A292673 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 3 X 3 block no symbol occurs twice.

1, 4, 9, 11, 13, 13, 13, 13, 14, 14, 15, 17, 18, 21, 22, 23, 25, 26, 27, 29, 30, 31, 32, 34, 35, 38, 39, 40, 42, 45, 47, 49, 51, 53, 54, 55, 55, 55, 58, 59, 60, 61, 62, 63, 64, 65, 66, 67, 68, 69, 71, 72, 74, 76, 78, 79, 83, 83, 85, 86, 88, 90, 91, 92, 93, 96

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 3 X 3 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 2 (having up to 5*5 = 25 grid points).

Examples

			For n = 4, the 4 X 4 grid is filled as follows (using hexadecimal digits):
   [1  2  3  4]
   [4  5  6  1]
   [7  8  9  A]
   [2  3  B  7], whence a(4) = # { 1, ..., 9, A, B} = 11.
For n = 8, the grid is filled as follows:
  [1  2  3  4  5  6  7  8]
  [4  5  6  1  2  3  9  A]
  [7  8  9  A  B  C  1  2]
  [2  3  C  7  4  5  6  B]
  [5  1  D  2  3  8  A  4]
  [6  4  8  9  1  D  2  3]
  [3  7  A  5  6  B  C  1]
  [9  B  2  3  7  4  5  6], whence a(8) = # { 1, ..., 9, A, B, C, D } = 13.
For n = 5, 6 and 7, the solution is just the upper left n X n part of the above grid: all of these also require 13 symbols.
		

Links

• Andrew Howroyd, Table of n, a(n) for n = 1..100
• Eric Weisstein's World of Mathematics, Moore Neighborhood

Crossrefs

Cf. A292670, A292671, A292672, ..., A292679.

Programs

PARI

a(n,m=3,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 A292673(n, b=3): # change b for A292672, ..., A292679
    m, S, N = 0, {1}, range(1, n+1)
    g = [[0 for j in range(n+b)] for i in range(n+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([A292673(n) for n in range(1, 101)]) # Michael S. Branicky, Apr 13 2023

Extensions

Terms a(40) and beyond from Andrew Howroyd, Feb 22 2020