Printing all solutions in N-Queen Problem

Difficulty Level : Hard
Last Updated : 06 Oct, 2020

The N Queen is the problem of placing N chess queens on an N×N chessboard so that no two queens attack each other. For example, following is a solution for 4 Queen problem.
The N Queen is the problem of placing N chess queens on an N×N chessboard so that no two queens attack each other. For example, following are two solutions for 4 Queen problem.

nQueen-solution2

In previous post, we have discussed an approach that prints only one possible solution, so now in this post the task is to print all solutions in N-Queen Problem. The solution discussed here is an extension of same approach.

Backtracking Algorithm
The idea is to place queens one by one in different columns, starting from the leftmost column. When we place a queen in a column, we check for clashes with already placed queens. In the current column, if we find a row for which there is no clash, we mark this row and column as part of the solution. If we do not find such a row due to clashes then we backtrack and return false.

1) Start in the leftmost column
2) If all queens are placed
    return true
3) Try all rows in the current column.  Do following
   for every tried row.
    a) If the queen can be placed safely in this row
       then mark this [row, column] as part of the 
       solution and recursively check if placing  
       queen here leads to a solution.
    b) If placing queen in [row, column] leads to a
       solution then return true.
    c) If placing queen doesn't lead to a solution 
       then unmark this [row, column] (Backtrack) 
       and go to step (a) to try other rows.
3) If all rows have been tried and nothing worked, 
   return false to trigger backtracking.

There is only a slight modification in above algorithm that is highlighted in the code.

C++

/* C/C++ program to solve N Queen Problem using
backtracking */
#include<bits/stdc++.h>
#define N 4
 
/* A utility function to print solution */
void printSolution(int board[N][N])
{
    static int k = 1;
    printf("%d-\n",k++);
    for (int i = 0; i < N; i++)
    {
        for (int j = 0; j < N; j++)
            printf(" %d ", board[i][j]);
        printf("\n");
    }
    printf("\n");
}
 
/* A utility function to check if a queen can
be placed on board[row][col]. Note that this
function is called when "col" queens are
already placed in columns from 0 to col -1.
So we need to check only left side for
attacking queens */
bool isSafe(int board[N][N], int row, int col)
{
    int i, j;
 
    /* Check this row on left side */
    for (i = 0; i < col; i++)
        if (board[row][i])
            return false;
 
    /* Check upper diagonal on left side */
    for (i=row, j=col; i>=0 && j>=0; i--, j--)
        if (board[i][j])
            return false;
 
    /* Check lower diagonal on left side */
    for (i=row, j=col; j>=0 && i<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 == N)
    {
        printSolution(board);
        return true;
    }
 
    /* Consider this column and try placing
    this queen in all rows one by one */
    bool res = false;
    for (int i = 0; i < 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;
 
            // Make result true if any placement
            // is possible
            res = solveNQUtil(board, col + 1) || res;
 
            /* 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 column col then return false */
    return res;
}
 
/* 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.*/
void solveNQ()
{
    int board[N][N];
    memset(board, 0, sizeof(board));
 
    if (solveNQUtil(board, 0) == false)
    {
        printf("Solution does not exist");
        return ;
    }
 
    return ;
}
 
// driver program to test above function
int main()
{
    solveNQ();
    return 0;
}

Java

/* Java program to solve N Queen  
Problem using backtracking */
 
class GfG 
{
 
static int N = 4; 
static int k = 1;
 
/* A utility function to print solution */
static void printSolution(int board[][]) 
{ 
    System.out.printf("%d-\n", k++); 
    for (int i = 0; i < N; i++) 
    { 
        for (int j = 0; j < N; j++) 
            System.out.printf(" %d ", board[i][j]); 
        System.out.printf("\n"); 
    } 
    System.out.printf("\n"); 
} 
 
/* A utility function to check if a queen can 
be placed on board[row][col]. Note that this 
function is called when "col" queens are 
already placed in columns from 0 to col -1. 
So we need to check only left side for 
attacking queens */
static boolean isSafe(int board[][], int row, int col) 
{ 
    int i, j; 
 
    /* Check this row on left side */
    for (i = 0; i < col; i++) 
        if (board[row][i] == 1) 
            return false; 
 
    /* Check upper diagonal on left side */
    for (i = row, j = col; i >= 0 && j >= 0; i--, j--) 
        if (board[i][j] == 1) 
            return false; 
 
