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

Author

M. F. Hasler, Sep 20 2017

Keywords

• nonn

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).

Examples

			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.
		

Links

• Eric Weisstein's World of Mathematics, Moore Neighborhood.

Crossrefs

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

Programs

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

Extensions

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