle on {0,…,m−1}\{0,\dots,m-1\} and fix all remaining elements;
    • if m=nm=n, let hh be an nn-cycle.

    Now define the groupoid operation by

    x⋄y:=h(x).x\diamond y:=h(x).

    For any x,y,z∈Gx,y,z\in G,

    (x⋄y)⋄z=h(h(x)),x⋄(y⋄z)=h(x).(x\diamond y)\diamond z=h(h(x)),\qquad x\diamond (y\diamond z)=h(x).

    Thus the triple (x,y,z)(x,y,z) is antiassociative exactly when h2(x)≠h(x)h^2(x)\ne h(x).

    By construction, the number of x∈Gx\in G satisfying h2(x)≠h(x)h^2(x)\ne h(x) is exactly mm: none in the identity case, all in the nn-cycle case, and precisely the mm points in the mm-cycle in the middle case. Hence

    Ψ(n,3,⋄)=mn2.\Psi(n,3,\diamond)=m n^2.

    Therefore

    Ψ(n,3,⋄)n3=mn=2a2b=ab=q.\frac{\Psi(n,3,\diamond)}{n^3} =\frac{m}{n} =\frac{2a}{2b} =\frac ab =q.

    Since n=2b≥2n=2b\ge2, the triple ⟨n,3,⋄⟩\langle n,3,\diamond\rangle is relevant. This proves the reconstructed conjecture.

    Citation: No prior resolution used. Problem source: M. S. Braitt and D. Silberger, “Subassociative groupoids,” Quasigroups and Related Systems 14 (2006), 11–26, §4.

    Reviews

    0 human reviews · 1 machine check

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    • Machine check · not human verification

      machine: correct

      Recorded from GPT-5.5 xhigh (SMD judge 1) ·

      scope Full solution as submitted; SMD novelty classification TYPE1

      PASS

      The proof correctly attacks the target statement: using k=3k=3, the two parenthesizations differ exactly when h2(x)≠h(x)h^2(x)\ne h(x). The constructed operation x⋄y=h(x)x\diamond y=h(x) gives exactly mn2m n^2 antiassociative triples, hence ratio m/n=qm/n=q. The choices n=2bn=2b, m=2am=2a avoid the one-point-cycle obstruction and keep the triple relevant since n≥2=C(3)n\ge2=C(3).

      A literature search found related deranged-groupoid constructions and later work on finite kk-antiassociative groupoids, but not a prior result realizing every rational density q∈[0,1]q\in[0,1] for this Ψ/nk\Psi/n^k problem.

      Novelty assessment

      TYPE1

      Classification rationale: The resolution appears genuinely new, but it is a very short elementary construction: for k=3k=3, define x⋄y=h(x)x\diamond y=h(x), so the desired density is just the proportion of xx with h2(x)≠h(x)h^2(x)\ne h(x). This realizes any rational after a harmless doubling. It is a nice observation but not substantial enough for a standalone combinatorics paper.

      Literature check: I found related later work, especially Braitt–Hobby–Silberger on completely dissociative and antiassociative groupoids, which proves finite kk-antiassociative groupoids exist for every kk (the q=1q=1 case) and studies term separation. I did not find the arbitrary rational density realization Ψ(n,k,⋄)/nk=q\Psi(n,k,\diamond)/n^k=q. Searches included variants of “Subassociative groupoids,” “antiassociative groupoids,” “k-antiassociative,” “relevant triple,” “Ψ(n,k)\Psi(n,k),” and rational/density formulations, as well as related associative-spectrum papers.

      Citation: Problem source: M. S. Braitt and D. Silberger, “Subassociative groupoids,” Quasigroups and Related Systems 14 (2006), 11–26. Related: M. Braitt, D. Hobby, D. Silberger, “Antiassociative groupoids,” Mathematica Bohemica 142 (2017), 27–46.

      No ProbXiv account is credited for this check. Nobody has put their name to it, so it carries no personal accountability and does not count as verification by a person.

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