4 Insert all the implicit `( )`s and <code>λ</code>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.
6 *In response to your feedback and questions, we refined the explanation of the conventions governing the use of the `.` shorthand. Thanks!*
9 <code><b>(((</b>x x<b>)</b> (<b>(</b>x x<b>)</b> x)<b>)</b> x<b>)</b></code>
11 <code><b>((</b>v w<b>)</b> (\x <b>(\</b>y <b>(</b>v x<b>))</b>)<b>)</b></code>
13 <code><b>((</b>(\x <b>(\</b>y x<b>)</b>) u<b>)</b> v<b>)</b></code>
14 4. `w (\x y z. x z (y z)) u v`
15 <code><b>(((</b>w (\x <b>(\</b>y <b>(\</b>z <b>((</b>x z<b>)</b> (y z)<b>)))</b>)<b>)</b> u<b>)</b> v<b>)</b></code>
17 Mark all occurrences of `(x y)` in the following terms:
19 5. <code>(\x y. <u>x y</u>) x y</code>
20 6. <code>(\x y. <u>x y</u>) (<u>x y</u>)</code>
21 7. <code>\x y. <u>x y</u> (<u>x y</u>)</code>
26 Find "normal forms" for the following---that is, reduce them until no more reductions are possible. As mentioned in the notes, we'll write <code>λx</code> as `\x`. If we ever say "reduce" without qualifications, we mean just "beta-reduce" (as opposed to "(beta + eta)-reduce").
28 8. `(\x \y. y x) z` ~~> `\y. y z`
29 9. `(\x (x x)) z` ~~> `z z`
30 10. `(\x (\x x)) z` ~~> `\x x`
31 11. `(\x (\z x)) z` ~~> `\y z`, be sure to change `\z` to a different variable so as not to "capture" `z`
32 12. `(\x (x (\y y))) (\z (z z))` ~~> `\y y`
33 13. `(\x (x x)) (\x (x x))` umm..., reductions will forever be possible, they just don't "do" much
34 14. `(\x (x x x)) (\x (x x x))` that's just mean
41 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.
43 Recall our definitions of true and false.
45 > **true** is defined to be `\t f. t`
46 > **false** is defined to be `\t f. f`
48 In Racket, these functions can be defined like this:
50 (define true (lambda (t) (lambda (f) t)))
51 (define false (lambda (t) (lambda (f) f)))
53 (Note that they are different from Racket's *primitive* boolean values `#t` and `#f`.)
56 15. Define a `neg` operator that negates `true` and `false`.
62 evaluates to `20`, and
68 (define neg (lambda (p) ((p false) true)))
70 16. Define an `or` operator.
72 (define or (lambda (p) (lambda (q) ((p p) q))))
76 (define or (lambda (p) (lambda (q) ((p true) q))))
79 17. Define an `xor` operator. If you haven't seen this term before, here's a truth table:
81 true xor true == false
82 true xor false == true
83 false xor true == true
84 false xor false == false
88 (define xor (lambda (p) (lambda (q) ((p (neg q)) q))))
94 Recall our definitions of ordered triples.
96 > the triple **(**a**, **b**, **c**)** is defined to be `\f. f a b c`
98 To extract the first element of a triple `t`, you write:
100 t (\fst snd trd. fst)
102 Here are some definitions in Racket:
104 (define make-triple (lambda (fst) (lambda (snd) (lambda (trd) (lambda (f) (((f fst) snd) trd))))))
105 (define fst_of_three (lambda (fst) (lambda (snd) (lambda (trd) fst))))
106 (define snd_of_three (lambda (fst) (lambda (snd) (lambda (trd) snd))))
110 (define t (((make-triple 10) 20) 30))
111 (t fst_of_three) ; will evaluate to 10
112 (t snd_of_three) ; will evaluate to 20
114 If you're puzzled by having the triple to the left and the function that
115 operates on it come second, think about why it's being done this way: the triple
116 is a package that takes a function for operating on its elements *as an
117 argument*, and returns *the result of* operating on its elements with that
118 function. In other words, the triple is a higher-order function.
