Help with writing a program that creates an odd n*n magic square where n is...
I am in a beginning C programming class and I have this assignment: Write a program that creates an N*N magic square, i.e., a square arrangement of the numbers 1,2,…,N^2 in which the sum of rows, columns, and diagonals are the same. The user will specify the size of the square matrix: N. The value N must be an odd number between 1 and 15.
I am using visual studio 2010. Also I am only allowed to use C89. My program will not compile and I have been trying to fix my problems, but I still have a problem with what the program prints out. My program prints out a vertical column of the magic square. For example if my magic square is size 3, it'll print:
8
1
6
3
5
7
4
9
2
Help please. Thank you!
Code:
#include <stdio.h>
#define array_size 15
int main (void)
{
int row, column, size, n = 2, magic_square [array_size][array_size];
for ( ; ; ) {
printf ("Enter size of magic square: ");
scanf ("%d", &size);
if (size % 2 == 0)
printf ("Enter an odd number!\n");
else if (size <= 0 || size > 15)
printf ("Enter a number between 1 and 15!\n");
else {
magic_square [array_size][array_size];
for (row = 0; row < size; row++) {
for (column = 0; column < size; column++) {
magic_square [row][column] = 0;
}
}
row = 0;
column = (size/2);
magic_square [row][column] = 1;
for (n; n<= size*size; n++) {
if (--row < 0)
row = (size -1);
if (++column > size -1)
column = 0;
if (magic_square [row][column] != 0) {
if (++row > (size -1))
row = 0;
if (--column < 0)
column = size - 1;
while (magic_square [row][column] != 0)
if (++row > (size -1))
row = 0;
}
magic_square [row][column] = n;
}
for (row = 0; row < size; row++) {
for (column = 0; column < size; column++) {
printf ("%4d\n", magic_square [row][column]);
}
printf ("\n");
}
break;
}
}
return 0;
}
