202 lines
5.2 KiB
Python
202 lines
5.2 KiB
Python
# 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.
|
|
|
|
"""
|
|
|
|
Global constraint regular in Google CP Solver.
|
|
|
|
This is a translation of MiniZinc's regular constraint (defined in
|
|
lib/zinc/globals.mzn). 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.
|
|
'''
|
|
|
|
It is, however, translated from the Comet model:
|
|
* Comet: http://www.hakank.org/comet/regular.co
|
|
|
|
Here we test with the following regular expression:
|
|
0*1{3}0+1{2}0+1{1}0*
|
|
using an array of size 10.
|
|
|
|
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 ortools.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.
|
|
|
|
d2 = []
|
|
for i in range(Q + 1):
|
|
for j in range(1, S + 1):
|
|
if i == 0:
|
|
d2.append((0, j, 0))
|
|
else:
|
|
d2.append((i, j, d[i - 1][j - 1]))
|
|
|
|
solver.Add(solver.TransitionConstraint(x, d2, q0, F))
|
|
|
|
#
|
|
# Make a transition (automaton) matrix from a
|
|
# single pattern, e.g. [3,2,1]
|
|
#
|
|
|
|
|
|
def make_transition_matrix(pattern):
|
|
|
|
p_len = len(pattern)
|
|
print 'p_len:', p_len
|
|
num_states = p_len + sum(pattern)
|
|
print 'num_states:', num_states
|
|
t_matrix = []
|
|
for i in range(num_states):
|
|
row = []
|
|
for j in range(2):
|
|
row.append(0)
|
|
t_matrix.append(row)
|
|
|
|
# convert pattern to a 0/1 pattern for easy handling of
|
|
# the states
|
|
tmp = [0 for i in range(num_states)]
|
|
c = 0
|
|
tmp[c] = 0
|
|
for i in range(p_len):
|
|
for j in range(pattern[i]):
|
|
c += 1
|
|
tmp[c] = 1
|
|
if c < num_states - 1:
|
|
c += 1
|
|
tmp[c] = 0
|
|
print 'tmp:', tmp
|
|
|
|
t_matrix[num_states - 1][0] = num_states
|
|
t_matrix[num_states - 1][1] = 0
|
|
|
|
for i in range(num_states):
|
|
if tmp[i] == 0:
|
|
t_matrix[i][0] = i + 1
|
|
t_matrix[i][1] = i + 2
|
|
else:
|
|
if i < num_states - 1:
|
|
if tmp[i + 1] == 1:
|
|
t_matrix[i][0] = 0
|
|
t_matrix[i][1] = i + 2
|
|
else:
|
|
t_matrix[i][0] = i + 2
|
|
t_matrix[i][1] = 0
|
|
|
|
print 'The states:'
|
|
for i in range(num_states):
|
|
for j in range(2):
|
|
print t_matrix[i][j],
|
|
print
|
|
print
|
|
|
|
return t_matrix
|
|
|
|
|
|
def main():
|
|
|
|
# Create the solver.
|
|
solver = pywrapcp.Solver('Regular test')
|
|
|
|
#
|
|
# data
|
|
#
|
|
|
|
this_len = 10
|
|
pp = [3, 2, 1]
|
|
|
|
transition_fn = make_transition_matrix(pp)
|
|
n_states = len(transition_fn)
|
|
input_max = 2
|
|
|
|
# Note: we use '1' and '2' (rather than 0 and 1)
|
|
# since 0 represents the failing state.
|
|
initial_state = 1
|
|
|
|
accepting_states = [n_states]
|
|
|
|
# declare variables
|
|
reg_input = [solver.IntVar(1, input_max, 'reg_input[%i]' % i)
|
|
for i in range(this_len)]
|
|
|
|
#
|
|
# constraints
|
|
#
|
|
regular(reg_input, n_states, input_max, transition_fn,
|
|
initial_state, accepting_states)
|
|
|
|
#
|
|
# solution and search
|
|
#
|
|
db = solver.Phase(reg_input,
|
|
solver.CHOOSE_MIN_SIZE_HIGHEST_MAX,
|
|
solver.ASSIGN_MIN_VALUE)
|
|
|
|
solver.NewSearch(db)
|
|
|
|
num_solutions = 0
|
|
while solver.NextSolution():
|
|
print 'reg_input:', [reg_input[i].Value() - 1 for i in range(this_len)]
|
|
num_solutions += 1
|
|
|
|
solver.EndSearch()
|
|
print
|
|
print 'num_solutions:', num_solutions
|
|
print 'failures:', solver.Failures()
|
|
print 'branches:', solver.Branches()
|
|
print 'WallTime:', solver.WallTime(), 'ms'
|
|
|
|
|
|
if __name__ == '__main__':
|
|
main()
|