📜  使用按位算法求解数独

📅  最后修改于: 2021-05-24 20:02:16             🧑  作者: Mango

给定部分填充的9×9矩阵,必须将数字(从1到9)分配给空单元格,以便大小为3×3的每一行,列和子矩阵都恰好包含一个1到9的数字实例。

这里描述了针对此问题的Pure backtracking解决方案。强烈建议读者在继续之前先了解纯回溯解决方案的工作原理。

在纯回溯解决方案中,我们遍历矩阵,每当找到一个空单元格(没有任何数字的单元格)时,我们就为该单元格分配一个数字,该数字在当前列,行和3×3中不存在子矩阵。将数字分配给当前单元格后,我们递归检查此分配是否导致有效的解决方案。如果分配没有找到有效的解决方案,则我们尝试使用当前空单元格的下一个有效数字。并且,如果没有一个数字导致有效的解决方案,则该实例是不可行的。

1. If there is no empty cell in the matrix M:
    return true
2. Let (i, j) be an empty cell in the matrix M
3. For i from 1 to 9:
    3.1. If i is not present in the row r, in column c, and the 3x3
    submatrix of (r, c):
        a) M(r, c) = i 
        b) recursively try fill in remaining empty cells
        c) If recursion was successful:
            return true
        d) M(r, c) = 0
4. return false
  • 可以通过遍历相应的行,列和3×3子矩阵来执行步骤(3.1)。但是,我们可以通过在回溯之前对这些数字进行预处理来加快此步骤,这是本文的重点。因此,让我们以下面的矩阵为例:

    我们可以跟踪整数位中的行,列和3×3子矩阵的数字,例如,考虑上一个矩阵的第一行,我们可以通过以下方式存储这些数字:

    bits order - 9 8 7 6 5 4 3 2 1
    bits       - 0 0 1 0 1 0 1 0 0
    

    然后在步骤(3.1)中,我们可以使用按位运算来确定数字i是否位于行,列和3×3子矩阵中。假设rowDigits [r]是包含第r行数字的整数,那么我们可以使用以下表达式检查数字i是否在第r行中:

    rowsDigits[r] & (1<<(i - 1))
    

    如果上面的表达式等于0,则在行r中不存在数字i。例如,如果r = 0,而i = 1,则:

    bits order                - 9 8 7 6 5 4 3 2 1
    rowDigits[r]              - 0 0 1 0 1 0 1 0 0
    1<<(i - 1)                - 0 0 0 0 0 0 0 0 1
    rowDigits[r]&(1<<(i - 1)) - 0 0 0 0 0 0 0 0 0
    
  • 一旦步骤(3.1)的条件为真,就执行步骤(3.1a),然后我们需要在rowDigits,columnDigits和subMatrixDigits中插入数字i,我们可以使用以下表达式进行操作:
    rowsDigits[r] | (1<<(i - 1))
    

    例如,如果r = 0,而i = 1,则:

    bits order                - 9 8 7 6 5 4 3 2 1
    rowDigits[r]              - 0 0 1 0 1 0 1 0 0
    1<<(i - 1)                - 0 0 0 0 0 0 0 0 1
    rowDigits[r]|(1<<(i - 1)) - 0 0 1 0 1 0 1 0 1
    
  • 如果步骤(3.1c)的条件为false,则执行步骤(3.1d),然后我们需要从rowDigits,columnDigits和subMatrixDigits中删除数字i,我们可以使用以下表达式进行操作:
    rowsDigits[r] & ~(1<<(i - 1))
    

    例如,如果r = 0,而i = 1,则:

    bits order                - 9 8 7 6 5 4 3 2 1
    rowDigits[r]              - 0 0 1 0 1 0 1 0 0
    1<<(i - 1)                - 0 0 0 0 0 0 0 0 1
    ~(1<<(i - 1))             - 1 1 1 1 1 1 1 1 0
    rowDigits[r]&~(1<<(i - 1) - 0 0 1 0 1 0 1 0 0
    

下面是上述方法的实现。

// C++ program to solve sudoku
#include 
#include 
  
// N is used for the size of Sudoku grid.  
// Size will be NxN  
#define N 9
  
using namespace std;
  
