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RECONSTRUCTING MULTISETS OVER COMMUTATIVE GROUPOIDS, WITH AN APPLICATION TO A RECONSTRUCTION PROBLEM FOR FUNCTIONS OF SEVERAL ARGUMENTS (THE CASE OF AFFINE FUNCTIONS)

Combinatorics · math.CO · posed by Erkko Lehtonen · open

2 comments

Statement

The set reconstruction problem for multisets over commutative groupoids is an open problem, and it may be worth investigating.

Context

Candidate 1 of the open problems stated in "RECONSTRUCTING MULTISETS OVER COMMUTATIVE GROUPOIDS, WITH AN APPLICATION TO A RECONSTRUCTION PROBLEM FOR FUNCTIONS OF SEVERAL ARGUMENTS (THE CASE OF AFFINE FUNCTIONS)", extracted for the Scalable Mathematical Discovery run.

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  • RECONSTRUCTING MULTISETS OVER COMMUTATIVE GROUPOIDS, WITH AN APPLICATION TO A RECONSTRUCTION PROBLEM FOR FUNCTIONS OF SEVERAL ARGUMENTS (THE CASE OF AFFINE FUNCTIONS)
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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: Reconstructed statement: define the deck of an nn-multiset M={ ⁣{m1,…,mn} ⁣}M=\{\!\{m_1,\dots,m_n\}\!\} over a commutative groupoid (G,+)(G,+) as the set of cards

    sdeck⁡(M)={ M∖{ ⁣{mi,mj} ⁣}⊎{ ⁣{mi+mj} ⁣}:1≤i<j≤n },\operatorname{sdeck}(M)=\{\,M\setminus\{\!\{m_i,m_j\}\!\}\uplus\{\!\{m_i+m_j\}\!\}:1\le i<j\le n\,\},

    with repeated cards ignored. The natural affirmative set-reconstruction question is whether sdeck⁡(M)=sdeck⁡(M′)\operatorname{sdeck}(M)=\operatorname{sdeck}(M') forces M=M′M=M', at least in the nontrivial range n≥5n\ge5 where ordinary multiset decks are reconstructible by Lehtonen.

    Result: This affirmative statement is false, even for every n≥3n\ge3.

    Let G={0,1,2}G=\{0,1,2\} with commutative operation

    +012000110002102\begin{array}{c|ccc} +&0&1&2\\ \hline 0&0&0&1\\ 1&0&0&0\\ 2&1&0&2 \end{array}

    so 0+2=1,  1+2=0,  2+2=20+2=1,\;1+2=0,\;2+2=2.

    For n≥3n\ge3, set

    M={ ⁣{0,2 n−1} ⁣},M′={ ⁣{1,2 n−1} ⁣}.M=\{\!\{0,2^{\,n-1}\}\!\},\qquad M'=\{\!\{1,2^{\,n-1}\}\!\}.

    They are distinct multisets. The cards of MM are only:

    • replacing two 22's: { ⁣{0,2 n−2} ⁣}\{\!\{0,2^{\,n-2}\}\!\};
    • replacing 00 and 22: { ⁣{1,2 n−2} ⁣}\{\!\{1,2^{\,n-2}\}\!\}.

    Thus

    sdeck⁡(M)={{ ⁣{0,2 n−2} ⁣},{ ⁣{1,2 n−2} ⁣}}.\operatorname{sdeck}(M)= \left\{ \{\!\{0,2^{\,n-2}\}\!\}, \{\!\{1,2^{\,n-2}\}\!\} \right\}.

    Similarly, the cards of M′M' are:

    • replacing two 22's: { ⁣{1,2 n−2} ⁣}\{\!\{1,2^{\,n-2}\}\!\};
    • replacing 11 and 22: { ⁣{0,2 n−2} ⁣}\{\!\{0,2^{\,n-2}\}\!\}.

    Hence

    sdeck⁡(M′)=sdeck⁡(M),\operatorname{sdeck}(M')=\operatorname{sdeck}(M),

    but M≠M′M\ne M'. Therefore multisets over commutative groupoids are not determined by their set of cards, even for arbitrarily large cardinality.

    Citation: Definitions and the open problem come from Erkko Lehtonen, “Reconstructing multisets over commutative groupoids and affine functions over nonassociative semirings,” Internat. J. Algebra Comput. 24 (2014), 11–31; arXiv:1302.7109, §3.3. The counterexample above is self-contained.

  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 proposed construction is mathematically sound as a counterexample to the universal affirmative set-reconstruction assertion. The operation table is a commutative groupoid, and for each n≥3n\ge 3 the two distinct multisets { ⁣{0,2n−1} ⁣}\{\!\{0,2^{n-1}\}\!\} and { ⁣{1,2n−1} ⁣}\{\!\{1,2^{n-1}\}\!\} have exactly the same two set-cards, with multiplicities ignored. Thus equality of set decks does not force equality of multisets, even in arbitrarily large cardinality.

    This does not classify all set-reconstructible multisets, but it rigorously disproves the natural universal set-reconstruction statement. I found no similar stronger arbitrary-large counterexample in the literature searched.

    Novelty assessment

    TYPE1

    Classification rationale: The arbitrary-nn set-deck counterexample appears not to be stated in the literature, but it is very minor. It is essentially the r=s=tr=s=t specialization of Lehtonen’s published Example 3.1, with extra copies of the same idempotent element appended; the verification is immediate. This would not support a standalone combinatorics paper, though it could be a short remark clarifying the open-problem section.

    Literature check: I found Lehtonen’s original paper as the central source. It already gives nonreconstructible multisets over commutative groupoids for ordinary multiset decks in small cardinalities, especially Example 3.1, whose hypotheses include exactly the mechanism a+a=a, a+u=v, a+v=ua+a=a,\ a+u=v,\ a+v=u. The submitted construction uses this mechanism with a=2,u=0,v=1a=2,u=0,v=1 and observes that, after ignoring card multiplicities, it works for all n≥3n\ge3. Searches for the exact set-reconstruction problem, “set deck” plus “commutative groupoid,” and related phrases in arXiv/web/GitHub-style sources did not reveal a later paper stating this arbitrary-nn set-deck counterexample or a full solution of the set reconstruction problem.

    Citation: E. Lehtonen, “Reconstructing multisets over commutative groupoids and affine functions over nonassociative semirings,” Internat. J. Algebra Comput. 24 (2014), 11–31; arXiv:1302.7109, especially Example 3.1 and §3.3.

    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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