3_jugs_regular.cs

//
// Copyright 2012 Hakan Kjellerstrand
//
// 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.

using System;
using System.Collections;
using System.Linq;
using System.Diagnostics;
using Google.OrTools.ConstraintSolver;

public class ThreeJugsRegular
{


  /*
   * 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
   *
   */
  static void MyRegular(Solver solver,
                        IntVar[] x,
                        int Q,
                        int S,
                        int[,] d,
                        int q0,
                        int[] F) {



    Debug.Assert(Q > 0, "regular: 'Q' must be greater than zero");
    Debug.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.
    int[][] d2 = new int[Q+1][];
    for(int i = 0; i <= Q; i++) {
      int[] row = new int[S];
      for(int j = 0; j < S; j++) {
        if (i == 0) {
          row[j] = 0;
        } else {
          row[j] = d[i-1,j];
        }
      }
      d2[i] = row;
    }

    int[] d2_flatten = (from i in Enumerable.Range(0, Q+1)
                        from j in Enumerable.Range(0, S)
                        select d2[i][j]).ToArray();

    // 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).
    int m = 0;
    int n = x.Length;

    IntVar[] a = solver.MakeIntVarArray(n+1-m, 0,Q+1, "a");
    // Check that the final state is in F
    solver.Add(a[a.Length-1].Member(F));
    // First state is q0
    solver.Add(a[m] == q0);

    for(int i = 0; i < n; i++) {
      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] == d2_flatten.Element(((a[i]*S)+(x[i]-1))));

    }

  }




  /**
   *
   * 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).
   *
   *
   * Also see http://www.hakank.org/or-tools/3_jugs_regular.py
   *
   */
  private static bool Solve(int n)
  {
    Solver solver = new Solver("ThreeJugProblem");

    //
    // Data
    //
    
    // the DFA (for regular)
    int n_states = 14;
    int input_max = 15;
    int initial_state = 1; // state 0 is for the failing state
    int[] accepting_states = {15};

    //
    // Manually crafted DFA
    // (from the adjacency matrix used in the other models)
    //
    /*
    int[,] 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.
    //
    int[][] states = {
      new int[] {2,9},  // state 1
      new int[] {3},    // state 2
      new int[] {4, 9}, // state 3
      new int[] {5},    // state 4
      new int[] {6,9},  // state 5
      new int[] {7},    // state 6
      new int[] {8,9},  // state 7
      new int[] {15},   // state 8
      new int[] {10},   // state 9
      new int[] {11},   // state 10
      new int[] {12},   // state 11
      new int[] {13},   // state 12
      new int[] {14},   // state 13
      new int[] {15}    // state 14
    };
    
    int[,] transition_fn = new int[n_states,input_max];
    for(int i = 0; i < n_states; i++) {
      for(int j = 1; j <= input_max; j++) {
        bool in_states = false;
        for(int s = 0; s < states[i].Length; s++) {
          if (j == states[i][s]) {
            in_states = true;
            break;
          }
        }
        if (in_states) {
          transition_fn[i,j-1] = j;
        } else {
          transition_fn[i,j-1] = 0;
        }
      }
    }

    //
    // The name of the nodes, for printing
    // the solution.
    //
    string[] 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
    };

    //
    // Decision variables
    //

    // Note: We use 1..2 (instead of 0..1) and subtract 1 in the solution
    IntVar[] x = solver.MakeIntVarArray(n, 1, input_max, "x");


    //
    // Constraints
    //
    MyRegular(solver, x, n_states, input_max, transition_fn,
            initial_state, accepting_states);



    //
    // Search
    //
    DecisionBuilder db = solver.MakePhase(x,
                                          Solver.CHOOSE_FIRST_UNBOUND,
                                          Solver.ASSIGN_MIN_VALUE);

    solver.NewSearch(db);

    bool found = false;
    while (solver.NextSolution()) {
      Console.WriteLine("\nFound a solution of length {0}", n+1);
      int[] x_val = new int[n];
      x_val[0] = 1;
      Console.WriteLine("{0} -> {1}", nodes[0], nodes[x_val[0]]);
      for(int i = 1; i < n; i++) {
        // Note: here we subtract 1 to get 0..1
        int val = (int)x[i].Value()-1;
        x_val[i] = val;
        Console.WriteLine("{0} -> {1}", nodes[x_val[i-1]], nodes[x_val[i]]);
      }
      Console.WriteLine();

      Console.WriteLine("\nSolutions: {0}", solver.Solutions());
      Console.WriteLine("WallTime: {0}ms", solver.WallTime());
      Console.WriteLine("Failures: {0}", solver.Failures());
      Console.WriteLine("Branches: {0} ", solver.Branches());
      
      found = true;

    }

    solver.EndSearch();

    return found;

  }

  public static void Main(String[] args)
  {
    
    for(int n = 1; n < 15; n++) {
      bool found = Solve(n);
      if (found) {
        break;
      }
    }
  }
}