Question

PYTHON: Implement the following algorithm to construct magic n × n squares; it works only if n is odd.

row ← n - 1

column ← n/2

for k = 1 . . . n2 do

Place k at [row][column]

Increment row and column

if the row or column is n then

replace it with 0

end if

if the element at [row][column] has already been filled then

Set row and column to their previous values

Decrement row end if end for

Here is the 5 × 5 square that you get if you follow this algorithm:

11 18 25 2 9

10 12 19 21 3

4 6 13 20 22

23 5 7 14 16

17 24 1 8 15

Write a program whose input is the number n and whose output is the magic square of order n if n is odd.

Please solve in Python

Answer 1

PYTHON Program:

n= int(input("Enter the value of n: "))
if n%2==0:
print("Even value entered. Expecting an odd number.")
else:
arr= [[0 for i in range(n)] for j in range(n)]
row = n-1
column = n//2
for k in range(1,n*n+1):
arr[row][column] = k
newrow = row + 1
newcolumn = column + 1
if newrow == n:
newrow = 0
if newcolumn == n:
newcolumn = 0
if arr[newrow][newcolumn] != 0:
row = row-1
else:
row = newrow
column = newcolumn
  
print("Magic Square of order",n,"x",n,":")
for i in range(0,n):
for j in range(0,n):
print(arr[i][j],end=" ")
print(" ")

Output: