Printing Simple Diamond Pattern in Python

Question

In python 3.5, I just want to print the following pattern

    *

   ***

  *****

 *******

*********

 *******

  *****

   ***

    *

But the problem is, I just know how to print the following pattern, I don't know how to make it as diamond

n = 5

print("Pattern 1")

for a1 in range (0,n):

    for a2 in range (a1):

        print("*", end="")

    print()

for a1 in range (n,0,-1):

    for a2 in range (a1):

        print("*", end="")

    print()

*

**

***

****

*****

****

***

**

*

Anyone, please help me?

Answer 1

When you look the diamond pattern, The middle and largest row has 9 starts, so you have to make n equal to 9. So, In each row we have to create number of spaces and stars.

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

From the above description,  You can understand that, for row 1 to (n+1)/2, the number of stars increases as the number of space decreases. From 1 to 5, The number of stars=(row number*2)-1 and number of spaces before stars=5-row number.

Now, For row(n+1)/2 +1 to 9th row, The number of stars decreases as the number of spaces increases. So, From row number 6 to n, the number of  stars=((n+1- row number)*2)-1 and number number of spaces before stars= row number -5

Look at the following code, The program look 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()

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