Reduction
---------
-Find "normal forms" for the following---that is, reduce them until no more reductions are possible. We'll write λ`x` as `\x`.
+Find "normal forms" for the following---that is, reduce them until no more reductions are possible. We'll write <code>λx</code> as `\x`.
1. `(\x \y. y x) z`
2. `(\x (x x)) z`
<LI>Define an `xor` operator.
-(If you haven't seen this term before, here's a truth table:
+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
-)
<LI>Inspired by our definition of boolean values, propose a data structure
capable of representing one of the two values `black` or `white`.
one of those values, call it a "black-or-white value", we should be able to
write:
- the-value if-black if-white
+ 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
To extract the first element of a pair p, you write:
- p (\fst \snd. fst)
+ 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)))
+ (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
+ (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 second, think about why it's being done this way: the pair
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)))
+ (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)
+ (lifted-get-first p)
instead of:
- (p get-first)
+ (p get-first)
However, the latter is still what's going on under the hood.
Expected behavior:
- (define p ((make-pair 10) 20))
- ((p swap) get-first) ; evaluates to 20
- ((p swap) get-second) ; evaluates to 10
+ (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.
whose elements are the same.
Expected behavior:
- ((dup 10) get-first) ; evaluates to 10
- ((dup 10) get-second) ; evaluates to 10
+ ((dup 10) get-first) ; evaluates to 10
+ ((dup 10) get-second) ; evaluates to 10
<LI>Define a `sixteen` function that makes
sixteen copies of its argument (and stores them in a data structure of
<LI>Inspired by our definition of ordered pairs, propose a data structure capable of representing ordered triples. That is,
- (((make-triple M) N) P)
+ (((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.
You can help yourself to the following definition:
- (define add (lambda (x) (lambda (y) (+ x y))))
+ (define add (lambda (x) (lambda (y) (+ x y))))
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