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# Some Comments on Philatelic Latin Squares from Pakistan

Some Comments on Philatelic Latin Squares from Pakistan

## 462

462 ong>Someong> ong>Commentsong> on Philatelic Latin Squares from Pakistan If we rearrange the rows in the solution in Figure 6.5.2 so that the entries in column 1 are in ascending order then we obtain the reduced-form 5 × 5 block-Latin square shown in Figure 6.5.3, which allows the generalized Kronecker-sum representation E5×5 ⊗ K2×2 + 2L5 ⊗ E2×2 − 2E5×5 ⊗ E2×2, (6.5.2) � � 1 2 where E has every element equal to 1, K2×2 = , and the 5 × 5 reduced-form Latin 2 1 square matrix ⎛ ⎞ 1 ⎜ ⎜2 ⎜ L5 = ⎜ ⎜3 ⎜ ⎝4 2 1 5 3 3 5 4 1 4 3 2 5 5 ⎟ 4 ⎟ 1 ⎟ ⎟, ⎟ 2⎟ ⎠ (6.5.3) 5 4 2 1 3 which we note is not a circulant, but is nonsingular (all 5 × 5 Latin square matrices are nonsingular). We find that the rank of the matrices in Figure 6.5.3 is equal to 7 and that a basis for its (column) null space is the set of columns of the matrix ⎛ ⎞ 1 ⎜ −1 ⎜ ⎜−1 ⎜ 1 ⎜ 0 ⎜ 0 ⎜ 0 ⎜ 0 ⎝ 0 0 0 0 0 1 −1 −1 1 0 1 ⎟ −1 ⎟ 0 ⎟ ⎛ ⎟ 1 0 ⎟ ⎜ ⎟ ⎜ 0 ⎟ ⎜ −1 ⎟ = ⎜ 0 ⎟ ⎜ 0 ⎟ ⎝ ⎟ 0 0 ⎟ 0 ⎟ 0 ⎟ −1 ⎠ 0 0 1 −1 0 ⎞ 1 ⎛ ⎞ ⎟ 0 ⎟ ⎜ 1 ⎟ 0 ⎟ ⊗ ⎝ ⎠. ⎟ ⎠ −1 0 −1 (6.5.4) 0 0 1

Ka Lok Chu et al. 463 6.6 Philatelic Sudoku puzzle S6: stamps from the USA featuring “American Advances in Aviation” FIGURE 6.6.1: USA 2007, Scott 3925v.

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