Reputation: 189
fact :: Int -> Int
fact n
|n < 0 = 0
|n == 0 = 1
|n > 0 && n `mod` 2 == 1 = fact (n-1) * n
|n > 0 && n `mod` 2 == 0 = n-1
When I enter an odd number for example: fact 5 will give 15, as it should 1 * 3 * 5 = 15. However I realized that if I do fact 7 or any other odd number, it only multiplies the first two odd numbers. How do I get the function to multiply all the odd numbers and not just the first 2. Eg. fact 7 = 35 (ie. 3 * 5). Also note, that if an even number is entered, it will work out the factorial of all the odd numbers up until and not including the even number.
Upvotes: 2
Views: 1142
Reputation: 456
This reminds me of the famous Evolution of a Haskell Programmer. Paraphrasing the tenured professor's answer:
factorialOfOdds :: Integer -> Integer
factorialOfOdds n = product [1,3..n]
Upvotes: 4
Reputation: 53901
You're problem is that your case for an even number is n-1
, which means that when you get to an odd number you're really just doing
n * (n - 1 - 1)
When what you want is
n * n-2 * n-4 ...
So try this instead
fact :: Integer -> Integer -- Overflows
fact n
|n < 0 = 0
|n == 0 || n == 1 = 1
|n `mod` 2 == 1 = fact (n-2) * n
|n `mod` 2 == 0 = fact (n-1)
I also took the liberty of removing some redundant logic. Here we decrement by two if it's odd, so 5 -> 3. And in the even case we decrement by one to end up on an odd number and that recurse on that.
Upvotes: 3