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

This C program generates and prints a doubly even magic square of a given order.

Input:

• The program takes an integer n as input from the command line, representing the order of the magic square.

Algorithm:

• The program uses a bit pattern bits = 38505 to determine the placement of numbers in the magic square.
• It calculates the size of the square as size = n * n.
• The mult variable is set to n / 4, which is used to determine the bit position in the bit pattern for each element in the square.
• The program creates a 2D array result of size n x n to store the elements of the magic square.
• It iterates through each row and column, calculating the bit position bitPos based on the current row and column indices.
• Using the bit pattern, it checks if (bits & (1 << bitPos)) is non-zero. If it is, the element is set to i + 1, where i is the current index. Otherwise, it is set to size - i.
• This bit pattern ensures that the elements in each row, column, and the two main diagonals sum up to the same magic constant.

Output:

• The program prints the generated magic square on the console.
• It displays the order of the magic square and the magic constant, which is (rows * rows + 1) * rows / 2.
• The magic square is printed with a consistent width for each element, determined based on the number of digits in the square's size.

Additional Functions:

• numDigits(int n): Calculates the number of digits in a given integer n.
• printMagicSquare(int** square, int rows): Prints the magic square in a user-friendly format.

Usage:

• To use this program, run it from the command line and provide an integer argument specifying the order of the magic square.
• For example, to generate a magic square of order 4, run:

  ./doublyEvenMagicSquare 4
  

#include<stdlib.h>
#include<ctype.h>
#include<stdio.h>

int** doublyEvenMagicSquare(int n) {
	if (n < 4 || n % 4 != 0)
		return NULL;

	int bits = 38505;
	int size = n * n;
	int mult = n / 4,i,r,c,bitPos;

	int** result = (int**)malloc(n*sizeof(int*));
	
	for(i=0;i<n;i++)
		result[i] = (int*)malloc(n*sizeof(int));

	for (r = 0, i = 0; r < n; r++) {
		for (c = 0; c < n; c++, i++) {
			bitPos = c / mult + (r / mult) * 4;
			result[r][c] = (bits & (1 << bitPos)) != 0 ? i + 1 : size - i;
		}
	}
	return result;
}

int numDigits(int n){
	int count = 1;
	
	while(n>=10){
		n /= 10;
		count++;
	}
	
	return count;
}

void printMagicSquare(int** square,int rows){
	int i,j,baseWidth = numDigits(rows*rows) + 3;
	
	printf("Doubly Magic Square of Order : %d and Magic Constant : %d\n\n",rows,(rows * rows + 1) * rows / 2);
	
	for(i=0;i<rows;i++){
		for(j=0;j<rows;j++){
			printf("%*s%d",baseWidth - numDigits(square[i][j]),"",square[i][j]);
		}
		printf("\n");
	}
}

int main(int argC,char* argV[])
{
	int n;
	
	if(argC!=2||isdigit(argV[1][0])==0)
		printf("Usage : %s <integer specifying rows in magic square>",argV[0]);
	else{
		n = atoi(argV[1]);
		printMagicSquare(doublyEvenMagicSquare(n),n);
	}
	return 0;
}