fix f = f (fix f)
First, f takes the result of fix f, and second, the return type of f must be the same as the return type of fix.
fix :: (a -> a) -> a
Research shows that fix is the Fixed point combinator (or Y-combinator). Apparently this allows the definition of anonymous recursive functions. So let’s try to make remainder into an anonymous function. A good way to start is to simply remove one of the arguments and see if we can return an anonymous function. Something tells me that a is the argument to remove, since b is used in the base case.
remanon b = (\f x -> if x < b then x
else f (x-b))
That's a good start. Note here that if we try to do something like:
(remanon 3) remanon
Haskell says "unification would give infinite type" - which is a good sign. This is what we would like to do with the help of fix. So now if we apply fix to this, we get:
fix (remanon b)
= (remanon b) (fix (remanon b))
= (\x -> if x < b then x
else (fix (remanon b)) (x-b))
Which looks good. So we need to fix remanon and apply it to a.
remainder' a b = fix (remanon b) a
And that works.
2 responses to “Exercise 9.9”
remainder’ = fix (\f a b -> if a < b then a else f (a-b) b)
thanks!