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Placement Counts Under Extreme Insertions

Abstract

Adding a final maximum, or inserting an interior maximum under a specified ordering condition, cannot decrease the rook-placement count.

Theorem 1.1 (Appending a maximum).

Lean statement: D5/S3/Combinatorics/CrosswordGrid/SkewMergedRookDeletions.corner_lift

Proof. Machine-checked in Lean as D5/S3/Combinatorics/CrosswordGrid/SkewMergedRookDeletions.corner_lift (✓ std3). ∎

Source. Repository-derived.

Acknowledgement. Joel Brewster Lewis, Robert Won (2026). Non-attacking rook placements on crossword grids. DOI: 10.48550/arXiv.2609.03081. URL: https://arxiv.org/abs/2609.03081v1.

Commentary.

Let n be positive and let u be a permutation of zero through n minus one. Let w be a permutation of zero through n with w(n) equal to n and w(i) equal to u(i) for every i less than n. The number of complete rook placements of the grid of u is at most the number for the grid of w.

Theorem 1.2 (Inserting an interior maximum).

Lean statement: D5/S3/Combinatorics/CrosswordGrid/SkewMergedRookDeletions.interior_lift

Proof. Machine-checked in Lean as D5/S3/Combinatorics/CrosswordGrid/SkewMergedRookDeletions.interior_lift (✓ std3). ∎

Source. Repository-derived.

Acknowledgement. Joel Brewster Lewis, Robert Won (2026). Non-attacking rook placements on crossword grids. DOI: 10.48550/arXiv.2609.03081. URL: https://arxiv.org/abs/2609.03081v1.

Commentary.

Let n be positive, let u be a permutation of zero through n minus one, and let w be a permutation of zero through n. Choose a position p strictly between zero and n with w(p) equal to n. Deleting that position from w, without changing the other values, gives u. If the position of value n minus one in w is less than p, the number of complete rook placements of the grid of u is at most the number for the grid of w.

References