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Erdős Problem 596

References:

    erdosproblems.com/596

    [Er87] Erdős, Some of my favourite problems in various branches of combinatorics, Mat. Lapok 1987.

    [NeRo75] Nešetřil and Rödl, The Ramsey property for graphs with forbidden complete subgraphs, J. Combin. Theory B 20 (1976), 243--249.

open SimpleGraph namespace Erdos596

Erdős Problem 596 (Erdős–Hajnal, [Er87]). For which graph pairs $(G_1, G_2)$ is it true that

(1) for every $n \geq 1$ there is a graph $H$ without a $G_1$ such that any $n$-colouring of $H$'s edges contains a monochromatic $G_2$, and yet (2) for every graph $H$ without a $G_1$ there is an $\aleph_0$-colouring of $H$'s edges with no monochromatic $G_2$?

Erdős and Hajnal originally conjectured that no such pair exists; but $(C_4, C_6)$ witnesses it (Nešetřil–Rödl + Erdős–Hajnal). The full question is to characterise the class of all such pairs, recorded here as answer(sorry).

See Problem 595 for the specific case $(G_1, G_2) = (K_4, K_3)$.

@[category research open, AMS 5] theorem declaration uses 'sorry'erdos_596 : {U₁ U₂ : Type} (G₁ : SimpleGraph U₁) (G₂ : SimpleGraph U₂), IsErdosHajnalExceptional G₁ G₂ (answer(sorry) : {U₁ U₂ : Type}, SimpleGraph U₁ SimpleGraph U₂ Prop) G₁ G₂ := {U₁ U₂ : Type} (G₁ : SimpleGraph U₁) (G₂ : SimpleGraph U₂), G₁.IsErdosHajnalExceptional G₂ (fun {U₁ U₂} => sorry) G₁ G₂ All goals completed! 🐙

Erdős–Hajnal exceptional pairs exist — recorded as a known direction of erdos_596.

@[category research solved, AMS 5] theorem declaration uses 'sorry'erdos_596.variants.exists_exceptional : (U₁ U₂ : Type) (G₁ : SimpleGraph U₁) (G₂ : SimpleGraph U₂), IsErdosHajnalExceptional G₁ G₂ := U₁ U₂ G₁ G₂, G₁.IsErdosHajnalExceptional G₂ All goals completed! 🐙

Every $C_4$-free graph is a countable union of trees (Erdős–Hajnal [Er87]); trees are acyclic, hence $C_6$-free, giving the countable Ramsey escape for $(C_4, C_6)$.

@[category research solved, AMS 5] theorem declaration uses 'sorry'erdos_596.variants.C4_free_countable_escape : HasCountableRamseyEscape (cycleGraph 4) (cycleGraph 6) := (cycleGraph 4).HasCountableRamseyEscape (cycleGraph 6) All goals completed! 🐙

Nešetřil–Rödl [NeRo75]: for every $n \geq 1$ there is a $C_4$-free graph whose edges cannot be $n$-coloured without a monochromatic $C_6$.

@[category research solved, AMS 5] theorem declaration uses 'sorry'erdos_596.variants.C4_C6_finite_ramsey : HasFiniteRamseyProperty (cycleGraph 4) (cycleGraph 6) := (cycleGraph 4).HasFiniteRamseyProperty (cycleGraph 6) All goals completed! 🐙

The pair $(C_4, C_6)$ is Erdős–Hajnal exceptional; combines C4_C6_finite_ramsey and C4_free_countable_escape.

@[category research solved, AMS 5] theorem erdos_596.variants.C4_C6_is_exceptional : IsErdosHajnalExceptional (cycleGraph 4) (cycleGraph 6) := erdos_596.variants.C4_C6_finite_ramsey, erdos_596.variants.C4_free_countable_escape

The original Erdős–Hajnal conjecture (that no exceptional pair exists) is false — witnessed by $(C_4, C_6)$ via C4_C6_is_exceptional.

@[category research solved, AMS 5] theorem declaration uses 'sorry'erdos_596.variants.original_conjecture_is_false : answer(False) {U₁ U₂ : Type} (G₁ : SimpleGraph U₁) (G₂ : SimpleGraph U₂), ¬IsErdosHajnalExceptional G₁ G₂ := False {U₁ U₂ : Type} (G₁ : SimpleGraph U₁) (G₂ : SimpleGraph U₂), ¬G₁.IsErdosHajnalExceptional G₂ All goals completed! 🐙

Whether $(K_4, K_3)$ is Erdős–Hajnal exceptional is precisely the content of Erdős Problem 595. The finite Ramsey property holds (Folkman 1970, Nešetřil–Rödl [NeRo75]); the open part is whether every $K_4$-free graph is a countable union of triangle-free graphs.

@[category research open, AMS 5] theorem declaration uses 'sorry'erdos_596.variants.K4_K3_exceptional_iff : answer(sorry) IsErdosHajnalExceptional (completeGraph (Fin 4)) (completeGraph (Fin 3)) := True (completeGraph (Fin 4)).IsErdosHajnalExceptional (completeGraph (Fin 3)) All goals completed! 🐙

Folkman 1970 / Nešetřil–Rödl [NeRo75]: for every $n \geq 1$ there is a $K_4$-free graph whose edges cannot be $n$-coloured without a monochromatic triangle.

@[category research solved, AMS 5] theorem declaration uses 'sorry'erdos_596.variants.K4_K3_finite_ramsey : HasFiniteRamseyProperty (completeGraph (Fin 4)) (completeGraph (Fin 3)) := (completeGraph (Fin 4)).HasFiniteRamseyProperty (completeGraph (Fin 3)) All goals completed! 🐙

The empty graph on Fin 0 is Free of any nontrivial subgraph (vacuous). This is the simplest non-trivial witness to G₁.Free H appearing in HasFiniteRamseyProperty.

@[category test, AMS 5] theorem erdos_596.test.empty_is_free : (cycleGraph 4).Free ( : SimpleGraph (Fin 0)) := fun f => Fin.elim0 (f.toEmbedding 0) end Erdos596