Syntax ------ Insert all the implicit `( )`s and `λ`s into the following abbreviated expressions. Don't just insert them *freely*; rather, provide the official expression, without any notational shortcuts, that is syntactically identical to the form presented. *In response to your feedback and questions, we refined the explanation of the conventions governing the use of the `.` shorthand. Thanks!* 1. `x x (x x x) x` `(((x x) ((x x) x)) x)` 2. `v w (\x y. v x)` `((v w) (\x (\y (v x))))` 3. `(\x y. x) u v` `(((\x (\y x)) u) v)` 4. `w (\x y z. x z (y z)) u v` `(((w (\x (\y (\z ((x z) (y z)))))) u) v)` Mark all occurrences of `(x y)` in the following terms: 5. `(\x y. x y) x y` 6. `(\x y. x y) (x y)` 7. `\x y. x y (x y)` Reduction --------- Find "normal forms" for the following---that is, reduce them until no more reductions are possible. As mentioned in the notes, we'll write `λx` as `\x`. If we ever say "reduce" without qualifications, we mean just "beta-reduce" (as opposed to "(beta + eta)-reduce"). 8. `(\x \y. y x) z` ~~> `\y. y z` 9. `(\x (x x)) z` ~~> `z z` 10. `(\x (\x x)) z` ~~> `\x x` 11. `(\x (\z x)) z` ~~> `\y z`, be sure to change `\z` to a different variable so as not to "capture" `z` 12. `(\x (x (\y y))) (\z (z z))` ~~> `\y y` 13. `(\x (x x)) (\x (x x))` umm..., reductions will forever be possible, they just don't "do" much 14. `(\x (x x x)) (\x (x x x))` that's just mean Booleans -------- For these questions, and the ones on triples below, we're setting them up so as to encourage you to experiment with Racket and to formulate your answer in Scheme/Racket syntax. But you can answer in Lambda Calculus syntax if you prefer. 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 functions can be defined like this: (define true (lambda (t) (lambda (f) t))) (define false (lambda (t) (lambda (f) f))) (Note that they are different from Racket's *primitive* boolean values `#t` and `#f`.) 15. 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`. (define neg (lambda (p) ((p false) true))) 16. Define an `or` operator. (define or (lambda (p) (lambda (q) ((p p) q)))) or: (define or (lambda (p) (lambda (q) ((p true) q)))) 17. 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 (define xor (lambda (p) (lambda (q) ((p (neg q)) q)))) Triples ------- Recall our definitions of ordered triples. > the triple **(**a**, **b**, **c**)** is defined to be `\f. f a b c` To extract the first element of a triple `t`, you write: t (\fst snd trd. fst) Here are some definitions in Racket: (define make-triple (lambda (fst) (lambda (snd) (lambda (trd) (lambda (f) (((f fst) snd) trd)))))) (define fst_of_three (lambda (fst) (lambda (snd) (lambda (trd) fst)))) (define snd_of_three (lambda (fst) (lambda (snd) (lambda (trd) snd)))) Now we can write: (define t (((make-triple 10) 20) 30)) (t fst_of_three) ; will evaluate to 10 (t snd_of_three) ; will evaluate to 20 If you're puzzled by having the triple to the left and the function that operates on it come second, think about why it's being done this way: the triple 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 triple is a higher-order function. 18. Define the `swap12` function that permutes the elements of a triple. Expected behavior: (define t (((make-triple 10) 20) 30)) ((t swap12) fst_of_three) ; evaluates to 20 ((t swap12) snd_of_three) ; evaluates to 10 Write out the definition of `swap12` in Racket. (define swap12 (lambda (x) (lambda (y) (lambda (z) (lambda (f) (((f y) x) z)))))) 19. Define a `dup3` function that duplicates its argument to form a triple whose elements are the same. Expected behavior: ((dup3 10) fst_of_three) ; evaluates to 10 ((dup3 10) snd_of_three) ; evaluates to 10 (define dup3 (lambda (x) (lambda (f) (((f x) x) x)))) 20. Define a `dup27` function that makes twenty-seven copies of its argument (and stores them in a data structure of your choice). OK, then we will store them in a triply-nested triple: (define dup27 (lambda (x) (dup3 (dup3 (dup3 x))))) Folds and Lists --------------- 21. Using Kapulet syntax, define `fold_left`. # fold_left (f, z) [a, b, c] == f (f (f z a) b) c letrec fold_left (f, z) xs = case xs of [] then z; x' & xs' then fold_left (f, f (z, x')) xs' end in fold_left 22. Using Kapulet syntax, define `filter` (problem 7 in last week's homework) in terms of `fold_right` and other primitive syntax like `lambda`, `&`, and `[]`. Don't use `letrec`! All the `letrec`-ing that happens should come from the one inside the definition of `fold_right`. let filter (p, xs) = fold_right ((lambda (y, ys). if p y then y & ys else ys), []) xs in filter 23. Using Kapulet syntax, define `&&` in terms of `fold_right`. (To avoid trickiness about the infix syntax, just call it `append`.) As with problem 22 (the previous problem), don't use `letrec`! let xs && ys = fold_right ((&), ys) xs # or append (xs, ys) = ... in (&&) 24. Using Kapulet syntax, define `head` in terms of `fold_right`. When applied to a non-empty list, it should give us the first element of that list. When applied to an empty list, let's say it should give us `'err`. As with problem 22, don't use `letrec`! let head xs = fold_right ((lambda (y, _). y), 'err) xs in head 25. We mentioned in the Encoding notes that `fold_left (flipped_cons, []) xs` would give us the elements of `xs` but in the reverse order. So that's how we can express `reverse` in terms of `fold_left`. How would you express `reverse` in terms of `fold_right`? As with problem 22, don't use `letrec`! *Here is a boring, inefficient answer* let append (ys, zs) = fold_right ((&), zs) ys; # aka (&&) f (y, prev) = append (prev, [y]); reverse xs = fold_right (f, []) xs in reverse or (same basic idea, just written differently): let f (y, prev) = fold_right ((&), [y]) prev; reverse xs = fold_right (f, []) xs in reverse *Here is an elegant, efficient answer following the [[hint|assignment2 hint]]* Suppose the list we want to reverse is `[10, 20, 30]`. Applying `fold_right` to this will begin by computing `f (30, z)` for some `f` and `z` that we specify. If we made the result of that be something like `30 & blah`, or any larger structure that contained something of that form, it's not clear how we could, using just the resources of `fold_right`, reach down into that structure and replace the `blah` with some other element, as we'd evidently need to, since after the next step we should get `30 & (20 & blah)`. What we'd like instead is something like this: 30 & < > Where `< >` isn't some *value* but rather a *hole*. Then with the next step, we want to plug into that hole `20 & < >`, which contains its own hole. Getting: 30 & (20 & < >) And so on. That is the key to the solution. The questions you need to answer, to turn this into something executable, are: 1. What is a hole? How can we implement it? A hole is a bound variable. `30 & < >` is `lambda x. 30 & x`. 2. What should `f` be, so that the result of the second step, namely `f (20, 30 & < >)`, is `30 & (20 & < >)`? let f (y, prev) = lambda x. prev (y & x) in ... 3. Given that choice of `f`, what should `z` be, so that the result of the first step, namely `f (30, z)` is `30 & < >`? The identity function: `f (30, (lambda y. y))` will reduce to `lambda x. (lambda y. y) (30 & x)`, which will reduce to `lambda x. 30 & x`. 4. At the end of the `fold_right`, we're going to end with something like `30 & (20 & (10 & < >))`. But what we want is `[30, 20, 10]`. How can we turn what we've gotten into what we want? Supply it with `[]` as an argument. 5. So now put it all together, and explain how to express `reverse xs` using `fold_right` and primitive syntax like `lambda`, `&`, and `[]`? let f (y, prev) = lambda x. prev (y & x); id match lambda y. y; reverse xs = (fold_right (f, id) xs) [] in reverse Numbers ------- 26. Given that we've agreed to Church's encoding of the numbers: `0 ≡ \f z. z` `1 ≡ \f z. f z` `2 ≡ \f z. f (f z)` `3 ≡ \f z. f (f (f z))` `...` How would you express the `succ` function in the Lambda Calculus? let succ = \n. \f z. f (n f z) in ... Compare the definition of `cons`, which has an additional element: `let cons = \d ds. \f z. f d (ds f z) in ...`