# Exercise 9.9

`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”

1. anonyoumous says:

remainder’ = fix (\f a b -> if a < b then a else f (a-b) b)

2. Dave says:

thanks!