# # Example using the Xpress Python interface # # Sudoku: place numbers from 1 to 9 into a 9x9 grid such that no # number repeats in any row, in any column, and in any 3x3 sub-grid. # # More generally, replace 3 with q and 9 with q^2 # from __future__ import print_function import xpress as xp # We model this problem as an assignment problem where certain # conditions must be met for all numbers in the columns, rows, and # sub-grids. # # These subgrids are lists of tuples with the coordinates of each # subgrid. In a 9x9 sudoku, for instance, subgrids[0,1] has the 9 # elements in the top square, i.e., the following: # # ___ ___ ___ # | |###| | # | |###| | # |___|###|___| # | | | | # | | | | # |___|___|___| # | | | | # | | | | # |___|___|___| # # while subgrids[2,2] has the 9 elements in the bottom right square. # The input is a starting grid where the unknown numbers are replaced by zero q = 3 starting_grid = \ [[8,0,0,0,0,0,0,0,0], [0,0,3,6,0,0,0,0,0], [0,7,0,0,9,0,2,0,0], [0,5,0,0,0,7,0,0,0], [0,0,0,0,4,5,7,0,0], [0,0,0,1,0,0,0,3,0], [0,0,1,0,0,0,0,6,8], [0,0,8,5,0,0,0,1,0], [0,9,0,0,0,0,4,0,0]] # Solve a 16x16 sudoku: Just uncomment the following lines. # # q = 4 # # starting_grid = \ # [[ 0, 0,12, 0, 0, 2, 0, 0, 0, 7, 3, 0,13,15, 0, 0], # [15, 0, 0, 0, 0, 3, 0, 0, 9, 0, 0, 0,12, 0, 0,10], # [ 0, 0, 0, 0, 9, 0, 6, 0, 0, 0,12, 0, 0, 0, 2, 5], # [ 6,11, 1, 0, 0,10, 5, 0, 0, 2, 0,15, 0, 0, 0, 0], # [ 4, 6, 3, 0, 0, 0,13,14, 0, 0, 0, 0, 0, 7, 0, 0], # [ 0,15,11, 0, 7, 0, 9, 0, 0, 0, 0, 0, 0, 0, 1, 0], # [ 0, 1, 0,10,15, 0, 0, 0,11, 3,14, 0, 6, 0, 0, 0], # [13, 0, 8, 7, 0, 5, 0, 0, 0, 1, 9,12, 0, 0, 0, 0], # [ 0, 0, 0, 6, 3, 7,15, 4, 0, 0, 0, 0, 0,14, 0, 0], # [ 0, 8, 0, 0, 0, 0, 0, 0, 0,11, 7, 0, 4, 0, 0, 0], # [ 0, 0, 0, 0, 0, 0, 0, 0,13, 0, 0, 6, 9, 0, 3, 0], # [ 0, 0, 0, 0, 2, 8,14, 0, 3, 0, 0,10, 0, 0,13, 7], # [ 0, 0, 0, 8, 0, 0, 0, 7,10, 0, 0, 0, 0, 0, 5, 1], # [ 0, 4,10, 1, 6, 0, 0, 0, 0,12, 0,14, 7, 3, 9,15], # [ 3, 0,15, 0, 0, 0, 0, 8, 0, 0, 1, 0,14,12, 0, 0], # [ 2, 0, 0, 9,12, 0, 0, 1, 0, 0, 0, 0, 0, 6, 8, 0]] n = q**2 # the size must be the square of the size of the subgrids N = range (n) x = {(i,j,k): xp.var (vartype = xp.binary, name='x{0}_{1}_{2}'.format (i,j,k)) for i in N for j in N for k in N} # define all q^2 subgrids subgrids = {(h,l): [(i,j) for i in range (q*h, q*h + q) for j in range (q*l, q*l + q)] for h in range (q) for l in range (q)} vertical = [xp.Sum (x[i,j,k] for i in N) == 1 for j in N for k in N] horizontal = [xp.Sum (x[i,j,k] for j in N) == 1 for i in N for k in N] subgrid = [xp.Sum (x[i,j,k] for (i,j) in subgrids[h,l]) == 1 for (h,l) in subgrids.keys() for k in N] # Assign exactly one number to each cell assign = [xp.Sum (x[i,j,k] for k in N) == 1 for i in N for j in N] # Fix those variables that are non-zero in the input grid init = [x[i,j,k] == 1 for k in N for i in N for j in N if starting_grid[i][j] == k+1] p = xp.problem() p.addVariable (x) p.addConstraint (vertical, horizontal, subgrid, assign, init) # we don't need an objective function, as long as a solution is found p.solve () print ('Solution:') for i in N: for j in N: l = [k for k in N if p.getSolution (x[i,j,k]) >= 0,5] assert (len (l) == 1) print ('{0:2d}'.format (1 + l[0]), end = '', sep='') print ('')