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.
+ ANSWER: Most of you just stated that `Ω` a fixed point for `ω`, but didn't explain why. The mere fact that `ω Ω` doesn't immediately reduce to `Ω` doesn't show that the two terms are not convertible. However, one can argue that `Ω` syntactically has two lambdas, and that upon reduction it always yields itself. So it suffices to show that neither `ω Ω` nor anything it can reduces to can have two lambdas. And indeed it has three lambdas, and anything it reduces to will either have three or four lambdas. For the last part of the question, `Y X` is a fixed point for any `X`, as we've already demonstrated in the notes.
+
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
question, begin by constructing `Y Ω`. Prove that
`Y Ω` is a fixed point for `Ω`.
+ ANSWER: Already demonstrated in the notes. We don't need `Ω ξ` to reduce to `ξ`, because `Y Ω` is a `ξ` that can do its own reducing.
+
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.)
+ ANSWER: Everything is a fixed point of `I`, so in particular `I` is. `I` is also a fixed point for `K I`. There are many other examples.
+
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)`.
-<!-- Proof: YY ~~> Y(YY) ~~> YY(Y(YY)); YY(YY) ~~> YY(Y(YY)) -->
+ ANSWER: By the definition of a fixed point operator, `Ψ Ψ` is a fixed point for `Ψ`; and `Ψ (Ψ Ψ)` is a fixed point for `Ψ Ψ`. That is: (a) `Ψ (Ψ Ψ) <~~> Ψ Ψ`; and (b) `Ψ Ψ (Ψ (Ψ Ψ)) <~~> Ψ (Ψ Ψ)`. Now a fact we did not discuss in class, but which Hankin and other readings do discuss, is that substitution of convertible subterms preserves convertibility (and as part of that, `<~~>` is transitive). Hence in the lhs of (b), we can substitute `Ψ Ψ` for the subterm `Ψ (Ψ Ψ)`, because of (a), getting (c) `Ψ Ψ (Ψ Ψ) <~~> Ψ (Ψ Ψ)`. But then by (a) and transitivity of `<~~>`, we get (d) `Ψ Ψ (Ψ Ψ) <~~> Ψ Ψ`. Which states that `Ψ Ψ` is a fixed point for itself.
## Writing recursive functions ##
fact 4
+ ANSWER: `let fact = Y (\fact n. (zero? n) 1 (mult n (fact (pred n))))`.
+
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
* The native Scheme function that most resembles the `mem?` you're defining is `memv`, though that is defined for more than just numbers, and when that `memv` finds a match it returns a *list* starting with the match, rather than `#t`.
+ ANSWER:
+
+ let not_equal? = \n m. neg (num_equal? n m) in
+ let mem? = \n xs. neg (empty? (drop_while (not_equal? n) xs)) in
+ let without = \n xs. append (take_while (not_equal? n) xs) (tail (drop_while (not_equal? n) xs)) in
+ let set_cons = \x xs. (mem? x xs) xs (cons x xs) in
+ let set_equal? = Y (\set_equal?. \xs ys. (empty? xs)
+ (empty? ys)
+ ; else when xs aren't empty
+ ((mem? (head xs) ys)
+ (set_equal? (tail xs) (without (head xs) ys))
+ ; else when head xs not in ys
+ false))
+
7. Linguists often analyze natural language expressions into **trees**. We'll need trees in future weeks, and tree structures provide good opportunities for learning how to write recursive functions. Making use of our current resources, we might approximate trees as follows. Instead of words or syntactic categories, we'll have the nodes of the tree labeled with Church numbers. We'll think of a tree as a list in which each element is itself a tree. For simplicity, we'll adopt the convention that a tree of length 1 must contain a number as its only element.
Then we have the following representations:
Your assignment is to write a Lambda Calculus function that expects a tree, encoded in the way just described, as an argument, and returns the sum of its leaves as a result. So for all of the trees listed above, it should return `1 + 2 + 3`, namely `6`. You can use any Lambda Calculus implementation of lists you like.
