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