The N–queens puzzle is the problem of placing N chess queens on an N × N chessboard so that no two queens threaten each other. Thus, the solution requires that no two queens share the same row, column, or diagonal.
N-Queens is a tricky problem. To solve this problem efficiently, it requires knowing the backtracking algorithm. Basically, the problem is to place N queens on an NxN chessboard. So in this article, we will discuss how to solve the N-Queens problem using a backtracking algorithm.
C# Code
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using System; using System.Collections.Generic; using System.Linq; using System.Text; using System.Threading.Tasks; namespace ConsoleApp1 { class Program { const int N =
4; static void
Main(string[] args) { int
count = 0; int[,]
board = new int[N, N]; //Initialize
the board array to 0 for (int i =
0; i < N; i++) { for (int j =
0; j < N; j++) {
board[i, j] = 0; } } //Initialize
the pointer array int[]
pointer = new int[N]; for (int i =
0; i < N; i++) { pointer[i]
= -1; } //Implementation
of Back Tracking Algorithm for (int j =
0; ;) {
pointer[j]++; //Reset
and move one column back if
(pointer[j] == N) {
board[pointer[j] - 1, j] = 0;
pointer[j] = -1; j--; if (j
== -1) {
Console.WriteLine("all possible solutions to N–Queens problem are done"); break; } } else {
board[pointer[j], j] = 1; if
(pointer[j] != 0) {
board[pointer[j] - 1, j] = 0; } if
(SolutionCheck(board)) {
j++;//move to next column if (j
== N) {
j--;
count++;
Console.WriteLine("Solution" +
count.ToString() + ":");
for (int p = 0; p < N; p++)
{ for (int q =
0; q < N; q++)
{
Console.Write(board[p, q] + " ");
}
Console.WriteLine(); } } } } }
Console.ReadLine(); } public static bool
SolutionCheck(int[,] board) { //Row
check for (int i =
0; i < N; i++) { int sum
= 0; for (int j =
0; j < N; j++) { sum =
sum + board[i, j]; } if
(sum > 1) { return false; } } //Main
diagonal check //above for (int i =
0, j = N - 2; j >= 0; j--) { int sum
= 0; for (int p =
i, q = j; q < N; p++, q++) { sum =
sum + board[p, q]; } if
(sum > 1) { return false; } } //below for (int i =
1, j = 0; i < N - 1; i++) { int sum
= 0; for (int p =
i, q = j; p < N; p++, q++) { sum =
sum + board[p, q]; } if
(sum > 1) { return false; } } //Minor
diagonal check //above for (int i =
0, j = 1; j < N; j++) { int sum
= 0; for (int p =
i, q = j; q >= 0; p++, q--) { sum =
sum + board[p, q]; } if
(sum > 1) { return false; } } //below for (int i =
1, j = N - 1; i < N - 1; i++) { int sum
= 0; for (int p =
i, q = j; p < N; p++, q--) { sum =
sum + board[p, q]; } if
(sum > 1) { return false; } } return true; } } } |
Output
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