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A292674
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Least number of symbols required to fill a grid of size n X n row by row in the greedy way such that in any row or column or rectangular 4 X 4 block no symbol occurs twice.
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4
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1, 4, 9, 16, 18, 18, 20, 20, 22, 22, 23, 23, 23, 24, 25, 26, 26, 26, 29, 32, 32, 34, 36, 38, 38, 38, 42, 42, 42, 44, 44, 45, 48, 49, 49, 49, 54, 54, 54, 56, 59, 59, 64, 65, 68, 69, 70, 73, 76, 78, 79, 79, 82, 82, 83, 86, 87, 89, 90, 92, 95, 95, 96, 96, 97, 97
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OFFSET
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1,2
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COMMENTS
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Consider the symbols as positive integers. By the greedy way we mean to fill the grid row by row from left to right always with the least possible positive integer such that the three constraints (on rows, columns and rectangular blocks) are satisfied.
In contrast to the sudoku case, the 4 X 4 rectangles have "floating" borders, so the constraint is actually equivalent to say that an element must be different from all neighbors in a Moore neighborhood of range 3 (having up to 7*7 = 49 grid points).
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LINKS
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EXAMPLE
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For n = 8, the grid is filled as follows:
[ 1 2 3 4 5 6 7 8]
[ 5 6 7 8 1 2 3 4]
[ 9 10 11 12 13 14 15 16]
[13 14 15 16 9 10 11 12]
[ 2 3 4 17 18 5 6 7]
[ 6 1 8 7 2 3 4 17]
[10 5 12 19 20 1 8 13]
[11 9 13 14 10 15 12 19]
whence a(8) = 20.
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PROG
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(PARI) a(n, m=4, g=matrix(n, n))={my(ok(g, k, i, j, m)=if(m, ok(g[i, ], k)&&ok(g[, j], k)&&ok(concat(Vec(g[max(1, i-m+1)..i, max(1, j-m+1)..min(#g, j+m-1)])), k), !setsearch(Set(g), k))); for(i=1, n, for(j=1, n, for(k=1, n^2, ok(g, k, i, j, m)&&(g[i, j]=k)&&break))); vecmax(g)} \\ without "vecmax" the program returns the full n X n board.
(Python) # uses function in A292673
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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