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Some Comments on Philatelic Latin Squares from Pakistan

Some Comments on Philatelic Latin Squares from Pakistan

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452 2 7 4 1 6 3 8 5 2 7 4 1 6 3 8 5 2 7 4 1 6 3 8 5 2 7 4 1 6 3 While 1 the 6 10 × 3 10 philatelic 8 5 Sudoku 2 4 puzzle 7 36 S1 based on the 6 × 5 sheetlet of mushroomstamps 2 from 7 Pakistan 4 1 is the 6 only 3 10×10 5 philatelic 8 36 Sudoku puzzle that we have found (from any country), we have found a sheetlet of 5×4 stamps (Figure 6.2.1) from the USA featur- 3 8 5 2 7 4 6 1 36 ing eight classical composers and conductors in the “Legends of American music series” and4which 1 yields 6 the 3 8 × 8 philatelic 5 7 Sudoku 2 36 puzzle S2 (Figure 6.2.1) in Maack–brickwall format 5 (Figure 2 7 6.1.2). 4 The 1 8 classical 6 8 composers 3 36 and conductors depicted in Figure 6.2.1 are (1) Leopold Stokowski (1882–1977), (2) Arthur Fiedler (1894–1979), (3) George Szell 6 3 8 5 2 7 1 4 36 (1897–1970); (4) Eugene Ormandy (1899–1985), (5) Samuel Barber (1910–1981), (6) 7 4 1 6 3 8 2 5 36 Ferde Grofé (1892–1972), (7) Charles Ives (1874–1954), (8) Louis Moreau Gottschalk (1829–1869). 8 5 2 7 4 1 3 6 36 36 36 36 36 36 36 36 288 1 2 3 4 5 6 7 8 3 4 1 2 7 8 5 6 4 3 2 1 ong>Someong> ong>Commentsong> on Philatelic Latin Squares from Pakistan 1 2 3 4 5 6 7 8 5 6 7 8 1 2 3 4 3 4 1 2 7 8 5 6 7 8 5 6 3 4 1 2 4 3 2 1 6 7 8 5 6 7 8 5 4 3 2 1 2 1 4 3 8 5 6 7 8 5 6 7 2 1 4 3 FIGURE 6.2.2: Philatelic Sudoku puzzle S2 based on Figure 6.2.1, with solution (right panel). The puzzle (Figure 6.2.2, left panel) is to create an 8 × 8 Latin square with the extra conditions that the numbers 1,2,3,4 appear, in some order, in each of the regions coloured pale blue, and that the numbers 5,6,7,8 appear, in some order, in each of the other regions. Our solution to puzzle S2, using the two 4 × 4 Latin squares in (6.2.2) below is shown in Figure 6.2.2 (right panel), the cells with “givens” are in brown. If the rows in Figure 6.2.2 (right panel) are rearranged so that the numbers in column 1 are in ascending order, and so the 8 × 8 Latin square is now in reduced-form, then it may be represented by the 8 × 8 Latin square matrix BS2 in block-Latin format, see also (6.1.2) above: ⎛ B2 = ⎝ L2 M2 M2 L2 ⎞ ⎠, (6.2.1)

Ka Lok Chu et al. 453 where the 4 × 4 matrices L2 and M2 are defined as: L2 = ⎛ ⎞ ⎛ ⎞ 1 ⎜ ⎜2 ⎜ ⎝3 2 1 4 3 4 1 4 ⎟ 3 ⎟ ⎟, 2 ⎟ ⎠ 5 ⎜ ⎜6 M2 = ⎜ ⎝7 6 7 8 7 8 5 8 ⎟ 5 ⎟ ⎟. 6 ⎟ ⎠ (6.2.2) 4 3 2 1 8 5 6 7 Moreover L2 is “criss-cross” in that all elements of the main-forwards diagonal are equal as are all elements of the main-backwards diagonal, and is “centro-symmetric” [12, p. 181] in that reversing the columns and the rows leaves L2 unchanged. The Latin square matrix M2 is the one-step backwards circulant. We also find it interesting that both matrices L2 and M2 in (6.2.2) are themselves 2 × 2 block-Latin (with 2 × 2 block submatrices). Using the 2 × 2 block-Latin rank formula (6.1.4) above repeatedly, it is easy to see that the matrix representing the solution (Figure 6.2.2, right panel) to the 8 × 8 philatelic Sudoku puzzle S2 is singular with rank � equal � to 6; moreover the (column) null space is spanned by X2 the (columns of the) matrix , where −X2 ⎛ ⎞ −1 0 ⎜ ⎟ ⎜ X2 = ⎜ 2 1 ⎟ ⎟. (6.2.3) ⎝ −1 0 ⎠ 0 −1 6.3 Philatelic Sudoku puzzle S3: transport-stamps from Hong Kong Our third philatelic Sudoku puzzle S3 is based on a 6 × 3 sheetlet (Figure 6.3.1, below) of stamps from Hong Kong featuring 6 different modes of government transport, which we code (Figure Kronecker-sum) as follows: (1) Correctional Services Department: Security Bus, (2) Fire Services Department: Hydraulic Platform, (3) Hong Kong Police Force: Versatile Traffic Patrol Motorcycle, (4) Customs and Excise Department: Mobile X-ray Vehicle Scanning System, (5) Government Flying Service: Super Puma Helicopter, (6) Immigration Department: Launch. Then we may represent the sheetlet as shown in Figure 6.3.2 (left panel).

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