A292670 Least number of symbols required to fill a grid of size n^2 X n^2 row by row in the greedy way such that in no row or column or n X n square any symbol occurs twice.

1, 6, 14, 26, 40, 53, 73, 114, 144, 161, 188, 210, 250, 279, 329, 482, 544, 577, 611, 656, 697, 752, 847, 906, 947, 973, 1049, 1113, 1210, 1284, 1425, 1986, 2112, 2177, 2242, 2326, 2408, 2473, 2632, 2748, 2818, 2886, 3004, 3125, 3334, 3390, 3539, 3661, 3806, 3870

Offset

1,2

Comments

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 n X n rectangles have "floating" borders, so the constraint is actually equivalent to say that any element must be different from all neighbors in a Moore neighborhood of range n-1 (having up to (2n-1)^2 grid points).

Links

Table of n, a(n) for n=1..50.

Eric Weisstein's World of Mathematics, Moore Neighborhood.

Example

For n = 2, the 4 X 4 grid is filled as follows:
   [1 2 3 4]
   [3 4 1 2]
   [2 5 6 3]
   [4 1 2 5], whence a(2) = 6.

Prog

(PARI) a(m, n=m^2, 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^2 X n^2 board.
(Python)
def a(n):
    s, b = n*n, n # side length, block size
    m, S, N = 0, {1}, range(1, s+1)
    g = [[0 for j in range(s+b)] for i in range(s+b)]
    row, col = {i:set() for i in N}, {j:set() for j in N}
    offsets = [(i, j) for i in range(-b+1, 1) for j in range(-b+1, 1)]
    offsets += [(i, j) for i in range(-b+1, 0) for j in range(1, b)]
    for i in N:
        for j in N:
            rect = set(g[i+o[0]][j+o[1]] for o in offsets)
            e = min(S - row[i] - col[j] - rect)
            g[i][j] = e
            if e > m:
                m = e
                S.add(m+1)
            row[i].add(e)
            col[j].add(e)
    return m
print([a(n) for n in range(1, 13)]) # Michael S. Branicky, Apr 13 2023

Crossrefs

Cf. A292671: grid size independent of block size; A292672, ..., A292679: A(m,n) for fixed n=2,...,9).

Sequence in context: A026055 A288939 A165986 * A131951 A168648 A093776

Adjacent sequences: A292667 A292668 A292669 * A292671 A292672 A292673

Keyword

nonn

Author

M. F. Hasler, Sep 20 2017

Extensions

a(9) and beyond from Michael S. Branicky, Apr 13 2023

Status

approved