تحديات برمجيةالتحدي السادس - حل التمرين الثالث بلغة بايثون

n = 0

while n <= 0 or n % 2 == 0:
    n = int(input("Enter and odd number as the number of lines: "))

for i in range(1, n + 1):

    for j in range(1, n + 1):

        if i <= n / 2 and i < j <= n - i or i > n / 2 and i > j > n - i + 1 or i == int(n / 2) + 1:
            print(' ', end='')

        else:
            print('*', end='')

    print()

Enter an odd number as the number of lines: 9
*       *
**     **
***   ***
**** ****
         
**** ****
***   ***
**     **
*       *