N Queen Problem Using Branch and Bound in C
Last Updated :
14 Jul, 2024
N Queen Problem is a classic combinatorial problem. The objective is to place N queens on an N×N chessboard such that no two queens threaten each other according to the rules of the chess game. No two queens can share the same row, column, or diagonal. This problem has a variety of applications in fields such as artificial intelligence, backtracking algorithms, etc.
In this article, we will learn to solve the N queen problem in C using the Branch and Bound technique which is a better method as compared to simple backtracking. The backtracking Algorithm for N-Queen is already discussed here.
Example:
Input:
N = 4
Output:
0 0 1 0
1 0 0 0
0 0 0 1
0 1 0 0
The Branch and Bound algorithm is a systematic method for solving optimization problems in computer programming. It solves the N-Queen problem efficiently after reducing the number of places we have to remove in order to check. Instead of exploring all possible positions for queens, Branch and Bound eliminates many possibilities by recognizing that placing a queen in a certain position will automatically exclude several other positions from consideration. In this way this algorithm provides us much faster and easier way to solve a problem.
Approach:
- Initialize the Board:
- Create a 2D array to represent the board.
- Initialize arrays to track columns and diagonals.
- Check Safety:
- Create a function isSafe to check if a position is safe by ensuring the current row, and both diagonals are free of other queens.
- Recursive Backtracking:
- Implement the recursive function solve that places queens column by column.
- For each column, try placing a queen in each row and check if it’s safe.
- If placing the queen is safe, mark the position and recursively try to place the next queen in the next column.
- If placing the queen is not safe, backtrack and try the next row.
- Print Solution:
- If a solution is found, print the board.
C Program to Solve N-Queen Problem
The below program implements the above approach to solve N-Queen Problem in C.
C
// C program to solve N queen problem using branch and bound
// algorithm
#include <stdbool.h>
#include <stdio.h>
#include <stdlib.h>
int N;
// Function to print the solution board
void printSolution(int** board)
{
for (int i = 0; i < N; i++) {
for (int j = 0; j < N; j++) {
// Print each cell of the board
printf("%d ", board[i][j]);
}
// Move to the next line after printing each row
printf("\n");
}
}
// Function to allocate memory for the 2D board
int** allocateBoard()
{
// Allocate memory for rows
int** board = (int**)malloc(N * sizeof(int*));
for (int i = 0; i < N; i++) {
// Allocate memory for columns in each row
board[i] = (int*)malloc(N * sizeof(int));
}
// Return the allocated board
return board;
}
// Function to allocate memory for a boolean array
bool* allocateBooleanArray(int size)
{
// Allocate memory for the array
bool* array = (bool*)malloc(size * sizeof(bool));
for (int i = 0; i < size; i++) {
// Initialize all elements of the array to false
array[i] = false;
}
// Return the allocated boolean array
return array;
}
// Optimized function to check if placing a queen at
// position (row, col) is safe
bool isSafe(int row, int col, bool* rows,
bool* leftDiagonals, bool* rightDiagonals)
{
if (rows[row] || leftDiagonals[row + col]
|| rightDiagonals[col - row + N - 1]) {
// If any of the row, left diagonal, or right
// diagonal is occupied, return false
return false;
}
// Otherwise, it's safe to place a queen at position
// (row, col)
return true;
}
// Recursive function to solve the N-Queen problem
bool solve(int** board, int col, bool* rows,
bool* leftDiagonals, bool* rightDiagonals)
{
if (col >= N) {
// If all queens are placed successfully, return
// true
return true;
}
// Try placing queen in each row of the current column
for (int i = 0; i < N; i++) {
if (isSafe(i, col, rows, leftDiagonals,
rightDiagonals)) {
// Mark the row and diagonals as occupied
rows[i] = true;
leftDiagonals[i + col] = true;
rightDiagonals[col - i + N - 1] = true;
// Place the queen at position (i, col)
board[i][col] = 1;
// Recur to place queens in subsequent columns
if (solve(board, col + 1, rows, leftDiagonals,
rightDiagonals)) {
// If placing queen in the next column leads
// to a solution, return true
return true;
}
// Backtrack: Unmark the row and diagonals, and
// remove the queen
rows[i] = false;
leftDiagonals[i + col] = false;
rightDiagonals[col - i + N - 1] = false;
board[i][col] = 0; // Remove the queen from
// position (i, col)
}
}
// If no position is safe in the current column, return
// false
return false;
}
int main()
{
// Taking input from the user for the size of the board
printf(
"Enter the number of rows for the square board: ");
scanf("%d", &N);
// Allocate memory for the board, and boolean arrays for
// rows and diagonals
int** board = allocateBoard();
bool* rows = allocateBooleanArray(N);
bool* leftDiagonals = allocateBooleanArray(2 * N - 1);
bool* rightDiagonals = allocateBooleanArray(2 * N - 1);
// Solve the N-Queen problem and print the solution
if (solve(board, 0, rows, leftDiagonals,
rightDiagonals)) {
printf("Solution found:\n");
// Print the board configuration
printSolution(board);
}
else {
printf("Solution does not exist\n");
}
// Free dynamically allocated memory
for (int i = 0; i < N; i++) {
free(board[i]);
}
free(board);
free(rows);
free(leftDiagonals);
free(rightDiagonals);
return 0;
}
Output
Enter the number of rows for the square board: 4
Solution found:
0 0 1 0
1 0 0 0
0 0 0 1
0 1 0 0
Time Complexity
The time complexity of the solver algorithm is (O(N!)), where (N) is the number of rows and columns in the square board. This is because for each column, the algorithm tries to place a queen in each row and then recursively tries to place the queens in the remaining columns. The number of possible combinations of queen placements in the board is (N!) since there can be only one queen in each row and each column.
Space Complexity
The space complexity of the solver algorithm is (O(N^2)), where (N) is the number of rows and columns in the square board. This is because we are using a 2D array to represent the board, which takes up (N^2) space. Additionally, we are using three boolean arrays to keep track of the occupied rows and diagonals, which take up (2N-1) space each. Therefore, the total space complexity is (O(N^2 + 6N - 3)), which is equivalent to (O(N^2)).
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