    /* Check lower diagonal on left side */
    for (i = row, j = col; j >= 0 && i < N; i++, j--) 
        if (board[i][j] == 1) 
            return false; 
 
    return true; 
} 
 
/* A recursive utility function  
to solve N Queen problem */
static boolean solveNQUtil(int board[][], int col) 
{ 
    /* base case: If all queens are placed 
    then return true */
    if (col == N) 
    { 
        printSolution(board); 
        return true; 
    } 
 
    /* Consider this column and try placing 
    this queen in all rows one by one */
    boolean res = false; 
    for (int i = 0; i < 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; 
 
            // Make result true if any placement 
            // is possible 
            res = solveNQUtil(board, col + 1) || res; 
 
            /* 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 column col then return false */
    return res; 
} 
 
/* 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.*/
static void solveNQ() 
{ 
    int board[][] = new int[N][N]; 
 
    if (solveNQUtil(board, 0) == false) 
    { 
        System.out.printf("Solution does not exist"); 
        return ; 
    } 
 
    return ; 
} 
 
// Driver code 
public static void main(String[] args)
{
    solveNQ();
}
}
 
// This code has been contributed
// by 29AjayKumar

Python3

''' Python3 program to solve N Queen Problem using 
backtracking '''
k = 1
 
# A utility function to print solution 
def printSolution(board): 
 
    global k
    print(k, "-\n")
    k = k + 1
    for i in range(4): 
        for j in range(4):
            print(board[i][j], end = " ")
        print("\n")
    print("\n") 
 
''' A utility function to check if a queen can 
be placed on board[row][col]. Note that this 
function is called when "col" queens are 
already placed in columns from 0 to col -1. 
So we need to check only left side for 
attacking queens '''
def isSafe(board, row, col) :
     
    # Check this row on left side 
    for i in range(col): 
        if (board[row][i]): 
            return False
 
    # Check upper diagonal on left side 
    i = row
    j = col
    while i >= 0 and j >= 0:
        if(board[i][j]):
            return False;
        i -= 1
        j -= 1
 
    # Check lower diagonal on left side 
    i = row
    j = col
    while j >= 0 and i < 4:
        if(board[i][j]):
            return False
        i = i + 1
        j = j - 1
 
    return True
 
''' A recursive utility function to solve N 
Queen problem '''
def solveNQUtil(board, col) :
     
    ''' base case: If all queens are placed 
    then return true '''
    if (col == 4): 
        printSolution(board) 
        return True
 
    ''' Consider this column and try placing 
    this queen in all rows one by one '''
    res = False
    for i in range(4):
     
        ''' 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; 
 
            # Make result true if any placement 
            # is possible 
            res = solveNQUtil(board, col + 1) or res; 
 
            ''' 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 column col then return false '''
    return res
 
''' 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.'''
def solveNQ() :
 
    board = [[0 for j in range(10)] 
                for i in range(10)]
 
    if (solveNQUtil(board, 0) == False): 
     
        print("Solution does not exist") 
        return
    return
 
# Driver Code 
solveNQ() 
 
# This code is contributed by YatinGupta

C#

/* C# program to solve N Queen 
Problem using backtracking */
using System;
 
class GfG 
{ 
 
static int N = 4; 
static int k = 1; 
 
/* A utility function to print solution */
static void printSolution(int [,]board) 
{ 
    Console.Write("{0}-\n", k++); 
    for (int i = 0; i < N; i++) 
    { 
        for (int j = 0; j < N; j++) 
            Console.Write(" {0} ", board[i, j]); 
        Console.Write("\n"); 
    } 
    Console.Write("\n"); 
} 
 
/* A utility function to check if a queen can 
be placed on board[row,col]. Note that this 
function is called when "col" queens are 
already placed in columns from 0 to col -1. 
So we need to check only left side for 
attacking queens */
static bool isSafe(int [,]board, int row, int col) 
{ 
    int i, j; 
 
    /* Check this row on left side */
    for (i = 0; i < col; i++) 
        if (board[row, i] == 1) 
            return false; 
 
    /* Check upper diagonal on left side */
    for (i = row, j = col; i >= 0 && j >= 0; i--, j--) 
        if (board[i, j] == 1) 
            return false; 
 
    /* Check lower diagonal on left side */
    for (i = row, j = col; j >= 0 && i < N; i++, j--) 
        if (board[i, j] == 1) 
            return false; 
 
    return true; 
} 
 
/* A recursive utility function 
to solve N Queen problem */
static bool solveNQUtil(int [,]board, int col) 
{ 
    /* base case: If all queens are placed 
    then return true */
    if (col == N) 
    { 
        printSolution(board); 
        return true; 
    } 
 
