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Last updated 29.03.04

Advanced Functional Programming [COMP4132]

Session 1, 2004

Exercises: Lambda Calculus

The following exercises should help you develop a better understanding of how the lambda calculus works.

A note on notation: In the following, I use the Haskell syntax \x -> e to denote lambda expressions .

Translation of let-Bindings into the Plain Lambda Calculus

In the lecture, we discussed that we can translate non-recusive and recursive let bindings into the plain lambda calculus as follows:

Non-recursive binding: let x = e1 in e2
=>  (\x -> e2) e1
Recursive binding: let x = e1 in e2
=>  let x = fix (\x -> e1) in e2

where the fixed point combinator fix is defined to be the function

fix :: (a -> a) -> a
fix f = f (fix f)

Your task is to find a translation rule for mutually recursive bindings of the form

let {x1 = e1; ...; xn = en} in e

where the xi may occur in the ei as well as in e. (Hint: It is sufficient if you turn the set of mutually-recursive bindings into a single recursive binding to which the previous rules can, then, be applied.)

Translation of Data Constructors and case into the Plain Lambda Calculus

In the lecture, we discussed that

pair = \a1 -> \a2 -> \sel -> sel a1 a2
fst  = \p -> p (\a1 -> \a2 -> a1)
snd  = \p -> p (\a1 -> \a2 -> a2)

encodes pair construction (pair) and decomposition (fst and snd). In particular, the following rules are satisfied:

fst (pair e1 e2) = e1
snd (pair e1 e2) = e2

Your tasks is to find an implementation of the constructors nil and cons, the destructors head and tail, and the conditional ifNull, which encode lists and have to satisfy the following rules:

head (cons e1 e2) = e1
tail (cons e1 e2) = e2

ifNull nil          e3 e4 = e3
ifNull (cons e1 e2) e3 e4 = e4

Afterwards, implement the function

map f []     = []
map f (x:xs) = x : map f xs

in terms of nil, cons, head, tail, and ifNull (without using case, [], (:), or pattern-matching).

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