A magic square of order n is an arrangement of n^2 numbers, usually distinct integers, in a square, such that the n numbers in all rows, all columns, and both diagonals sum to the same constant. A magic square contains the integers from 1 to n^2.

The constant sum in every row, column and diagonal is called the magic constant or magic sum, M. The magic constant of a normal magic square depends only on n and has the following value:

M = n(n^2+1)/2

For normal magic squares of order n = 3, 4, 5, ..., the magic constants are: 15, 34, 65, 111, 175, 260, ...

In this post, we will discuss how programmatically we can generate a magic square of size n. Before we go further, consider the below examples:

Magic Square of size 3 ----------------------- 2 7 6 9 5 1 4 3 8 Sum in each row & each column = 3*(3^2+1)/2 = 15 Magic Square of size 5 ---------------------- 9 3 22 16 15 2 21 20 14 8 25 19 13 7 1 18 12 6 5 24 11 10 4 23 17 Sum in each row & each column = 5*(5^2+1)/2 = 65 Magic Square of size 7 ---------------------- 20 12 4 45 37 29 28 11 3 44 36 35 27 19 2 43 42 34 26 18 10 49 41 33 25 17 9 1 40 32 24 16 8 7 48 31 23 15 14 6 47 39 22 21 13 5 46 38 30 Sum in each row & each column = 7*(7^2+1)/2 = 175

Did you find any pattern in which the numbers are stored?

In any magic square, the first number i.e. 1 is stored at position (n/2, n-1). Let this position be (i,j). The next number is stored at position (i-1, j+1) where we can consider each row & column as circular array i.e. they wrap around.

**Three conditions hold:**

1. The position of next number is calculated by decrementing row number of previous number by 1, and incrementing the column number of previous number by 1. At any time, if the calculated row position becomes -1, it will wrap around to n-1. Similarly, if the calculated column position becomes n, it will wrap around to 0.

2. If the magic square already contains a number at the calculated position, calculated column position will be decremented by 2, and calculated row position will be incremented by 1.

3. If the calculated row position is -1 & calculated column position is n, the new position would be: (0, n-2).

Example: Magic Square of size 3 ---------------------- 2 7 6 9 5 1 4 3 8Steps:1. position of number 1 = (3/2, 3-1) = (1, 2) 2. position of number 2 = (1-1, 2+1) = (0, 0) 3. position of number 3 = (0-1, 0+1) = (3-1, 1) = (2, 1) 4. position of number 4 = (2-1, 1+1) = (1, 2) Since, at this position, 1 is there. So, apply condition 2. new position=(1+1,2-2)=(2,0) 5. position of number 5=(2-1,0+1)=(1,1) 6. position of number 6=(1-1,1+1)=(0,2) 7. position of number 7 = (0-1, 2+1) = (-1,3) // this is tricky, see condition 3 new position = (0, 3-2) = (0,1) 8. position of number 8=(0-1,1+1)=(-1,2)=(2,2) //wrap around 9. position of number 9=(2-1,2+1)=(1,3)=(1,0) //wrap around

Based on the above approach, following is the working code:

## C/C++

// C program to generate odd sized magic squares #include<stdio.h> #include<string.h> // A function to generate odd sized magic squares void generateSquare(int n) { int magicSquare[n][n]; // set all slots as 0 memset(magicSquare, 0, sizeof(magicSquare)); // Initialize position for 1 int i = n/2; int j = n-1; // One by one put all values in magic square for (int num=1; num <= n*n; ) { if (i==-1 && j==n) //3rd condition { j = n-2; i = 0; } else { //1st condition helper if next number // goes to out of square's right side if (j == n) j = 0; //1st condition helper if next number // is goes to out of square's upper side if (i < 0) i=n-1; } if (magicSquare[i][j]) //2nd condition { j -= 2; i++; continue; } else magicSquare[i][j] = num++; //set number j++; i--; //1st condition } // Print magic square printf("The Magic Square for n=%d:\nSum of each row or column %d:\n\n", n, n*(n*n+1)/2); for(i=0; i<n; i++) { for(j=0; j<n; j++) printf("%3d ", magicSquare[i][j]); printf("\n"); } } // Driver program to test above function int main() { int n = 7; // Works only when n is odd generateSquare (n); return 0; }

## Java

// Java program to generate odd sized magic squares import java.io.*; class GFG { // Function to generate odd sized magic squares static void generateSquare(int n) { int[][] magicSquare = new int[n][n]; // Initialize position for 1 int i = n/2; int j = n-1; // One by one put all values in magic square for (int num=1; num <= n*n; ) { if (i==-1 && j==n) //3rd condition { j = n-2; i = 0; } else { //1st condition helper if next number // goes to out of square's right side if (j == n) j = 0; //1st condition helper if next number is // goes to out of square's upper side if (i < 0) i=n-1; } //2nd condition if (magicSquare[i][j] != 0) { j -= 2; i++; continue; } else //set number magicSquare[i][j] = num++; //1st condition j++; i--; } // print magic square System.out.println("The Magic Square for "+n+":"); System.out.println("Sum of each row or column "+n*(n*n+1)/2+":"); for(i=0; i<n; i++) { for(j=0; j<n; j++) System.out.print(magicSquare[i][j]+" "); System.out.println(); } } // driver program public static void main (String[] args) { // Works only when n is odd int n = 7; generateSquare(n); } } // Contributed by Pramod Kumar

## Python

# Python program to generate odd sized magic squares # A function to generate odd sized magic squares def generateSquare(n): # 2-D array with all slots set to 0 magicSquare = [[0 for x in range(n)]for y in range(n)] # initialize position of 1 i = n/2 j = n-1 # Fill the magic square by placing values num = 1 while num <= (n*n): if i == -1 and j == n: # third condition j = n-2 i = 0 else: # next number goes out of right side of square if j == n: j = 0 # next number goes out of upper side if i < 0: i = n-1 if magicSquare[i][j]: # 2nd condition j = j-2 i = i+1 continue else: magicSquare[i][j] = num num = num+1 j = j+1 i = i-1 #1st condition # Printing magic square print "Magic Squre for n = ",n print "Sum of each row or column",n*(n*n+1)/2 for i in range(0,n): for j in range(0,n): print '%2d ' %(magicSquare[i][j]), print # Driver program # Works only when n is odd n = 7 generateSquare(n) #Contributed by Harshit Agrawal

Output:

The Magic Square for n=7: Sum of each row or column 175: 20 12 4 45 37 29 28 11 3 44 36 35 27 19 2 43 42 34 26 18 10 49 41 33 25 17 9 1 40 32 24 16 8 7 48 31 23 15 14 6 47 39 22 21 13 5 46 38 30

NOTE: This approach works only for odd values of n.

References:

http://en.wikipedia.org/wiki/Magic_square

This article is compiled by **Aashish Barnwal** and reviewed by GeeksforGeeks team. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above