Ali R. - 2 months ago 21

Python Question

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

`*`

***

*****

*******

*********

*******

*****

***

*

But I only know how to print the following using the code below, but not sure how to invert it to make it a complete 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()

*

**

***

****

*****

****

***

**

*

Any help would be appreciated!

Answer

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.