1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 154 155 156 157 158 159 160 161 162 163 164 165 166 167 168 169 170 171 172 173 174 175 176 177 178 179 180 181 182 183 184 185 186 187 188 189 190 191 192 193 194 195 196 197 198 199 200 201 202 203 204 205 206 207 208 209 210 211 212 213 214 215 216 217 218 219 220 221 222 223 224 225 226 | # Copyright 2010 Hakan Kjellerstrand hakank@gmail.com, lperron@google.com # # Licensed under the Apache License, Version 2.0 (the 'License'); # you may not use this file except in compliance with the License. # You may obtain a copy of the License at # # # Unless required by applicable law or agreed to in writing, software # distributed under the License is distributed on an 'AS IS' BASIS, # WITHOUT WARRANTIES OR CONDITIONS OF ANY KIND, either express or implied. # See the License for the specific language governing permissions and # limitations under the License. """ Nonogram (Painting by numbers) in Google CP Solver. ''' Nonograms or Paint by Numbers are picture logic puzzles in which cells in a grid have to be colored or left blank according to numbers given at the side of the grid to reveal a hidden picture. In this puzzle type, the numbers measure how many unbroken lines of filled-in squares there are in any given row or column. For example, a clue of '4 8 3' would mean there are sets of four, eight, and three filled squares, in that order, with at least one blank square between successive groups. ''' See problem 12 at http://www.csplib.org/. Haskell solution: Brunetti, Sara & Daurat, Alain (2003) 'An algorithm reconstructing convex lattice sets' http://geodisi.u-strasbg.fr/~daurat/papiers/tomoqconv.pdf """ import sys from constraint_solver import pywrapcp # # Make a transition (automaton) list of tuples from a # single pattern, e.g. [3,2,1] # def make_transition_tuples(pattern): p_len = len (pattern) num_states = p_len + sum (pattern) tuples = pywrapcp.IntTupleSet( 3 ) # this is for handling 0-clues. It generates # just the minimal state if num_states = = 0 : tuples.Insert3( 1 , 0 , 1 ) return (tuples, 1 ) # convert pattern to a 0/1 pattern for easy handling of # the states tmp = [ 0 ]; c = 0 for pattern_index in range (p_len): tmp.extend([ 1 ] * pattern[pattern_index]) tmp.append( 0 ) for i in range (num_states): state = i + 1 if tmp[i] = = 0 : tuples.Insert3(state, 0 , state) tuples.Insert3(state, 1 , state + 1 ) else : if i < num_states - 1 : if tmp[i + 1 ] = = 1 : tuples.Insert3(state, 1 , state + 1 ) else : tuples.Insert3(state, 0 , state + 1 ) tuples.Insert3(num_states, 0 , num_states) return (tuples, num_states) # # check each rule by creating an automaton and transition constraint. # def check_rule(rules, y): cleaned_rule = [rules[i] for i in range ( len (rules)) if rules[i] > 0 ] (transition_tuples, last_state) = make_transition_tuples(cleaned_rule) initial_state = 1 accepting_states = [last_state] solver = y[ 0 ].solver() solver.Add(solver.TransitionConstraint(y, transition_tuples, initial_state, accepting_states)) def main(rows, row_rule_len, row_rules, cols, col_rule_len, col_rules): # Create the solver. solver = pywrapcp.Solver( 'Nonogram' ) # # variables # board = {} for i in range (rows): for j in range (cols): board[i, j] = solver.IntVar( 0 , 1 , 'board[%i, %i]' % (i, j)) board_flat = [board[i, j] for i in range (rows) for j in range (cols)] # Flattened board for labeling. # This labeling was inspired by a suggestion from # Pascal Van Hentenryck about my (hakank's) Comet # nonogram model. board_label = [] if rows * row_rule_len < cols * col_rule_len: for i in range (rows): for j in range (cols): board_label.append(board[i, j]) else : for j in range (cols): for i in range (rows): board_label.append(board[i, j]) # # constraints # for i in range (rows): check_rule(row_rules[i], [board[i, j] for j in range (cols)]) for j in range (cols): check_rule(col_rules[j], [board[i, j] for i in range (rows)]) # # solution and search # parameters = pywrapcp.DefaultPhaseParameters() parameters.heuristic_period = 200000 db = solver.DefaultPhase(board_label, parameters) print 'before solver, wall time = ' , solver.WallTime(), 'ms' solver.NewSearch(db) num_solutions = 0 while solver.NextSolution(): print num_solutions + = 1 for i in range (rows): row = [board[i, j].Value() for j in range (cols)] row_pres = [] for j in row: if j = = 1 : row_pres.append( '#' ) else : row_pres.append( ' ' ) print ' ' , ''.join(row_pres) print print ' ' , '-' * cols if num_solutions > = 2 : print '2 solutions is enough...' break solver.EndSearch() print print 'num_solutions:' , num_solutions print 'failures:' , solver.Failures() print 'branches:' , solver.Branches() print 'WallTime:' , solver.WallTime(), 'ms' # # Default problem # # The lambda picture # rows = 12 row_rule_len = 3 row_rules = [ [ 0 , 0 , 2 ], [ 0 , 1 , 2 ], [ 0 , 1 , 1 ], [ 0 , 0 , 2 ], [ 0 , 0 , 1 ], [ 0 , 0 , 3 ], [ 0 , 0 , 3 ], [ 0 , 2 , 2 ], [ 0 , 2 , 1 ], [ 2 , 2 , 1 ], [ 0 , 2 , 3 ], [ 0 , 2 , 2 ] ] cols = 10 col_rule_len = 2 col_rules = [ [ 2 , 1 ], [ 1 , 3 ], [ 2 , 4 ], [ 3 , 4 ], [ 0 , 4 ], [ 0 , 3 ], [ 0 , 3 ], [ 0 , 3 ], [ 0 , 2 ], [ 0 , 2 ] ] if __name__ = = '__main__' : if len (sys.argv) > 1 : file = sys.argv[ 1 ] execfile ( file ) main(rows, row_rule_len, row_rules, cols, col_rule_len, col_rules) |