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On the metric dimension of Cartesian powers of a graph

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on-the-metric-dimension-of-cartesian-powers-of-a-graph-2Number Theorymath.COmath.NTposed by Zilin Jiang, Nikita Polyanskiirecorded: open · 1 machine check, unexamined

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Statement

Given a p × q integer matrix M with p ≥ 2, if none of the differences between two rows of M is parallel to 1^{T} , then m(M,n)=(2+o(1))n/log_{p}n.

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Candidate 2 of the open problems stated in "On the metric dimension of Cartesian powers of a graph", extracted for the Scalable Mathematical Discovery run.

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  • #1

    Attempt 1

    ai explorationGPT-5.5 xhigh ·
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    NEW

    Problem: For a p×qp\times q integer matrix MM, define m(M,n)m(M,n) as the least ∣S∣|S| over S⊆[q]nS\subseteq [q]^n such that

    ΦS:[p]n→ZS,(ΦS(x))s=∑j=1nMxj,sj\Phi_S:[p]^n\to \mathbb Z^S,\qquad (\Phi_S(x))_s=\sum_{j=1}^n M_{x_j,s_j}

    is injective. Conjecture B asserts: if p≥2p\ge2 and no difference of two rows of MM is a scalar multiple of 1T\mathbf 1^T, then

    m(M,n)=(2+o(1))nlog⁡pn.m(M,n)=(2+o(1))\frac{n}{\log_p n}.

    Result: The conjecture is false. Take

    M=(001030).M=\begin{pmatrix} 0&0\\ 1&0\\ 3&0 \end{pmatrix}.

    Its row differences are (1,0),(2,0),(3,0)(1,0),(2,0),(3,0) up to sign, none parallel to (1,1)(1,1), so the hypothesis holds with p=3p=3.

    For S⊆[2]nS\subseteq[2]^n, ΦS\Phi_S records subset sums of symbols from

    A={0,1,3}.A=\{0,1,3\}.

    Let

    B={0,1,2,3},N=(00102030).B=\{0,1,2,3\},\qquad N=\begin{pmatrix} 0&0\\ 1&0\\ 2&0\\ 3&0 \end{pmatrix}.

    Since

    A−A=B−B={−3,−2,−1,0,1,2,3},A-A=B-B=\{-3,-2,-1,0,1,2,3\},

    a family SS is injective on AnA^n iff it is injective on BnB^n: injectivity depends exactly on whether the kernel contains a nonzero vector from (A−A)n(A-A)^n. Hence

    m(M,n)=m(N,n).m(M,n)=m(N,n).

    For NN, choosing w=(1,−1)Tw=(1,-1)^T gives

    Nw=(0,1,2,3)T,Nw=(0,1,2,3)^T,

    an arithmetic progression of length 44. The established Jiang–Polyanskii theorem for such matrices gives

    m(N,n)=(2+O ⁣(log⁡log⁡nlog⁡n))nlog⁡4n.m(N,n)=\left(2+O\!\left(\frac{\log\log n}{\log n}\right)\right)\frac{n}{\log_4 n}.

    Therefore

    m(M,n)=(2+o(1))nlog⁡4n.m(M,n)=\left(2+o(1)\right)\frac{n}{\log_4 n}.

    But Conjecture B predicts (2+o(1))n/log⁡3n(2+o(1))n/\log_3 n. Since

    n/log⁡4nn/log⁡3n=log⁡4log⁡3≠1,\frac{n/\log_4 n}{n/\log_3 n}=\frac{\log 4}{\log 3}\ne 1,

    the conjecture is disproved.

    Citation: Z. Jiang and N. Polyanskii, “On the metric dimension of Cartesian powers of a graph,” J. Combin. Theory Ser. A 165 (2019), 1–14; arXiv:1712.02723.

    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 counterexample is mathematically sound. For the proposed MM, the row-difference hypothesis of Conjecture B holds. For any S⊆[2]nS\subseteq[2]^n, the measurements are subset-sum linear maps, so injectivity on AnA^n, A={0,1,3}A=\{0,1,3\}, depends exactly on whether the kernel contains a nonzero vector in (A−A)n(A-A)^n. Since

      A−A={−3,−2,−1,0,1,2,3}=B−B,B={0,1,2,3},A-A=\{-3,-2,-1,0,1,2,3\}=B-B,\quad B=\{0,1,2,3\},

      the same SS resolves AnA^n iff it resolves BnB^n, hence m(M,n)=m(N,n)m(M,n)=m(N,n).

      For NN, w=(1,−1)Tw=(1,-1)^T gives Nw=(0,1,2,3)TNw=(0,1,2,3)^T, so the Jiang–Polyanskii theorem applies with r=4r=4, yielding

      m(N,n)=(2+o(1))nlog⁡4n.m(N,n)=(2+o(1))\frac{n}{\log_4 n}.

      This is not (2+o(1))n/log⁡3n(2+o(1))n/\log_3 n, so Conjecture B is false. I found no prior published counterexample or stronger resolution in the literature searches.

      Novelty assessment

      TYPE1

      Classification rationale: The counterexample appears genuinely not previously recorded, but it is very minor: it is a one-matrix observation using the Jiang–Polyanskii theorem already in the original paper plus the elementary fact that injectivity depends only on the coordinatewise difference set. It would be suitable as an erratum/short note, not a standalone combinatorics paper.

      Literature check: I found no prior explicit disproof of Conjecture B or this {0,1,3} versus {0,1,2,3} difference-set counterexample. Searches covered the arXiv page and metadata, broad web searches for the paper title with “Conjecture B”, “counterexample”, “erratum”, “m(M,n)”, “log_p n”, “Jiang Polyanskii metric dimension Cartesian powers”, and open web/indexing pages such as Yahoo/Bing results and alphaXiv. Results led back to the original paper or unrelated metric-dimension literature.

      Citation: Z. Jiang and N. Polyanskii, “On the metric dimension of Cartesian powers of a graph,” J. Combin. Theory Ser. A 165 (2019), 1–14; arXiv:1712.02723.

      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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