FormalConjectures/ErdosProblems/659.lean

GitHub source file

1 declarations, 1 contain sorry, 0 declared as axioms

1. theorem Erdos659.erdos_659 Documented declaration: erdos_659 sorry line 58 source

   Source documentation

   Is there a set of $n$ points in $\mathbb{R}^2$ such that every subset of $4$ points determines at least $3$ distances, yet the total number of distinct distances is $\ll \frac{n}{\sqrt{\log n}}$?

   There does exist such a set: a suitable truncation of the lattice $\{(a,b\sqrt{2}): a,b\in\mathbb{Z}\}$ suffices. This construction appears to have been first considered by Moree and Osburn \cite{MoOs06}, who proved that it has $\ll \frac{n}{\sqrt{\log n}}$ many distinct distances. This construction was independently found by Lund and Sheffer, who further noted that this configuration contains no squares or equilateral triangles.

   There are only six possible configurations of $4$ points which determine only $2$ distances (first noted by Erdős and Fishburn [ErFi96]), and five of them contain either a square or an equilateral triangle. The remaining configuration contains four points from a regular pentagon, and Grayzel [Gr26] (using Gemini) has noted in the comments that this configuration can also be ruled out, thus giving a complete solution to this problem.

   Boris Alexeev provides a formalisation of the reduction, which is conditional on Bernays' theorem (assumed as an axiom in the proof to obtain the $O(n/\sqrt{\log n})$ bound). See the formal proof.

   L4DEL explanation

   Is there a set of $n$ points in $\mathbb{R}^2$ such that every subset of $4$ points determines at least $3$ distances, yet the total number of distinct distances is $\ll \frac{n}{\sqrt{\log n}}$? There does exist such a set: a suitable truncation of the lattice $\{(a,b\sqrt{2}): a,b\in\mathbb{Z}\}$ suffices. This construction appears to have been first considered by Moree and Osburn \cite{MoOs06}, who proved that it has $\ll \frac{n}{\sqrt{\log n}}$ many distinct distances. This construction was independently found by Lund and Sheffer, who further noted that this configuration contains no squares or equilateral triangles. There are only six possible configurations of $4$ points which determine only $2$ distances (first noted by Erdős and Fishburn [ErFi96]), and five of them contain either a square or an equilateral triangle. The remaining configuration contains four points from a regular pentagon, and Grayzel [Gr26] (using Gemini) has noted in the comments that this configuration can also be ruled out, thus giving a complete solution to this problem. Boris Alexeev provides a formalisation of the reduction, which is conditional on Bernays' theorem (assumed as an axiom in the proof to obtain the $O(n/\sqrt{\log n})$ bound). See the formal proof.

   Sources and attribution: Pinned Lean source (1646ca16afd6)

   depends on sorry what this means

   Statement uses: Iff, True, Exists, Nat, Finset, EuclideanSpace, Real, Fin, OfNat.ofNat, And, Eq, Finset.card, LE.le, distinctDistances, ENNReal, Asymptotics.IsBigO, Filter.atTop, HDiv.hDiv, Real.sqrt, Real.log

   Proof uses: sorryAx, Lean.Name, Bool.false, Lean.Name.num, Lean.Name.str, Lean.Name.anonymous

   42 instances, coercions and auxiliary constants left out.

   Tactic trace: 1 steps

   Open to load the checked before-and-after goal states.

   theorem erdos_659 : answer(True) ↔ ∃ A : ℕ → Finset ℝ²,
      (∀ n, #(A n) = n ∧ ∀ S ⊆ A n, #S = 4 → 3 ≤ distinctDistances S) ∧
       (fun n ↦ distinctDistances (A n)) ≪ fun n ↦ n / sqrt (log n) := by
   …

   Mentions: distinctDistances