N Queen Problem | Backtracking-3

We have discussed Knight’s tour and Rat in a Maze problems in Set 1 and Set 2 respectively. Let us discuss N Queen as another example problem that can be solved using Backtracking.

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 expected output is a binary matrix which has 1s for the blocks where queens are placed. For example, following is the output matrix for above 4 queen solution.

              { 0,  1,  0,  0}
              { 0,  0,  0,  1}
              { 1,  0,  0,  0}
              { 0,  0,  1,  0}

Recommended: Please solve it on “PRACTICE ” first, before moving on to the solution.

Naive Algorithm
Generate all possible configurations of queens on board and print a configuration that satisfies the given constraints.

while there are untried configurations
{
   generate the next configuration
   if queens don't attack in this configuration then
   {
      print this configuration;
   }
}

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 the 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.

Implementation of Backtracking solution

C/C++

/* C/C++ program to solve N Queen Problem using 
   backtracking */
#define N 4 
#include <stdbool.h> 
#include <stdio.h> 
  
/* A utility function to print solution */
void printSolution(int board[N][N]) 
{ 
    for (int i = 0; i < N; i++) { 
        for (int j = 0; j < N; j++) 
            printf(" %d ", board[i][j]); 
        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) 
        return true; 
  
    /* Consider this column and try placing 
       this queen in all rows one by one */
    for (int i = 0; i < N; i++) { 
        /* Check if the 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 the queen cannot be placed 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; 
} 
  
// driver program to test above function 
int main() 
{ 
    solveNQ(); 
    return 0; 
} 

Java

/* Java program to solve N Queen Problem using 
   backtracking */
public class NQueenProblem { 
    final int N = 4; 
  
    /* A utility function to print solution */
    void printSolution(int board[][]) 
    { 
        for (int i = 0; i < N; i++) { 
            for (int j = 0; j < N; j++) 
                System.out.print(" " + board[i][j] 
                                 + " "); 
            System.out.println(); 
        } 
    } 
  
    /* 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 
       placeed in columns from 0 to col -1. So we need 
       to check only left side for attacking queens */
    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 */
    boolean solveNQUtil(int board[][], int col) 
    { 
        /* base case: If all queens are placed 
           then return true */
        if (col >= N) 
            return true; 
  
        /* Consider this column and try placing 
           this queen in all rows one by one */
        for (int i = 0; i < N; i++) { 
            /* Check if the 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) == true) 
                    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 the queen can not be placed 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.*/
    boolean solveNQ() 
    { 
        int board[][] = { { 0, 0, 0, 0 }, 
                          { 0, 0, 0, 0 }, 
                          { 0, 0, 0, 0 }, 
                          { 0, 0, 0, 0 } }; 
  
        if (solveNQUtil(board, 0) == false) { 
            System.out.print("Solution does not exist"); 
            return false; 
        } 
  
        printSolution(board); 
        return true; 
    } 
  
    // driver program to test above function 
    public static void main(String args[]) 
    { 
        NQueenProblem Queen = new NQueenProblem(); 
        Queen.solveNQ(); 
    } 
} 
// This code is contributed by Abhishek Shankhadhar 

Python3

# Python3 program to solve N Queen  
# Problem using backtracking 
global N 
N = 4
  
def printSolution(board): 
    for i in range(N): 
        for j in range(N): 
            print (board[i][j], end = " ") 
        print() 
  
# 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] == 1: 
            return False
  
    # Check upper diagonal on left side 
    for i, j in zip(range(row, -1, -1),  
                    range(col, -1, -1)): 
        if board[i][j] == 1: 
            return False
  
    # Check lower diagonal on left side 
    for i, j in zip(range(row, N, 1),  
                    range(col, -1, -1)): 
        if board[i][j] == 1: 
            return False
  
    return True
  
def solveNQUtil(board, col): 
      
    # base case: If all queens are placed 
    # then return true 
    if col >= N: 
        return True
  
    # Consider this column and try placing 
    # this queen in all rows one by one 
    for i in range(N): 
  
        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) == True: 
                return True
  
            # If placing queen in board[i][col 
            # doesn't lead to a solution, then 
            # queen from board[i][col] 
            board[i][col] = 0
  
    # if the queen can not be placed 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 
# placement of queens in the form of 1s. 
# note that there may be more than one 
# solutions, this function prints one of the 
# feasible solutions. 
def solveNQ(): 
    board = [ [0, 0, 0, 0], 
              [0, 0, 0, 0], 
              [0, 0, 0, 0], 
              [0, 0, 0, 0] ] 
  
    if solveNQUtil(board, 0) == False: 
        print ("Solution does not exist") 
        return False
  
    printSolution(board) 
    return True
  
# Driver Code 
solveNQ() 
  
# This code is contributed by Divyanshu Mehta 

C#

// C# program to solve N Queen Problem  
// using backtracking  
using System; 
      
class GFG  
{ 
    readonly int N = 4; 
  
