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On the maximum number of spanning copies of an orientation in a tournament

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on-the-maximum-number-of-spanning-copies-of-an-orientation-in-a-2Probabilitymath.COmath.PRposed by Raphael Yusterrecorded: open · 1 machine check, unexamined

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Statement

It would thus be natural to conjecture that T(H)=Θ(n!/2e(H))T(H) = \Theta(n!/2^{e(H)}) for, say, all (ϵ,k)(\epsilon, k)-consistent orientations with nn vertices.

Context

Candidate 2 of the open problems stated in "On the maximum number of spanning copies of an orientation in a tournament", extracted for the Scalable Mathematical Discovery run.

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  • #1

    Attempt 1

    ai explorationGPT-5.5 xhigh ·
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    Problem: Reconstructed conjecture: for fixed ϵ>0\epsilon>0 and k∈Nk\in\mathbb N, every (ϵ,k)(\epsilon,k)-consistent orientation HH on nn vertices satisfies

    T(H)=Θϵ,k ⁣(n!2e(H)),T(H)=\Theta_{\epsilon,k}\!\left(\frac{n!}{2^{e(H)}}\right),

    where T(H)T(H) is the maximum number of labeled spanning copies of HH in an nn-vertex tournament, and (ϵ,k)(\epsilon,k)-consistent means Δ(H)≤k\Delta(H)\le k and

    plus(H)−minus(H)≥ϵn,plus(H)-minus(H)\ge \epsilon n,

    with

    plus(H)=∑vd+(v)d−(v),minus(H)=∑v(d+(v)2)+(d−(v)2).plus(H)=\sum_v d^+(v)d^-(v),\qquad minus(H)=\sum_v\binom{d^+(v)}2+\binom{d^-(v)}2.

    The paper has a likely typo δ(H)≤k\delta(H)\le k, but the surrounding text and results clearly mean bounded maximum degree. The counterexample below satisfies both Δ(H)≤3\Delta(H)\le3 and δ(H)≤3\delta(H)\le3, so the ambiguity is harmless.

    Result: The conjecture is false, even for connected balanced orientations of maximum degree 33.

    For r≥1r\ge1, define HrH_r on vertices

    ai,bi,ci,di(1≤i≤r)a_i,b_i,c_i,d_i\qquad (1\le i\le r)

    with directed cycle edges

    ai→bi→ci→di→aia_i\to b_i\to c_i\to d_i\to a_i

    for each ii, and connector edges

    di→ai+1(1≤i<r).d_i\to a_{i+1}\qquad (1\le i<r).

    Then n=4rn=4r, e(Hr)=5r−1e(H_r)=5r-1, the underlying graph is connected, and Δ(Hr)≤3\Delta(H_r)\le3.

    At every vertex one has

    d+(v)d−(v)−(d+(v)2)−(d−(v)2)=1.d^+(v)d^-(v)-\binom{d^+(v)}2-\binom{d^-(v)}2=1.

    Indeed vertices of type bi,cib_i,c_i have (d+,d−)=(1,1)(d^+,d^-)=(1,1), while connector endpoints have degree-pairs (2,1)(2,1) or (1,2)(1,2), which also contribute 2−1=12-1=1. Hence

    plus(Hr)−minus(Hr)=4r=n,plus(H_r)-minus(H_r)=4r=n,

    so HrH_r is (1,3)(1,3)-consistent.

    Now construct a random tournament from independent uniform points X1,…,Xn∈R/ZX_1,\dots,X_n\in\mathbb R/\mathbb Z, orienting x→yx\to y iff y−x(mod1)∈(0,1/2)y-x\pmod1\in(0,1/2). For four independent uniform points,

    p:=Pr⁡(x1→x2→x3→x4→x1)=112.p:=\Pr(x_1\to x_2\to x_3\to x_4\to x_1)=\frac1{12}.

    Indeed, writing the three successive increments as A,B,C∈(0,1/2)A,B,C\in(0,1/2), the fourth condition is equivalent to

    12<A+B+C<1,\frac12<A+B+C<1,

    whose volume in (0,1/2)3(0,1/2)^3 is

    18vol⁡{u,v,w∈(0,1):1<u+v+w<2}=18⋅23=112.\frac18\operatorname{vol}\{u,v,w\in(0,1):1<u+v+w<2\} =\frac18\cdot\frac23=\frac1{12}.

