تحديات برمجيةالتحدي السابع - 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 (j == 1 or j == n or i == int(n / 2) or i == int(n / 2) + 2 or i == j or j == n - i + 1)\
                and i != int(n / 2) + 1 and j != int(n / 2) + 1:
            print('*', end='')

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

    print()

Enter an odd number as the number of lines: 11
*         *
**       **
* *     * *
*  *   *  *
***** *****
           
***** *****
*  *   *  *
* *     * *
**       **
*         *