    /* A utility function to print solution */
    void printSolution(int [,]board) 
    { 
        for (int i = 0; i < N; i++)  
        { 
            for (int j = 0; j < N; j++) 
                Console.Write(" " + board[i, j] 
                                  + " "); 
            Console.WriteLine(); 
        } 
    } 
  
    /* 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 
    placeed in columns from 0 to col -1. So we need 
    to check only left side for attacking queens */
    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 */
    bool solveNQUtil(int [,]board, int col) 
    { 
        /* base case: If all queens are placed 
        then return true */
        if (col >= N) 
            return true; 
  
        /* Consider this column and try placing 
        this queen in all rows one by one */
        for (int i = 0; i < N; i++)  
        { 
            /* Check if the 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) == true) 
                    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 the queen can not be placed 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 = {{ 0, 0, 0, 0 }, 
                        { 0, 0, 0, 0 }, 
                        { 0, 0, 0, 0 }, 
                        { 0, 0, 0, 0 }}; 
  
        if (solveNQUtil(board, 0) == false) 
        { 
            Console.Write("Solution does not exist"); 
            return false; 
        } 
  
        printSolution(board); 
        return true; 
    } 
  
    // Driver Code 
    public static void Main(String []args) 
    { 
        GFG Queen = new GFG(); 
        Queen.solveNQ(); 
    } 
} 
  
// This code is contributed by Princi Singh 

Output: The 1 values indicate placements of queens

Optimization in is_safe() function
The idea is not to check every element in right and left diagonal instead use property of diagonals:
1.The sum of i and j is constant and unique for each right diagonal where i is the row of element and j is the
column of element.
2.The difference of i and j is constant and unique for each left diagonal where i and j are row and column of element respectively.

Implementation of Backtracking solution(with optimization)

C/C++

/* C/C++ program to solve N Queen Problem using 
   backtracking */
#define N 4 
#include <stdbool.h> 
#include <stdio.h> 
/* ld is an array where its indices indicate row-col+N-1 
 (N-1) is for shifting the difference to store negative  
 indices */
int ld[30] = { 0 }; 
/* rd is an array where its indices indicate row+col 
   and used to check whether a queen can be placed on  
   right diagonal or not*/
int rd[30] = { 0 }; 
/*column array where its indices indicates column and  
  used to check whether a queen can be placed in that 
    row or not*/
int cl[30] = { 0 }; 
/* A utility function to print solution */
void printSolution(int board[N][N]) 
{ 
    for (int i = 0; i < N; i++) { 
        for (int j = 0; j < N; j++) 
            printf(" %d ", board[i][j]); 
        printf("\n"); 
    } 
} 
  
/* 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) 
        return true; 
  
    /* Consider this column and try placing 
       this queen in all rows one by one */
    for (int i = 0; i < N; i++) { 
        /* Check if the queen can be placed on 
          board[i][col] */
        /* A check if a queen can be placed on  
           board[row][col].We just need to check 
           ld[row-col+n-1] and rd[row+coln] where 
           ld and rd are for left and right  
           diagonal respectively*/
        if ((ld[i - col + N - 1] != 1 && 
                  rd[i + col] != 1) && cl[i] != 1) { 
            /* Place this queen in board[i][col] */
            board[i][col] = 1; 
            ld[i - col + N - 1] = 
                          rd[i + col] = cl[i] = 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 
            ld[i - col + N - 1] = 
                         rd[i + col] = cl[i] = 0; 
        } 
    } 
  
    /* If the queen cannot be placed 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; 
} 
  
// driver program to test above function 
int main() 
{ 
    solveNQ(); 
    return 0; 
} 

Java

/* Java program to solve N Queen Problem  
using backtracking */
import java.util.*; 
  
class GFG  
{ 
static int N = 4; 
  
/* ld is an array where its indices indicate row-col+N-1 
(N-1) is for shifting the difference to store negative  
indices */
static int []ld = new int[30]; 
  
/* rd is an array where its indices indicate row+col 
and used to check whether a queen can be placed on  
right diagonal or not*/
static int []rd = new int[30]; 
  
/*column array where its indices indicates column and  
used to check whether a queen can be placed in that 
    row or not*/
static int []cl = new int[30]; 
  
/* A utility function to print solution */
static void printSolution(int board[][]) 
{ 
    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"); 
    } 
} 
  
/* 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) 
        return true; 
  
    /* Consider this column and try placing 
    this queen in all rows one by one */
    for (int i = 0; i < N; i++) 
    { 
          
        /* Check if the queen can be placed on 
        board[i][col] */
        /* A check if a queen can be placed on  
        board[row][col].We just need to check 
        ld[row-col+n-1] and rd[row+coln] where 
        ld and rd are for left and right  
        diagonal respectively*/
        if ((ld[i - col + N - 1] != 1 && 
             rd[i + col] != 1) && cl[i] != 1) 
        { 
            /* Place this queen in board[i][col] */
            board[i][col] = 1; 
            ld[i - col + N - 1] = 
            rd[i + col] = cl[i] = 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 
            ld[i - col + N - 1] = 
            rd[i + col] = cl[i] = 0; 
        } 
    } 
  
