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8 Queens problem with a SAT solver
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| import pycosat | |
| import numpy | |
| import itertools | |
| WORLDS_HARDEST_RIDDLE_ACCORDING_TO_TELEGRAPH = """\ | |
| 8........ | |
| ..36..... | |
| .7..9.2.. | |
| .5...7... | |
| ....457.. | |
| ...1...3. | |
| ..1....68 | |
| ..85...1. | |
| .9....4..""" | |
| def get_cnf(riddle): | |
| # * add one because 0 is reserved in picosat | |
| # * object type because pycosat expects 'int's | |
| # * 9^3 vars x_ij^d where (i, j) == row and col, d == digit | |
| vars = (numpy.arange(9 * 9 * 9).reshape(9, 9, 9) + 1).astype('object') | |
| cnf = [] | |
| # At least one digit per square | |
| for i in xrange(9): | |
| for j in xrange(9): | |
| cnf.append(vars[i, j, :].tolist()) | |
| # Only one digit per square | |
| for i in xrange(9): | |
| for j in xrange(9): | |
| cnf += list(itertools.combinations(-vars[i, j, :], 2)) | |
| # Each 3x3 block must contain 9 differrent digits | |
| for i in xrange(3): | |
| for j in xrange(3): | |
| for d in xrange(9): | |
| cnf += list(itertools.combinations(-vars[i*3:i*3+3, j*3:j*3+3, d].ravel(), 2)) | |
| # Each row and each column must contain 9 different digits | |
| for i in xrange(9): | |
| for d in xrange(9): | |
| cnf += list(itertools.combinations(-vars[i,:,d].ravel(), 2)) | |
| cnf += list(itertools.combinations(-vars[:,i,d].ravel(), 2)) | |
| # Tranform riddle board to CNF | |
| for i, x in enumerate(riddle.split()): | |
| for j, y in enumerate(x): | |
| if y == '.': | |
| continue | |
| d = int(y) - 1 | |
| cnf.append([vars[i, j, d]]) | |
| return [list(x) for x in cnf] | |
| def print_solution(solution): | |
| solution_a = numpy.array(solution).reshape(9, 9, 9) | |
| for i in xrange(9): | |
| for j in xrange(9): | |
| for d in xrange(9): | |
| if solution_a[i, j, d] > 0: | |
| print d + 1, | |
| print "" | |
| print_solution(pycosat.solve(get_cnf(WORLDS_HARDEST_RIDDLE_ACCORDING_TO_TELEGRAPH))) |
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