4 Insert all the implicit `( )`s and <code>λ</code>s into the following abbreviated expressions.
7 <code><b>(((</b>x x<b>)</b> (<b>(</b>x x<b>)</b> x)<b>)</b> x<b>)</b></code>
9 <code><b>((</b>v w<b>)</b> (\x <b>(\</b>y <b>(</b>v x<b>))</b>)<b>)</b></code>
11 <code><b>((</b>(\x <b>(\</b>y x<b>)</b>) u<b>)</b> v<b>)</b></code>
12 4. `w (\x y z. x z (y z)) u v`
13 <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>
15 Mark all occurrences of `(x y)` in the following terms:
17 5. <code>(\x y. <u>x y</u>) x y</code>
18 6. <code>(\x y. <u>x y</u>) (<u>x y</u>)</code>
19 7. <code>\x y. <u>x y</u> (<u>x y</u>)</code>
24 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").
26 8. `(\x \y. y x) z` ~~> `\y. y z`
27 9. `(\x (x x)) z` ~~> `z z`
28 10. `(\x (\x x)) z` ~~> `\x x`
29 11. `(\x (\z x)) z` ~~> `\y z`, be sure to change `\z` to a different variable so as not to "capture" `z`
30 12. `(\x (x (\y y))) (\z (z z))` ~~> `\y y`
31 13. `(\x (x x)) (\x (x x))` umm..., reductions will forever be possible, they just don't "do" much
32 14. `(\x (x x x)) (\x (x x x))` that's just mean
39 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.
41 Recall our definitions of true and false.
43 > **true** is defined to be `\t f. t`
44 > **false** is defined to be `\t f. f`
46 In Racket, these functions can be defined like this:
48 (define true (lambda (t) (lambda (f) t)))
49 (define false (lambda (t) (lambda (f) f)))
51 (Note that they are different from Racket's *primitive* boolean values `#t` and `#f`.)
54 15. Define a `neg` operator that negates `true` and `false`.
60 evaluates to `20`, and
66 (define neg (lambda (p) ((p false) true)))
68 16. Define an `or` operator.
70 (define or (lambda (p) (lambda (q) ((p p) q))))
74 (define or (lambda (p) (lambda (q) ((p true) q))))
77 17. Define an `xor` operator. If you haven't seen this term before, here's a truth table:
79 true xor true == false
80 true xor false == true
81 false xor true == true
82 false xor false == false
86 (define xor (lambda (p) (lambda (q) ((p (neg q)) q))))
92 Recall our definitions of ordered triples.
94 > the triple **(**a**, **b**, **c**)** is defined to be `\f. f a b c`
96 To extract the first element of a triple `t`, you write:
100 Here are some definitions in Racket:
102 (define make-triple (lambda (fst) (lambda (snd) (lambda (trd) (lambda (f) (((f fst) snd) trd))))))
103 (define fst_of_three (lambda (fst) (lambda (snd) (lambda (trd) fst))))
104 (define snd_of_three (lambda (fst) (lambda (snd) (lambda (trd) snd))))
108 (define t (((make-triple 10) 20) 30))
109 (t fst_of_three) ; will evaluate to 10
110 (t snd_of_three) ; will evaluate to 20
112 If you're puzzled by having the triple to the left and the function that
113 operates on it come second, think about why it's being done this way: the triple
114 is a package that takes a function for operating on its elements *as an
115 argument*, and returns *the result of* operating on its elements with that
116 function. In other words, the triple is a higher-order function.
119 18. Define the `swap12` function that permutes the elements of a triple. Expected behavior:
121 (define t (((make-triple 10) 20) 30))
122 ((t swap12) fst_of_three) ; evaluates to 20
123 ((t swap12) snd_of_three) ; evaluates to 10
125 Write out the definition of `swap12` in Racket.
127 (define swap12 (lambda (x) (lambda (y) (lambda (z)
128 (lambda (f) (((f y) x) z))))))
131 19. Define a `dup3` function that duplicates its argument to form a triple
132 whose elements are the same. Expected behavior:
134 ((dup3 10) fst_of_three) ; evaluates to 10
135 ((dup3 10) snd_of_three) ; evaluates to 10
139 (define dup3 (lambda (x)
140 (lambda (f) (((f x) x) x))))
142 20. Define a `dup27` function that makes
143 twenty-seven copies of its argument (and stores them in a data structure of
146 OK, then we will store them in a triply-nested triple:
148 (define dup27 (lambda (x) (dup3 (dup3 (dup3 x)))))
153 21. Using Kapulet syntax, define `fold_left`.
155 # fold_left (f, z) [a, b, c] == f (f (f z a) b) c
157 fold_left (f, z) xs = case xs of
159 x' & xs' then fold_left (f, f (z, x')) xs'
164 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`.
167 filter (p, xs) = fold_right ((lambda (y, ys). if p y then y & ys else ys), []) xs
170 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`!
173 xs && ys = fold_right ((&), ys) xs
174 # or append (xs, ys) = ...
177 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`!
180 head xs = fold_right ((lambda (y, _). y), 'err) xs
183 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`!
185 *Here is a boring, inefficient answer*
188 append (ys, zs) = fold_right ((&), zs) ys; # aka (&&)
189 f (y, prev) = append (prev, [y]);
190 reverse xs = fold_right (f, []) xs
193 or (same basic idea, just written differently):
196 f (y, prev) = fold_right ((&), [y]) prev;
197 reverse xs = fold_right (f, []) xs
200 *Here is an elegant, efficient answer following the [[hint|assignment2 hint]]*
202 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:
206 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:
210 And so on. That is the key to the solution. The questions you need to answer, to turn this into something executable, are:
212 1. What is a hole? How can we implement it?
214 A hole is a bound variable. `30 & < >` is `lambda x. 30 & x`.
216 2. What should `f` be, so that the result of the second step, namely `f (20, 30 & < >)`, is `30 & (20 & < >)`?
219 f (y, prev) = lambda x. prev (y & x)
222 3. Given that choice of `f`, what should `z` be, so that the result of the first step, namely `f (30, z)` is `30 & < >`?
224 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`.
226 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?
228 Supply it with `[]` as an argument.
230 5. So now put it all together, and explain how to express `reverse xs` using `fold_right` and primitive syntax like `lambda`, `&`, and `[]`?
233 f (y, prev) = lambda x. prev (y & x);
234 id match lambda y. y;
235 reverse xs = (fold_right (f, id) xs) []
241 26. Given that we've agreed to Church's encoding of the numbers:
243 <code>0 ≡ \f z. z</code>
244 <code>1 ≡ \f z. f z</code>
245 <code>2 ≡ \f z. f (f z)</code>
246 <code>3 ≡ \f z. f (f (f z))</code>
249 How would you express the `succ` function in the Lambda Calculus?
251 let succ = \n. \f z. f (n f z) in ...
253 Compare the definition of `cons`, which has an additional element:
255 <code>let cons = \<u>d</u> ds. \f z. f <u>d</u> (ds f z) in ...</code>