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A Bounded Capacity Generalization of the Classic n-Queens Problem #4177

@LucDuponcheelAtGitHub

Description

@LucDuponcheelAtGitHub

What is the conjecture

We introduce a generalization of the classic $n$-queens problem, parameterized by a capacity bound $0 \le c \le n$. The puzzle requires placing exactly $n . c$ coins on an $n \times n$ board such that no row, column, or diagonal contains more than $c$ coins. When $c=1$, this reduces precisely to the canonical $n$-queens formulation. We formulate this problem mathematically and present extensive empirical evidence verifying the conjecture up to $n = 1500$. Using a highly optimized local search framework based on the min-conflicts heuristic, we observe smooth polynomial execution scaling and a structural transition from highly geometric permutation-like arrangements at low $c$ to maximum-entropy, uniform distribution profiles (``noisy TV screen'') in the dense middle-spectrum where $c \approx n/2$. This smooth behavior strongly suggests that the solution space remains dense and free of hard phase transitions across all valid parameter ranges.

AMS categories

ams-05

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  • I plan on adding this conjecture to the repository
  • This issue is up for grabs: I would like to see this conjecture added by somebody else

Lean code

import Mathlib.Data.Matrix.Basic
import Mathlib.Data.Fin.Basic

namespace DuponcheelConjecture

variable {n : ℕ} (hn : n ≥ 4) {c : ℕ} (hc : c ≤ n)

abbrev BoolBoard (n : ℕ) := Matrix (Fin n) (Fin n) Bool

abbrev NatBoard (n : ℕ) := Matrix (Fin n) (Fin n) ℕ

theorem n_c_coins_placement_conjecture :
    ∃ (B : BoolBoard n),
      let N := B.map (fun b ↦ if b then 1 else 0) ;
      -- Each row, column, upward diagonal, and downward diagonal
      -- contains at most `c` coins.
      ∀ i : Fin n, ∑ j : Fin n, N i j ≤ c ∧
      ∀ j : Fin n, ∑ i : Fin n, N i j ≤ c ∧
      ∀ k : ℕ, ∑ i : Fin n, ∑ j : Fin n, (if i + j = k then N i j else 0) ≤ c ∧
      ∀ k : ℤ, ∑ i : Fin n, ∑ j : Fin n, (if i - j = k then N i j else 0) ≤ c ∧
      -- Total number of coins placed is `n * c`.
      ∑ i : Fin n, ∑ j : Fin n, N i j = n * c

  := by sorry

end DuponcheelConjecture

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