erdos_1080

Let $G$ be a bipartite graph on $n$ vertices such that one part has $\lfloor n^{2/3}\rfloor$ vertices. Is there a constant $c>0$ such that if $G$ has at least $cn$ edges then $G$ must contain a $C_6$? The answer is no, as shown by De Caen and Székely [DeSz92], who in fact show a stronger result. Let $f(n,m)$ be the maximum number of edges of a bipartite graph between $n$ and $m$ vertices which does not contain either a $C_4$ or $C_6$. A positive answer to this question would then imply $f(n,\lfloor n^{2/3}\rfloor)\ll n$. De Caen and Székely prove $n^{10/9}\gg f(n,\lfloor n^{2/3}\rfloor) \gg n^{58/57+o(1)}$ for $m\sim n^{2/3}$. They also prove more generally that, for $n^{1/2}\leq m\leq n$, $f(n,m) \ll (nm)^{2/3},$ which was also proved by Faudree and Simonovits. This was formalized in Lean by Alexeev using Aristotle.

theorem erdos_1080 :
    answer(False) ↔
    ∃ c > (0 : ℝ), ∀ (V : Type) [Fintype V] [Nonempty V] (G : SimpleGraph V) (X Y : Set V),
      IsBipartition G X Y → X.ncard = ⌊(Fintype.card V : ℝ) ^ (2/3 : ℝ)⌋₊ →
      G.edgeSet.ncard ≥ c * Fintype.card V →
        ∃ (v : V) (walk : G.Walk v v), walk.IsCycle ∧ walk.length = 6

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