    /* If the queen cannot be placed 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.*/
static boolean solveNQ() 
{ 
    int board[][] = {{ 0, 0, 0, 0 }, 
                     { 0, 0, 0, 0 }, 
                     { 0, 0, 0, 0 }, 
                     { 0, 0, 0, 0 }}; 
  
    if (solveNQUtil(board, 0) == false)  
    { 
        System.out.printf("Solution does not exist"); 
        return false; 
    } 
  
    printSolution(board); 
    return true; 
} 
  
// Driver Code 
public static void main(String[] args) 
{ 
    solveNQ(); 
} 
} 
  
// This code is contributed by Princi Singh 

Python3

""" Python3 program to solve N Queen Problem using  
backtracking """
N = 4
  
""" ld is an array where its indices indicate row-col+N-1  
(N-1) is for shifting the difference to store negative  
indices """
ld = [0] * 30
  
""" rd is an array where its indices indicate row+col  
and used to check whether a queen can be placed on  
right diagonal or not"""
rd = [0] * 30
  
"""column array where its indices indicates column and  
used to check whether a queen can be placed in that  
    row or not"""
cl = [0] * 30
  
""" A utility function to print solution """
def printSolution(board):  
    for i in range(N): 
        for j in range(N): 
            print(board[i][j], end = " ") 
        print()  
  
""" 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 >= N): 
        return True
          
    """ Consider this column and try placing 
        this queen in all rows one by one """
    for i in range(N): 
          
        """ Check if the queen can be placed on board[i][col] """
        """ A check if a queen can be placed on board[row][col]. 
        We just need to check ld[row-col+n-1] and rd[row+coln]  
        where ld and rd are for left and right diagonal respectively"""
        if ((ld[i - col + N - 1] != 1 and 
             rd[i + col] != 1) and cl[i] != 1): 
                   
            """ Place this queen in board[i][col] """
            board[i][col] = 1
            ld[i - col + N - 1] = rd[i + col] = cl[i] = 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  
            ld[i - col + N - 1] = rd[i + col] = cl[i] = 0
              
            """ If the queen cannot be placed 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."""
def solveNQ(): 
    board = [[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
      
# Driver Code 
solveNQ()  
  
# This code is contributed by SHUBHAMSINGH10 

C#

/* C# program to solve N Queen Problem  
using backtracking */
using System; 
      
class GFG  
{ 
static int N = 4; 
  
/* ld is an array where its indices indicate row-col+N-1 
(N-1) is for shifting the difference to store negative  
indices */
static int []ld = new int[30]; 
  
/* rd is an array where its indices indicate row+col 
and used to check whether a queen can be placed on  
right diagonal or not*/
static int []rd = new int[30]; 
  
/*column array where its indices indicates column and  
used to check whether a queen can be placed in that 
    row or not*/
static int []cl = new int[30]; 
  
/* A utility function to print solution */
static void printSolution(int [,]board) 
{ 
    for (int i = 0; i < N; i++) 
    { 
        for (int j = 0; j < N; j++) 
            Console.Write(" {0} ", board[i, j]); 
        Console.Write("\n"); 
    } 
} 
  
/* 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) 
        return true; 
  
    /* Consider this column and try placing 
    this queen in all rows one by one */
    for (int i = 0; i < N; i++) 
    { 
          
        /* Check if the queen can be placed on 
        board[i,col] */
        /* A check if a queen can be placed on  
        board[row,col].We just need to check 
        ld[row-col+n-1] and rd[row+coln] where 
        ld and rd are for left and right  
        diagonal respectively*/
        if ((ld[i - col + N - 1] != 1 && 
             rd[i + col] != 1) && cl[i] != 1) 
        { 
            /* Place this queen in board[i,col] */
            board[i, col] = 1; 
            ld[i - col + N - 1] = 
            rd[i + col] = cl[i] = 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 
            ld[i - col + N - 1] = 
            rd[i + col] = cl[i] = 0; 
        } 
    } 
  
    /* If the queen cannot be placed 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.*/
static bool solveNQ() 
{ 
    int [,]board = {{ 0, 0, 0, 0 }, 
                    { 0, 0, 0, 0 }, 
                    { 0, 0, 0, 0 }, 
                    { 0, 0, 0, 0 }}; 
  
    if (solveNQUtil(board, 0) == false)  
    { 
        Console.Write("Solution does not exist"); 
        return false; 
    } 
  
    printSolution(board); 
    return true; 
} 
  
// Driver Code 
public static void Main(String[] args) 
{ 
    solveNQ(); 
} 
} 
  
// This code is contributed by Rajput-Ji 

Output: The 1 values indicate placements of queens

Printing all solutions in N-Queen Problem

Sources:
http://see.stanford.edu/materials/icspacs106b/H19-RecBacktrackExamples.pdf
http://en.literateprograms.org/Eight_queens_puzzle_%28C%29
http://en.wikipedia.org/wiki/Eight_queens_puzzle

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

My Personal Notes

Recommended: Please solve it on “PRACTICE ” first, before moving on to the solution.

C/C++

Java

Python3

C#

C/C++

Java

Python3

C#

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