121 18. Define the `swap12` function that permutes the elements of a triple. Expected behavior:
123 (define t (((make-triple 10) 20) 30))
124 ((t swap12) fst_of_three) ; evaluates to 20
125 ((t swap12) snd_of_three) ; evaluates to 10
127 Write out the definition of `swap12` in Racket.
129 (define swap12 (lambda (x) (lambda (y) (lambda (z)
130 (lambda (f) (((f y) x) z))))))
133 19. Define a `dup3` function that duplicates its argument to form a triple
134 whose elements are the same. Expected behavior:
136 ((dup3 10) fst_of_three) ; evaluates to 10
137 ((dup3 10) snd_of_three) ; evaluates to 10
141 (define dup3 (lambda (x)
142 (lambda (f) (((f x) x) x))))
144 20. Define a `dup27` function that makes
145 twenty-seven copies of its argument (and stores them in a data structure of
148 OK, then we will store them in a triply-nested triple:
150 (define dup27 (lambda (x) (dup3 (dup3 (dup3 x)))))
155 21. Using Kapulet syntax, define `fold_left`.
157 # fold_left (f, z) [a, b, c] == f (f (f z a) b) c
159 fold_left (f, z) xs = case xs of
161 x' & xs' then fold_left (f, f (z, x')) xs'
166 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`.
169 filter (p, xs) = fold_right ((lambda (y, ys). if p y then y & ys else ys), []) xs
172 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`!
175 xs && ys = fold_right ((&), ys) xs
176 # or append (xs, ys) = ...
179 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`!
182 head xs = fold_right ((lambda (y, _). y), 'err) xs
185 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`!
187 *Here is a boring, inefficient answer*
190 append (ys, zs) = fold_right ((&), zs) ys; # aka (&&)
191 f (y, prev) = append (prev, [y]);
192 reverse xs = fold_right (f, []) xs
195 or (same basic idea, just written differently):
198 f (y, prev) = fold_right ((&), [y]) prev;
199 reverse xs = fold_right (f, []) xs
202 <a id=cps-reverse></a>
203 *Here is an elegant, efficient answer following the [[hint|assignment2 hint]]*
205 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:
209 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:
213 And so on. That is the key to the solution. The questions you need to answer, to turn this into something executable, are:
215 1. What is a hole? How can we implement it?
217 A hole is a bound variable. `30 & < >` is `lambda x. 30 & x`.
219 2. What should `f` be, so that the result of the second step, namely `f (20, 30 & < >)`, is `30 & (20 & < >)`?
222 f (y, prev) = lambda x. prev (y & x)
225 3. Given that choice of `f`, what should `z` be, so that the result of the first step, namely `f (30, z)` is `30 & < >`?
227 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`.
229 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?
231 Supply it with `[]` as an argument.
233 5. So now put it all together, and explain how to express `reverse xs` using `fold_right` and primitive syntax like `lambda`, `&`, and `[]`?
236 f (y, prev) = lambda x. prev (y & x);
237 id match lambda y. y;
238 reverse xs = (fold_right (f, id) xs) []
241 The ideas here are explored further in Chapter 8 of *The Little Schemer*. There they first introduce the idea of passing function as arguments to other functions, and having functions be the return values from functions. Then the `multirember&co` function discussed on pp. 137--140 (and the other `...&co` functions discussed in that chapter) are more complex examples of the kind of strategy used here to define `reverse`. We will be returning to these ideas and considering them more carefully later in the term.
247 26. Given that we've agreed to Church's encoding of the numbers:
249 <code>0 ≡ \f z. z</code>
250 <code>1 ≡ \f z. f z</code>
251 <code>2 ≡ \f z. f (f z)</code>
252 <code>3 ≡ \f z. f (f (f z))</code>
255 How would you express the `succ` function in the Lambda Calculus?
257 let succ = \n. \f z. f (n f z) in ...
259 Compare the definition of `cons`, which has an additional element:
261 <code>let cons = \<u>d</u> ds. \f z. f <u>d</u> (ds f z) in ...</code>