## Fixed points ##
1. Recall that `ω ≡ \x. x x`, and `Ω ≡ ω ω`. Is `Ω` a fixed point for `ω`? Find a fixed point for `ω`, and prove that it is a fixed point.
2. Consider `Ω ξ` for an arbitrary term `ξ`.
`Ω` is so busy reducing itself (the eternal narcissist) that it
never gets around to noticing whether it has an argument, let alone
doing anything with that argument. If so, how could `Ω` have a
fixed point? That is, how could there be an `ξ` such that
`Ω ξ <~~> ξ`? To answer this
question, begin by constructing `Y Ω`. Prove that
`Y Ω` is a fixed point for `Ω`.
3. Find two different terms that have the same fixed point. That is,
find terms `F`, `G`, and `ξ` such that `F ξ <~~> ξ` and `G ξ
<~~> ξ`. (If you need a hint, reread the notes on fixed
points.)
4. Assume that `Ψ` is some fixed point combinator; we're not telling you which one. (You can just write `Psi` in your homework if you don't know how to generate the symbol `Ψ`.) Prove that `Ψ Ψ` is a fixed point of itself, that is, that `Ψ Ψ <~~> Ψ Ψ (Ψ Ψ)`.
## Writing recursive functions ##
5. Helping yourself to the functions given below,
write a recursive function called `fact` that computes the factorial.
The factorial `n! = n * (n - 1) * (n - 2) * ... * 3 * 2 * 1`.
For instance, `fact 0 ~~> 1`, `fact 1 ~~> 1`, `fact 2 ~~> 2`, `fact 3 ~~>
6`, and `fact 4 ~~> 24`.
let true = \y n. y in
let false = \y n. n in
let pair = \a b. \v. v a b in
let fst = \a b. a in ; aka true
let snd = \a b. b in ; aka false
let zero = \s z. z in
let succ = \n s z. s (n s z) in
let zero? = \n. n (\p. false) true in
let pred = \n. n (\p. p (\a b. pair (succ a) a)) (pair zero zero) snd in
let add = \l r. r succ l in
let mult = \l r. r (add l) 0 in
let Y = \h. (\u. h (u u)) (\u. h (u u)) in
let fact = ... in
fac 4
6. For this question, we want to implement **sets** of numbers in terms of lists of numbers, where we make sure as we construct those lists that they never contain a single number more than once. (It would be even more efficient if we made sure that the lists were always sorted, but we won't try to implement that refinement here.) To enforce the idea of modularity, let's suppose you don't know the details of how the lists are implemented. You just are given the functions defined below for them (but pretend you don't see the actual definitions). These define lists in terms of [[one of the new encodings discussed last week|/topics/week3_lists#v5-lists]].
; all functions from the previous question, plus
; `num_cmp x y lt eq gt` returns lt when xy
let num_cmp = (\base build consume. \l r. r consume (l build base) fst)
; where base is
(pair (\a b c. b) (K (\a b c. a)))
; and build is
(\p. pair (\a b c. c) p)
; and consume is
(\p. p fst p (p snd) (p snd)) in
let num_equal? = \x y. num_cmp x y false true false in
let neg = \b y n. b n y in
let empty = \f n. n in
let cons = \x xs. \f n. f x xs in
let take_while = Y (\take_while. \p xs. xs (\y ys. (p y) (cons y (take_while p ys)) empty) empty) in
let drop_while = Y (\drop_while. \p xs. xs (\y ys. (p y) (drop_while p ys) xs) empty) in
...
The functions `take_while` and `drop_while` work as described in [[Week 1's homework|assignment1]].
Using those resources, define a `set_cons` and a `set_equal?` function. The first should take a number argument `x` and a set argument `xs` (implemented as a list of numbers assumed to have no repeating elements), and return a (possibly new) set argument which contains `x`. (But make sure `x` doesn't appear in the result twice!) The `set_equal?` function should take two set arguments `xs` and `ys` and say whether they represent the same set. (Be careful, the lists `[1, 2]` and `[2, 1]` are different lists but do represent the same set. Hence, you can't just use the `list_equal?` function you defined in last week's homework.)
