//
// 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.Collections.Generic;
using System.Linq;
using Google.OrTools.ConstraintSolver;


public class Kakuro
{

  /**
   * Ensure that the sum of the segments
   * in cc == res
   *
   */
  public static void  calc(Solver solver,
                           int[] cc,
                           IntVar[,] x,
                           int res)
  {

    // ensure that the values are positive
    int len = cc.Length / 2;
    for(int i = 0; i < len; i++) {
      solver.Add(x[cc[i*2]-1,cc[i*2+1]-1] >= 1);
    }

    // sum the numbers
    solver.Add( (from i in Enumerable.Range(0, len)
                 select x[cc[i*2]-1,cc[i*2+1]-1])
                .ToArray().Sum() == res);
  }



  /**
   *
   * Kakuru puzzle.
   *
   * http://en.wikipedia.org/wiki/Kakuro
   * """
   * The object of the puzzle is to insert a digit from 1 to 9 inclusive
   * into each white cell such that the sum of the numbers in each entry
   * matches the clue associated with it and that no digit is duplicated in
   * any entry. It is that lack of duplication that makes creating Kakuro
   * puzzles with unique solutions possible, and which means solving a Kakuro
   * puzzle involves investigating combinations more, compared to Sudoku in
   * which the focus is on permutations. There is an unwritten rule for
   * making Kakuro puzzles that each clue must have at least two numbers
   * that add up to it. This is because including one number is mathematically
   * trivial when solving Kakuro puzzles; one can simply disregard the
   * number entirely and subtract it from the clue it indicates.
   * """
   *
   * This model solves the problem at the Wikipedia page.
   * For a larger picture, see
   * http://en.wikipedia.org/wiki/File:Kakuro_black_box.svg
   *
   * The solution:
   *  9 7 0 0 8 7 9
   *  8 9 0 8 9 5 7
   *  6 8 5 9 7 0 0
   *  0 6 1 0 2 6 0
   *  0 0 4 6 1 3 2
   *  8 9 3 1 0 1 4
   *  3 1 2 0 0 2 1
   *
   * Also see http://www.hakank.org/or-tools/kakuro.py
   * though this C# model has another representation of
   * the problem instance.
   *
   */
  private static void Solve()
  {

    Solver solver = new Solver("Kakuro");

    // size of matrix
    int n = 7;

    // segments:
    //  sum, the segments
    // Note: this is 1-based
    int[][] problem =
      {
        new int[] {16,  1,1, 1,2},
        new int[] {24,  1,5, 1,6, 1,7},
        new int[] {17,  2,1, 2,2},
        new int[] {29,  2,4, 2,5, 2,6, 2,7},
        new int[] {35,  3,1, 3,2, 3,3, 3,4, 3,5},
        new int[] { 7,  4,2, 4,3},
        new int[] { 8,  4,5, 4,6},
        new int[] {16,  5,3, 5,4, 5,5, 5,6, 5,7},
        new int[] {21,  6,1, 6,2, 6,3, 6,4},
        new int[] { 5,  6,6, 6,7},
        new int[] { 6,  7,1, 7,2, 7,3},
        new int[] { 3,  7,6, 7,7},

        new int[] {23,  1,1, 2,1, 3,1},
        new int[] {30,  1,2, 2,2, 3,2, 4,2},
        new int[] {27,  1,5, 2,5, 3,5, 4,5, 5,5},
        new int[] {12,  1,6, 2,6},
        new int[] {16,  1,7, 2,7},
        new int[] {17,  2,4, 3,4},
        new int[] {15,  3,3, 4,3, 5,3, 6,3, 7,3},
        new int[] {12,  4,6, 5,6, 6,6, 7,6},
        new int[] { 7,  5,4, 6,4},
        new int[] { 7,  5,7, 6,7, 7,7},
        new int[] {11,  6,1, 7,1},
        new int[] {10,  6,2, 7,2}

      };


    int num_p = 24; // Number of segments

    // The blanks
    // Note: 1-based
    int[,] blanks = {
      {1,3}, {1,4},
      {2,3},
      {3,6}, {3,7},
      {4,1}, {4,4}, {4,7},
      {5,1}, {5,2},
      {6,5},
      {7,4}, {7,5}
    };

    int num_blanks = blanks.GetLength(0);


    //
    // Decision variables
    //
    IntVar[,] x =  solver.MakeIntVarMatrix(n, n, 0, 9, "x");
    IntVar[] x_flat = x.Flatten();

    //
    // Constraints
    //

    // fill the blanks with 0
    for(int i = 0; i < num_blanks; i++) {
      solver.Add(x[blanks[i,0]-1,blanks[i,1]-1]==0);
    }

    for(int i = 0; i < num_p; i++) {
      int[] segment = problem[i];

      // Remove the sum from the segment
      int[] s2 = new int[segment.Length-1];
      for(int j = 1; j < segment.Length; j++) {
        s2[j-1] = segment[j];
      }

      // sum this segment
      calc(solver, s2, x, segment[0]);

      // all numbers in this segment must be distinct
      int len = segment.Length / 2;
      solver.Add( (from j in Enumerable.Range(0, len)
                   select x[s2[j * 2] - 1, s2[j * 2 + 1] - 1])
                  .ToArray().AllDifferent());
    }

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

    solver.NewSearch(db);

    while (solver.NextSolution()) {
      for(int i = 0; i < n; i++) {
        for(int j = 0; j < n; j++) {
          int v = (int)x[i,j].Value();
          if (v > 0) {
            Console.Write(v + " ");
          } else {
            Console.Write("  ");
          }
        }
        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());

    solver.EndSearch();

  }


  public static void Main(String[] args)
  {

    Solve();
  }
}