/*
<longKPathEval.h>
Copyright (C) DOLPHIN Project-Team, INRIA Lille - Nord Europe, 2006-2010

Sébastien Verel

This software is governed by the CeCILL license under French law and
abiding by the rules of distribution of free software.  You can  ue,
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*/

#ifndef __longKPathEval_h
#define __longKPathEval_h

#include <eoEvalFunc.h>

/**
 * Full evaluation function for long k-path problem
 */
template< class EOT >
class LongKPathEval : public eoEvalFunc<EOT>
{
private:
  // parameter k of the problem
  unsigned k;

  // tempory variable if the solution is in the long path
  bool inPath;

  /**
   * compute the number k j in solution = u1^kO^jw between i and i - k + 1 with |u1^kO^j| = i and k+j <= k
   *
   * @param solution the solution to evaluate
   * @param l last position in the bit string
   * @param n0 number of consecutive 0
   * @param n1 number of consecutive 1
   */
  void nbOnesZeros(EOT & solution, unsigned l, unsigned & n0, unsigned & n1) {
    n0 = 0;

    unsigned ind = l - 1;

    while (n0 < k && solution[ind - n0] == 0) 
      n0++;

    n1 = 0;

    ind = ind - n0;
    while (n0 + n1 < k && solution[ind - n1] == 1)
      n1++;
  }

  /**
   * true if the solution is the last solution of the path of bitstring length l
   *
   * @param solution the solution to evaluate
   * @param l size of the path, last position in the bit string
   * @return true if the solution is the solution of the path
   */
  bool final(EOT & solution, unsigned l) {
    if (l == 1)
      return (solution[0] == 1);
    else {
      int i = 0;

      while (i < l - k && solution[i] == 0)
	i++;

      if (i < l - k)
	return false;
      else {
	while (i < l && solution[i] == 1)
	  i++;
	
	return (i == l);
      }
    }
  }

  /**
   * position in the long path
   *
   * @param solution the solution to evaluate
   * @param l size of the path, last position in the bit string
   * @return position in the path
   */
  unsigned rank(EOT & solution, unsigned int l) { 
    if (l == 1) { // long path l = 1
      inPath = true;

      if (solution[0] == 0) 
	return 0;
      else
	return 1;
    } else { // long path for l>1
      unsigned n0, n1;

      // read the k last bits, and count the number of last successive 0 follow by the last successive 1
      nbOnesZeros(solution, l, n0, n1);

      if (n0 == k) // first part of the path
	return rank(solution, l - k);
      else
	if (n1 == k) { // last part of the path
	  return (k+1) * (1 << ((l-1) / k)) - k - rank(solution, l - k);
	} else 
	  if (n0 + n1 == k) {
	    if (final(solution, l - k)) {
	      inPath = true;
	      return (k+1) * (1 << ((l-k-1) / k)) - k + n1;
	    } else {
	      inPath = false;
	      return 0;
	    }
	  } else {
	    inPath = false;
	    return 0;
	  }
    }
  }

  /**
   * compute the number of zero of the bit string
   *
   * @param solution the solution to evaluate
   * @return number of zero in the bit string
   */
  unsigned int nbZero(EOT & solution){
    unsigned int res = 0;

    for(unsigned int i=0; i < solution.size(); i++)
      if (solution[i] == 0)
	res++;

    return res;
  }

public:
  /**
   * Default constructor
   *
   * @param _k parameter k of the long K-path problem
   */
  LongKPathEval(unsigned _k) : k(_k) {
  };

  /**
   * default destructor
   */
  ~LongKPathEval(void) {} ;
    
  /**
   * compute the fitnes of the solution
   *
   * @param solution the solution to evaluate
   * @return fitness of the solution
   */
  void operator()(EOT & solution) {
    inPath = true;

    unsigned r = rank(solution, solution.size());

    if (inPath)
      solution.fitness(solution.size() + r);
    else
      solution.fitness(nbZero(solution));
  }

};

#endif