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Statement

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

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  • On coloring numbers of graph powers
  • FAR
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No person has examined this. Everything below was judged by machines. say whether it holds →

  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: 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)≤2k−m(C)\operatorname{mad}(G)\le 2k-m(C)

    satisfies

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

    Here mad⁡(G)=max⁡H⊆G, H≠∅2∣E(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 χℓ\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)≤2−m<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

    χℓ(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)≤2−m\operatorname{mad}(G)\le 2-m satisfies

    χℓ(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

    χℓ(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)<2k−m(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.

  2. Read by a language model on #1 · not a proof

    model says: correctGPT-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.

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