/* A utility function to print grid */
void printGrid(int grid[N][N])  
{  
    for (int row = 0; row < N; row++)  
    {  
    for (int col = 0; col < N; col++)  
            cout << grid[row][col] << " ";  
        cout << endl; 
    }  
}
/* Takes a partially filled-in grid and attempts  
to assign values to all unassigned locations in  
such a way to meet the requirements for 
Sudoku solution (non-duplication across rows, 
columns, and boxes) */
bool solve(int r, int c, int board[9][9], 
           int submatrixDigits[3][3], 
           int rowDigits[9], 
           int columnDigits[9])
{
    if (r == 9)
    {
          
        return true;
    }
    if (c == 9)
    {
        return solve(r + 1, 0, board, submatrixDigits, 
                     rowDigits, columnDigits);
    }
      
    if (board[r] == 0) {
        for (int i = 1; i <= 9; i++)
        {
            int digit = 1 << (i - 1);
              
            if (!((submatrixDigits[r / 3] & digit) 
                  || (rowDigits[r] & digit) 
                  || (columnDigits & digit)))
            {
                // set digit
                submatrixDigits[r / 3] |= digit;
                rowDigits[r] |= digit;
                columnDigits |= digit;
                board[r] = i;
                  
                if (solve(r, c + 1, board, submatrixDigits,
                          rowDigits, columnDigits))
                {
                    return true;
                }
                else
                {
                    submatrixDigits[r / 3] &= ~digit;
                    rowDigits[r] &= ~digit;
                    columnDigits &= ~digit;
                    board[r] = 0;
                }
            }
        }
        return false;
    }
    return solve(r, c + 1, board, submatrixDigits, 
                 rowDigits, columnDigits);
}
  
// Function checks if Sudoku can be
// solved or not
bool SolveSudoku(int board[9][9])
{
    int submatrixDigits[3][3];
    int columnDigits[9];
    int rowDigits[9];
      
    for (int i = 0; i < 3; i++)
        memset(submatrixDigits[i], 0, 3 * sizeof(int));
      
      
    memset(rowDigits, 0, 9 * sizeof(int));
    memset(columnDigits, 0, 9 * sizeof(int));
      
    // get 3x3 submatrix, row and column digits
    for (int i = 0; i < 9; i++)
        for (int j = 0; j < 9; j++)
            if (board[i][j] > 0)
            {
                int value = 1 << (board[i][j] - '1');
                submatrixDigits[i / 3][j / 3] |= value;
                rowDigits[i] |= value;
                columnDigits[j] |= value;
            }
    // Backtrack
    if (solve(0, 0, board, submatrixDigits,
              rowDigits, columnDigits))
        return true;
    else
        return false;
}
  
  
// Driver Code 
int main()  
{  
    // 0 means unassigned cells  
    int grid[N][N] = {{3, 0, 6, 5, 0, 8, 4, 0, 0},  
                      {5, 2, 0, 0, 0, 0, 0, 0, 0},  
                      {0, 8, 7, 0, 0, 0, 0, 3, 1},  
                      {0, 0, 3, 0, 1, 0, 0, 8, 0},  
                      {9, 0, 0, 8, 6, 3, 0, 0, 5},  
                      {0, 5, 0, 0, 9, 0, 6, 0, 0},  
                      {1, 3, 0, 0, 0, 0, 2, 5, 0},  
                      {0, 0, 0, 0, 0, 0, 0, 7, 4},  
                      {0, 0, 5, 2, 0, 6, 3, 0, 0}};  
    if (SolveSudoku(grid) == true)  
        printGrid(grid);  
    else
        cout << "No solution exists";  
    
    return 0;  
}  
输出:
3 1 6 5 2 8 4 3 4 
5 2 2 1 3 4 5 6 7 
3 8 7 5 6 7 1 3 1 
1 2 3 3 1 5 4 8 6 
9 3 4 8 6 3 2 1 5 
5 5 6 2 9 1 6 7 3 
1 3 1 4 5 2 2 5 8 
2 4 3 6 1 8 7 7 4 
6 5 5 2 7 6 3 2 1