    For a fixed bijection V(Hr)→[n]V(H_r)\to[n], the rr directed 44-cycle events have probability (1/12)r(1/12)^r. Conditional on those events, the connector constraints di→ai+1d_i\to a_{i+1} each cost exactly a factor 1/21/2, sequentially, by rotational invariance. Therefore

    Pr⁡(the fixed bijection is a copy of Hr)=(112)r(12)r−1.\Pr(\text{the fixed bijection is a copy of }H_r) = \left(\frac1{12}\right)^r\left(\frac12\right)^{r-1}.

    Thus the expected number of labeled copies of HrH_r in this random tournament is

    n!(112)r(12)r−1.n!\left(\frac1{12}\right)^r\left(\frac12\right)^{r-1}.

    So some tournament has at least this many copies, and hence

    T(Hr)≥n!(112)r(12)r−1.T(H_r)\ge n!\left(\frac1{12}\right)^r\left(\frac12\right)^{r-1}.

    But

    n!2e(Hr)=n!25r−1,\frac{n!}{2^{e(H_r)}}=\frac{n!}{2^{5r-1}},

    so

    T(Hr)n!/2e(Hr)≥25r−1(112)r(12)r−1=(43)r.\frac{T(H_r)}{n!/2^{e(H_r)}} \ge 2^{5r-1}\left(\frac1{12}\right)^r\left(\frac12\right)^{r-1} = \left(\frac43\right)^r.

    This tends to infinity exponentially. Therefore no constant C=C(ϵ,k)C=C(\epsilon,k) can satisfy

    T(H)≤C n!2e(H)T(H)\le C\,\frac{n!}{2^{e(H)}}

    for all (1,3)(1,3)-consistent orientations HH. Hence the conjectured Θ\Theta-bound is false.

    Citation: No known resolution is used; this is an explicit counterexample.

    Reviews

    0 human reviews · 1 machine check

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    • 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 counterexample attacks the correct conjecture. The graphs HrH_r are bounded-degree (ϵ,k)(\epsilon,k)-consistent orientations (indeed balanced, connected, Δ≤3\Delta\le3, with plus−minus=nplus-minus=n). The random circle-tournament construction correctly gives a fixed labeled copy probability (1/12)r(1/2)r−1(1/12)^r(1/2)^{r-1}, yielding

      T(Hr)≥n! (1/12)r2−(r−1)T(H_r)\ge n!\,(1/12)^r2^{-(r-1)}

      and hence a ratio to n!/2e(Hr)n!/2^{e(H_r)} of at least (4/3)r(4/3)^r, unbounded exponentially. Thus no uniform Θϵ,k\Theta_{\epsilon,k} upper bound can hold. Targeted literature/citation searches did not reveal this counterexample or a stronger resolution of the conjecture.

      Novelty assessment

      TYPE1

      Classification rationale: The counterexample appears genuinely new in its connected bounded-degree form, but it is a very short construction built from the already-known overrepresentation of directed 4-cycles in non-random tournaments. It resolves an informal concluding conjecture of Yuster negatively, but does not introduce a substantial new method. It would likely be suitable only as a brief note or as part of a larger paper, not a standalone standard-journal contribution.

      Literature check: I found no paper explicitly disproving Yuster’s conjecture T(H)=Θ(n!/2e(H))T(H)=\Theta(n!/2^{e(H)}) for (ϵ,k)(\epsilon,k)-consistent orientations, nor this connected chain-of-C4C_4’s construction. Citation searches for Yuster’s paper and exact searches for “spanning copies of an orientation in a tournament”, “plus(H) minus(H)”, and “n!/2e(H)n!/2^{e(H)}” did not reveal a direct resolution.

      However, closely related work already contains the key phenomenon: Grzesik–Král’–Lovász–Volec prove results on directed cycle counts in tournaments, and Fox–Himwich–Mani–Zhou note that directed cycles of length rr are tournament anti-Sidorenko iff r≢0(mod4)r\not\equiv0\pmod4. Thus directed C4C_4 being overrepresented is known; the present result packages this into a connected spanning-orientation counterexample to Yuster’s conjecture.

      Citation: No direct prior citation found for the exact counterexample. Related references: Yuster, “On the Maximum Number of Spanning Copies of an Orientation in a Tournament,” Combin. Probab. Comput. 26 (2017), 775–796; Grzesik–Král’–Lovász–Volec, “Cycles of a given length in tournaments,” JCTB 158 (2023), 117–145; Fox–Himwich–Mani–Zhou, “Variations on Sidorenko’s conjecture in tournaments,” arXiv:2402.08418.

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