## Reduction

Find "normal forms" for the following---that is, reduce them until no more reductions are possible. We'll write λx as \x.

1. (\x \y. y x) z
2. (\x (x x)) z
3. (\x (\x x)) z
4. (\x (\z x)) z
5. (\x (x (\y y))) (\z (z z))
6. (\x (x x)) (\x (x x))
7. (\x (x x x)) (\x (x x x))

## Booleans

Recall our definitions of true and false.

true is defined to be \t \f. t
false is defined to be \t \f. f

In Racket, these can be defined like this:

(define true (lambda (t) (lambda (f) t)))
(define false (lambda (t) (lambda (f) f)))


1. Define a neg operator that negates true and false.

Expected behavior:

(((neg true) 10) 20)


evaluates to 20, and

(((neg false) 10) 20)


evaluates to 10.

2. Define an and operator.

3. Define an xor operator. If you haven't seen this term before, here's a truth table:

true xor true = false
true xor false = true
false xor true = true
false xor false = false


4. Inspired by our definition of boolean values, propose a data structure capable of representing one of the two values black or white. If we have one of those values, call it a "black-or-white value", we should be able to write:

the-value if-black if-white


(where if-black and if-white are anything), and get back one of if-black or if-white, depending on which of the black-or-white values we started with. Give a definition for each of black and white. (Do it in both lambda calculus and also in Racket.)

5. Now propose a data structure capable of representing one of the three values red green or blue, based on the same model. (Do it in both lambda calculus and also in Racket.)

## Pairs

Recall our definitions of ordered pairs.

the pair (x,y) is defined to be \f. f x y

To extract the first element of a pair p, you write:

p (\fst \snd. fst)


Here are some definitions in Racket:

(define make-pair (lambda (fst) (lambda (snd) (lambda (f) ((f fst) snd)))))
(define get-first (lambda (fst) (lambda (snd) fst)))
(define get-second (lambda (fst) (lambda (snd) snd)))


Now we can write:

(define p ((make-pair 10) 20))
(p get-first)   ; will evaluate to 10
(p get-second)  ; will evaluate to 20


If you're puzzled by having the pair to the left and the function that operates on it come second, think about why it's being done this way: the pair is a package that takes a function for operating on its elements as an argument, and returns the result of operating on its elements with that function. In other words, the pair is a higher-order function. (Consider the similarities between this definition of a pair and a generalized quantifier.)

If you like, you can disguise what's going on like this:

(define lifted-get-first (lambda (p) (p get-first)))
(define lifted-get-second (lambda (p) (p get-second)))


Now you can write:

(lifted-get-first p)


(p get-first)


However, the latter is still what's going on under the hood. (Remark: (lifted-f ((make-pair 10) 20)) stands to (((make-pair 10) 20) f) as (((make-pair 10) 20) f) stands to ((f 10) 20).)

1. Define a swap function that reverses the elements of a pair. Expected behavior:

(define p ((make-pair 10) 20))
((p swap) get-first) ; evaluates to 20
((p swap) get-second) ; evaluates to 10


Write out the definition of swap in Racket.

2. Define a dup function that duplicates its argument to form a pair whose elements are the same. Expected behavior:

((dup 10) get-first) ; evaluates to 10
((dup 10) get-second) ; evaluates to 10


3. Define a sixteen function that makes sixteen copies of its argument (and stores them in a data structure of your choice).

4. Inspired by our definition of ordered pairs, propose a data structure capable of representing ordered triples. That is,

(((make-triple M) N) P)


should return an object that behaves in a reasonable way to serve as a triple. In addition to defining the make-triple function, you have to show how to extract elements of your triple. Write a get-first-of-triple function, that does for triples what get-first does for pairs. Also write get-second-of-triple and get-third-of-triple functions.

5. Write a function second-plus-third that when given to your triple, returns the result of adding the second and third members of the triple.

(define add (lambda (x) (lambda (y) (+ x y))))