# 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.
"""
3 jugs problem using regular constraint in Google CP Solver.
A.k.a. water jugs problem.
Problem from Taha 'Introduction to Operations Research',
page 245f .
For more info about the problem, see:
http://mathworld.wolfram.com/ThreeJugProblem.html
This model use a regular constraint for handling the
transitions between the states. This version differs
from the original version of this model
(http://www.hakank.org/google_or_tools/3_jugs_regular.py)
in the following aspects:
- the last state (15) can now be entered (looped) repeated times
- it use minimization of the costs of not being in the
last state. This is when a 'fix-point' is reached.
This means that we don't have to call main() many times
with increasing length of the array of decision variables
(states).
Compare with other models that use MIP/CP approach,
as a shortest path problem:
* Comet: http://www.hakank.org/comet/3_jugs.co
* Comet: http://www.hakank.org/comet/water_buckets1.co
* MiniZinc: http://www.hakank.org/minizinc/3_jugs.mzn
* MiniZinc: http://www.hakank.org/minizinc/3_jugs2.mzn
* SICStus: http://www.hakank.org/sicstus/3_jugs.pl
* ECLiPSe: http://www.hakank.org/eclipse/3_jugs.ecl
* ECLiPSe: http://www.hakank.org/eclipse/3_jugs2.ecl
* Gecode: http://www.hakank.org/gecode/3_jugs2.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
from collections import defaultdict
#
# 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():
# Create the solver.
solver = pywrapcp.Solver('3 jugs problem using regular constraint')
#
# data
#
# the DFA (for regular)
# n_states = 14
n_states = 15
input_max = 15
initial_state = 1 # 0 is for the failing state
accepting_states = [15]
##
## Manually crafted DFA
## (from the adjacency matrix used in the other models)
##
# transition_fn = [
# # 1 2 3 4 5 6 7 8 9 0 1 2 3 4 5
# [0, 2, 0, 0, 0, 0, 0, 0, 9, 0, 0, 0, 0, 0, 0], # 1
# [0, 0, 3, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0], # 2
# [0, 0, 0, 4, 0, 0, 0, 0, 9, 0, 0, 0, 0, 0, 0], # 3
# [0, 0, 0, 0, 5, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0], # 4
# [0, 0, 0, 0, 0, 6, 0, 0, 9, 0, 0, 0, 0, 0, 0], # 5
# [0, 0, 0, 0, 0, 0, 7, 0, 0, 0, 0, 0, 0, 0, 0], # 6
# [0, 0, 0, 0, 0, 0, 0, 8, 9, 0, 0, 0, 0, 0, 0], # 7
# [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 15], # 8
# [0, 0, 0, 0, 0, 0, 0, 0, 0, 10, 0, 0, 0, 0, 0], # 9
# [0, 2, 0, 0, 0, 0, 0, 0, 0, 0, 11, 0, 0, 0, 0], # 10
# [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 12, 0, 0, 0], # 11
# [0, 2, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 13, 0, 0], # 12
# [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 14, 0], # 13
# [0, 2, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 15], # 14
# # 15
# ]
#
# However, the DFA is easy to create from adjacency lists.
#
states = [
[2,9], # state 1
[3], # state 2
[4, 9], # state 3
[5], # state 4
[6,9], # state 5
[7], # state 6
[8,9], # state 7
[15], # state 8
[10], # state 9
[11], # state 10
[12], # state 11
[13], # state 12
[14], # state 13
[15], # state 14
[15] # state 15: loop ("fix-point")
]
transition_fn = []
for i in range(n_states):
row = []
for j in range(1,input_max+1):
if j in states[i]:
row.append(j)
else:
row.append(0)
transition_fn.append(row)
#
# The name of the nodes, for printing
# the solution.
#
nodes = [
'8,0,0', # 1 start
'5,0,3', # 2
'5,3,0', # 3
'2,3,3', # 4
'2,5,1', # 5
'7,0,1', # 6
'7,1,0', # 7
'4,1,3', # 8
'3,5,0', # 9
'3,2,3', # 10
'6,2,0', # 11
'6,0,2', # 12
'1,5,2', # 13
'1,4,3', # 14
'4,4,0' # 15 goal
]
#
# declare variables
#
# x = [solver.IntVar(1, input_max, 'x[%i]'% i) for i in range(n)]
x = [solver.IntVar(0, input_max, 'x[%i]'% i) for i in range(input_max)]
cost = solver.IntVar(0, input_max, 'cost')
#
# constraints
#
regular(x, n_states, input_max, transition_fn,
initial_state, accepting_states)
sum_b = [solver.IsLessCstVar(x[i], input_max-1) for i in range(input_max)]
solver.Add(cost == solver.Sum(sum_b))
# objective
objective = solver.Minimize(cost, 1)
#
# solution and search
#
db = solver.Phase(x,
solver.INT_VAR_DEFAULT,
solver.INT_VALUE_DEFAULT)
solver.NewSearch(db, [objective])
num_solutions = 0
x_val = []
while solver.NextSolution():
num_solutions += 1
print 'cost:', cost.Value()
x_val = [1] + [x[i].Value() for i in range(input_max)]
print 'x:', x_val
for i in range(1, input_max+1):
node_from, node_to = nodes[x_val[i-1]-1], nodes[x_val[i]-1]
if node_from == node_to:
break
print '%s -> %s' % (node_from, node_to)
solver.EndSearch()
print
print 'num_solutions:', num_solutions
print 'failures:', solver.failures()
print 'branches:', solver.branches()
print 'wall_time:', solver.wall_time(), 'ms'
if __name__ == '__main__':
main()