The following code I wrote in haskell
```hs
factorial(1)=1
factorial(n)=n*(n-1)*factorial(n-2)
```
gives me an error when I try to run it. Is there anything I did wrong?
user
The code you provided looks like a recursive implementation of the factorial function in Haskell. However, it appears that you have missed a base case for the recursion. When n=1, the recursive call will be factorial(n-2), which will cause an infinite recursion.

To fix this, you need to provide a base case for n=0 or n=1, which should return 1. Here's a corrected version of your code:
"factorial(0) = 1
factorial(1) = 1
factorial(n) = n * factorial(n-1)
"
assistant
Can you break down what happens in the code you provided step-by-step for a given value of n, such as n=5?
user
For n = 5, here are the steps of the code above :

`factorial(5)` is called, it returns `5 * factorial(4)` so the return value of `factorial(4)` needs to be calculated and inserted.

`factorial(4)` returns `4 * factorial(3)` so `factorial(3)` needs to be calculated and inserted.

`factorial(3)` returns `3 * factorial(2)` so `factorial(2)` needs to be calculated and inserted.

`factorial(2)` returns `2 * factorial(1)` so `factorial(1)` needs to be calculated and inserted.

`factorial(1)` returns `1` as it we defined it to.

Now that we know `factorial(1) = 1` we can calculate the return of `factorial(2) = 2 * 1 = 2`.

Now that we know `factorial(2) = 2` we can calculate the return of `factorial(3) = 3 * 2 =6`.

Now that we know `factorial(3) = 6` we can calculate the return of `factorial(4) = 4 * 6 = 24`.

And finally, now that we know `factorial(4) = 24` we can calculate the return of `factorial(5) = 5 * 24 = 120`.
assistant