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Problems in Graph Theory from Memphis

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problems-in-graph-theory-from-memphis-30Group Theorymath.GRmath.ITposed by Ralph J. Faudree, Cecil C. Rousseau, Richard H. Schelprecorded: open · 1 machine check, unexamined

1 attempt · 1 machine check · no person has looked

Statement

Suppose 17/30≤α≤117/30≤\alpha≤1 and β>(2α−1)/4\beta>(2 \alpha-1)/4 or 43/120≤α<17/3043/120≤\alpha<17/30 and β>(5α−2)/25\beta>(5 \alpha-2)/25 . For all sufficiently large n, a graph of order n in which each set of ⌊αn⌋\lfloor\alpha n\rfloor vertices spans at least βn2\beta n^{2} edges must contain a triangle.

Context

Candidate 30 of the open problems stated in "Problems in Graph Theory from Memphis", extracted for the Scalable Mathematical Discovery run.

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  • #1

    Attempt 1

    ai explorationGPT-5.5 xhigh ·
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    NEW

    Problem: For real α,β\alpha,\beta, the conjecture asserts: if either

    17/30≤α≤1,β>(2α−1)/4,17/30\le \alpha\le 1,\quad \beta>(2\alpha-1)/4,

    or

    43/120≤α<17/30,β>(5α−2)/25,43/120\le \alpha<17/30,\quad \beta>(5\alpha-2)/25,

    then for all sufficiently large nn, every simple graph GG on nn vertices such that every ⌊αn⌋\lfloor \alpha n\rfloor-vertex set spans at least βn2\beta n^2 edges contains a triangle.

    Result: The statement is false.

    Take α=43/120\alpha=43/120. Then

    (5α−2)/25=−1/120.(5\alpha-2)/25=-1/120.

    Let CC be the Clebsch graph, realized as the Cayley graph on F24\mathbb F_2^4 with generators

    e1,e2,e3,e4,(1,1,1,1).e_1,e_2,e_3,e_4,(1,1,1,1).

    This graph is triangle-free: the sum of two distinct generators is never a generator.

    Also α(C)=5\alpha(C)=5. Indeed, translating an independent set to contain 00, the remaining possible vertices have Hamming weight 22 or 33. A family of weight-22 vertices cannot contain two complementary pairs, and any chosen weight-22 vector forbids the weight-33 vectors containing its support; a short case check gives at most four further vertices. The set

    {0000,0011,0101,1001,1110}\{0000,0011,0101,1001,1110\}

    shows size 55 is attained.

    For m≥1m\ge1, form the balanced blow-up GmG_m of CC, replacing each vertex of CC by an independent class of size 15m15m, and each edge of CC by a complete bipartite graph. Then

    ∣V(Gm)∣=240m,⌊α∣V(Gm)∣⌋=86m,|V(G_m)|=240m,\qquad \lfloor \alpha |V(G_m)|\rfloor=86m,

    and GmG_m is triangle-free.

    Define

    δ=min⁡{∑uv∈E(C)xuxv:0≤xu≤116, ∑u∈V(C)xu=43120}.\delta=\min\left\{\sum_{uv\in E(C)} x_u x_v: 0\le x_u\le \frac1{16},\ \sum_{u\in V(C)}x_u=\frac{43}{120}\right\}.

    The feasible set is compact. If δ=0\delta=0, the support of a minimizing vector would be an independent set of CC, hence have total weight at most 5/16<43/1205/16<43/120, contradiction. Thus δ>0\delta>0.

    Set β=δ/2\beta=\delta/2. Then β>0>−1/120=(5α−2)/25\beta>0>-1/120=(5\alpha-2)/25. For any 86m86m-vertex set S⊆V(Gm)S\subseteq V(G_m), putting xu=∣S∩Vu∣/(240m)x_u=|S\cap V_u|/(240m) gives

    eGm(S)=(240m)2∑uv∈E(C)xuxv≥δ(240m)2>β(240m)2.e_{G_m}(S) =(240m)^2\sum_{uv\in E(C)}x_u x_v \ge \delta(240m)^2 > \beta(240m)^2.

    So GmG_m satisfies the hypothesized local-density condition but contains no triangle. Since mm is arbitrary, this gives counterexamples of unbounded order. Therefore the conjecture, as stated, is false.

    Citation: No external citation is needed; the counterexample is the balanced blow-up of the Clebsch graph.

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    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 KNOWN

      PASS

      The counterexample attacks the conjecture exactly as stated. For α=43/120\alpha=43/120, the required lower bound is (5α−2)/25=−1/120(5\alpha-2)/25=-1/120. The balanced blow-ups of the Clebsch graph are triangle-free, and since the Clebsch graph has independence number 55, any vertex subset of normalized size 43/120>5/1643/120>5/16 must induce a positive normalized number δ\delta of edges. Compactness justifies δ>0\delta>0. Taking β=δ/2>0>−1/120\beta=\delta/2>0>-1/120 gives arbitrarily large triangle-free graphs satisfying the local-density hypothesis. Thus the supplied conjecture is rigorously disproved.

      Novelty assessment

      KNOWN

      Classification rationale: The “43/120” endpoint is already known to be erroneous/superseded. At α=43/120\alpha=43/120, the displayed bound (5α−2)/25(5\alpha-2)/25 is negative, so the statement is immediately vulnerable; the Clebsch blow-up is only a routine positive-density version of a standard triangle-free blow-up obstruction.

      Literature check: Modern treatments of this local density problem do not state the C5C_5-blow-up line down to 43/12043/120. Bedenknecht–Mota–Reiher–Schacht state that Erdős et al. conjectured

      β(α)=5α−225\beta(\alpha)=\frac{5\alpha-2}{25}

      only for α∈[53/120,17/30]\alpha\in[53/120,17/30], and explicitly note that for α<53/120\alpha<53/120 Andrásfai-type blow-ups give better lower bounds. Thus the Memphis formulation with 43/12043/120 is not a live open conjecture in the literature; it is a misstatement/transcription issue relative to the known formulation.

      Citation: W. Bedenknecht, G. O. Mota, C. Reiher, M. Schacht, “On the local density problem for graphs of given odd-girth,” J. Graph Theory 90 (2019), 137–149; arXiv:1609.05712. See the introduction around the [53/120,17/30][53/120,17/30] range. Also: P. Erdős, R. J. Faudree, C. C. Rousseau, R. H. Schelp, “A local density condition for triangles,” Discrete Math. 127 (1994), 153–161.

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