--- /dev/null
+*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))))
+