Printing Simple Diamond Pattern in Python

Question

I would like to print the following pattern in Python 3.5 (I\'m new to coding):

    *
   ***
  *****
 *******
*********
 *******
  *****
   ***
    *
         


        

Answer 1

Since the middle and largest row of stars has 9 stars, you should make n equal to 9. You were able to print out half of the diamond, but now you have to try to make a function that prints a specific number of spaces, then a specific number of stars. So try to develop a pattern with the number of spaces and stars in each row,

Row1: 4 spaces, 1 star, 4 spaces
Row2: 3 spaces, 3 stars, 3 spaces
Row3: 2 spaces, 5 stars, 2 spaces
Row4: 1 space, 7 stars, 1 space
Row5: 0 spaces, 9 stars, 0 spaces
Row6: 1 space, 7 stars, 1 space
Row7: 2 spaces, 5 stars, 2 spaces
Row8: 3 spaces, 3 stars, 3 spaces
Row9: 4 spaces, 1 star, 4 spaces

So what can you deduce? From row 1 to (n+1)/2, the number of spaces decreases as the number of stars increase. So from 1 to 5, the # of stars = (row number * 2) - 1, while # of spaces before stars = 5 - row number.

Now from row (n+1)/2 + 1 to row 9, the number of spaces increase while the number of stars decrease. So from 6 to n, the # of stars = ((n+1 - row number) * 2) - 1, while # of spaces before stars = row number - 5.

From this information, you should be able to make a program that looks like this,

n = 9
print("Pattern 1")
for a1 in range(1, (n+1)//2 + 1): #from row 1 to 5
    for a2 in range((n+1)//2 - a1):
        print(" ", end = "")
    for a3 in range((a1*2)-1):
        print("*", end = "")
    print()

for a1 in range((n+1)//2 + 1, n + 1): #from row 6 to 9
    for a2 in range(a1 - (n+1)//2):
        print(" ", end = "")
    for a3 in range((n+1 - a1)*2 - 1):
        print("*", end = "")
    print()

Note that you can replace n with any odd number to create a perfect diamond of that many lines.