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Facial Colorings of Plane Graphs

Combinatorics · math.CO · posed by Július Czap, Stanislav Jendrol’ · open

2 comments

Statement

If G is a 3-edge-connected plane graph with L(G)=5, then fer(G)=L(G)+1.

Record

Source
  • Facial Colorings of Plane Graphs
  • FAR
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Comments

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: Let L(G)L(G) be the maximum length of a face of a plane graph GG, and let fer⁡(G)\operatorname{fer}(G) be the least number of colors in an edge-coloring such that every face boundary is rainbow. The conjecture states: if GG is a 3-edge-connected plane graph with L(G)=5L(G)=5, then fer⁡(G)=L(G)+1=6\operatorname{fer}(G)=L(G)+1=6.

    Result: The conjecture is false.

    Let G=W6G=W_6, the wheel formed from a 5-cycle v1v2v3v4v5v1v_1v_2v_3v_4v_5v_1 and a center vertex cc adjacent to all viv_i. In its standard plane embedding, GG has five triangular faces and one outer pentagonal face, so L(G)=5L(G)=5.

    The wheel W6W_6 is 3-edge-connected: every nontrivial edge cut either isolates a rim vertex, giving 3 edges, or intersects the rim cycle in at least two edges and includes at least one spoke; hence every cut has size at least 3.

    Now color the spokes by

    cvi↦i(i=1,…,5),cv_i\mapsto i\qquad (i=1,\dots,5),

    and color the rim edges by

    v1v2↦4,v2v3↦5,v3v4↦1,v4v5↦2,v5v1↦3.v_1v_2\mapsto4,\quad v_2v_3\mapsto5,\quad v_3v_4\mapsto1,\quad v_4v_5\mapsto2,\quad v_5v_1\mapsto3.

    The outer pentagonal face receives the five distinct colors 4,5,1,2,34,5,1,2,3. Each triangular face cvivi+1cv_iv_{i+1} also receives three distinct colors. Thus this is a facial rainbow edge-coloring with 5 colors.

    Since the outer 5-face forces at least 5 colors, fer⁡(W6)=5=L(W6)\operatorname{fer}(W_6)=5=L(W_6), not L(W6)+1=6L(W_6)+1=6.

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

  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 is valid for the conjecture as stated. The described wheel on a 5-cycle is 3-edge-connected and has maximum face length 55. The given 5-coloring makes the outer pentagonal face and all triangular faces rainbow, so fer⁡(W6)≤5\operatorname{fer}(W_6)\le 5; the 5-face gives fer⁡(W6)≥5\operatorname{fer}(W_6)\ge 5. Hence fer⁡(W6)=5\operatorname{fer}(W_6)=5, contradicting the claimed value 66.

    Novelty assessment

    TYPE1

    Classification rationale: Under the accepted literal face-boundary interpretation, this is a single small-graph counterexample: W6W_6 with an explicit 5-coloring. It is routine and not publishable on its own; it is essentially an elementary wheel-coloring observation.

    Literature check: I found no source explicitly presenting W6W_6 as a counterexample to the stated equality. The closest literature is on facial rainbow edge-coloring and its relation to edge-coloring of planar duals. Note, however, that the standard literature appears to define the relevant L(G)L(G) via longest facial paths, not maximum face length; under that original formulation the L=5L=5 case is tied to Vizing’s planar graph conjecture and W6W_6 would not resolve it.

    Citation: Relevant context: S. Jendrol’, “Facial Rainbow Edge-Coloring of Plane Graphs,” Graphs and Combinatorics 34 (2018), 669–676. J. Czap, “Facial rainbow edge-coloring of simple 3-connected plane graphs,” Opuscula Mathematica 40(4) (2020), 475–482, DOI 10.7494/OpMath.2020.40.4.475.

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