Chromatic number of random graphs
DefinitionProbability of a large chromatic–cochromatic gap#✓fully proved
For natural numbers \(C\) and \(n\), the proportion of simple graphs \(G\) on the vertex set \(\{0, \dots, n - 1\}\) with \(\chi(G) - \zeta(G) > C\), where \(\chi\) is the chromatic number and \(\zeta\) the cochromatic number; equivalently, the probability of this event for a uniformly random labelled graph, \(G \sim G(n, 1/2)\).
noncomputable def probDiffGT (C : ℕ) (n : ℕ) : ℝ :=
(Fintype.card { G : SimpleGraph (Fin n) //
G.chromaticNumber.toNat - G.cochromaticNumber.toNat > C
} : ℝ) /
(Fintype.card (SimpleGraph (Fin n)) : ℝ)
Uses 1 · Used by 1
DefinitionErdős Problem 625#✓fully proved
The statement that for every natural number \(C\), the probability that a uniformly random simple graph \(G\) on the vertex set \(\{0, \dots, n - 1\}\) (that is, \(G \sim G(n, 1/2)\)) satisfies \(\chi(G) - \zeta(G) > C\) tends to \(1\) as \(n \to \infty\), where \(\chi\) is the chromatic number and \(\zeta\) the cochromatic number. In other words, \(\chi(G) - \zeta(G) \to \infty\) with high probability.
def conjecture : Prop :=
∀ C : ℕ, Filter.Tendsto (fun n => probDiffGT C n) Filter.atTop (𝓝 (1 : ℝ))
noncomputable def probDiffGT (C : ℕ) (n : ℕ) : ℝ :=
(Fintype.card { G : SimpleGraph (Fin n) //
G.chromaticNumber.toNat - G.cochromaticNumber.toNat > C
} : ℝ) /
(Fintype.card (SimpleGraph (Fin n)) : ℝ)
WantedExt/Combinatorics/ErdosProblem625Wanted.lean:27
Uses 1 · Used by 1
TheoremChromatic minus cochromatic number of a random graph tends to infinity#●ready to prove
For every natural number \(C\), the probability that a uniformly random simple graph \(G\) on the vertex set \(\{0, \dots, n - 1\}\) (that is, \(G \sim G(n, 1/2)\)) satisfies \(\chi(G) - \zeta(G) > C\) tends to \(1\) as \(n \to \infty\), where \(\chi\) is the chromatic number and \(\zeta\) the cochromatic number.
Petkov and GPT-5.6 improved Heckel's bounds to \(\chi(G) - \zeta(G) \gg n / (\log n)^3\) with high probability for \(G \sim G(n, 1/2)\), as Heckel predicted; this answers Erdős Problem 625 in the affirmative.
public theorem_wanted conjecture_holds : conjecture
def conjecture : Prop :=
∀ C : ℕ, Filter.Tendsto (fun n => probDiffGT C n) Filter.atTop (𝓝 (1 : ℝ))
WantedExt/Combinatorics/ErdosProblem625Wanted.lean:35
Sources
- T. F. Bloom, Erdős Problem #625, erdosproblems.com (live page, LaTeX source and forum threads read 2026-10-22), https://www.erdosproblems.com/625; S. Petkov (with GPT-5.6), proof for Erdős Problem #625, https://github.com/SamPetkov/Erdos/blob/main/625/arxiv_625.pdf.
Uses 1
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