回溯N皇后算法

Question

I was seeing the below code and trying to understand how it works. I got stuck in backtracking . could some one explain how the r value is going from 2 to 1. when it failed to find the suitable position for the queen to place . ` Below is the debugging section :

nQueen (mat=0x7fffffffddb0, r=2) at N_Queens_problem.cpp:47
47      for (int i = 0; i < N; i++) 
(gdb) print i
$1 = 3
(gdb) n
62  }
(gdb) print i
No symbol "i" in current context.
(gdb) print i
No symbol "i" in current context.
**(gdb) print r
$2 = 2**
(gdb) n
nQueen (mat=0x7fffffffddb0, r=1) at N_Queens_problem.cpp:47
47      for (int i = 0; i < N; i++) 
**(gdb) print r
$3 = 1**
(gdb)

Below is the nqueen function :

void nQueen(char mat[][N], int r)
{
    // if N queens are placed successfully, print the solution
    if (r == N)
    {
        for (int i = 0; i < N; i++) 
        {
            for (int j = 0; j < N; j++)
                cout << mat[i][j] << " ";
            cout << endl;
        }
        cout << endl;

        return;
    }

    // place Queen at every square in current row r
    // and recur for each valid movement    
    for (int i = 0; i < N; i++) 
    {
        // if no two queens threaten each other
        if (isSafe(mat, r, i)) 
        {
            // place queen on current square
            mat[r][i] = 'Q';

            // recur for next row
            nQueen(mat, r + 1);

            // backtrack and remove queen from current square
            mat[r][i] = '-';
        }
    }
}

Below is the function to check whether the queen is safe to place :

bool isSafe(char mat[][N], int r, int c)
{
    // return false if two queens share the same column
    for (int i = 0; i < r; i++)
        if (mat[i][c] == 'Q')
            return false;

    // return false if two queens share the same \ diagonal
    for (int i = r, j = c; i >= 0 && j >= 0; i--, j--)
        if (mat[i][j] == 'Q')
            return false;

    // return false if two queens share the same / diagonal
    for (int i = r, j = c; i >= 0 && j < N; i--, j++)
        if (mat[i][j] == 'Q')
            return false;

    return true;
}

Answer 1

Say you have placed two queens and going to place the 3rd queen.

1. Till now there will be two recursive calls to nQueens() in the stack. One for the queen 0 and one for the queen 3.
2. nQueen(mat, r + 1); will be called with nQueen(mat, 2) and third nQueens() will be pushed to stack.

considering that it is not safe to place the queen at any of the N locations of 3rd row and

3. if (isSafe(mat, r, i)) will return false for all columns 0 to N.
4. Ultimately the loop for (int i = 0; i < N; i++) will end for r == 2 and stack will unwind.
5. Stack will remove the topmost call and we will be left function calls in stack(bcak to step 1).
6. And loop for (int i = 0; i < N; i++) will try to reposition the 2nd queen.