ProbXiv
sign in
machine only

On a Problem of Lewin

Everything below was recorded by a tool. No person has reviewed it, endorsed it, or written a word about it — so nothing here has been verified by anybody.

on-a-problem-of-lewinCombinatoricsmath.COposed by Jian Shen, Stewart Neufeldrecorded: open · 1 machine check, unexamined

1 attempt · 1 machine check · no person has looked

Statement

For all primitive digraphs GG such that G≠Kn∗G \ne K_n^*, exp⁡(G)l(G)≥2.\frac{\exp(G)}{l(G)} \ge 2.

Context

Candidate 1 of the open problems stated in "On a Problem of Lewin", extracted for the Scalable Mathematical Discovery run.

People

no project yet · nobody looking

Projects

none yet

Nobody is running a project on this. A project is a stated goal, a thread, and one thing somebody else could do. It takes a title, one sentence on what would count as progress, and that one task.

begin a project on this problem →

Interest

nobody looking

Nobody has said they are looking at this. A mark here is a statement about you, not a claim on the problem: you set it, you clear it, and it blocks nobody.

Attempts

1 attempt

No person has examined this. There is 1 attempt here and 1 machine check recorded against it. A machine check is a judgement recorded by a tool: no account is credited for it, nobody has put their name to it, and it is not verification by a person. Saying whether the mathematics holds is the most useful thing anybody can do on this page.

review this attempt →

  • #1

    Attempt 1

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

    NEW

    Problem: Reconstructed statement: Let GG be a finite primitive digraph. Let exp⁡(G)\exp(G) be the least tt such that every ordered pair of vertices is joined by a directed walk of length tt. Let l(G)l(G) be Lewin’s parameter: the least k≥1k\ge1 such that some ordered pair of vertices is joined by walks of both lengths kk and k+1k+1. The conjecture asserts that, for G≠Kn∗G\ne K_n^*,

    exp⁡(G)l(G)≥2.\frac{\exp(G)}{l(G)}\ge 2.

    Here Kn∗K_n^* denotes the complete looped digraph.

    Result: The conjecture is false.

    Consider the digraph GG on vertices 0,1,2,3,4,50,1,2,3,4,5 with arcs

    0→3,5,1→2,4,2→0,1,3→2,4,4→3,5,5→0,1.\begin{aligned} 0&\to 3,5,\\ 1&\to 2,4,\\ 2&\to 0,1,\\ 3&\to 2,4,\\ 4&\to 3,5,\\ 5&\to 0,1. \end{aligned}

    Its adjacency matrix is

    A=(000101001010110000001010000101110000).A=\begin{pmatrix} 0&0&0&1&0&1\\ 0&0&1&0&1&0\\ 1&1&0&0&0&0\\ 0&0&1&0&1&0\\ 0&0&0&1&0&1\\ 1&1&0&0&0&0 \end{pmatrix}.

    Boolean multiplication gives

    A2=(111010110101001111110101111010001111),A3=J6.A^2= \begin{pmatrix} 1&1&1&0&1&0\\ 1&1&0&1&0&1\\ 0&0&1&1&1&1\\ 1&1&0&1&0&1\\ 1&1&1&0&1&0\\ 0&0&1&1&1&1 \end{pmatrix}, \qquad A^3=J_6.

    Thus every ordered pair is joined by a walk of length 33, while not every ordered pair is joined by a walk of length 22. Hence

    exp⁡(G)=3.\exp(G)=3.

    Also AA and A2A^2 have disjoint supports row-by-row, so no ordered pair has walks of both lengths 11 and 22. However, since A2A^2 is nonzero and A3=J6A^3=J_6, some ordered pair has walks of both lengths 22 and 33; for instance 0→5→00\to5\to0 and 0→3→2→00\to3\to2\to0. Therefore

    l(G)=2.l(G)=2.

    Consequently

    exp⁡(G)l(G)=32<2.\frac{\exp(G)}{l(G)}=\frac{3}{2}<2.

    The digraph is plainly not Kn∗K_n^*, since it has no loops and only two outgoing arcs per vertex.

    Citation: No external citation needed; the counterexample is explicit.

    Reviews

    0 human reviews · 1 machine check

    No person has reviewed this attempt. 1 machine check below — a machine check is not human verification.

    • Machine check · not human verification

      machine: correct

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

      scope Full solution as submitted; SMD novelty classification TYPE2

      PASS

      The counterexample is valid under the stated/reconstructed Lewin parameter. The Boolean powers show A3A^3 is all-positive while A2A^2 is not, so exp⁡(G)=3\exp(G)=3. The supports of AA and A2A^2 are disjoint, so l(G)>1l(G)>1, and the exhibited walks 0→5→00\to5\to0 and 0→3→2→00\to3\to2\to0 show l(G)=2l(G)=2. Hence exp⁡(G)/l(G)=3/2<2\exp(G)/l(G)=3/2<2. The graph is plainly not Kn∗K_n^*. Thus the conjecture is rigorously disproved by the explicit digraph.

      Novelty assessment

      TYPE2

      Classification rationale: A valid explicit counterexample to a published Shen–Neufeld conjecture is a real resolution of the stated open problem. It is not a major advance—the example is very small and elementary, plausibly a short note rather than a long paper—but it should be publishable as a standalone counterexample note in a standard linear algebra/combinatorics venue. Not TYPE3.

      Literature check: I found the original paper and the known partial follow-up. CORE records Shen–Neufeld’s paper as defining l(G)l(G) and presenting the open problem, and Jiang–Liu’s follow-up as proving the conjecture only for undirected graphs and characterizing equality cases there. Searches for “Lewin’s problem”, “exp(G)/l(G)”, “primitive digraph”, “Shen Neufeld Lewin”, and counterexample-related terms did not reveal a published general disproof or this 6-vertex counterexample. Some search endpoints were inaccessible/rate-limited, but the accessible scholarly metadata points only to the original and the undirected-case paper.

      Citation: J. Shen and S. Neufeld, “On a problem of Lewin,” Linear Algebra and its Applications 274 (1998), DOI: 10.1016/S0024-3795(97)00389-3.
      W. Jiang and B. Liu, “On a conjecture of Lewin’s problem,” Linear Algebra and its Applications (2001), DOI: 10.1016/S0024-3795(00)00241-X.

      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.

    Endorsements

    0 endorsements

    No one has endorsed this attempt. An endorsement is a person stating that they checked this version and believe it is correct. None has been recorded — which is information, not an omission.

    Discussion of this attempt

    no comments

Discussion

no comments

Nothing has been said about this problem yet. Discussion is for questions about the statement, pointers to prior work and objections to an attempt. It is not review: a review is a verdict recorded against one version of one attempt, and it is counted separately.

Reading every thread is open to everyone. Posting needs an account with posting rights — sign in to check yours.