title: Time complexity analysis of backtracking classic n-queens
body: <p>I have been reading some code of solving the classic n-queens problem. It is just to find one solution(not all of them or to count the # of solutions). You can find complete code on <a href="http://www.geeksforgeeks.org/backtracking-set-3-n-queen-problem/" rel="nofollow">geeksforgeeks website</a>, which is summed up below. <strong>The question is, what's the time complexity of this code, really</strong>?</p>
  
  <pre><code>bool isSafe(int board[N][N], int row, int col)
  {
      int i, j;
  
      /* Check this row on left side */
      for (i = 0; i &lt; col; i++)
          if (board[row][i])
              return false;
  
      /* Check upper diagonal on left side */
      for (i=row, j=col; i&gt;=0 &amp;&amp; j&gt;=0; i--, j--)
          if (board[i][j])
              return false;
  
      /* Check lower diagonal on left side */
      for (i=row, j=col; j&gt;=0 &amp;&amp; i&lt;N; i++, j--)
          if (board[i][j])
              return false;
  
      return true;
  }
  
  /* A recursive utility function to solve N
     Queen problem */
  bool solveNQUtil(int board[N][N], int col)
  {
      /* base case: If all queens are placed
        then return true */
      if (col &gt;= N)
          return true;
  
      /* Consider this column and try placing
         this queen in all rows one by one */
      for (int i = 0; i &lt; N; i++)
      {
          /* Check if queen can be placed on
            board[i][col] */
          if ( isSafe(board, i, col) )
          {
              /* Place this queen in board[i][col] */
              board[i][col] = 1;
  
              /* recur to place rest of the queens */
              if ( solveNQUtil(board, col + 1) )
                  return true;
  
              /* If placing queen in board[i][col]
                 doesn't lead to a solution, then
                 remove queen from board[i][col] */
              board[i][col] = 0; // BACKTRACK
          }
      }
  
       /* If queen can not be place in any row in
          this colum col  then return false */
      return false;
  }
  
  /* This function solves the N Queen problem using
     Backtracking. It mainly uses solveNQUtil() to
     solve the problem. It returns false if queens
     cannot be placed, otherwise return true and
     prints placement of queens in the form of 1s.
     Please note that there may be more than one
     solutions, this function prints one  of the
     feasible solutions.*/
  bool solveNQ()
  {
      int board[N][N] = { {0, 0, 0, 0},
          {0, 0, 0, 0},
          {0, 0, 0, 0},
          {0, 0, 0, 0}
      };
  
      if ( solveNQUtil(board, 0) == false )
      {
        printf("Solution does not exist");
        return false;
      }
  
      printSolution(board);
      return true;
  }
  </code></pre>
  
  <p>If you went through the history of comments, some said it's <code>O(n!)</code>, or even exponential. But I think neither of them is correct.</p>
  
  <p>E.g., for claim of <code>O(n!)</code>, one gave <code>T(n)= n*(T(n-1) + O(n))</code> which leads to <code>O(n!)</code>.</p>
  
  <h2><strong>Why I think it's wrong(not <code>O(n!)</code>)?</strong></h2>
  
  <p>1.The problem is, the <code>for</code> loop in <code>solveNQUtil</code> always runs <code>N</code> times. It doesn't decrease with problem scope <code>n</code>. So the multiplier <code>n</code> in the above formula is not correct. It should be replaced with a fixed number <code>N</code>.</p>
  
  <p>2.<code>col</code> in <code>isSafe</code> increases down the recursion tree, which means the <code>for</code> loop in <code>isSafe</code> has more and more iterations. The # of iterations is <code>N - n</code>.</p>
  
  <p>So basically, the recursion should be <code>T(n)= N *(T(n-1) + O(N - n))</code> (N is fixed). Not sure how to solve this one, but it's should be at least <code>O(N^N)</code>. <strong>Any idea?</strong></p>
  
  <h2>Recursion tree method</h2>
  
  <p>If using a recursion tree, there are <code>O(n^n)</code> different paths down to the leaves, and each path takes <code>O(1+2+..n)</code> operations for conflict check. So total time should be <code>O(n^(n+2))</code>. <strong>Is this right?</strong></p>
  
  <p>Could anyone point out if this is correct and give appropriate reasonings?</p>