RESEARCH / Focused target
Every Natural Number Counts the Rook Placements of a Crossword Grid
RELEASED FOUNDATIONS / PROPOSED CONNECTION
Research connections
All source anchors (1)
- D5/S3/Combinatorics/CrosswordRookCounts.resultAbsent from this graph
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
- For
r ≥ 1take the grid of sidemax(5, 2r − 1)whose white cells are the first2r − 1cells of column 2 together with, for0 ≤ 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 eveniand(1, 0)for oddi. - The cells
(2i + 1, t_i)are down words of length one, so every placement contains them and no other cell of row2i + 1. - Hence column 2 carries exactly one rook, at some row
2kwithk < r, and the three-cell side words are then forced to their upper endpoint before blockkand to their lower endpoint from blockkon. - Each of these
rforced choices is a placement, so the grid has exactlyrplacements. - 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.