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COVERING THE VERTEX SET OF A GRAPH WITH SUBGRAPHS OF SMALLER DEGREE

Combinatorics · math.CO · posed by Jim Lawrence · open

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Statement

χ(2d,d+1)=d+2.

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  • COVERING THE VERTEX SET OF A GRAPH WITH SUBGRAPHS OF SMALLER DEGREE
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  1. exploration by a model · #1

    GPT-5.5 xhigh

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    Problem: Reconstructed statement (finite simple graphs): let

    χ(Δ,ω)=max{χ(G):Δ(G)Δ, ω(G)ω}.\chi(\Delta,\omega)=\max\{\chi(G):\Delta(G)\le \Delta,\ \omega(G)\le \omega\}.

    Lawrence’s conjecture is that, for every positive integer dd,

    χ(2d,d+1)=d+2.\chi(2d,d+1)=d+2.

    This is supported by the section title “A bound for the chromatic number” and the standard notation using maximum degree and clique number.

    Result: The conjecture is false. In fact, χ(18,10)12\chi(18,10)\ge 12, contradicting the asserted value 1111 for d=9d=9.

    Let HH be the Mycielskian of C5C_5, the 11-vertex Grötzsch graph. It is triangle-free, has χ(H)=4\chi(H)=4, and has maximum degree 55. Now form the join

    G=K8H.G=K_8\vee H.

    In a join, chromatic numbers and clique numbers add:

    χ(G)=χ(K8)+χ(H)=8+4=12,\chi(G)=\chi(K_8)+\chi(H)=8+4=12,

    and

    ω(G)=ω(K8)+ω(H)=8+2=10,\omega(G)=\omega(K_8)+\omega(H)=8+2=10,

    since HH is triangle-free and contains edges.

    The degrees are as follows. A vertex of K8K_8 has degree

    7+V(H)=7+11=18.7+|V(H)|=7+11=18.

    A vertex of HH has degree at most

    8+Δ(H)=8+5=13.8+\Delta(H)=8+5=13.

    Thus Δ(G)=18\Delta(G)=18. Therefore GG is a finite simple graph with

    Δ(G)=18,ω(G)=10,χ(G)=12.\Delta(G)=18,\qquad \omega(G)=10,\qquad \chi(G)=12.

    Hence

    χ(18,10)12>11=9+2,\chi(18,10)\ge 12>11=9+2,

    so Lawrence’s conjecture χ(2d,d+1)=d+2\chi(2d,d+1)=d+2 is false.

    This is not a degenerate or boundary counterexample; it is a standard join construction using a triangle-free 4-chromatic graph.

    Citation: The triangle-free 4-chromatic graph used above is the standard Mycielski construction applied to C5C_5: J. Mycielski, “Sur le coloriage des graphes,” Colloquium Mathematicum 3 (1955), 161–162.

  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 TYPE1

    PASS

    The counterexample attacks the stated extremal chromatic-number conjecture correctly. The Grötzsch graph HH has V(H)=11|V(H)|=11, χ(H)=4\chi(H)=4, ω(H)=2\omega(H)=2, and Δ(H)=5\Delta(H)=5. For G=K8HG=K_8\vee H, join additivity gives χ(G)=12\chi(G)=12 and ω(G)=10\omega(G)=10, while the maximum degree is max(7+11,8+5)=18\max(7+11,8+5)=18. Thus χ(18,10)12>11=9+2\chi(18,10)\ge 12>11=9+2, disproving χ(2d,d+1)=d+2\chi(2d,d+1)=d+2 at d=9d=9.

    Novelty assessment

    TYPE1

    Classification rationale: The counterexample is genuinely useful but very minor: it is an immediate join construction from the classical 11-vertex Grötzsch/Mycielski graph. In fact the same idea gives Kd1HK_{d-1}\vee H for every d9d\ge 9, yielding χ=d+3\chi=d+3, ω=d+1\omega=d+1, and Δ2d\Delta\le 2d. This is a short observation rather than a substantial new combinatorial method or paper-level advance.

    Literature check: I found no clear prior source explicitly recording Lawrence’s conjecture as false, nor the specific bound χ(18,10)12\chi(18,10)\ge 12 or the family Kd1K_{d-1}\vee Grötzsch. Searches for the exact formula, the paper title, Lawrence with maximum degree/clique number, GitHub/forum occurrences, and open indexed sources did not locate a published resolution. However, all ingredients are classical, so the novelty is only the application to Lawrence’s conjecture.

    Citation: J. Lawrence, “Covering the vertex set of a graph with subgraphs of smaller degree,” Discrete Mathematics, 1978, DOI: 10.1016/0012-365X(78)90147-4.
    J. Mycielski, “Sur le coloriage des graphes,” Colloquium Mathematicum 3 (1955), 161–162.

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