Added nonogram_default_search.py Nonogram solver using DefaultSearch
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python/nonogram_default_search.py
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225
python/nonogram_default_search.py
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# Copyright 2010 Hakan Kjellerstrand hakank@bonetmail.com, lperron@google.com
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#
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# Licensed under the Apache License, Version 2.0 (the 'License');
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# you may not use this file except in compliance with the License.
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# You may obtain a copy of the License at
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#
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# http://www.apache.org/licenses/LICENSE-2.0
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#
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# Unless required by applicable law or agreed to in writing, software
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# distributed under the License is distributed on an 'AS IS' BASIS,
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# WITHOUT WARRANTIES OR CONDITIONS OF ANY KIND, either express or implied.
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# See the License for the specific language governing permissions and
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# limitations under the License.
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"""
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Nonogram (Painting by numbers) in Google CP Solver.
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http://en.wikipedia.org/wiki/Nonogram
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'''
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Nonograms or Paint by Numbers are picture logic puzzles in which cells in a
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grid have to be colored or left blank according to numbers given at the
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side of the grid to reveal a hidden picture. In this puzzle type, the
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numbers measure how many unbroken lines of filled-in squares there are
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in any given row or column. For example, a clue of '4 8 3' would mean
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there are sets of four, eight, and three filled squares, in that order,
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with at least one blank square between successive groups.
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'''
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See problem 12 at http://www.csplib.org/.
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http://www.puzzlemuseum.com/nonogram.htm
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Haskell solution:
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http://twan.home.fmf.nl/blog/haskell/Nonograms.details
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Brunetti, Sara & Daurat, Alain (2003)
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'An algorithm reconstructing convex lattice sets'
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http://geodisi.u-strasbg.fr/~daurat/papiers/tomoqconv.pdf
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"""
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import sys
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from constraint_solver import pywrapcp
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#
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# Make a transition (automaton) list of tuples from a
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# single pattern, e.g. [3,2,1]
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#
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def make_transition_tuples(pattern):
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p_len = len(pattern)
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num_states = p_len + sum(pattern)
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# this is for handling 0-clues. It generates
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# just the minimal state
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if num_states == 0:
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return [(1, 0, 1)], 1
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tuples = []
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# convert pattern to a 0/1 pattern for easy handling of
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# the states
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tmp = [0];
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c = 0
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for pattern_index in range(p_len):
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tmp.extend([1] * pattern[pattern_index])
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tmp.append(0)
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for i in range(num_states):
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state = i + 1
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if tmp[i] == 0:
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tuples.append((state, 0, state))
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tuples.append((state, 1, state + 1))
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else:
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if i < num_states - 1:
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if tmp[i + 1] == 1:
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tuples.append((state, 1, state + 1))
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else:
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tuples.append((state, 0, state + 1))
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tuples.append((num_states, 0, num_states))
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return (tuples, num_states)
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#
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# check each rule by creating an automaton and transition constraint.
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#
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def check_rule(rules, y):
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cleaned_rule = [rules[i] for i in range(len(rules)) if rules[i] > 0]
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(transition_tuples, last_state) = make_transition_tuples(cleaned_rule)
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initial_state = 1
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accepting_states = [last_state]
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solver = y[0].solver()
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solver.Add(solver.TransitionConstraint(y,
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transition_tuples,
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initial_state,
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accepting_states))
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def main(rows, row_rule_len, row_rules, cols, col_rule_len, col_rules):
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# Create the solver.
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solver = pywrapcp.Solver('Nonogram')
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#
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# variables
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#
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board = {}
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for i in range(rows):
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for j in range(cols):
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board[i, j] = solver.IntVar(0, 1, 'board[%i, %i]' % (i, j))
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board_flat = [board[i, j] for i in range(rows) for j in range(cols)]
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# Flattened board for labeling.
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# This labeling was inspired by a suggestion from
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# Pascal Van Hentenryck about my (hakank's) Comet
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# nonogram model.
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board_label = []
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if rows * row_rule_len < cols * col_rule_len:
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for i in range(rows):
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for j in range(cols):
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board_label.append(board[i, j])
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else:
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for j in range(cols):
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for i in range(rows):
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board_label.append(board[i, j])
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#
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# constraints
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#
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for i in range(rows):
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check_rule(row_rules[i], [board[i, j] for j in range(cols)])
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for j in range(cols):
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check_rule(col_rules[j], [board[i, j] for i in range(rows)])
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#
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# solution and search
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#
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parameters = pywrapcp.DefaultPhaseParameters()
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parameters.heuristic_period = 200000
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db = solver.DefaultPhase(board_label, parameters)
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print 'before solver, wall time = ', solver.wall_time(), 'ms'
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solver.NewSearch(db)
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num_solutions = 0
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while solver.NextSolution():
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print
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num_solutions += 1
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for i in range(rows):
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row = [board[i, j].Value() for j in range(cols)]
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row_pres = []
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for j in row:
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if j == 1:
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row_pres.append('#')
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else:
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row_pres.append(' ')
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print ' ', ''.join(row_pres)
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print
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print ' ', '-' * cols
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if num_solutions >= 2:
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print '2 solutions is enough...'
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break
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solver.EndSearch()
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print
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print 'num_solutions:', num_solutions
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print 'failures:', solver.failures()
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print 'branches:', solver.branches()
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print 'wall_time:', solver.wall_time(), 'ms'
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#
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# Default problem
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#
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# From http://twan.home.fmf.nl/blog/haskell/Nonograms.details
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# The lambda picture
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#
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rows = 12
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row_rule_len = 3
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row_rules = [
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[0,0,2],
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[0,1,2],
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[0,1,1],
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[0,0,2],
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[0,0,1],
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[0,0,3],
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[0,0,3],
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[0,2,2],
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[0,2,1],
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[2,2,1],
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[0,2,3],
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[0,2,2]
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]
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cols = 10
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col_rule_len = 2
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col_rules = [
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[2,1],
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[1,3],
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[2,4],
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[3,4],
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[0,4],
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[0,3],
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[0,3],
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[0,3],
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[0,2],
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[0,2]
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]
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if __name__ == '__main__':
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if len(sys.argv) > 1:
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file = sys.argv[1]
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execfile(file)
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main(rows, row_rule_len, row_rules, cols, col_rule_len, col_rules)
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