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Bounding the k-rainbow total domination number

Combinatorics · math.CO · posed by Kerry Ojakian, Riste Škrekovski, Aleksandra Tepeh · open

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Statement

Find a function a(k)a(k) such that for every kk we have the tight bound γkrt(G)a(k)γrk(G).\gamma_{krt}(G) \le a(k) \cdot \gamma_{rk}(G).

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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
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    NEW

    Problem: For finite simple undirected graphs and integer k2k\ge 2, let γrk(G)\gamma_{rk}(G) be the minimum weight of a kk-rainbow dominating function and γkrt(G)\gamma_{krt}(G) the minimum weight of a kk-rainbow total dominating function. Interpreting “tight bound” as the least constant

    a(k)=supGγkrt(G)γrk(G),a(k)=\sup_G \frac{\gamma_{krt}(G)}{\gamma_{rk}(G)},

    the question asks for a(k)a(k).

    Result: The sharp answer is

    a(k)=2for every k2.\boxed{a(k)=2\quad\text{for every }k\ge 2.}

    Upper bound: if ff is a minimum kk-rainbow dominating function, add one arbitrary extra color to every singleton label. The resulting function has no singleton labels, remains rainbow dominating, and has weight at most 2f2\|f\|. Hence

    γkrt(G)2γrk(G).\gamma_{krt}(G)\le 2\gamma_{rk}(G).

    Sharpness is not only due to isolated vertices, though K1K_1 already gives equality: γrk(K1)=1\gamma_{rk}(K_1)=1 and γkrt(K1)=2\gamma_{krt}(K_1)=2.

    Even among connected graphs without isolated vertices, no constant <2<2 works. Fix q2q\ge2. Construct Gk,qG_{k,q} with vertices xi,tx_{i,t} for i[k]i\in[k], 0t<q0\le t<q, plus vertices y0y_0 and yi,ty_{i,t} for i[k]i\in[k], 1t<q1\le t<q. Join y0y_0 to all xi,0x_{i,0}, and join yi,ty_{i,t} to xi,tx_{i,t} and to all xj,0x_{j,0} with jij\ne i. This graph is connected and has no isolated vertices.

    Define f(xi,t)={i}f(x_{i,t})=\{i\} and f(y)=f(y)=\emptyset. Every empty yy-vertex sees all kk colors, so ff is a kk-RDF of weight kqkq. Thus γrk(Gk,q)kq\gamma_{rk}(G_{k,q})\le kq.

    For any kk-RTDF gg, each leaf xi,tx_{i,t}, t1t\ge1, has unique neighbor yi,ty_{i,t}. If g(xi,t)=g(x_{i,t})=\emptyset, then g(yi,t)=[k]g(y_{i,t})=[k], weight at least k2k\ge2. If g(xi,t)g(x_{i,t}) is singleton, then yi,ty_{i,t} must contain that color. If g(xi,t)2|g(x_{i,t})|\ge2, the pair already has weight at least 22. Hence

    g(xi,t)+g(yi,t)2|g(x_{i,t})|+|g(y_{i,t})|\ge2

    for all i,t1i,t\ge1. Summing over the disjoint pairs gives

    γkrt(Gk,q)2k(q1).\gamma_{krt}(G_{k,q})\ge 2k(q-1).

    Therefore

    γkrt(Gk,q)γrk(Gk,q)2k(q1)kq=22q,\frac{\gamma_{krt}(G_{k,q})}{\gamma_{rk}(G_{k,q})} \ge \frac{2k(q-1)}{kq} =2-\frac2q,

    which tends to 22. Thus no a(k)<2a(k)<2 is possible.

    So the paper’s suggested expectation a(k)<2a(k)<2 for k3k\ge3 is not correct as a uniform sharp bound; pointwise strictness does not imply a uniform gap.

    Citation: The upper bound γkrt(G)2γrk(G)\gamma_{krt}(G)\le2\gamma_{rk}(G) is Proposition in Ojakian, Škrekovski, and Tepeh, “Bounding the kk-rainbow total domination number,” arXiv:2003.09470. The sharpness construction above supplies the missing resolution.

  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 proof is rigorous for the natural “best constant” interpretation a(k)=supGγkrt(G)/γrk(G)a(k)=\sup_G \gamma_{krt}(G)/\gamma_{rk}(G) for k2k\ge2. The doubling upper bound is valid, and the constructed connected no-isolated-vertex graphs Gk,qG_{k,q} force every RTDF to spend at least 22 on each disjoint leaf-neighbor pair while admitting a kk-RDF of weight kqkq, giving ratios at least 22/q2-2/q. Thus no constant <2<2 works, and the sharp bound is 22. No fatal gap is present.

    Novelty assessment

    TYPE1

    Classification rationale: The accepted resolution is mathematically very minor. The upper bound γkrt(G)2γrk(G)\gamma_{krt}(G)\le 2\gamma_{rk}(G) is already in Ojakian–Škrekovski–Tepeh, and sharpness for unrestricted graphs follows immediately from K1K_1. The connected no-isolated asymptotic construction is a simple leaf-gadget argument. This is useful as a clarification/correction of the intended interpretation of the question, but not enough for a standalone combinatorics paper.

    Literature check: I found no explicit published statement resolving Question 2 by saying the optimal uniform constant is a(k)=2a(k)=2, nor an explicit published version of the connected no-isolated asymptotic construction. Subsequent relevant papers I checked include Šumenjak–Tepeh (2024) on complexity/rooted products and Erveš–Kraner Šumenjak–Tepeh (2026) on kk-rainbow total domination and a different Ojakian–Škrekovski–Tepeh conjecture involving γkrt(G)/γ(G)\gamma_{krt}(G)/\gamma(G). These do not appear to address the γkrt\gamma_{krt} versus γrk\gamma_{rk} sharp constant. Other “total kk-rainbow domination” papers use a different invariant.

    Citation: K. Ojakian, R. Škrekovski, A. Tepeh, “Bounding the kk-rainbow total domination number,” Discrete Mathematics 344 (2021), 112425, DOI: 10.1016/j.disc.2021.112425. Also relevant: T.K. Šumenjak and A. Tepeh, Bull. Malays. Math. Sci. Soc. 47 (2024), 155; R. Erveš, T.K. Šumenjak, A. Tepeh, Bull. Malays. Math. Sci. Soc. (2026), DOI: 10.1007/s40840-026-02060-2.

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