    /* Consider this column and try placing 
    this queen in all rows one by one */
    bool res = false; 
    for (int i = 0; i < 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; 
 
            // Make result true if any placement 
            // is possible 
            res = solveNQUtil(board, col + 1) || res; 
 
            /* 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 column col then return false */
    return res; 
} 
 
/* 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.*/
static void solveNQ() 
{ 
    int [,]board = new int[N, N]; 
 
    if (solveNQUtil(board, 0) == false) 
    { 
        Console.Write("Solution does not exist"); 
        return ; 
    } 
 
    return ; 
} 
 
// Driver code 
public static void Main() 
{ 
    solveNQ(); 
} 
} 
 
/* This code contributed by PrinciRaj1992 */

Output

Output:

This article is contributed by Sahil Chhabra. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.
Efficient Backtracking Approach Using Bit-Masking
Algorithm:
There is always only one queen in each row and each column, so idea of backtracking is to start placing queen from leftmost column of each row and find a column where queen could be placed without collision with previously placed queens. It is repeated from first row till last row. While placing a queen, it is tracked as if it is not making a collision (row-wise, column-wise and diagonally) with queens placed in previous rows. Once it is found that queen can’t be placed at particular column index in a row, algorithm backtracks and change the position of queen placed in previous row then moves forward to place the queen in next row.
1. Start with three bit vector which is used to track safe place for queen placement row-wise, column-wise and diagonally in each iteration.
2. Three bit vector will contain information as bellow:
rowmask: set bit index (i) of this bit vector will indicate, queen can’t be placed at ith column of next row.
ldmask: set bit index (i) of this bit vector will indicate, queen can’t e placed at ith column of next row. It represents the unsafe column index for next row falls under left diagonal of queens placed in previous rows.
rdmask: set bit index (i) of this bit vector will indicate, queen can’t be placed at ith column of next row. It represents the unsafe column index for next row falls right diagonal of queens placed in previous rows.
3. There is a 2-D (NxN) matrix (board), which will have ‘ ‘ character at all indexes in beginning and it gets filled by ‘Q’ row-by-row. Once all rows are filled by ‘Q’, this board matrix is displayed and counter variable to count number of ways to successful queen placement is increased. At last, counter variable is displayed as total number of ways to place the queens.

C++

// CPP program for above approach
#include <iostream>
#include <vector>
#include <math.h>
using namespace std;
 
// Program to print board
void printBoard(vector< vector<char> >& board)
{
  for (auto row : board) 
  {
    cout<<"|";
    for (auto ch : row) 
    {
      cout<<ch<<"|";
    }
    cout<<endl;
  }
}
 
// Program to solve N Quuens problem
void solveBoard(vector<vector<char> >& board, int row,
                int rowmask, int ldmask, int rdmask, int& ways)
{
 
  int n = board.size();
 
  // All_rows_filled is a bit mask having all N bits set
  int all_rows_filled = (1 << n) - 1;
 
  // If rowmask will have all bits set, means queen has been
  // placed successfully in all rows and board is diplayed
  if (rowmask == all_rows_filled) 
  {
    ways++;
    cout<<"=====================\n";
    cout<<"Queen placement - "<<ways<<endl;
    cout<<"=====================\n";
    printBoard(board);
    return;
  }
 
  // We extract a bit mask(safe) by rowmask,
  // ldmask and rdmask. all set bits of 'safe' 
  // indicates the safe column index for queen
  // placement of this iteration for row index(row).
  int safe = all_rows_filled & (~(rowmask | 
                                  ldmask | rdmask));
  while (safe) 
  {
 
    // Extracts the right-most set bit
    // (safe column index) where queen 
    // can be placed for this row
    int p = safe & (-safe);
    int col = (int)log2(p);
    board[row][col] = 'Q';
 
    // This is very important:
    // we need to update rowmask, ldmask and rdmask 
    // for next row as safe index for queen placement
    // will be decided by these three bit masks.
 
