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# Some Comments on Philatelic Latin Squares from Pakistan

Some Comments on Philatelic Latin Squares from Pakistan

## 454

454 ong>Someong> ong>Commentsong> on Philatelic Latin Squares from Pakistan FIGURE 6.3.1: Hong Kong 2006, Scott 1219v.

Hong Kong transport: 6x3 splits into two 3x3 Latin squares 1 2 3 4 5 6 2 3 1 5 6 4 3 1 2 6 4 5 1 2 3 4 4 5 1 6 2 5 1 3 5 6 4 2 5 3 4 1 1� 2� 4 X3 3 6 4 5 3 1 2 −X3 6 7 9 10 6 8 1 2 3 4 8 5 10 6 6 7 1 9 4 5 6 1 1 2 2 3 3B3 = ⎝ 4 2 5 2 3 1 5 2 6 3 4 5 1 3 4 L3 M3 M3 L3 5 6 4 2 3 3 1 1 4 5 4 2 3 1 2 6 4 5 5 L3 = 4 5 2 3 1 6 4 5 3 1 2 6 5 4 1 2 3 3 3 3 3 3 3 3 3 3 6 7 8 9 10 5 4 1 2 3 7 9 10 6 8 4 5 2 3 1 8 10 6 7 9 1 2 3 4 5 6 1 1 2 3 4 5 6 1 2 3 1 5 6 4 4 3 5 6 1 2 3 2 3 1 2 6 4 5 2 5 3 1 5 6 4 3 4 5 6 1 2 3 5 2 6 4 2 3 1 4 5 6 4 2 3 1 3 4 1 2 6 4 5 5 6 4 5 3 1 2 6 6 4 5 3 1 2 6 FIGURE 6.3.2: Philatelic 2 5 Sudoku 4 puzzle 3 1S3 (left panel) with solution (right panel). 6 7 8 9 10 8 10 6 7 9 7 9 5 B3 = E2×2 9 ⊗ 8K3×3 7+ 3K2×2 10 ⊗6E3×3 − 3E2×2 ⊗ E3×3 = B (2,3) 2 3 1 5 6 4 3 3 1 2 6 4 5 5 4 5 6 1 2 3 2 5 6 4 2 3 1 4 6 4 5 3 1 2 6 Ka Lok Chu et al. 455 2To solve 3 this 1 puzzle 5 S3 6 we4first sort so3 that the elements in column 1 are in ascending 3 4 5 1 2 5 7 9 order (Figure 6.3.2, center panel, shaded part). We then form a 6 × 6 Latin square which 3 1 2 6 1 4 2 5 3 4 5 5 7 9 5 is 2 × 2 block-Latin with 3 × 3 blocks (Figure 6.3.2, center panel). We then rearrange the rows 4 back 5 into 6 the 1original 6 2 7sequence 3 8 to9 find 2 10the solution (Figure 9 5 6.3.2, 7 right panel). It is easy to see that the matrix representing the solution has rank equal to 4 and (column) null space spanned by the columns of , where X3 = 9 5 7 say, where ⊗ denotes 10 Kronecker 6 product, 9 8 Ep×q 7 is the p×q matrix with every element equal L-M equals minus ⎛ ⎞ L+M � L-M � 1 1 −1 0 . 0 3−1 3 3 equals minus The reduced-form solution to puzzle S3 (Figure 6.3.2, center panel) may be represented by the matrix 4 5 2 3 1 3 3 3 ⎛ ⎞ where 3 3 3 ⎠, (6.3.1) ⎛ ⎞ ⎜ 1 ⎜ ⎜2 ⎝ 2 3 3 ⎟ 1 ⎟ ⎠ 3 1 2 , M3 ⎛ ⎞ ⎜ 4 ⎜ = ⎜ ⎜5 ⎝ 5 6 6 ⎟ 4 ⎟ ⎠ 6 4 5 = L3 ⎛ ⎞ ⎜ 3 ⎜ + ⎜ ⎜3 ⎝ 3 3 3 ⎟ 3 ⎟ ⎟. ⎠ (6.3.2) 3 3 3 The 2×2 block-Latin representation (6.3.1) and the fact that all the elements of M3 −L3 L+M are equal (to 3) allow7us to9rewrite 10 the6matrix 8 B3 in (6.3.1) as a “generalized Kroneckersum” 5 7 9 to 1 and K2×2 = 3 , (6.3.3) ⎛ ⎞ ⎜ 1 2 3 ⎟ 1 2 ⎜ ⎟ ⎝ ⎠, K3×3 = ⎜ ⎜2 3 1 ⎟ 2 1 ⎝ ⎠ 3 1 2 = L3. (6.3.4)

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