c++ - 从理论上使用回溯递归解决数独难题

Question

回答:我的伪代码在递归方面是模糊的，但是这个视频和下面的资源很有帮助

http://www.youtube.com/watch?v=p-gpaIGRCQI

http://norvig.com/sudoku.html

无法掌握这种回溯递归算法在数独游戏中的实现。

我正在尝试使用递归回溯来解决数独难题。考虑到我正在处理的问题领域，我仍然无法在脑海中形成通用算法。

我尝试使用的回溯算法似乎是标准的，但我无法遵循逻辑并知道下面发生了什么。

包括回溯算法及其定义。

编辑:“同样，取出类定义，留下声明，放上伪代码”

这是我利用它的伪代码。

伪代码(C++实现)回溯游戏 (81,9)//表示游戏的所有可能的输入和值组合

    //All info is loading into a vector of size 81 with the initial state 
    //puzzle = the initial state 9x9 grid from left to right of integers  
        vector <int> puzzle
while(!not solved && not the end of the vector){
     for(int i =puzzle.begin::iterator i , puzzle.end()) //from 0-80 of the vector until end
        if puzzle[i] != 0
             //leave alone, original state of board
        else
              if (valid move) //a guess is allowed in this column/row/square of the board  
                   solution[i] = puzzle_guess[i]  //guess a move and insert 
              else  // one of previous guesses were wrong
                   game.prune(i); //backtracks, or reverses guesses until valid move
}

//游戏初始状态

    4 0 0  6 0 5  2 0 3
    0 0 0  0 4 9  0 7 5
    0 0 0  1 0 7  6 0 0

    6 0 1  0 0 0  4 8 7
    0 8 0  0 0 0  0 3 0
    2 7 4  0 0 0  5 0 6

    0 0 8  7 0 3  0 0 0
    3 1 0  9 6 0  0 0 0
    7 0 9  2 0 8  0 0 1

谢谢

唯一知道的线索是使用回溯游戏 (81,9) 的声明//表示 81 个可能的数字和 9 个不同的选项的 9。

#ifndef BACKTRACK_H
#define BACKTRACK_H

#include <vector>
#include <algorithm>

class BackTrack {
public:
  typedef std::vector<unsigned>::const_iterator const_iterator;
  typedef std::vector<unsigned>::const_iterator iterator;

  BackTrack (unsigned nVariables, unsigned arity=2);
  // Create a backtracking state for a problem with
  // nVariables variables, each of which has the same
  // number of possible values (arity).

  template <class Iterator>
  BackTrack (Iterator arityBegin,
         Iterator arityEnd);
  // Create a backtracking state in which each variable may have
  // a different number of possible values. The values are obtained
  // as integers stored in positions arityBegin .. arityEnd as per
  // the usual conventions for C++ iterators. The number of
  // variables in the system are inferred from the number of
  // positions in the given range.

  unsigned operator[] (unsigned variableNumber) const;
  // Returns the current value associated with the indicated
  // variable.

  unsigned numberOfVariables() const;
  // Returns the number of variables in the backtracking system.

  unsigned arity (unsigned variableNumber) const;
  // Returns the number of potential values that can be assigned
  // to the indicated variable.

  bool more() const;
  // Indicates whether additional candidate solutions exist that
  // can be reached by subsequent ++ or prune operaations.

  void prune (unsigned level);
  // Indicates that the combination of values associated with
  // variables 0 .. level-1 (inclusive) has been judged unacceptable
  // (regardless of the values that could be given to variables
  // level..numberOfVariables()-1.  The backtracking state will advance
  // to the next solution in which at least one of the values in the
  // variables 0..level-1 will have changed.

  BackTrack& operator++();
  // Indicates that the combination of values associated with
  // variables 0 .. nVariables-1 (inclusive) has been judged unacceptable.
  // The backtracking state will advance
  // to the next solution in which at least one of the values in the
  // variables 0..level-1 will have changed.

  BackTrack operator++(int);
  // Same as other operator++, but returns a copy of the old backtrack state


  // Iterator operations for easy access to the currently assigned values
  const_iterator begin() const {return values.begin();}
  iterator begin()             {return values.begin();}

  const_iterator end() const {return values.end();}
  iterator       end()       {return values.end();}

private:
  bool done;
  std::vector<unsigned> arities;
  std::vector<unsigned> values;

};
inline
unsigned BackTrack::operator[] (unsigned variableNumber) const
  // Returns the current value associated with the indicated
  // variable.
{
  return values[variableNumber];
}

inline
unsigned BackTrack::numberOfVariables() const
  // Returns the number of variables in the backtracking system.
{
  return values.size();
}

inline
unsigned BackTrack::arity (unsigned variableNumber) const
  // Returns the number of potential values that can be assigned
  // to the indicated variable.
{
  return arities[variableNumber];
}


inline
bool BackTrack::more() const
  // Indicates whether additional candidate solutions exist that
  // can be reached by subsequent ++ or prune operaations.
{
  return !done;
}

template <class Iterator>
BackTrack::BackTrack (Iterator arityBegin,
              Iterator arityEnd):
  // Create a backtracking state in which each variable may have
  // a different number of possible values. The values are obtained
  // as integers stored in positions arityBegin .. arityEnd as per
  // the usual conventions for C++ iterators. The number of
  // variables in the system are inferred from the number of
  // positions in the given range.
  arities(arityBegin, arityEnd), done(false) 
{
  fill_n (back_inserter(values), arities.size(), 0);
}


#endif

Answer 1

这是一个简单的伪代码，可以帮助您理解递归和回溯:

solve(game):
    if (game board is full)
        return SUCCESS
    else
        next_square = getNextEmptySquare()
        for each value that can legally be put in next_square
            put value in next_square (i.e. modify game state)
            if (solve(game)) return SUCCESS
            remove value from next_square (i.e. backtrack to a previous state)
    return FAILURE

一旦您理解了这一点，接下来就是了解 getNextEmptySquare() 的各种实现如何通过以不同方式修剪状态空间图来影响性能。

我在你的原始伪代码中没有看到任何递归或有条不紊的搜索，虽然它并不完全清楚，它似乎只是一遍又一遍地尝试随机排列？