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}

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 umark 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 #include    /* 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(" ");     } }    /* 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)         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)             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

Python

 # Python 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],         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 program to test above function solveNQ()    # This code is contributed by Divyanshu Mehta

Output: The 1 values indicate placements of queens

0  0  1  0
1  0  0  0
0  0  0  1
0  1  0  0

Printing all solutions in N-Queen Problem