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Do Almost All Trees Have No Perfect Dominating Set?

Analysis · math.CO · math.SP · posed by Bill Quan Yue · open

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Can we find the limit?

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  • Do Almost All Trees Have No Perfect Dominating Set?
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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: Reconstructed statement: for finite non-plane unlabeled rooted and unrooted trees, with perfect dominating set S⊆V(T)S\subseteq V(T) meaning

    ∣N[v]∩S∣=1for every v∈V(T),|N[v]\cap S|=1\quad\text{for every }v\in V(T),

    find

    lim⁡n→∞average number of perfect dominating sets among n-vertex treesn.\lim_{n\to\infty}\sqrt[n]{\text{average number of perfect dominating sets among }n\text{-vertex trees}}.

    As is standard for unlabeled-tree generating functions, I interpret “tree with a perfect dominating set” as a two-coloured tree (T,S)(T,S) counted up to the relevant isomorphism. Thus the numerator is the number of isomorphism classes of trees equipped with such an SS. This matches the paper’s rooted/unrooted generating-function context.

    Result: Let τ\tau be Otter’s rooted-tree singularity, determined by

    T(z)=zexp⁡ ⁣(∑k≥1T(zk)k),T(τ)=1,T(z)=z\exp\!\left(\sum_{k\ge1}\frac{T(z^k)}k\right),\qquad T(\tau)=1,

    so

    τ=0.3383218568992077….\tau=0.3383218568992077\ldots .

    For rooted trees with a perfect dominating set, use three rooted classes:

    • BB: root is in SS;
    • FF: root is not in SS, is dominated by its parent in SS, and has no child in SS;
    • GG: root is not in SS, is not dominated by its parent, and is dominated by exactly one child in SS.

    Then

    B=zexp⁡ ⁣(∑k≥1F(zk)k),B=z\exp\!\left(\sum_{k\ge1}\frac{F(z^k)}k\right), F=zexp⁡ ⁣(∑k≥1G(zk)k),F=z\exp\!\left(\sum_{k\ge1}\frac{G(z^k)}k\right), G=BF.G=BF.

    The rooted perfect-dominating-set generating function is

    P(z)=B(z)+G(z)=B(z)(1+F(z)).P(z)=B(z)+G(z)=B(z)(1+F(z)).

    The dominant singularity ρ\rho of P(z)P(z) is the unique positive solution at which the Perron eigenvalue of the k=1k=1 Jacobian equals 11. Since

    J=(0BF2BF),J=\begin{pmatrix} 0&B\\ F^2&BF \end{pmatrix},

    this condition is

    BF(1+F)=1.BF(1+F)=1.

    Solving the above system gives

    ρ=0.381104495337003….\rho=0.381104495337003\ldots .

    Standard Pólya-tree singularity theory gives

    [zn]P(z)∼Cρ−nn−3/2,[zn]T(z)∼CTτ−nn−3/2,[z^n]P(z)\sim C\rho^{-n}n^{-3/2},\qquad [z^n]T(z)\sim C_T\tau^{-n}n^{-3/2},

    with positive constants C,CTC,C_T. Hence the rooted average An(r)A_n^{(r)} satisfies

    lim⁡n→∞An(r)n=τρ=0.8877403993884578….\lim_{n\to\infty}\sqrt[n]{A_n^{(r)}}=\frac{\tau}{\rho} =0.8877403993884578\ldots .

    For unrooted trees, rooting an nn-vertex tree-with-SS gives between 11 and nn rooted isomorphism classes, and the same is true for uncoloured trees. Therefore rooted and unrooted numerator and denominator have the same exponential growth constants. Thus

    lim⁡n→∞An(r)n=lim⁡n→∞An(u)n=0.8877403993884578…\boxed{ \lim_{n\to\infty}\sqrt[n]{A_n^{(r)}} = \lim_{n\to\infty}\sqrt[n]{A_n^{(u)}} = 0.8877403993884578\ldots }

    In particular the average number decays exponentially.

    Citation: No known source appears to give this exact constant for Yue’s problem. The analytic ingredients are standard: Otter, “The number of trees,” Ann. of Math. 49 (1948), and the Drmota–Lalley–Woods/Pólya-tree singularity theory as presented, for example, in Drmota, Random Trees, Springer, 2009.

  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 solution attacks the correct Yue open problem under the paper’s unlabeled/Pólya-tree counting convention. Its three classes are exactly Yue’s R,B,CR,B,C system in different notation, and the functional equations are correct. The critical condition det⁡(I−J)=0\det(I-J)=0, i.e. BF(1+F)=1BF(1+F)=1, is the standard smooth implicit/Pólya-tree singularity condition, giving the stated exponential growth ρ−n\rho^{-n}. Comparing with Otter’s rooted-tree constant then yields the limit τ/ρ\tau/\rho. The rooted/unrooted transfer via the factor between 1 and nn is valid for exponential growth. I found no prior source giving this exact constant.

    Novelty assessment

    TYPE1

    Classification rationale: The result appears genuinely new, but it is a routine analytic-combinatorics completion of Yue’s own generating-function setup. Yue already derives essentially the same rooted classes and functional equations; the accepted solution applies standard Pólya-tree singularity theory and the usual Perron/Jacobian critical condition to extract the exponential growth constant. This answers the stated open question, but introduces no new method and is likely too narrow for a standalone standard-journal paper unless substantially expanded.

    Literature check: I found no prior source giving the limit or the constants ρ=0.381104495337003…\rho=0.381104495337003\ldots and τ/ρ=0.8877403993884578…\tau/\rho=0.8877403993884578\ldots. Searches of the exact title, Yue’s conjecture wording, “perfect dominating set(s)” with rooted/unlabeled/random trees, “efficient dominating set,” and “perfect code(s) in trees” found only Yue’s paper or unrelated domination/algorithmic/Roman-domination papers. The SCIRP citation page lists no Google Scholar or CrossRef citing articles. OEIS searches for the title and “perfect dominating set” did not reveal a relevant sequence. The closest literature is Yue’s 2018 paper, which gives the functional equations and numerical evidence but explicitly asks for convergence and the limit.

    Citation: Closest source: Bill Quan Yue, “Do Almost All Trees Have No Perfect Dominating Set?”, Open Journal of Discrete Mathematics 8 (2018), 1–13, doi:10.4236/ojdm.2018.81001. Standard background: R. Otter, “The Number of Trees,” Ann. of Math. 49 (1948), 583–599; M. Drmota, Random Trees, Springer, 2009.

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