trureturingGitHub MATHEMATICAL ATLAS

RESEARCH / Focused target

Every Natural Number Counts the Rook Placements of a Crossword Grid

Literature status: not recheckedRepository record: proved

RELEASED FOUNDATIONS / PROPOSED CONNECTION

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Problem

The question

Joel Brewster Lewis and Robert Won, Non-attacking rook placements on crossword grids, arXiv:2609.03081v1, Question 4.2:

For which positive integers r does there exist a crossword grid which admits exactly r rook placements? Can the crossword grids which admit exactly r rook placements be characterized?

The paper observes that block-diagonal juxtaposition multiplies counts, so that it suffices to treat prime r, and it conjectures that permutation grids miss infinitely many values.

Motivation

Our foothold

The frozen theorem D5/S3/Combinatorics/CrosswordRookCounts.result shows that every natural number is attained by a square crossword grid. This answers the first part of Question 4.2; the characterization asked in the second part is not addressed.

Gap

Missing bridges

Issue 10086 records the screen made before the work: the arXiv record has only v1, and neither citing papers, the paper's public review page, the searched formal-conjecture catalogues nor this repository answer the question.

Route

Proposed approach

  1. For r ≥ 1 take the grid of side max(5, 2r − 1) whose white cells are the first 2r − 1 cells of column 2 together with, for 0 ≤ i ≤ r − 2, the cells (2i, s_i), (2i + 1, s_i), (2i + 2, s_i), (2i + 1, t_i), where (s_i, t_i) = (3, 4) for even i and (1, 0) for odd i.
  2. The cells (2i + 1, t_i) are down words of length one, so every placement contains them and no other cell of row 2i + 1.
  3. Hence column 2 carries exactly one rook, at some row 2k with k < r, and the three-cell side words are then forced to their upper endpoint before block k and to their lower endpoint from block k on.
  4. Each of these r forced choices is a placement, so the grid has exactly r placements.
  5. For r = 0, the 3 × 3 grid with white cells (0,1), (0,2), (1,0), (2,0) has two length-one down words inside one across word.

Falsifier

What would falsify this route

The statement would fail if some grid of the family had a placement avoiding a length-one down word or more than one rook in column 2.

Evidence

Evidence to collect

Exhaustive enumeration shows that 2 × 2, 3 × 3 and 4 × 4 grids attain {1, 2}, {0, 1, 2, 3, 4, 6} and {0, …, 8, 10, 12, 14, 18, 24}; the family was recounted independently for 1 ≤ r ≤ 120.

Triage

Scope assessment

theorem; Tier 1 open question stated in a 2026 paper, preregistered in issue 10086 before the work. The computational use is none: the central theorems hold for every r.

ASSUMED-UNVERIFIED

Unverified assumptions

The literature screen is bounded to the sources named above; no worldwide priority claim is made.

Dossier & anchor history

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