    // We have all three rowmask, ldmask and 
    // rdmask as 0 in beginning. Suppose, we are placing
    // queen at 1st column index at 0th row. rowmask, ldmask
    // and rdmask will change for next row as below:
 
    // rowmask's 1st bit will be set by OR operation
    // rowmask = 00000000000000000000000000000010
 
    // ldmask will change by setting 1st
    // bit by OR operation  and left shifting 
    // by 1 as it has to block the next column
    // of next row because that will fall on left diagonal.
    // ldmask = 00000000000000000000000000000100 
 
    // rdmask will change by setting 1st bit 
    // by OR operation and right shifting by 1
    // as it has to block the previous column
    // of next row because that will fall on right diagonal.
    // rdmask = 00000000000000000000000000000001
 
    // these bit masks will keep updated in each 
    // iteration for next row
    solveBoard(board, row+1, rowmask|p, (ldmask|p) << 1, 
               (rdmask|p) >> 1, ways);
     
    // Reset right-most set bit to 0 so, 
    // next iteration will continue by placing the queen
    // at another safe column index of this row
    safe = safe & (safe-1);
     
    // Backtracking, replace 'Q' by ' '
    board[row][col] = ' ';  
  }
  return;
}
 
// Driver Code
int main()
{
  // Board size
  int n = 4;  
  int ways = 0;
 
  vector< vector<char> > board;
  for (int i = 0; i < n; i++)
  {
    vector<char> tmp;
    for (int j = 0; j < n; j++) 
    {
      tmp.push_back(' ');
    }
    board.push_back(tmp);
  }
 
  int rowmask = 0, ldmask = 0, rdmask = 0;
  int row = 0;
   
  // Function Call
  solveBoard(board, row, rowmask, ldmask, rdmask, ways);
  cout<<endl<<"Number of ways to place "<<n<<" queens : "
    <<ways<<endl;
 
  return 0;
}  
// This code is contributed by Nikhil Vinay

Java

// Java Program for aove approach
public class NQueenSolution 
{
  static private int ways = 0;
  
  // Program to solve N-Queens Problem
  public void solveBoard(char[][] board, int row,
                         int rowmask, int ldmask, 
                         int rdmask)
  {
    int n = board.length;
 
    // All_rows_filled is a bit mask 
    // having all N bits set
    int all_rows_filled = (1 << n) - 1;
     
    // If rowmask will have all bits set, 
    // means queen has been placed successfully
    // in all rows and board is diplayed
    if (rowmask == all_rows_filled)
    {
      ways++;
      System.out.println("=====================");
      System.out.println("Queen placement - " + ways);
      System.out.println("=====================");
      printBoard(board);
    }
     
    // We extract a bit mask(safe) by rowmask,
    // ldmask and rdmask. all set bits of 'safe' 
    // indicates the safe column index for queen
    // placement of this iteration for row index(row).
    int safe = all_rows_filled & (~(rowmask | 
                                    ldmask | rdmask));
    while (safe > 0) 
    {
       
      // Extracts the right-most set bit
      // (safe column index) where queen 
      // can be placed for this row
      int p = safe & (-safe);
      int col = (int)(Math.log(p) / Math.log(2));
      board[row][col] = 'Q';
       
      // This is very important:
      // we need to update rowmask, ldmask and rdmask 
      // for next row as safe index for queen placement
      // will be decided by these three bit masks.
 
      // We have all three rowmask, ldmask and 
      // rdmask as 0 in beginning. Suppose, we are placing
      // queen at 1st column index at 0th row. rowmask, ldmask
      // and rdmask will change for next row as below:
  
      // rowmask's 1st bit will be set by OR operation
      // rowmask = 00000000000000000000000000000010
 
      // ldmask will change by setting 1st
      // bit by OR operation  and left shifting 
      // by 1 as it has to block the next column
      // of next row because that will fall on left diagonal.
      // ldmask = 00000000000000000000000000000100 
 
      // rdmask will change by setting 1st bit 
      // by OR operation and right shifting by 1
      // as it has to block the previous column
      // of next row because that will fall on right diagonal.
      // rdmask = 00000000000000000000000000000001
 
      // these bit masks will keep updated in each 
      // iteration for next row
      solveBoard(board, row+1, rowmask|p, 
                          (ldmask|p) << 1, (rdmask|p) >> 1);
       
      // Reset right-most set bit to 0 so, 
      // next iteration will continue by placing the queen
      // at another safe column index of this row
      safe = safe & (safe-1);
       
      // Backtracking, replace 'Q' by ' '
      board[row][col] = ' ';  
    }
  }
 
  // Program to print board
  public void printBoard(char[][] board)
  {
    for (int i = 0; i < board.length; i++) 
    {
      System.out.print("|");
      for (int j = 0; j < board[i].length; j++) 
      {
        System.out.print(board[i][j] + "|");
      }
      System.out.println();
    }
  }
 
