`λ`

s into the following abbreviated expressions.
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`!
This problem does have an elegant and concise solution, but it may be hard for you to figure it out. We think it will a useful exercise for you to try, anyway. We'll give a [[hint|assignment2 hint]]. Don't look at the hint until you've gotten really worked up about the problem. Before that, it probably will just be baffling. If your mind has really gotten its talons into the problem, though, the hint might be just what you need to break it open.
There are also other, less cool answers. Perhaps you'll find one of them first.
Even if you don't get any answer, we think the experience of working on the problem, and then understanding the answer when we reveal it, will be satisfying and worthwhile. It also fits our pedagogical purposes for some of the recurring themes of the class.
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)