All possible solution of the n-Queen's algorithm

Question

When implementing an algorithm for all possible solution of an n-Queen problem, i found that the same solution is reached by many branches. Is there any good way to generate every unique solutions to the n-Queens problem? How to avoid the duplicate solutions generated by the different branches (except store and compare)?

Here is what i have tried, for the first solution: http://www.ideone.com/hDpr3

Code:

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

/* crude */

#define QUEEN 'Q'
#define BLANK '.'

int is_valid (char **board, int n, int a, int b)
{
  int i, j;

  for (i=0; i<n; i++)
  {
    if (board[a][i] == QUEEN)
      return 0;
    if (board[i][b] == QUEEN)
      return 0;
  }

  for (i=a, j=b; (i>=0) && (j>=0); i--, j--)
  {
    if (board[i][j] == QUEEN)
      return 0;
  }

  for (i=a, j=b; (i<n) && (j<n); i++, j++)
  {
    if (board[i][j] == QUEEN)
      return 0;
  }

  for (i=a, j=b; (i>=0) && (j<n); i--, j++)
  {
    if (board[i][j] == QUEEN)
      return 0;
  }

  for (i=a, j=b; (i<n) && (j>=0); i++, j--)
  {
    if (board[i][j] == QUEEN)
      return 0;
  }
  return 1;
}

void show_board (char **board, int n)
{
  int i, j;

  for (i=0; i<n; i++)
  {
    printf ("\n");
    for (j=0; j<n; j++)
    {
      printf (" %c", board[i][j]);
    }
  }
}

int nqdfs_all (char **board, int n, int d)
{
  int i, j, ret = 0;

  /* the last queen was placed on the last depth
   * therefore this dfs branch in the recursion 
   * tree is a solution, return 1
   */
  if (d == n)
  {
    /* Print whenever we find one solution */
    printf ("\n");
    show_board (board, n);
    return 1;
  }

  /* check all position */
  for (i=0; i<n; i++)
  {
    for (j=0; j<n; j++)
    {
      if (is_valid (board, n, i, j))
      {
    board[i][j] = QUEEN;
    nqdfs_all (board, n, d + 1);
    board[i][j] = BLANK;
      }
    }
  }

  return ret;  
}

int nqdfs_first (char **board, int n, int d)
{
  int i, j, ret = 0;

  /* the last queen was placed on the last depth
   * therefore this dfs branch in the recursion 
   * tree is a solution, return 1
   */
  if (d == n)
    return 1;

  /* check all position */
  for (i=0; i<n; i++)
  {
    for (j=0; j<n; j++)
    {
      if (is_valid (board, n, i, j))
      {
    board[i][j] = QUEEN;
    ret = nqdfs_first (board, n, d + 1);
    if (ret)
    {
      /* if the first branch is found, tell about its 
       * success to its parent, we will not look in other
       * solutions in this function.
       */
      return ret;
    }
    else
    {
      /* this was not a successful path, so replace the
       * queen with a blank, and prepare board for next
       * pass
       */
      board[i][j] = BLANK;
    }
      }
    }
  }

  return ret;
}

int main (void)
{
  char **board;
  int n, i, j, ret;

  printf ("\nEnter \"n\": ");
  scanf ("%d", &n);

  board = malloc (sizeof (char *) * n);
  for (i=0; i<n; i++)
  {
    board[i] = malloc (sizeof (char) * n);
    memset (board[i], BLANK, n * sizeof (char));
  }

  nqdfs_first (board, n, 0);
  show_board (board, n);

  printf ("\n");
  return 0;
}

To generate all possible solution i have used the same code nqdfs_all () function, but did not return the control to the parent, instead continued enumerating and checking. A call to this function displays the duplicate results reached by different branches.

Answer 1

You should make use of the fact that each queen must be placed in a different column. If you have placed already k queens in the first k columns, recursively place queen number k+1 in column k+1 and go through rows 1 to n (and not through all n*n cells, as you algo currently does). Continue with k:=k+1 for each valid placement. That will avoid any duplicate results, since this algo does not generate any duplicate boards at all.

EDIT: to your question about avoiding of symmetries: a part of those can be avoided beforehand, for example, by restricting queen 1 in column 1 to rows 1,...n/2. If you want to avoid the output of symmetric solutions completely, you will have to store every found solution in a list and whenever you find a new solution, before printing it out, test if one of it's symmetric equivalents is already there in the list.

To make this more efficient, you can work with a "canoncial representation" of each board, defined as follows. Generate all symmetric boards of a given one, pack each one of it into a byte array, and among those arrays keep the array which, interpreted as a big number, has the minimum value. This packed represention is a unique identifier of the symmetry class of each board and can be easily put in a dictionary / hash table, which makes testing if that symmetry class already appeared very efficient.