*Reduction*
Find "normal forms" for the following (that is, reduce them as far as it's possible to reduce
them):
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" defined to be `\t \f. t`
"false" 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)))
8. Define a "neg" operator that negates "true" and "false".
Expeceted behavior: (((neg true) 10) 20) evaluates to 20,
(((neg false) 10) 20) evaluates to 10.
9. Define an "and" operator.
10. 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
)
11. 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-black-or-white-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.)
12. 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 as `\f. f x y`
To extract the first element of a pair p, you write:
p (\fst \snd. fst)
Here are some defintions in Racket:
(define make-pair (lambda (fst) (lambda (snd) (lambda (f) ((f fst) snd)))))
(define get-first (lamda (fst) (lambda (snd) fst)))
(define get-second (lamda (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 bothered by having the pair to the left and the function that operates on it come seco\
nd, think about why it's being done this way: the pair is a package that takes a function for op\
erating on its elements as an argument, and returns the result of operating on its elemens with \
that function. In other words, the pair is also a function.
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.
13. 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.
14. 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
15. Define a "sixteen" function that makes
sixteen copies of its argument (and stores them in a data structure of
your choice).
16. Inspired by our definition of ordered pairs, propose a data structure capable of representin\
g ordered tripes. 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 de\
fining the make-triple function, you have to show how to extraxt 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.
> I expect some to come back with the lovely
> (\f. f first second third)
> and others, schooled in a certain mathematical perversion, to come back
> with:
> (\f. f first (\g. g second third))
17. Write a function second-plus-third that when given to your triple, returns the result of add\
ing the second and third members of the triple.
You can help yourself to the following definition:
(define add (lambda (x) (lambda (y) (+ x y))))