# Week 4

### Lecture slides¶

### Recommended reading¶

- Textbook: new Chapter 6 on types.
- Type systems for programming languages by Didier Rémy

### Type tetris¶

In our parametricity explorations we've explored some problems that not only
asked you to implement parametric functions, but do so in terms of other
parametric functions. This can be quite tricky, but also fun. Below is a worked
out example of the `hog`

function from the homework using only the `(.)`

function:

-------------------------------------------------------------- -- Simplified example: -------------------------------------------------------------- foo0 :: (b -> c) -> (a -> b) -> a -> c foo0 f g x = f (g x) -- Can be rewritten with the function composition operator (.) as: foo1 :: (b -> c) -> (a -> b) -> a -> c foo1 f g x = (f . g) x -- We can get rid of the x (eta conversion): foo2 :: (b -> c) -> (a -> b) -> a -> c foo2 f g = f . g -- Let's rewrite it in prefix form now: foo3 :: (b -> c) -> (a -> b) -> a -> c foo3 f g = (.) f g -- Get rid of g (eta again): foo4 :: (b -> c) -> (a -> b) -> a -> c foo4 f = (.) f -- Get rid of f (eta again): foo5 :: (b -> c) -> (a -> b) -> a -> c foo5 = (.) -------------------------------------------------------------- -------------------------------------------------------------- hog :: (c -> d) -> (a -> b -> c) -> a -> b -> d hog = undefined -- Let's write this with explicit arguments hog0 :: (c -> d) -> (a -> b -> c) -> a -> b -> d hog0 fCD fABC vA vB = undefined -- Note that we want to produce a d at the end of the day -- What gives us a d? fCD when called with a c -- How do we get a c? Well fABC gives us c when called with an a and b hog1 :: (c -> d) -> (a -> b -> c) -> a -> b -> d hog1 fCD fABC vA vB = fCD (fABC vA vB) -- Another way to look at the type is to write all the parantheses: How can you -- think of this? fABC is almost the thing you want to return. You only need to -- change that c into a d. And the fCD function can be used to do exacly that! hog2 :: (c -> d) -> (a -> (b -> c)) -> (a -> (b -> d)) hog2 fCD fABC vA = (.) fCD {- :: (c -> d) -} (fABC vA {- :: (b -> c) -}) -- We can now get rid of the vA: hog3 :: (c -> d) -> (a -> (b -> c)) -> (a -> (b -> d)) hog3 fCD fABC = ((.) fCD) . fABC -- To get rid of fABC, let's first rewrite it in prefix: hog4 :: (c -> d) -> (a -> (b -> c)) -> (a -> (b -> d)) hog4 fCD fABC = (.) ((.) fCD) fABC -- Now you can just get rid of fABC (eta conversion): hog5 :: (c -> d) -> (a -> (b -> c)) -> (a -> (b -> d)) hog5 fCD = (.) ((.) fCD) -- We can now rewrite hog5 by moving fCD to the outside (look at foo1, the (.) -- are just functions like f and g): hog6 :: (c -> d) -> (a -> (b -> c)) -> (a -> (b -> d)) hog6 fCD = ((.) . (.)) fCD -- Finally we can just get rid of fCD (eta) and redundant parantheses: hog7 :: (c -> d) -> (a -> (b -> c)) -> (a -> (b -> d)) hog7 = (.) . (.)