FormalConjectures/ErdosProblems/617.lean
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Project sources: Pinned project documentation, Formal Conjectures paper, arXiv v1
4 declarations, 4 contain sorry, 0 declared as axioms
Erdős Problem 617
References:
- erdosproblems.com/617
- [ErGy99] Erdős, Paul and Gyárfás, András, Split and balanced colorings of complete graphs. Discrete Math. (1999), 79-86.
Imports (1)
Imported by 1 file
- theorem
Erdos617.erdos_617Let $r\geq 3$ sorry line 42 sourceSource documentation
Let $r\geq 3$. If the edges of $K_{r^2+1}$ are $r$-coloured then there exist $r+1$ vertices with at least one colour missing on the edges of the induced $K_{r+1}$.
In other words, there is no balanced colouring.
A conjecture of Erdős and Gyárfás [ErGy99].
depends on sorry what this means
Statement uses:
Nat,GE.ge,OfNat.ofNat,Fintype,DecidableEq,Eq,Fintype.card,HAdd.hAdd,HPow.hPow,Sym2,Fin,Exists,Finset,And,Finset.card,Membership.mem,Ne,Sym2.mkProof uses:
sorryAx,Lean.Name,Bool.false,Lean.Name.num,Lean.Name.str,Lean.Name.anonymous20 instances, coercions and auxiliary constants left out.
Tactic trace: 1 steps
Open to load the checked before-and-after goal states.
theorem erdos_617 (r : ℕ) (hr : r ≥ 3) {V : Type} [Fintype V] [DecidableEq V] (hV : Fintype.card V = r^2 + 1) (coloring : Sym2 V → Fin r) : ∃ (S : Finset V) (k : Fin r), S.card = r + 1 ∧ ∀ u ∈ S, ∀ v ∈ S, u ≠ v → coloring s(u, v) ≠ k := by … - theorem
Erdos617.erdos_617.variants.r_eq_3Erdős and Gyárfás [ErGy99] proved the conjecture for $r=3$ sorry line 53 sourceSource documentation
Erdős and Gyárfás [ErGy99] proved the conjecture for $r=3$.
depends on sorry what this means
Statement uses:
Nat,GE.ge,OfNat.ofNat,Fintype,DecidableEq,Eq,Fintype.card,HAdd.hAdd,HPow.hPow,Sym2,Fin,Exists,Finset,And,Finset.card,Membership.mem,Ne,Sym2.mkProof uses:
sorryAx,Lean.Name,Bool.false,Lean.Name.num,Lean.Name.str,Lean.Name.anonymous20 instances, coercions and auxiliary constants left out.
Tactic trace: 1 steps
Open to load the checked before-and-after goal states.
theorem erdos_617.variants.r_eq_3 (r : ℕ) (hr : r ≥ 3) {V : Type} [Fintype V] [DecidableEq V] (hV : Fintype.card V = 3^2 + 1) (coloring : Sym2 V → Fin 3) : ∃ (S : Finset V) (k : Fin 3), S.card = 3 + 1 ∧ ∀ u ∈ S, ∀ v ∈ S, u ≠ v → coloring s(u, v) ≠ k := by … - theorem
Erdos617.erdos_617.variants.r_eq_4Erdős and Gyárfás [ErGy99] proved the conjecture for $r=4$ sorry line 64 sourceSource documentation
Erdős and Gyárfás [ErGy99] proved the conjecture for $r=4$.
depends on sorry what this means
Statement uses:
Nat,GE.ge,OfNat.ofNat,Fintype,DecidableEq,Eq,Fintype.card,HAdd.hAdd,HPow.hPow,Sym2,Fin,Exists,Finset,And,Finset.card,Membership.mem,Ne,Sym2.mkProof uses:
sorryAx,Lean.Name,Bool.false,Lean.Name.num,Lean.Name.str,Lean.Name.anonymous20 instances, coercions and auxiliary constants left out.
Tactic trace: 1 steps
Open to load the checked before-and-after goal states.
theorem erdos_617.variants.r_eq_4 (r : ℕ) (hr : r ≥ 3) {V : Type} [Fintype V] [DecidableEq V] (hV : Fintype.card V = 4^2 + 1) (coloring : Sym2 V → Fin 4) : ∃ (S : Finset V) (k : Fin 4), S.card = 4 + 1 ∧ ∀ u ∈ S, ∀ v ∈ S, u ≠ v → coloring s(u, v) ≠ k := by … - theorem
Erdos617.erdos_617.variants.r2Erdős and Gyárfás [ErGy99] showed this property fails for infinitely many $r$ if we replace $r^2+1$ by $r^2$ sorry line 76 sourceSource documentation
Erdős and Gyárfás [ErGy99] showed this property fails for infinitely many $r$ if we replace $r^2+1$ by $r^2$.
depends on sorry what this means
Statement uses:
Set.Infinite,Nat,Set.ofPred,Exists,Fintype,DecidableEq,And,Eq,Fintype.card,HPow.hPow,OfNat.ofNat,Sym2,Fin,Finset,Finset.card,HAdd.hAdd,Membership.mem,Ne,Sym2.mkProof uses:
sorryAx,Lean.Name,Bool.false,Lean.Name.num,Lean.Name.str,Lean.Name.anonymous18 instances, coercions and auxiliary constants left out.
Tactic trace: 1 steps
Open to load the checked before-and-after goal states.
theorem erdos_617.variants.r2 : {r : ℕ | ∃ (V : Type) (_ : Fintype V) (_ : DecidableEq V), Fintype.card V = r^2 ∧ ∃ (coloring : Sym2 V → Fin r), ∀ (S : Finset V), S.card = r + 1 → ∀ (k : Fin r), ∃ u ∈ S, ∃ v ∈ S, u ≠ v ∧ coloring s(u, v) = k}.Infinite := by …