A function that prints an xo square

I have a school proejct. I'm having a bit of trouble. I was thinking of implementing a 2D lists in python and then using a while loop to update the variables and index the lists to reassign the index to an x. The details are

> Go to main.py write a function called xo that accepts an integer size as a parameter and prints a square of size by size characters,
where all characters are "o" except that an “x” pattern of "x" characters has been drawn from the corners of the square.
On the first line, the first and last characters are "x";
on the second line, the second and second-from-last characters are "x"; and so on.

> For example, the call of xo(5) should produce the following outputs:

xooox
oxoxo
ooxoo
oxoxo
xooox

my code so far was

def xo(num):
  square=[]
  section=["o"*num]
  for i in range(1,num+1):
    square.append(section)

  for row in square:
    print("".join(row))

xo(5)

and it prints

ooooo
ooooo
ooooo
ooooo
ooooo

vertically

It helps to look at this type of exercise as a math problem before trying to turn it into code. We can start with code that prints every square as an "o" (as you've already done), but it'll be easier to take it to the next step if each time we print an "o" we have a particular coordinate on the grid (x and y) to work with:

def xo(n):
	for y in range(n):
		for x in range(n):
			print("o", end="")
		print()

xo(5)

This gets us the output:

ooooo
ooooo
ooooo
ooooo
ooooo

We know we want some of those os to be xs, so let's look at what the x and y values in our code will look like when we want that to happen:

y  x
0: 0, 4
1: 1, 3
2: 2, 2
3: 1, 3
4: 0, 4

This is the math problem -- how do we come up with a mathematical way to express that relationship between x and y?

We can see that in every case we want to change what we print when x == y, or x == 4 - y. Remembering that in this case n == 5, we might change our code to do:

def xo(n):
	for y in range(n):
		for x in range(n):
			print("x" if x in (y, n - 1 - y) else "o", end="")
		print()

which for xo(5) gets us:

xooox
oxoxo
ooxoo
oxoxo
xooox

and since we wrote our code in terms of n instead of hardcoding 4 in there as a magic number, we can also run, say, xo(20) and get:

xoooooooooooooooooox
oxooooooooooooooooxo
ooxooooooooooooooxoo
oooxooooooooooooxooo
ooooxooooooooooxoooo
oooooxooooooooxooooo
ooooooxooooooxoooooo
oooooooxooooxooooooo
ooooooooxooxoooooooo
oooooooooxxooooooooo
oooooooooxxooooooooo
ooooooooxooxoooooooo
oooooooxooooxooooooo
ooooooxooooooxoooooo
oooooxooooooooxooooo
ooooxooooooooooxoooo
oooxooooooooooooxooo
ooxooooooooooooooxoo
oxooooooooooooooooxo
xoooooooooooooooooox