+ ANSWER:
+
+ The tricky thing about defining these functions is that it's easy to unwittingly violate the conventions about how we're encoding trees. Thus `Leaf1 ≡ [1]` is a tree, and `[Leaf1, Leaf1, Leaf1]` is also a tree, but `[Leaf1]` is not a tree. (It's a singleton whose content is not a number, but rather a leaf, a list containing a number.) Thus when we are recursively processing `[Leaf1, Leaf1, Leaf1]`, we have to be careful not to just blindly call our function with the tail of our input, since when we get to the end the tail `[Leaf1]` is not itself a tree, on the conventions we've adopted. And our function is likely to rely on those conventions in such a way that it chokes when they are violated.
+
+ That understood, here is a recursive implementation of `sum_leaves`:
+
+ let singleton? = \xs. num_equal? one (length xs) in
+ let singleton = \x. cons x empty in
+ let doubleton = \x y. cons x (singleton y) in
+ let second = \xs. head (tail xs) in
+
+ let sum_leaves = Y (\sum_leaves. \t.
+ (singleton? t)
+ (head t)
+ ; else if t is a tree, it contains two or more subtrees
+ (add
+ (sum_leaves (head t))
+ ; don't recurse if (tail t) is a singleton
+ ((singleton? (tail t))
+ (sum_leaves (second t))
+ ; else it's ok to recurse
+ (sum_leaves (tail t))))) in
+ ...
8. The **fringe** of a leaf-labeled tree is the list of values at its leaves, ordered from left-to-right. For example, the fringe of all three trees displayed above is the same list, `[1, 2, 3]`. We are going to return to the question of how to tell whether trees have the same fringe several times this course. We'll discover more interesting and more efficient ways to do it as our conceptual toolboxes get fuller. For now, we're going to explore the straightforward strategy. Write a function that expects a tree as an argument, and returns the list which is its fringe. Next write a function that expects two trees as arguments, converts each of them into their fringes, and then determines whether the two lists so produced are equal. (Convert your `list_equal?` function from last week's homework into the Lambda Calculus for this last step.)
+ ANSWER using right-fold lists:
+
+ ; are xs strictly longer than ys?
+ let longer? = \xs ys. neg (leq? (length xs) (length ys)) in
+
+ ; uncons xs f ~~> f (head xs) (tail xs)
+ let uncons = \xs f. f (head xs) (tail xs) in
+
+ let check = \x p. p (\bool ys. uncons ys (\y ys. pair (and (num_equal? x y) bool) ys)) in
+ let finish = \bool ys. (empty? ys) bool false in
+ let list_equal? = \xs ys. (longer? xs ys) false (xs check (pair true (rev ys)) finish) in
+
+ let get_fringe = Y (\get_fringe. \t.
+ ; this uses a similar pattern to previous problem
+ (singleton? t)
+ t
+ ; else if t is a tree, it contains two or more subtrees
+ (append
+ (get_fringe (head t))
+ ; don't recurse if (tail t) is a singleton
+ ((singleton? (tail t))
+ (get_fringe (second t))
+ ; else it's ok to recurse
+ (get_fringe (tail t))))) in
+
+ Here is some test data:
+
+ let leaf1 = singleton 1 in
+ let leaf2 = singleton 2 in
+ let leaf3 = singleton 3 in
+ let t12 = doubleton leaf1 leaf2 in
+ let t23 = doubleton leaf2 leaf3 in
+ let alpha = cons leaf1 t23 in
+ let beta = doubleton t12 leaf3 in
+ let gamma = doubleton leaf1 t23 in
+ list_equal? (get_fringe gamma) (get_fringe alpha)
+
+ And here are some cleverer implementations of some of the functions used above:
+
+ let box = \a. \v. v a in
+ let singleton? = \xs. xs (\x b. box (b (K true))) (K false) not in
+
+ ; this function works by first converting [x1,x2,x3] into (true,(true,(true,(K false))))
+ ; then each element of ys unpacks that stack by applying its fst to its snd and itself
+ ; so long as we've not gotten to the end, this will have the result of selecting the snd each time
+ ; when we get to the end of the stack, ((K false) fst) ((K false) snd) (K false) ~~> K false
+ ; after ys are done iterating, we apply the result to fst, which will give us either true or ((K false) fst) ~~> false
+ let longer? = \xs ys. ys (\y p. (p fst) (p snd) p) (xs (\x. pair true) (K false)) fst in
+
+ let shift = \x t. t (\a b c. triple (cons x a) a (pair x))) in
+ let uncons = \xs. xs shift (triple empty empty (K err_head)) (\a b c. c b) in
+ ...