  // Driver Code
  public static void main(String args[]) 
  {
     
    // Board size
    int n = 4;  
 
    char board[][] = new char[n][n];
    for (int i = 0; i < n; i++) 
    {
      for (int j = 0; j < n; j++) 
      {
        board[i][j] = ' ';
      }
    }
 
    int rowmask = 0, ldmask = 0, rdmask = 0;
    int row = 0;
 
    NQueenSolution solution = new
      NQueenSolution();
     
    // Function Call
    solution.solveBoard(board, row, rowmask, 
                        ldmask, rdmask);
    System.out.println();
    System.out.println("Number of ways to place " + 
                       n + " queens : " + ways);
  }
}
 
// This code is contributed by Nikhil Vinay

Python3

# Python program for above approach
import math
ways = 0
 
# Program to solve N-Queens Problem
def solveBoard(board, row, rowmask, 
                              ldmask, rdmask):
   
    n = len(board)
     
    # All_rows_filled is a bit mask 
    # having all N bits set
    all_rows_filled = (1 << n) - 1
     
    # If rowmask will have all bits set, means
    # queen has been placed successfully
    # in all rows and board is diplayed
    if (rowmask == all_rows_filled):
        global ways
        ways = ways + 1
        print("=====================")
        print("Queen placement - " + (str)(ways))
        print("=====================")
        printBoard(board)
         
    # We extract a bit mask(safe) by rowmask,
    # ldmask and rdmask. all set bits of 'safe' 
    # indicates the safe column index for queen
    # placement of this iteration for row index(row).
    safe = all_rows_filled & (~(rowmask | 
                                   ldmask | rdmask))
     
    while (safe > 0):
       
        # Extracts the right-most set bit
        # (safe column index) where queen 
        # can be placed for this row
        p = safe & (-safe)
        col = (int)(math.log(p)/math.log(2))
        board[row][col] = 'Q'
         
        # This is very important:
        # we need to update rowmask, ldmask and rdmask 
        # for next row as safe index for queen placement
        # will be decided by these three bit masks.
 
        # We have all three rowmask, ldmask and 
        # rdmask as 0 in beginning. Suppose, we are placing
        # queen at 1st column index at 0th row. rowmask, ldmask
        # and rdmask will change for next row as below:
  
        # rowmask's 1st bit will be set by OR operation
        # rowmask = 00000000000000000000000000000010
 
        # ldmask will change by setting 1st
        # bit by OR operation  and left shifting 
        # by 1 as it has to block the next column
        # of next row because that will fall on left diagonal.
        # ldmask = 00000000000000000000000000000100 
 
        # rdmask will change by setting 1st bit 
        # by OR operation and right shifting by 1
        # as it has to block the previous column
        # of next row because that will fall on right diagonal.
        # rdmask = 00000000000000000000000000000001
 
        # these bit masks will keep updated in each 
        # iteration for next row
        solveBoard(board, row+1, rowmask|p, 
                           (ldmask|p) << 1, (rdmask|p) >> 1)
         
        # Reset right-most set bit to 0 so, next 
        # iteration will continue by placing the queen
        # at another safe column index of this row
        safe = safe & (safe-1)
         
        # Backtracking, replace 'Q' by ' '
        board[row][col] = ' '
 
# Program to print board
def printBoard(board):
    for row in board:
        print("|" + "|".join(row) + "|")
 
# Driver Code        
def main():
   
    n = 4;  # board size
    board = []
     
    for i in range (n):
        row = []
        for j in range (n):
            row.append(' ')
        board.append(row)
 
    rowmask = 0
    ldmask = 0
    rdmask = 0
    row = 0
     
    # Function Call
    solveBoard(board, row, rowmask, ldmask, rdmask)
     
    # creates a new line
    print() 
    print("Number of ways to place " + (str)(n) +
                          " queens : " + (str)(ways))
     
if __name__== "__main__":
    main()
 
#This code is contributed by Nikhil Vinay

Output

=====================
Queen placement - 1
=====================
| |Q| | |
| | | |Q|
|Q| | | |
| | |Q| |
=====================
Queen placement - 2
=====================
| | |Q| |
|Q| | | |
| | | |Q|
| |Q| | |

Number of ways to place 4 queens : 2

Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.

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