# Copyright 2010 Hakan Kjellerstrand hakank@gmail.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. Instead of minimizing the cost in a cost matrix (as shortest path problem), we here call the model with increasing length of the sequence array (x). 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@gmail.com) Also see my other Google CP Solver models: http://www.hakank.org/google_or_tools/ """ from __future__ import print_function from ortools.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 = list(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): # Create the solver. solver = pywrapcp.Solver('3 jugs problem using regular constraint') # # data # # the DFA (for regular) n_states = 14 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 ] 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)] # # constraints # regular(x, n_states, input_max, transition_fn, initial_state, accepting_states) # # solution and search # db = solver.Phase(x, solver.INT_VAR_DEFAULT, solver.INT_VALUE_DEFAULT) solver.NewSearch(db) num_solutions = 0 x_val = [] while solver.NextSolution(): num_solutions += 1 x_val = [1] + [x[i].Value() for i in range(n)] print('x:', x_val) for i in range(1, n + 1): print('%s -> %s' % (nodes[x_val[i - 1] - 1], nodes[x_val[i] - 1])) solver.EndSearch() if num_solutions > 0: print() print('num_solutions:', num_solutions) print('failures:', solver.Failures()) print('branches:', solver.Branches()) print('WallTime:', solver.WallTime(), 'ms') # return the solution (or an empty array) return x_val # Search for a minimum solution by increasing # the length of the state array. if __name__ == '__main__': for n in range(1, 15): result = main(n) result_len = len(result) if result_len: print() print('Found a solution of length %i:' % result_len, result) print() break