## Arithmetic infinity? ##
(Note, though, that with *some* notions of infinite numbers, like [[!wikipedia ordinal numbers]], operations like `+` are defined in such a way that `inf + n` is different from `n + inf`, and does exceed `inf`; similarly for `*` and `^`. With other notions of infinite numbers, like the [[!wikipedia cardinal numbers]], even less familiar arithmetic operations are employed.)
+ ANSWER: Let `H ≡ \u. succ (u u)`, and `X ≡ H H ≡ (\u. succ (u u)) H`. Note that `X ≡ H H ~~> succ (H H)`. Hence `X` is a fixed point for `succ`.
+
10. Prove that `add ξ 1 <~~> ξ`, where `ξ` is the fixed
point you found in (1). What about `add ξ 2 <~~> ξ`?
-Comment: a fixed point for the successor function is an object such that it
+ 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` and `(+n)` (recall this is shorthand for `\x. add x n`) may not be a fixed point for `(*n)`, for example.
+ ANSWER: Prove that `add X 1 <~~> X`:
+
+ add X 1 == (\m n. n succ m) X 1
+ ~~> 1 succ X
+ == (\s z. s z) succ X
+ ~~> succ X
+
+ Which by the previous problem is convertible with `X`. (In particular, `X ~~> succ X`.) What about `add X 2`?
+
+ add X 2 == (\m n. n succ m) X 2
+ ~~> 2 succ X
+ == (\s z. s (s z)) succ X
+ ~~> succ (succ X)
+
+ And we know the inner term will be convertible with `X`, and hence we get that the whole result is convertible with `succ X`. Which we already said is convertible with `X`. We can readily see that `add X n <~~> X` for all (finite) natural numbers `n`.
+
+
## Mutually-recursive functions ##
If we gave you such a `Y1` and `Y2`, how would you implement the above
definitions of `even?` and `odd?`?
+ ANSWER:
+
+ <pre>
+ let proto_even = \even odd. <span class="ul">\n. (zero? n) true (odd (pred n))</span> in
+ let proto_odd = \even odd. <span class="ul">\n. (zero? n) false (even (pred n))</span> in
+ let even = Y1 proto_even proto_odd in
+ let odd = Y2 proto_even proto_odd in
+ ...
+ </pre>
+
+ By the definitions of `Y1` and `Y2`, we know that `even <~~> proto_even even odd`, and `odd <~~> proto_odd even odd`. Hence the bound variables `even` and `odd` inside the `proto_...` functions have the values we want. `even` will be bound to (something convertible with) the underlined portion of `proto_even`, and `odd` will be bound to (something convertible with) the underlined portion of `proto_odd`.
12. (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`. [[Here are some hints|assignment4_hint]].
+
+ ANSWER: One solution is given in the hint. Here is another:
+
+ let Y1 = \f g . (\u v . f (u u v)(v v u))
+ (\u v . f (u u v)(v v u))
+ (\v u . g (v v u)(u u v)) in
+ let Y2 = \f g . Y1 g f in
+ ...
+