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On coloring numbers of graph powers

Combinatorics · math.CO · posed by H. A. Kierstead, Daqing Yang, Junjun Yi · open

1 attempt · 1 machine check

Statement

What is, for a given integer k1k \ge 1 and any C (if k=1k = 1, then C1C \ge 1), the minimum m(C)m(C) such that any graph G with mad(G)2km(C)\text{mad}(G) \le 2k - m(C) satisfies χl(G2)kΔ(G)+C\chi_l(G^2) \le k\Delta(G) + C.

Context

Candidate 4 of the open problems stated in "On coloring numbers of graph powers", extracted for the Scalable Mathematical Discovery run.

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Attempts

1 attempt

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  • #1

    Attempt 1

    ai explorationGPT-5.5 xhigh ·
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    NEW

    Problem: For finite simple graphs, reconstruct Question 3.6 as asking whether, for each allowed pair (k,C)(k,C), there is a minimum real number m(C)m(C) such that every graph GG with

    mad(G)2km(C)\operatorname{mad}(G)\le 2k-m(C)

    satisfies

    χ(G2)kΔ(G)+C.\chi_\ell(G^2)\le k\Delta(G)+C.

    Here mad(G)=maxHG,H2E(H)/V(H)\operatorname{mad}(G)=\max_{H\subseteq G,\,H\ne\emptyset}2|E(H)|/|V(H)|, G2G^2 joins vertices at distance at most 22, and χ\chi_\ell is list chromatic number.

    Result: The literal question has no answer as stated: the requested minimum need not exist. Already for the allowed case k=1, C=1k=1,\ C=1, there is no minimum mm.

    Indeed, if m>0m>0, then mad(G)2m<2\operatorname{mad}(G)\le 2-m<2, so GG is a forest. For a forest TT, T2T^2 is chordal and has clique number Δ(T)+1\Delta(T)+1, hence

    χ(T2)=χ(T2)=ω(T2)=Δ(T)+1.\chi_\ell(T^2)=\chi(T^2)=\omega(T^2)=\Delta(T)+1.

    Thus every graph GG with mad(G)2m\operatorname{mad}(G)\le 2-m satisfies

    χ(G2)Δ(G)+1.\chi_\ell(G^2)\le \Delta(G)+1.

    So every m>0m>0 works.

    But m=0m=0 does not work: take G=C5G=C_5. Then mad(C5)=2\operatorname{mad}(C_5)=2, Δ(C5)=2\Delta(C_5)=2, and C52=K5C_5^2=K_5, so

    χ(C52)=5>3=Δ(C5)+1.\chi_\ell(C_5^2)=5>3=\Delta(C_5)+1.

    Hence the admissible set of mm's is (0,)(0,\infty), which has no minimum.

    The defect is an endpoint/minimum issue caused by using \le in the maximum-average-degree condition while asking for a minimum deficit. A natural repair is to ask for the infimum of such mm, or to use a strict inequality mad(G)<2km(C)\operatorname{mad}(G)<2k-m(C). For (k,C)=(1,1)(k,C)=(1,1), the repaired value is 00.

    Citation: No external citation is needed for the counterexample. The facts used are standard: forests have square chordal with clique number Δ+1\Delta+1, and C52=K5C_5^2=K_5.

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    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 KNOWN

      PASS

      The argument rigorously shows that for the allowed pair (k,C)=(1,1)(k,C)=(1,1), every m>0m>0 works, while m=0m=0 fails via C5C_5. Hence the admissible set is (0,)(0,\infty), which has no minimum. This correctly attacks the literal statement with “\le” in the mad condition.

      Novelty assessment

      KNOWN

      Classification rationale: The accepted resolution is an immediate endpoint reformulation of a known small-case threshold. Bonamy–Lévêque–Pinlou already state m(1)=m(2)=2m(1)=m(2)=2: C5C_5 gives the obstruction at mad=2\operatorname{mad}=2, while every graph with mad<2\operatorname{mad}<2 is a forest and has square list chromatic number Δ+1\Delta+1. Translating to Question 3.6 with k=C=1k=C=1, every m>0m>0 works and m=0m=0 fails, so no minimum exists.

      Literature check: I checked the Kierstead–Yang–Yi paper text around Questions 3.5–3.6 and the cited Bonamy–Lévêque–Pinlou paper. I also searched for the exact “Question 3.6” phrasing; I did not find a later explicit note about the minimum/endpoint issue, but the stronger underlying threshold statement and the exact proof ingredients are already in the cited literature.

      Citation: M. Bonamy, B. Lévêque, A. Pinlou, “List coloring the square of sparse graphs with large degree,” European Journal of Combinatorics 41 (2014), 128–137; see Conclusion, where they note m(1)=m(2)=2m(1)=m(2)=2 using C5C_5 and forests.

      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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