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Statement

Let k ≥5 be an odd integer and G be a (n,d,λ)(n,d,\lambda) -graph satisfying dk−1≫λk−2d^{k-1}\gg \lambda^{k-2} . Then G has global resilience (1 / 4+o(1)) n d with respect to being CkC_{k} -free.

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  • Some problems in the theory of pseudo-random graphs
  • FAR
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No person has examined this. Everything below was judged by machines. say whether it holds →

  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 odd k≥5k\ge 5, interpret an (n,d,λ)(n,d,\lambda)-graph as a finite simple dd-regular graph on nn vertices whose nontrivial adjacency eigenvalues have absolute value at most λ\lambda. The conjecture asserts that if

    dk−1≫λk−2,d^{k-1}\gg \lambda^{k-2},

    then the minimum number of edges to delete from GG to make it CkC_k-free is (1/4+o(1))nd(1/4+o(1))nd.

    Result: The statement is false.

    Let g=k+2g=k+2, still odd. For t→∞t\to\infty, let GtG_t be the balanced blow-up of the cycle CgC_g: replace each vertex of CgC_g by an independent set of size tt, and replace each edge of CgC_g by a complete bipartite graph between the corresponding parts.

    Then

    n=gt,d=2t.n=gt,\qquad d=2t.

    The adjacency matrix is A(Cg)⊗JtA(C_g)\otimes J_t, so the nonzero eigenvalues are

    2tcos⁡(2πj/g),j=0,…,g−1.2t\cos(2\pi j/g),\qquad j=0,\dots,g-1.

    Thus the largest nontrivial absolute eigenvalue is

    λ=2tcos⁡(π/g)<2t=d.\lambda=2t\cos(\pi/g)<2t=d.

    Hence

    dk−1λk−2=(2t)k−1(2tcos⁡(π/g))k−2=2tcos⁡(π/g)k−2→∞,\frac{d^{k-1}}{\lambda^{k-2}} = \frac{(2t)^{k-1}}{(2t\cos(\pi/g))^{k-2}} = \frac{2t}{\cos(\pi/g)^{k-2}} \to\infty,

    so the hypothesis holds.

    But GtG_t is CkC_k-free. Indeed, any CkC_k in GtG_t would project to a closed walk of length kk in CgC_g. Such a walk has steps ±1\pm1 modulo gg, so the sum of its kk steps is congruent to 0(modg)0\pmod g. Since k<gk<g, the only possible multiple of gg is 00; but a sum of kk many ±1\pm1’s is odd, hence cannot be 00. Contradiction.

    Therefore GtG_t is already CkC_k-free, so its deletion resilience toward becoming CkC_k-free is 00, not (1/4+o(1))nd(1/4+o(1))nd. Equivalently, ex⁡(Gt,Ck)=e(Gt)=nd/2\operatorname{ex}(G_t,C_k)=e(G_t)=nd/2, not (1/4+o(1))nd(1/4+o(1))nd.

    The defect is not merely bipartiteness: these GtG_t are connected and non-bipartite with λ<d\lambda<d. The hypothesis dk−1≫λk−2d^{k-1}\gg\lambda^{k-2} is simply too weak; a natural repair would need an additional density/pseudorandomness condition involving nn, such as the commonly appearing stronger form dk−1/n≫λk−2d^{k-1}/n\gg\lambda^{k-2}.

    Citation: No external citation is needed; the counterexample is explicit.

  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 KNOWN

    PASS

    The construction is a valid counterexample to the conjecture as stated. The balanced blow-up of Ck+2C_{k+2} is an (n,d,λ)(n,d,\lambda)-graph with d=2td=2t, λ=2tcos⁡(π/(k+2))\lambda=2t\cos(\pi/(k+2)), so dk−1/λk−2→∞d^{k-1}/\lambda^{k-2}\to\infty. It is CkC_k-free because any CkC_k would project to a closed walk of odd length k<k+2k<k+2 in Ck+2C_{k+2}, impossible. Hence the deletion resilience is 00, contradicting (1/4+o(1))nd(1/4+o(1))nd.

    Novelty assessment

    KNOWN

    Classification rationale: The counterexample is not a new combinatorial contribution. The key defect is the missing factor of nn in the standard pseudorandomness condition. Stronger known counterexamples already exist: Alon–Kahale constructions of C2r+1C_{2r+1}-free pseudorandom (n,d,λ)(n,d,\lambda)-graphs show the sharpness of the condition

    λ2r−1≪d2r/n.\lambda^{2r-1}\ll d^{2r}/n.

    Such examples satisfy d2r/λ2r−1→∞d^{2r}/\lambda^{2r-1}\to\infty, so they already refute the weaker hypothesis with no 1/n1/n factor.

    Literature check: The relevant literature states the conjecture with the nn denominator. Aigner-Horev–Hàn–Schacht prove the odd-cycle Turán/resilience statement under

    λℓ−2≪dℓ−1nlog⁡(n)−(ℓ−2)(ℓ−3),\lambda^{\ell-2}\ll \frac{d^{\ell-1}}{n}\log(n)^{-(\ell-2)(\ell-3)},

    and explicitly note that Alon–Kahale constructions show this range is best possible up to polylogarithmic factors. Berger–Lee–Schacht later remove the polylogarithmic loss and state that the Alon–Kahale C2k+1C_{2k+1}-free pseudorandom graphs make the result asymptotically best possible. These are stronger known obstructions than the submitted blow-up-of-a-cycle example.

    Citation: E. Aigner-Horev, H. Hàn, M. Schacht, “Extremal results for odd cycles in sparse pseudorandom graphs,” Combinatorica 34 (2014), 379–406; arXiv:1602.03663.
    S. Berger, J. Lee, M. Schacht, “Odd cycles in subgraphs of sparse pseudorandom graphs,” arXiv:1906.05100.

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