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Graeko-Latin Squares
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Arrange the playing cards in a 4 by 4 grid such that each value occurs once in each row and once in each column and each suit occurs once in each row and once in each column (problem is due to Jacques Ozanam 1725).
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0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 Numbering
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0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 (0,0) (0,1) (0,2) (0,3) (1,0) (1,1) (1,2) (1,3) (2,0) (2,1) (2,2) (2,3) (3,0) (3,1) (3,2) (3,3) (Z[i][j]/n,Z[i][j]%n) Numbering
![(0,0) (0,1) (0,2) (0,3) (1,0) (1,1) (1,2) (1,3) (2,0) (2,1) (2,2) (2,3) (3,0) (3,1) (3,2) (3,3) (Z[i][j]/n,Z[i][j]%n) Numbering](http://images.slideplayer.com./33/10121135/slides/slide_10.jpg "(0,0) (0,1) (0,2) (0,3) (1,0) (1,1) (1,2) (1,3) (2,0) (2,1) (2,2) (2,3) (3,0) (3,1) (3,2) (3,3) (Z[i][j]/n,Z[i][j]%n) Numbering")
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0 5 10 15 6 3 12 9 11 14 1 4 13 8 7 2 Solution n=4
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0 5 10 15 6 3 12 9 11 14 1 4 13 8 7 2 (0,0) (1,1) (2,2) (3,3) (1,2) (0,3) (3,0) (2,1) (2,3) (3,2) (0,1) (1,0) (3,1) (2,0) (1,3) (0,2) Solution n=4
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0 5 10 15 6 3 12 9 11 14 1 4 13 8 7 2 (0,0) (1,1) (2,2) (3,3) (1,2) (0,3) (3,0) (2,1) (2,3) (3,2) (0,1) (1,0) (3,1) (2,0) (1,3) (0,2) Solution n=4
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0123 1032 2301 3210 X Latin All different in a row All different in a column Model 0
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0123 1032 2301 3210 0123 2301 3210 1032 XY Greek All different in a row All different in a column Model 0
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0123 1032 2301 3210 0123 2301 3210 1032 XY Graeco-Latin (0,0)(1,1)(2,2)(3,3) (1,2)(0,3)(3,0)(2,1) (2,3)(3,2)(0,1)(1,0) (3,1)(2,0)(1,3)(0,2) Model 0
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0123 1032 2301 3210 0123 2301 3210 1032 XY (0,0)(1,1)(2,2)(3,3) (1,2)(0,3)(3,0)(2,1) (2,3)(3,2)(0,1)(1,0) (3,1)(2,0)(1,3)(0,2) Z Model 0
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0123 1032 2301 3210 0123 2301 3210 1032 XY 051015 63129 111414 13872 Z Z[i][j] = 4.X[i][j] + Y[i][j Model 0
![XY Z Z[i][j] = 4.X[i][j] + Y[i][j Model 0](http://images.slideplayer.com./33/10121135/slides/slide_19.jpg "XY Z Z[i][j] = 4.X[i][j] + Y[i][j Model 0")
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0123 1032 2301 3210 0123 2301 3210 1032 XY 051015 63129 111414 13872 Z Z[i][j] = 4.X[i][j] + Y[i][j Can fix first row Symmetry Model 1
![XY Z Z[i][j] = 4.X[i][j] + Y[i][j Can fix first row Symmetry Model 1](http://images.slideplayer.com./33/10121135/slides/slide_22.jpg "XY Z Z[i][j] = 4.X[i][j] + Y[i][j Can fix first row Symmetry Model 1")
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0123 1032 2301 3210 0123 2301 3210 1032 XY 051015 63129 111414 13872 Z Z[i][j] = 4.X[i][j] + Y[i][j Can fix first row Can fix first column Symmetry Model 1
![XY Z Z[i][j] = 4.X[i][j] + Y[i][j Can fix first row Can fix first column Symmetry Model 1](http://images.slideplayer.com./33/10121135/slides/slide_23.jpg "XY Z Z[i][j] = 4.X[i][j] + Y[i][j Can fix first row Can fix first column Symmetry Model 1")
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051015 63129 111414 13872 Z Z[i][j] = 4.X[i][j] + Y[i][j Observe sum of each row and sum of each column Magic? Model 2
![Z Z[i][j] = 4.X[i][j] + Y[i][j Observe sum of each row and sum of each column Magic.](http://images.slideplayer.com./33/10121135/slides/slide_25.jpg "Model 2.")
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051015 63129 111414 13872 Z Flatten Z, Z’ 0510156312911141413872 Channeling Model BMS
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051015 63129 111414 13872 Z 0510156312911141413872 010155111414 1372861293 location Channeling Model BMS
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051015 63129 111414 13872 Z 0510156312911141413872 010155111414 1372861293 Channeling constraint Z’[i] = x ↔ loc[x] = i Channeling Model BMS
![Z Channeling constraint Z’[i] = x ↔ loc[x] = i Channeling Model BMS](http://images.slideplayer.com./33/10121135/slides/slide_29.jpg "Z Channeling constraint Z’[i] = x ↔ loc[x] = i Channeling Model BMS")
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051015 63129 111414 13872 Z 0510156312911141413872 010155111414 1372861293 Channeling constraint Z’[i] = x ↔ loc[x] = i Can then do away with an allDiff on Z’ Model BMS
![Z Channeling constraint Z’[i] = x ↔ loc[x] = i Can then do away with an allDiff on Z’ Model BMS](http://images.slideplayer.com./33/10121135/slides/slide_30.jpg "Z Channeling constraint Z’[i] = x ↔ loc[x] = i Can then do away with an allDiff on Z’ Model BMS")
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