Reduction
---------
-Find "normal forms" for the following (that is, reduce them as far as it's possible to reduce
-them):
+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
- 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))
+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`
+> **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)))
-* Define a "neg" operator that negates "true" and "false".
+<OL start=8>
+<LI>Define a `neg` operator that negates `true` and `false`.
Expected behavior:
evaluates to 10.
-* Define an "and" operator.
+<LI>Define an `and` operator.
-* Define an "xor" operator.
-(If you haven't seen this term before, here's a truth table:
+<LI>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
-)
-* Inspired by our definition of boolean values, propose a data structure
-capable of representing one of the two values "black" or "white".
+<LI>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
+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
+ 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
+(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.)
-* 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
+<LI>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.)
+</OL>
Recall our definitions of ordered pairs.
- the pair (x,y) is defined as `\f. f x y`
+> 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)
+ p (\fst \snd. fst)
-Here are some defintions in Racket:
+Here are some definitions 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)))
+ (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 is a package that takes a function for operating 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'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)))
+ (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.
+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)`.)
-13. Define a "swap" function that reverses the elements of a pair.
-Expected behavior:
+<OL start=13>
+<LI>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
+ (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.
+Write out the definition of `swap` in Racket.
-14. Define a "dup" function that duplicates its argument to form a pair
+<LI>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
+ ((dup 10) get-first) ; evaluates to 10
+ ((dup 10) get-second) ; evaluates to 10
-15. Define a "sixteen" function that makes
+<LI>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 representing ordered tripes. That is,
+<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 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.
+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.
-17. 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.
+<LI>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))))
+ (define add (lambda (x) (lambda (y) (+ x y))))
+
+<!-- Write a function that reverses the order of the elements in a list. [Only attempt this problem if you're feeling frisky, it's super hard unless you have lots of experience programming.] -->
+
+</OL>
-18. [Super hard, unless you have lots of experience programming] Write a function that reverses the order of the elements in a list.