Here are some tips for getting started. Use `drop_while` and `num_equal?` to define a `mem?` function that returns `true` if number `x` is a member of a list of numbers `xs`, else returns `false`. Also use `take_while` and `drop_while` to define a `without` function that returns a copy of a list of numbers `xs` that omits the first occurrence of a number `x`, if there be such. You may find these functions `mem?` and `without` useful in defining `set_cons` and `set_equal?`. Also, for `set_equal?`, you are probably going to want to define the function recursively... as now you know how to do.
7. Questions about trees.
## Arithmetic infinity? ##
The next few questions involve reasoning about Church arithmetic and
infinity. Let's choose some arithmetic functions:
succ = \n s z. s (n s z)
add = \l r. r succ l in
mult = \l r. r (add l) 0 in
exp = \base r. r (mult base) 1 in
There is a pleasing pattern here: addition is defined in terms of
the successor function, multiplication is defined in terms of
addition, and exponentiation is defined in terms of multiplication.
8. Find a fixed point `ξ` for the successor function. Prove it's a fixed
point, i.e., demonstrate that `succ ξ <~~> ξ`.
We've had surprising success embedding normal arithmetic in the lambda
calculus, modeling the natural numbers, addition, multiplication, and
so on. But one thing that some versions of arithmetic supply is a
notion of infinity, which we'll write as `inf`. This object usually
satisfies the following constraints, for any finite natural number `n`:
n + inf == inf
n * inf == inf
n ^ inf == inf
leq n inf == true
(Note, though, that with some notions of infinite numbers, operations like `+` and `*` are defined in such a way that `inf + n` is different from `n + inf`, and does exceed `inf`.)
9. Prove that `add 1 ξ <~~> ξ`, where `ξ` is the fixed
point you found in (1). What about `add 2 ξ <~~> ξ`?
Comment: a fixed point for the successor function is an object such that it
is unchanged after adding 1 to it. It makes a certain amount of sense
to use this object to model arithmetic infinity. For instance,
depending on implementation details, it might happen that `leq n ξ` is
true for all (finite) natural numbers `n`. However, the fixed point
you found for `succ` may not be a fixed point for `mult n` or for
`exp n`.
## Mutually-recursive functions ##
10. (Challenging.) One way to define the function `even?` is to have it hand off
part of the work to another function `odd?`:
let even? = \x. (zero? x)
; if x == 0 then result is
true
; else result turns on whether x-1 is odd
(odd? (pred x))
At the same tme, though, it's natural to define `odd?` in such a way that it
hands off part of the work to `even?`:
let odd? = \x. (zero? x)
; if x == 0 then result is
false
; else result turns on whether x-1 is even
(even? (pred x))
Such a definition of `even?` and `odd?` is called **mutually recursive**. If you
trace through the evaluation of some sample numerical arguments, you can see
that eventually we'll always reach a base step. So the recursion should be
perfectly well-grounded:
even? 3
~~> (zero? 3) true (odd? (pred 3))
~~> odd? 2
~~> (zero? 2) false (even? (pred 2))
~~> even? 1
~~> (zero? 1) true (odd? (pred 1))
~~> odd? 0
~~> (zero? 0) false (even? (pred 0))
~~> false
But we don't yet know how to implement this kind of recursion in the Lambda
Calculus.
The fixed point operators we've been working with so far worked like this:
let ξ = Y h in
ξ <~~> h ξ
Suppose we had a pair of fixed point operators, `Y1` and `Y2`, that operated on
a *pair* of functions `h` and `g`, as follows:
let ξ1 = Y1 h g in
let ξ2 = Y2 h g in
ξ1 <~~> h ξ1 ξ2 and
ξ2 <~~> g ξ1 ξ2
If we gave you such a `Y1` and `Y2`, how would you implement the above
definitions of `even?` and `odd?`?
11. (More challenging.) Using our derivation of `Y` from [[this week's notes|topics/week4_fixed_point_combinators#deriving-y]] as a model, construct a pair `Y1` and `Y2` that behave in the way described above.
Here is one hint to get you started: remember that in the notes, we constructed a fixed point for `h` by evolving it into `H` and using `H H` as `h`'s fixed point. We suggested the thought exercise, how might you instead evolve `h` into some `T` and then use `T T T` as `h`'s fixed point. Try solving this problem first. It may help give you the insights you need to define a `Y1` and `Y2`.