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/****   , [ MatrixGame.cpp ], 
Copyright (c) 2007 Universite d'Orleans - Arnaud Lallouet 

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 *************************************************************************/

#include <cstdlib> /* pour srand, rand et RAND_MAX */
#include <ctime> /* pour time */
#include <math.h> /* pour pow */

#include <iostream>

#include "gecode/minimodel.hh"
#include "gecode/int/element.hh"

#include "qsolver.hh"
#include "implicative.hh"
#include "SDFVariableHeuristic.hh"
#include "NaiveValueHeuristics.hh"

#define UNIVERSAL true
#define EXISTENTIAL false

// The Matrix game consists in a square boolean matrix of size 2^depth. First player cuts it vertically in two parts and removes one half,
// while secodn player do the same, but cutting the matrix horizontally. The game ends when there are only one cell left in the matrix. 
// If this last cell has value 1, the first player wins. If it has value 0, the second player wins. 

// The present model of this game is pure QBF, that QeCode can handle (though not as fast as QBF solvers...)

using namespace MiniModel;

int main (int argc, char * const argv[]) {
    
    int depth = 6;  // Size of the matrix is 2^depth. Larger values may take long to solve...
    int nbDecisionVar = 2*depth;
    int nbScope = nbDecisionVar+1;
    int boardSize = (int)pow((double)2,(double)depth);
    
    std::srand(std::time(NULL));
    
    IntArgs board(boardSize*boardSize);
    for (int i=0; i<boardSize; i++)
        for (int j=0; j<boardSize; j++)
            board[j*boardSize+i] = (int)( (double)rand()  /  ((double)RAND_MAX + 1) * 50 ) < 25 ? 0:1;
    
    IntArgs access(nbDecisionVar);
    access[nbDecisionVar-1]=1;
    access[nbDecisionVar-2]=boardSize;
    for (int i=nbDecisionVar-3; i>=0; i--)
        access[i]=access[i+2]*2;
         
    // debug
    for (int i=0; i<boardSize; i++)
    {
        for (int j=0; j<boardSize; j++)
            cout << board[j*boardSize+i] << " ";
        cout << endl;
         }
    cout  << endl;
    for (int i=0; i<nbDecisionVar; i++)
        cout << access[i] << " ";
    cout << endl;
    // end debug
    
    int *scopesSize = new int[nbScope];
    for (int i=0; i<nbScope-1; i++)
        scopesSize[i]=1;
    scopesSize[nbScope-1]=2;
    
    Implicative p(nbScope, QECODE_EXISTENTIAL, scopesSize);
    
    // Defining the variable of the n first scopes ...
    for (int i=0; i<nbDecisionVar; i++)
    {
        p.QIntVar(i, 0, 1);
        p.nextScope();
    }
         
    // Declaring last scope variables ...
    
    p.QIntVar(nbDecisionVar, 0, 1);
    p.QIntVar(nbDecisionVar+1, 0, boardSize*boardSize);
    p.nextScope();
    // Body
    
    post(p.space(), p.var(nbDecisionVar) == 1);
    
    IntVarArgs X(nbDecisionVar);
    for (int i=0; i<nbDecisionVar; i++)
        X[i]=p.var(i);
    
    linear(p.space(), access, X, IRT_EQ, p.var(nbDecisionVar+1));
    //  MiniModel::LinRel R(E, IRT_EQ, MiniModel::LinExpr(p.var(nbDecisionVar+1)));
    element(p.space(), board, p.var(nbDecisionVar+1), p.var(nbDecisionVar));
    
    // When every variables and constraints have been declared, the makeStructure method
    // must be called in order to lead the problem ready for solving.
    p.makeStructure();
    
    /*
     cout << (p.quantification(0)==UNIVERSAL? "universal" : "existential") << "-0\n";
     cout << "  " << (p.quantification(1)==UNIVERSAL? "universal" : "existential") << "-1\n";
     cout << "    " << (p.quantification(2)==UNIVERSAL? "universal" : "existential") << "-2\n";
     cout << "      " << (p.quantification(3)==UNIVERSAL? "universal" : "existential") << "-3\n";
     cout << "        " << (p.quantification(4)==UNIVERSAL? "universal" : "existential") << "-4\n";
     
     */
    
    // We will use a Smallest Domain First branching heuristic for solving this problem.
    SmallestDomainFirst heur;
    SmallestValueFirst v_heur;
    
    // So, we build a quantified solver for our problem p, using the heuristic we just created.
    QSolver solver(&p,&heur, &v_heur);
    
    unsigned long int nodes=0;
    unsigned long int steps=0;
    
    // then we solve the problem. Nodes and Steps will contain the number of nodes encountered and
    // of propagation steps achieved during the solving.
    bool outcome = solver.solve(nodes,steps);
    
    cout << "  outcome: " << ( outcome? "TRUE" : "FALSE") << endl;
    cout << "  nodes visited: " << nodes << " " << steps << endl;
    
    return outcome? 10:20;
    
}

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