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Vertex Turán problems for the oriented hypercube

Combinatorics · math.CO · posed by Dániel Gerbner, Abhishek Methuku, Dániel T. Nagy, Balázs Patkós, Máté Vizer · open

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Statement

We conjecture that exv(V⃗r,Q⃗n)=2n−1+Θ(nr−2)ex_v(\vec{V}_r, \vec{Q}_n) = 2^{n-1} + \Theta(n^{r-2}) holds for every r≥3r \ge 3.

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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 fixed r≥3r\ge 3, Q⃗n\vec Q_n is the Boolean lattice 2[n]2^{[n]} with edges A→A∪{i}A\to A\cup\{i\}. V⃗r\vec V_r is the out-star with one source and rr leaves. Thus a family A⊆2[n]\mathcal A\subseteq 2^{[n]} is V⃗r\vec V_r-free iff every A∈AA\in\mathcal A has at most r−1r-1 selected immediate supersets. The conjecture says

    exv(V⃗r,Q⃗n)=2n−1+Θ(nr−2).ex_v(\vec V_r,\vec Q_n)=2^{n-1}+\Theta(n^{r-2}).

    Result: The conjecture is false. In fact, for every fixed r≥3r\ge3,

    exv(V⃗r,Q⃗n)≥2n−1+crnr−1.ex_v(\vec V_r,\vec Q_n)\ge 2^{n-1}+c_r n^{r-1}.

    Proof. Put k=r−1k=r-1, and partition [n][n] into kk parts P1,…,PkP_1,\dots,P_k as equally as possible. Let

    H={{x1,…,xk}:xi∈Pi}\mathcal H=\{\{x_1,\dots,x_k\}:x_i\in P_i\}

    be the complete kk-partite kk-uniform hypergraph, so

    ∣H∣=∏i=1k∣Pi∣=Ωr(nk)=Ωr(nr−1).|\mathcal H|=\prod_{i=1}^k |P_i|=\Omega_r(n^k)=\Omega_r(n^{r-1}).

    No r=k+1r=k+1 element set has all its kk-subsets in H\mathcal H: among k+1k+1 vertices two lie in the same part, and some kk-subset containing them is not transversal.

    Now work with complements. Define

    F={X⊆[n]:∣X∣≡r(mod2)}∪H,\mathcal F=\{X\subseteq[n]: |X|\equiv r\pmod 2\}\cup \mathcal H,

    and let

    A={[n]∖X:X∈F}.\mathcal A=\{[n]\setminus X:X\in\mathcal F\}.

    Then

    ∣A∣=∣F∣=2n−1+∣H∣=2n−1+Ωr(nr−1).|\mathcal A|=|\mathcal F|=2^{n-1}+|\mathcal H| =2^{n-1}+\Omega_r(n^{r-1}).

    It remains to check V⃗r\vec V_r-freeness. For X∈FX\in\mathcal F, selected immediate supersets of [n]∖X[n]\setminus X correspond exactly to selected immediate subsets X∖{x}∈FX\setminus\{x\}\in\mathcal F.

    If X∈HX\in\mathcal H, then ∣X∣=r−1|X|=r-1, and all its immediate subsets have parity rr, so exactly r−1r-1 of them lie in F\mathcal F. If XX lies in the parity class, then its immediate subsets can lie in F\mathcal F only when ∣X∣=r|X|=r, in which case they are the (r−1)(r-1)-subsets of XX belonging to H\mathcal H. By the previous paragraph, not all rr of them belong to H\mathcal H. Hence every selected vertex has outdegree at most r−1r-1, so A\mathcal A is V⃗r\vec V_r-free.

    Thus the excess over 2n−12^{n-1} can be order nr−1n^{r-1}, contradicting the claimed O(nr−2)O(n^{r-2}) upper bound. This is not a boundary case; it holds for every fixed r≥3r\ge3 and all large nn.

    Citation: No external resolution used. The conjecture and notation are from Gerbner–Methuku–Nagy–Patkós–Vizer, “Vertex Turán problems for the oriented hypercube,” arXiv:1807.06866.

  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 construction is a rigorous counterexample to the stated conjecture. In the paper, V⃗r\vec V_r is indeed the out-star with rr leaves, so V⃗r\vec V_r-freeness is exactly the condition that no selected vertex has rr selected immediate supersets.

    The complement/parity construction is valid: H\mathcal H has Ωr(nr−1)\Omega_r(n^{r-1}) sets, is disjoint from the chosen parity class, and every X∈FX\in\mathcal F has at most r−1r-1 selected immediate subsets. The key point is correct: no rr-set has all its (r−1)(r-1)-subsets in the complete (r−1)(r-1)-partite (r−1)(r-1)-uniform hypergraph. Thus the resulting family is V⃗r\vec V_r-free and has size

    2n−1+Ωr(nr−1),2^{n-1}+\Omega_r(n^{r-1}),

    contradicting the conjectured 2n−1+O(nr−2)2^{n-1}+O(n^{r-2}).

    Targeted arXiv/web searches did not reveal a prior resolution of this specific conjecture.

    Novelty assessment

    TYPE1

    Classification rationale: The construction appears genuinely new, but it is a very short elementary counterexample rather than a substantial resolution of the underlying extremal problem. It refutes the stated conjecture, but does not determine the true asymptotic value. This would more plausibly be an erratum/comment or part of a broader note than a standalone combinatorics paper.

    Literature check: I found no evidence that this counterexample or the stronger lower bound 2n−1+Ωr(nr−1)2^{n-1}+\Omega_r(n^{r-1}) has appeared in the literature. Searches around the exact title, “oriented hypercube,” “directed cherry,” exvex_v, V⃗r\vec V_r, and related vertex Turán hypercube work returned the original Gerbner–Methuku–Nagy–Patkós–Vizer paper and unrelated/unoriented hypercube Turán papers, but no correction or later resolution of this conjecture.

    Citation: D. Gerbner, A. Methuku, D. T. Nagy, B. Patkós, M. Vizer, “Vertex Turán problems for the oriented hypercube,” arXiv:1807.06866.

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