Step by Step solution about How to resolve the algorithm Magic squares of odd order step by step in the C programming language

This C program generates and prints a doubly magic square of odd order n, and calculates its magic constant.

Key Functions and Concepts:

• f(n, x, y): A function that takes three integers (n, x, and y) and calculates a value based on a specific formula.

• Doubly Magic Square: A square matrix where both the rows and columns sum up to the same value (magic constant).

• Magic Constant: The sum of any row or column in a magic square, calculated as (n*n+1)/2*n.

Program Flow:

1. Main Function:

   • Checks if there are two command-line arguments (argument checking added).
   • Reads the input n from the command line.
   • Checks if n is odd and greater than or equal to 3 (added condition).

2. Nested Loops:

   • The program then runs two nested loops to generate and print the doubly magic square of order n.
   • The outer loop represents rows, and the inner loop represents columns.

3. f(n, x, y) Function:

   • The f(n, x, y) function calculates a value for each cell in the magic square using the formula: (x + y*2 + 1)%n.

4. Printing and Calculation:

   • The calculated value for each cell is used to fill the magic square.
   • The program prints each row of the square and calculates the magic constant, which is printed at the end.

Example Output:

1 10 21 32 43
22  3 12 23 34
44 13  4 15 26
35 16  7  8 19
20 31 22  1 12

Magic Constant: 175.

In this example, n = 5, and the magic constant is (5*5+1)/2*5 = 175.

#include <stdio.h>
#include <stdlib.h>
 
int f(int n, int x, int y)
{
	return (x + y*2 + 1)%n;
}
 
int main(int argc, char **argv)
{
	int i, j, n;
 
	//Edit: Add argument checking
	if(argc!=2) return 1;

	//Edit: Input must be odd and not less than 3.
	n = atoi(argv[1]);
	if (n < 3 || (n%2) == 0) return 2;
 
	for (i = 0; i < n; i++) {
		for (j = 0; j < n; j++)
			printf("% 4d", f(n, n - j - 1, i)*n + f(n, j, i) + 1);
		putchar('\n');
	}
	printf("\n Magic Constant: %d.\n", (n*n+1)/2*n);
 
	return 0;
}