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) instead of: (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. You can help yourself to the following definition: (define add (lambda (x) (lambda (y) (+ x y))))