A292672 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 2 X 2 block no symbol occurs twice.

1, 4, 6, 6, 7, 8, 10, 10, 13, 15, 16, 17, 19, 20, 21, 22, 23, 25, 28, 30, 31, 32, 33, 35, 35, 37, 38, 39, 40, 41, 43, 44, 45, 47, 50, 52, 53, 55, 57, 58, 60, 60, 61, 63, 64, 65, 67, 68, 70, 71, 72, 73, 74, 76, 78, 78, 79, 80, 82, 84, 85, 87, 89, 90, 92, 93, 94

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 2 X 2 rectangles have "floating" borders, so the constraint is actually equivalent to saying that any element must be different from all neighbors in a Moore neighborhood of range 1 (having up to 3*3=9 grid points).

Examples

			For n = 4, 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(4) = 6.
For n = 3 the result would be the upper 3 X 3 part of the above grid, showing that also a(3) = 6.
		

Links

• Michael S. Branicky, Table of n, a(n) for n = 1..1000 (terms 1..100 from Andrew Howroyd)
• Eric Weisstein's World of Mathematics, Moore Neighborhood

Crossrefs

Cf. A292670, A292671, A292673, ..., A292679.

Programs

PARI

a(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 X n board.

Python

def a(n):
    m, s, N = 0, {1}, range(1, n+1)
    g = [[0 for j in range(n+2)] for i in range(n+2)]
    row, col = {i:set() for i in N}, {j:set() for j in N}
    for i in N:
        for j in N:
            rect = {g[i-1][j-1], g[i-1][j], g[i][j-1], g[i-1][j+1]}
            e = min(s - row[i] - col[j] - rect)
            g[i][j] = e
            row[i].add(e)
            col[j].add(e)
            if e > m:
                m = e
                s.add(m+1)
    return m
print([a(n) for n in range(1, 71)]) # Michael S. Branicky, Apr 13 2023

Extensions

Terms a(60) and beyond from Andrew Howroyd, Feb 22 2020