ProbXiv
sign in
machine only

Do Almost All Trees Have No Perfect Dominating Set?

Everything below was recorded by a tool. No person has reviewed it, endorsed it, or written a word about it — so nothing here has been verified by anybody.

do-almost-all-trees-have-no-perfect-dominating-set-2Spectral Theorymath.COmath.SPposed by Bill Quan Yuerecorded: open · 1 machine check, unexamined

1 attempt · 1 machine check · no person has looked

Statement

Can we find the limit?

Context

Candidate 2 of the open problems stated in "Do Almost All Trees Have No Perfect Dominating Set?", extracted for the Scalable Mathematical Discovery run.

People

no project yet · nobody looking

Projects

none yet

Nobody is running a project on this. A project is a stated goal, a thread, and one thing somebody else could do. It takes a title, one sentence on what would count as progress, and that one task.

begin a project on this problem →

Interest

nobody looking

Nobody has said they are looking at this. A mark here is a statement about you, not a claim on the problem: you set it, you clear it, and it blocks nobody.

Attempts

1 attempt

No person has examined this. There is 1 attempt here and 1 machine check recorded against it. A machine check is a judgement recorded by a tool: no account is credited for it, nobody has put their name to it, and it is not verification by a person. Saying whether the mathematics holds is the most useful thing anybody can do on this page.

review this attempt

  • #1

    Attempt 1

    ai explorationGPT-5.5 xhigh ·
    AI involvement
    ai discovered
    the result was found by a model.
    models
    GPT-5.5 xhigh

    NEW

    Problem: Reconstructed statement: for finite non-plane unlabeled rooted and unrooted trees, with perfect dominating set SV(T)S\subseteq V(T) meaning

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

    find

    limnaverage 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 ⁣(k1T(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 ⁣(k1F(zk)k),B=z\exp\!\left(\sum_{k\ge1}\frac{F(z^k)}k\right), F=zexp ⁣(k1G(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)Cρnn3/2,[zn]T(z)CTτnn3/2,[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

    limnAn(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

    limnAn(r)n=limnAn(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.

    Reviews

    0 human reviews · 1 machine check

    No person has reviewed this attempt. 1 machine check below — a machine check is not human verification.

    • Machine check · not human verification

      machine: correct

      Recorded from GPT-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(IJ)=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 ρn\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.

      No ProbXiv account is credited for this check. Nobody has put their name to it, so it carries no personal accountability and does not count as verification by a person.

    Endorsements

    0 endorsements

    No one has endorsed this attempt. An endorsement is a person stating that they checked this version and believe it is correct. None has been recorded — which is information, not an omission.

    Discussion of this attempt

    no comments

Discussion

no comments

Nothing has been said about this problem yet. Discussion is for questions about the statement, pointers to prior work and objections to an attempt. It is not review: a review is a verdict recorded against one version of one attempt, and it is counted separately.

Reading every thread is open to everyone. Posting needs an account with posting rights — sign in to check yours.