# -*- coding: utf-8 -*-
# Copyright 2010 Hakan Kjellerstrand hakank@bonetmail.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
#
# http://www.apache.org/licenses/LICENSE-2.0
#
# 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.
"""
Simple regular expression in Google CP Solver.
My last name (Kjellerstrand) is quite often misspelled
in ways that this regular expression shows:
k(je|ä)ll(er|ar)?(st|b)r?an?d
This model generates all the words that can be construed
by this regular expression.
Compare with the following model:
* Gecode: http://www.hakank.org/gecode/all_regexp.cpp
This model was created by Hakan Kjellerstrand (hakank@bonetmail.com)
Also see my other Google CP Solver models: http://www.hakank.org/google_or_tools/
"""
from constraint_solver import pywrapcp
#
# Global constraint regular
#
# This is a translation of MiniZinc's regular constraint (defined in
# lib/zinc/globals.mzn), via the Comet code refered above.
# All comments are from the MiniZinc code.
# '''
# The sequence of values in array 'x' (which must all be in the range 1..S)
# is accepted by the DFA of 'Q' states with input 1..S and transition
# function 'd' (which maps (1..Q, 1..S) -> 0..Q)) and initial state 'q0'
# (which must be in 1..Q) and accepting states 'F' (which all must be in
# 1..Q). We reserve state 0 to be an always failing state.
# '''
#
# x : IntVar array
# Q : number of states
# S : input_max
# d : transition matrix
# q0: initial state
# F : accepting states
def regular(x, Q, S, d, q0, F):
solver = x[0].solver()
assert Q > 0, 'regular: "Q" must be greater than zero'
assert S > 0, 'regular: "S" must be greater than zero'
# d2 is the same as d, except we add one extra transition for
# each possible input; each extra transition is from state zero
# to state zero. This allows us to continue even if we hit a
# non-accepted input.
# Comet: int d2[0..Q, 1..S]
d2 = []
for i in range(Q+1):
row = []
for j in range(S):
if i == 0:
row.append(0)
else:
row.append(d[i-1][j])
d2.append(row)
d2_flatten = [d2[i][j] for i in range(Q+1) for j in range(S)]
# If x has index set m..n, then a[m-1] holds the initial state
# (q0), and a[i+1] holds the state we're in after processing
# x[i]. If a[n] is in F, then we succeed (ie. accept the
# string).
x_range = range(0,len(x))
m = 0
n = len(x)
a = [solver.IntVar(0, Q+1, 'a[%i]' % i) for i in range(m, n+1)]
# Check that the final state is in F
solver.Add(solver.MemberCt(a[-1], F))
# First state is q0
solver.Add(a[m] == q0)
for i in x_range:
solver.Add(x[i] >= 1)
solver.Add(x[i] <= S)
# Determine a[i+1]: a[i+1] == d2[a[i], x[i]]
solver.Add(a[i+1] == solver.Element(d2_flatten, ((a[i])*S)+(x[i]-1)))
def main(n, res):
# Create the solver.
solver = pywrapcp.Solver('Regular expression')
#
# data
#
# the DFA (for regular)
n_states = 11
input_max = 12
initial_state = 1 # 0 is for the failing state
accepting_states = [12]
# The DFA
transition_fn = [
# 1 2 3 4 5 6 7 8 9 0 1 2 #
[0,2,3,0,0,0,0,0,0,0,0,0], # 1 k
[0,0,0,4,0,0,0,0,0,0,0,0], # 2 je
[0,0,0,4,0,0,0,0,0,0,0,0], # 3 ä
[0,0,0,0,5,6,7,8,0,0,0,0], # 4 ll
[0,0,0,0,0,0,7,8,0,0,0,0], # 5 er
[0,0,0,0,0,0,7,8,0,0,0,0], # 6 ar
[0,0,0,0,0,0,0,0,9,10,0,0], # 7 st
[0,0,0,0,0,0,0,0,9,10,0,0], # 8 b
[0,0,0,0,0,0,0,0,0,10,0,0], # 9 r
[0,0,0,0,0,0,0,0,0,0,11,12], # 10 a
[0,0,0,0,0,0,0,0,0,0,0,12], # 11 n
# 12 d
]
s = ['k','je','ä','ll','er','ar','st','b','r','a','n','d']
print 'n:', n
#
# declare variables
#
x = [solver.IntVar(1, 12, 'x[%i]'% i) for i in range(n)]
#
# constraints
#
regular(x, n_states, input_max, transition_fn,
initial_state, accepting_states)
#
# solution and search
#
db = solver.Phase(x,
solver.CHOOSE_FIRST_UNBOUND,
solver.ASSIGN_MIN_VALUE)
solver.NewSearch(db)
num_solutions = 0
while solver.NextSolution():
num_solutions += 1
# Note: 1 is the start state which is not included in the
# state array (x)
x_val = [1] + [x[i].Value() for i in range(n)]
# print 'x:', x_val
ss = ''.join([str(s[i-1]) for i in x_val])
res.append(ss)
# print
solver.EndSearch()
print 'num_solutions:', num_solutions
print 'failures:', solver.failures()
print 'branches:', solver.branches()
print 'wall_time:', solver.wall_time(), 'ms'
print
if __name__ == '__main__':
res = []
for n in range(4,9+1):
main(n, res)
print 'The following %i words where generated:' % len(res)
# res.sort(key=len)
for r in res:
print r