Tricky factorial in Haskell

Question

Is there is a way to create function to calculate factorial in Haskell, using the following function:

fix f = f (fix f)

Yes, this is possible. In fact it's a pretty standard thing. Care to show any research effort on your own part? — leftaroundabout, Jan 11 '16 at 23:06
@leftaroundabout I know how to calculate factorial in a different way, but it is kind of confusing to do it with function ''fix'' — Jack, Jan 11 '16 at 23:08
Have you considered to use the following [solution](https://en.wikipedia.org/wiki/Fixed-point_combinator)? — Rodrigo Ribeiro, Jan 11 '16 at 23:50
The article @RodrigoRibeiro mentions even has a specific example for factorial: https://en.wikipedia.org/wiki/Fixed-point_combinator#The_factorial_function — Louis Wasserman, Jan 12 '16 at 00:04

score 4 · Answer 1 · answered Jan 12 '16 at 04:21

4

To be original

fac = (fix ((1:) . zipWith (*) [1..]) !!)

answered Jan 12 '16 at 04:21

user3237465

12,116
2
15
33

score 2 · Answer 2 · answered Jan 12 '16 at 01:04

Yes. The trivial way:

factorial = fix go 1 where
  go f acc n = fact acc n

Where fact acc n === acc * n!. This is, of course, an altogether ridiculous definition.

Can you see how to modify the ridiculous definition so that it uses the f parameter? This will, hopefully, mimic the structure of the explicitly recursive definition you've already written.

Tricky factorial in Haskell

2 Answers2