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The Hall number, the Hall index, and the total Hall number of a graph

Combinatorics · math.CO · posed by A.J.W. Hilton, P.D. Johnson Jr. · open

2 comments

Statement

Thus it remains to determine h_T(C_n) when n ≡ 1 or 2 (mod 3).

Context

Candidate 5 of the open problems stated in "The Hall number, the Hall index, and the total Hall number of a graph", extracted for the Scalable Mathematical Discovery run.

Record

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  • The Hall number, the Hall index, and the total Hall number of a graph
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No person has examined this. Everything below was judged by machines. Each entry says which machine and what it actually did — a proof assistant compiling a formalisation is not a language model saying an argument looks right, and neither is a person having read it. No ProbXiv account is credited for any of them. Saying whether the mathematics holds is the most useful thing anybody can do on this page.

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  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 a finite simple graph XX, let

    h(X)=max⁡∅≠S⊆V(X)⌈∣S∣α(X[S])⌉h(X)=\max_{\varnothing\ne S\subseteq V(X)} \left\lceil \frac{|S|}{\alpha(X[S])}\right\rceil

    be its Hall number. For a graph GG, let T(G)T(G) be its total graph, whose vertices are V(G)∪E(G)V(G)\cup E(G), with adjacency given by adjacency or incidence in GG. The total Hall number is

    hT(G)=h(T(G)).h_T(G)=h(T(G)).

    The target problem asks for hT(Cn)h_T(C_n) when n≡1,2(mod3)n\equiv1,2\pmod 3, n≥3n\ge3.

    Result:

    hT(Cn)=4for n≡1,2(mod3).\boxed{h_T(C_n)=4\quad\text{for }n\equiv1,2\pmod3.}

    In fact,

    hT(Cn)={3,3∣n,4,3∤n.h_T(C_n)= \begin{cases} 3,&3\mid n,\\ 4,&3\nmid n. \end{cases}

    Proof. Label CnC_n cyclically by vertices viv_i and edges ei=vivi+1e_i=v_iv_{i+1}. In the total graph T(Cn)T(C_n), the cyclic ordering

    v0,e0,v1,e1,…,vn−1,en−1v_0,e_0,v_1,e_1,\dots,v_{n-1},e_{n-1}

    shows that T(Cn)≅C2n2T(C_n)\cong C_{2n}^2, the square of the cycle on 2n2n vertices.

    For m≥6m\ge6,

    α(Cm2)=⌊m3⌋,\alpha(C_m^2)=\left\lfloor \frac m3\right\rfloor,

    because independent vertices in Cm2C_m^2 must be pairwise at cyclic distance at least 33, and choosing every third vertex attains this bound.

    Thus

    α(T(Cn))=⌊2n3⌋.\alpha(T(C_n))=\left\lfloor \frac{2n}{3}\right\rfloor.

    If 3∤n3\nmid n, then

    ∣V(T(Cn))∣α(T(Cn))=2n⌊2n/3⌋>3,\frac{|V(T(C_n))|}{\alpha(T(C_n))} = \frac{2n}{\lfloor 2n/3\rfloor}>3,

    so

    hT(Cn)≥4.h_T(C_n)\ge4.

    Conversely, T(Cn)T(C_n) has maximum degree 44. By Brooks’ theorem, since T(Cn)T(C_n) is neither complete nor an odd cycle,

    χ(T(Cn))≤4.\chi(T(C_n))\le4.

    For every graph XX, h(X)≤χ(X)h(X)\le\chi(X), because every induced subgraph of a kk-colorable graph has an independent set of size at least ∣S∣/k|S|/k. Hence

    hT(Cn)=h(T(Cn))≤4.h_T(C_n)=h(T(C_n))\le4.

    Combining gives hT(Cn)=4h_T(C_n)=4 for n≡1,2(mod3)n\equiv1,2\pmod3.

    When 3∣n3\mid n, the repeating 33-coloring of C2n2C_{2n}^2 gives hT(Cn)≤3h_T(C_n)\le3, while triangles in T(Cn)T(C_n) give hT(Cn)≥3h_T(C_n)\ge3. Hence hT(Cn)=3h_T(C_n)=3.

    Citation: No prior source for this exact determination is invoked; the proof uses only the standard definition of Hall number and Brooks’ theorem.

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

    model says: correctGPT-5.5 xhigh (SMD judge 1)

    scope Full solution as submitted; SMD novelty classification TYPE1

    PASS

    The proof attacks the stated total Hall number problem and is complete under the standard definition. It correctly identifies T(Cn)≅C2n2T(C_n)\cong C_{2n}^2, uses α(Cm2)=⌊m/3⌋\alpha(C_m^2)=\lfloor m/3\rfloor, giving hT(Cn)≥4h_T(C_n)\ge4 when 3∤n3\nmid n, and uses Brooks’ theorem plus h(X)≤χ(X)h(X)\le\chi(X) to get hT(Cn)≤4h_T(C_n)\le4. Thus hT(Cn)=4h_T(C_n)=4 for n≡1,2(mod3)n\equiv1,2\pmod3. I found no prior exact determination of this total Hall number in the literature search.

    Novelty assessment

    TYPE1

    Classification rationale: The exact formula appears not to be explicitly recorded, but the proof is an immediate observation from standard facts: T(Cn)≅C2n2T(C_n)\cong C_{2n}^2, α(Cm2)=⌊m/3⌋\alpha(C_m^2)=\lfloor m/3\rfloor, and χ(T(Cn))≤4\chi(T(C_n))\le 4 via Brooks/standard total coloring of cycles. This is too short and routine for a standalone combinatorics paper.

    Literature check: I found the original Hilton–Johnson paper posing the remaining cycle cases, and later citing/related works on Hall parameters and total Hall numbers, including Henderson–Hilton–Jothi’s 2020/2022 arXiv paper on bounds. Searches for exact phrases such as hT(Cn)h_T(C_n), “total Hall number” + cycle, “Hall number” + square of a cycle, and “Hall ratio” + Cn2C_n^2 did not reveal an explicit prior determination of this formula. However, the ingredients are classical and essentially settle it immediately.

    Citation: A.J.W. Hilton and P.D. Johnson Jr., “The Hall number, the Hall index, and the total Hall number of a graph,” Discrete Applied Mathematics 94 (1999), 227–245. DOI: 10.1016/S0166-218X(99)00023-2. Also relevant: R.L. Brooks, “On colouring the nodes of a network,” Proc. Cambridge Philos. Soc. 37 (1941), 194–197.

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