What is N Queen Problem?
We are given N chess’s queens and is asked to place them in the chessboard of size NxN so that no two queens attack/eliminate each other.
A queen can only attack if another queen is either on:
- Same Row
- Same Column
- Same Diagonal (left or right)
Example: For 4 Queens problem, we will have a 4×4 board. If a queen is placed at (2,2), then the places at which no queen can be placed is crossed by red line as shown in the figure:
N Queen Algorithm
Now the question is how to solve N Queen Problem?
There are two approaches to solve this problem:
Naive Algorithm
In this approach, we try the brute force method to find all possible solutions to the problem by randomly choosing the position.
1 2 3 4 5 6 7 8 | while some configurations are unchecked { Generate configuration of queen position if in current configuration queens don't attack { print the configuration } } |
Backtracking Algorithm
In this approach, we place a queen in one column and then we proceed to the next column (since a single column cannot contain more than one queen). Then, we check the possible safe row to place next queen, so that we can proceed to the next column.
If there is no safe row in the current column, we go back to the previous column and change the queen’s position to the next possible safe row. If found any then we proceed to the next column again, else we go back further and repeat the same task.
If we reached the last column and found the possible safe position (i.e row), then we print the configuration of queens (i.e solution), else if we came back to the first column and none of the position (i.e row) is safe then we assume that there is no possible solution.
So, In backtracking, if no further solution is possible, we go one step back and proceed with next available option.
Check N-Queens Visualizer to get visual working of backtrack algorithm.
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 | find_solution(row, column) { if (column == lastColumn + 1) { passed all columns, therefore successfully found a solution; print board; return true; } else { for(all rows of column) { if([row, column] is safe for placing queen) { //place queen, i.e board[row][column] = 1; //check for next column if(find_solution(0,column + 1)) { return true; } //No safe place in column+1, remove queen, i.e board[row][column] = 0; } move to next row } checked all rows of the column, no safe place; return false; } } |
N Queen in C|Java
This C program gives all possible safe configuration for N Queen Problem as output by giving the value of N during program execution.
C
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 | #include <stdio.h> #include <stdlib.h> #include <stdbool.h> int queens, count =0; int chessBoard[20][20]; bool hasSolution(int,int); bool isValidPlace(int , int); void displayBoard(); void solveNqueen(); bool flag = false; int main() { //Enter value of N printf("Enter number of Queens \n"); scanf("%d", &queens); solveNqueen(); return 0; } void solveNqueen(){ int i; //starting from (0,0) of the board hasSolution(0,0); //After running hasSolution() still flag == false //It means no Solution if(!flag) printf("No Solution \n"); } //to reset value to 0 void reset(){ int i, j; for(i=0; i<queens; i++){ for(j=0; j<queens; j++){ chessBoard[i][j] = 0; } } } //finding all solution by backtracking //row == ctr //column == colQueen bool hasSolution(int ctr,int colQueen){ //Reached beyond last column //Means solution (configuration) found if(colQueen == queens){ flag = true; displayBoard(); //intentionally returning false to find more possible solution return false; } int i,j; for(i=ctr; i<queens; i++){ //checking if position is safe if(isValidPlace(i,colQueen)){ //setting current value to 1 means placing a queen chessBoard[i][colQueen] = 1; //moving to next column's 1st row if(hasSolution(0,colQueen+1)) return true; //backtrack //reset to 0 means removing queen chessBoard[i][colQueen] = 0; } } //When no row is safe in the current column return false; } //To check whether the particular position is safe or not bool isValidPlace(int row, int col){ int i,j; //Checking horizontally for(i=col; i>=0; i--){ if(chessBoard[row][i] == 1) return false; } //checking left diagonal for(i=row, j=col; i>=0 && j>=0; i--,j--){ if(chessBoard[i][j] == 1) return false; } //checking right diagonal for(i=row, j=col; i<queens && j>=0; i++,j--){ if(chessBoard[i][j] == 1) return false; } return true; } //To display chess board with queen configuration // In place of 1 we print * // In place of 0 we print - void displayBoard(){ int i, j; for(i=0; i<queens; i++){ for(j=0; j<queens; j++){ if(chessBoard[i][j] == 1) printf(" * "); else printf(" - "); } printf("\n"); } printf("\n"); printf("\n"); } |
Java
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 | import java.util.Scanner; public class Board { int chessBoard[][]; int queens; public Board(int queens) { chessBoard = new int[20][20]; this.queens = queens; } void displayBoard(){ int i, j; for(i=0; i<queens; i++){ for(j=0; j<queens; j++){ if(chessBoard[i][j] == 1) System.out.print(" * "); else System.out.print(" - "); } System.out.print("\n"); } System.out.print("\n"); System.out.print("\n"); } void reset(){ for(int i=0; i<queens; i++){ for(int j=0; j<queens; j++){ chessBoard[i][j] = 0; } } } boolean isValidPlace(int row, int col){ int i,j; //Checking horizontally for(i=col; i>=0; i--){ if(chessBoard[row][i] == 1) return false; } //checking left diagonal for(i=row, j=col; i>=0 && j>=0; i--,j--){ if(chessBoard[i][j] == 1) return false; } //checking right diagonal for(i=row, j=col; i<queens && j>=0; i++,j--){ if(chessBoard[i][j] == 1) return false; } return true; } } public class NQueen { int queens; boolean flag; Board board; public NQueen(int queens) { this.flag = false; this.board = new Board(queens);; this.queens = queens; } void solveNqueen(){ int i; //starting from (0,0) of the board hasSolution(0,0); //After running hasSolution() still flag == false //It means no Solution if(!flag) System.out.println("No Solution"); } boolean hasSolution(int ctr, int colQueen){ //Reached beyond last column //Means solution (configuration) found if(colQueen == queens){ flag = true; board.displayBoard(); //intentionally returning false to find more possible solution return false; } int i,j; for(i=ctr; i<queens; i++){ //checking if position is safe if(board.isValidPlace(i,colQueen)){ //setting current value to 1 means placing a queen board.chessBoard[i][colQueen] = 1; //moving to next column's 1st row if(hasSolution(0,colQueen+1)) return true; //backtrack //reset to 0 means removing queen board.chessBoard[i][colQueen] = 0; } } //When no row is safe in current column return false; } } public class App { static Scanner in = new Scanner(System.in); public static void main(String args[]){ System.out.println("Enter number of Queens"); int queens = in.nextInt(); NQueen nQueen = new NQueen(queens); nQueen.solveNqueen(); } } |
Output
When N is 4:
There are total 92 safe configurations possible to place 8 queens in the 8×8 board, so it is not possible to show output here. Check the output yourself by giving the value of N during program execution. If you have any doubts then do comment below.
wow,so helpful
thank you so much for saving me.
one request sir
how to solve sum of subsets problem using backtracking.
please help 👨🦱
Check out I have posted the solution of Subset problem.