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Competition Numbers and Phylogeny Numbers of Connected Graphs and Hypergraphs*

Combinatorics · math.CO · posed by Yanzhen Xiong, Soesoe Zaw, Yinfeng Zhu · open

2 comments

Statement

Is it true that for every nonnegative integer k, there exists a connected hypergraph H satisfying φ_ST(H) − κ_ST(H) + 1 = k?

Context

Candidate 2 of the open problems stated in "Competition Numbers and Phylogeny Numbers of Connected Graphs and Hypergraphs*", extracted for the Scalable Mathematical Discovery run.

Record

Source
  • Competition Numbers and Phylogeny Numbers of Connected Graphs and Hypergraphs*
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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 every kN0k\in\mathbb N_0, there is a finite connected hypergraph HH such that

    ϕST(H)κST(H)+1=k,\phi_{ST}(H)-\kappa_{ST}(H)+1=k,

    where κST\kappa_{ST} is the Sonntag--Teichert competition number and ϕST\phi_{ST} is the corresponding ST-phylogeny number: for an acyclic digraph DD,

    CST(D): E={ND(v):ND(v)2},C_{ST}(D):\ E=\{N^-_D(v): |N^-_D(v)|\ge2\},

    and

    PST(D): E={ND[v]:ND[v]2},P_{ST}(D):\ E=\{N^-_D[v]: |N^-_D[v]|\ge2\},

    with phylogeny realization taken by the induced trace on the original vertex set.

    Result: The statement is true.

    For k=0k=0, take the one-edge hypergraph HH with vertex set {a,b}\{a,b\} and edge {a,b}\{a,b\}. Then κST(H)=1\kappa_{ST}(H)=1, since the edge is all of V(H)V(H) and so cannot be an in-neighborhood of an original vertex; one new sink realizes it. Also ϕST(H)=0\phi_{ST}(H)=0, realized by the acyclic digraph aba\to b. Thus ϕST(H)κST(H)+1=0\phi_{ST}(H)-\kappa_{ST}(H)+1=0.

    Now let k1k\ge1, and put t=k+2t=k+2. Let

    C={c1,,ct},R={r1,,rt},V=CR.C=\{c_1,\dots,c_t\},\qquad R=\{r_1,\dots,r_t\},\qquad V=C\cup R.

    Define a hypergraph HtH_t on VV with edges

    Fi=C{ci}(1it),andV.F_i=C\setminus\{c_i\}\quad(1\le i\le t), \qquad\text{and}\qquad V.

    It is connected because VV itself is an edge.

    First, κST(Ht)=1\kappa_{ST}(H_t)=1. One new vertex zz suffices: add arcs xrix\to r_i for every xFix\in F_i, and arcs vzv\to z for every vVv\in V. With order C<R<zC<R<z, the digraph is acyclic, and its nontrivial in-neighborhoods are exactly F1,,Ft,VF_1,\dots,F_t,V. Conversely, the edge VV cannot be the in-neighborhood of any original vertex, since no vertex belongs to its own in-neighborhood. Hence at least one added vertex is necessary.

    Second, ϕST(Ht)=t2\phi_{ST}(H_t)=t-2. For the upper bound, use t2t-2 new vertices w2,,wt1w_2,\dots,w_{t-1}. Realize F1F_1 at ctc_t, realize FtF_t at ct1c_{t-1}, realize VV at rtr_t, and realize each FiF_i for 2it12\le i\le t-1 at wiw_i. This is done by putting arcs from every other vertex of the desired edge into the representing vertex; the order

    c1,,ct2,ct1,ct,r1,,rt1,rt,w2,,wt1c_1,\dots,c_{t-2},c_{t-1},c_t,r_1,\dots,r_{t-1},r_t,w_2,\dots,w_{t-1}

    is acyclic. The induced trace of the ST-phylogeny hypergraph is exactly HtH_t.

    For the lower bound, consider any acyclic phylogeny realization and restrict a topological order to VV. If an edge of HtH_t is represented by an original vertex, that vertex must be the maximum element of that edge in this restricted order. Among the core edges Fi=C{ci}F_i=C\setminus\{c_i\}, only two vertices can occur as such maxima: the latest and second-latest vertices of CC. The large edge VV contributes at most one further maximum. Hence at most three edges of HtH_t can be represented by original vertices. Since HtH_t has t+1t+1 edges, at least (t+1)3=t2(t+1)-3=t-2 added vertices are necessary. Therefore ϕST(Ht)=t2\phi_{ST}(H_t)=t-2.

    Thus

    ϕST(Ht)κST(Ht)+1=(t2)1+1=t2=k.\phi_{ST}(H_t)-\kappa_{ST}(H_t)+1=(t-2)-1+1=t-2=k.

    So every nonnegative integer occurs.

    Citation: No prior resolution is used here. The problem and notation are from Xiong, Zaw, and Zhu, “Competition Numbers and Phylogeny Numbers of Connected Graphs and Hypergraphs,” Problem 2(ii).

  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 construction attacks the stated hypergraph problem and is mathematically sound. The competition-number argument is correct, and the phylogeny-number lower bound validly uses the topological-order maximum condition: among the FiF_i's only the two latest CC-vertices can serve as original representatives, and VV can add at most one more original representative, so at least t2t-2 added vertices are necessary. The proposed acyclic realization attains this bound. The k=0k=0 case is also correct. Targeted searches did not reveal an existing stronger resolution.

    Novelty assessment

    TYPE1

    Classification rationale: Genuinely new as far as I can determine, but minor. The proof is a short explicit construction plus an elementary topological-order counting argument in a very specialized hypergraph variant. It resolves a stated problem, but likely only as a short note or part of a larger paper, not a standalone standard-journal contribution.

    Literature check: I searched for the exact title, exact Problem 2(ii) wording, formula variants involving ϕST\phi_{ST}, κST\kappa_{ST}, ASCII variants “phi_ST/kappa_ST,” “ST competition number,” “ST phylogeny number,” “Sonntag Teichert” with phylogeny/competition hypergraphs, author-name combinations, and citation/metadata pages. The results found the original Xiong–Zaw–Zhu paper and bibliographic mirrors, plus related work on ordinary competition/phylogeny numbers, but no subsequent resolution or stronger exact-spectrum theorem for the ST hypergraph invariant.

    Citation: Y. Xiong, S. Zaw, and Y. Zhu, “Competition Numbers and Phylogeny Numbers of Connected Graphs and Hypergraphs,” Algebra Colloquium 27(1) (2020), 79–86, DOI: 10.1142/S1005386720000073, Problem 2(ii).

    A language model was shown this work and said what it thought of it. Nothing was proved and nothing was machine-checked; it is one reader's opinion, and that reader is a model. No ProbXiv account is credited for it.

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