ω ≡ \x. x x, and
Ω ≡ ω ω. Is
Ωa 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
Ωa fixed point for
ω, but didn't explain why. The mere fact that
ω Ωdoesn't immediately reduce to
Ωdoesn't show that the two terms are not convertible. However, one can argue that
Ωsyntactically has two lambdas, and that upon reduction it always yields itself. So it suffices to show that neither
ω Ωnor 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 Xis a fixed point for any
X, as we've already demonstrated in the notes.
Ω ξfor an arbitrary term
Ωis 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
Ωhave a fixed point? That is, how could there be an
Ω ξ <~~> ξ? To answer this question, begin by constructing
Y Ω. Prove that
Y Ωis a fixed point for
ANSWER: Already demonstrated in the notes. We don't need
Ω ξto reduce to
Y Ωis a
ξthat can do its own reducing.
Find two different terms that have the same fixed point. That is, find terms
F ξ <~~> ξand
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
Iis also a fixed point for
K I. There are many other examples.
Ψis some fixed point combinator; we're not telling you which one. (You can just write
Psiin your homework if you don't know how to generate the symbol
Ψ.) Prove that
Ψ Ψis a fixed point of itself, that is, that
Ψ Ψ <~~> Ψ Ψ (Ψ Ψ).
ANSWER: By the definition of a fixed point operator,
Ψ Ψis a fixed point for
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 Ψfor 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 Ψis a fixed point for itself.
Writing recursive functions
Helping yourself to the functions given below, write a recursive function called
factthat 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 fact 4
let fact = Y (\fact n. (zero? n) 1 (mult n (fact (pred n)))).
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.
; all functions from the previous question, plus ; `num_cmp x y lt eq gt` returns lt when x<y, eq when x==y, gt when x>y 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 empty? = \xs. xs (\y ys. false) true in let head = \xs. xs (\y ys. y) err in let tail = \xs. xs (\y ys. ys) empty in let append = Y (\append. \xs zs. xs (\y ys. (cons y (append ys zs))) zs) 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 ...
drop_whilework as described in Week 1's homework.
Using those resources, define a
set_equal?function. The first should take a number argument
xand 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
xdoesn't appear in the result twice!) The
set_equal?function should take two set arguments
ysand say whether they represent the same set. (Be careful, the lists
[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
empty?to define a
mem?function that returns
xis a member of a list of numbers
xs, else returns
false. Also use
appendto define a
withoutfunction that returns a copy of a list of numbers
xsthat omits the first occurrence of a number
x, if there be such. You may find these functions
withoutuseful in defining
set_equal?. Also, for
set_equal?, you are probably going to want to define the function recursively... as now you know how to do.
Some comments comparing this exercise to The Little Schemer, and Scheme more generally:
set_equal?you're trying to define here is like
eqset?in Chapter 7 of The Little Schemer, and
set_cons x xswould be like
(makeset (cons x xs)), from that same chapter.
withoutare like the
remberfunctions defined in Chapter 2 and 3 of The Little Schemer, though those functions are defined for lists of symbolic atoms, and here you are instead defining them for lists of numbers. The Little Schemer also defines
multirember, which removes all occurrences of a match rather than just the first; and
rember*in Chapter 5, that operate on lists that may contain other, embedded lists.
- 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
memvfinds a match it returns a list starting with the match, rather than
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))
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:
. /|\ / | \ 1 2 3 [, , ] . / \ /\ 3 1 2 [[, ], ] . / \ 1 /\ 2 3 [, [, ]]
Some limitations of this scheme: there is no easy way to label an inner, branching node (for example with a syntactic category like VP), and there is no way to represent a tree in which a mother node has a single daughter.
When processing a tree, you can test for whether the tree is a leaf node (that is, contains only a single number), by testing whether the length of the list is 1. This will be your base case for your recursive definitions that work on these trees. (You'll probably want to write a function
leaf?that encapsulates this check.)
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.
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 ≡ 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
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 ...
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 ...
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.
Find a fixed point
ξfor 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 sometimes satisfies the following constraints, for any finite natural number
n + inf == inf n * inf == inf n ^ inf == inf leq n inf == true
(Note, though, that with some notions of infinite numbers, like ordinal numbers, operations like
+are defined in such a way that
inf + nis different from
n + inf, and does exceed
inf; similarly for
^. With other notions of infinite numbers, like the cardinal numbers, even less familiar arithmetic operations are employed.)
H ≡ \u. succ (u u), and
X ≡ H H ≡ (\u. succ (u u)) H. Note that
X ≡ H H ~~> succ (H H). Hence
Xis a fixed point for
add ξ 1 <~~> ξ, where
ξis 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 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 ξis true for all (finite) natural numbers
n. However, the fixed point you found for
(+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 <~~> Xfor all (finite) natural numbers
(Challenging.) One way to define the function
even?is to have it hand off part of the work to another function
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
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
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,
Y2, that operated on a pair of functions
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
Y2, how would you implement the above definitions of
let proto_even = \even odd. \n. (zero? n) true (odd (pred n)) in let proto_odd = \even odd. \n. (zero? n) false (even (pred n)) in let even = Y1 proto_even proto_odd in let odd = Y2 proto_even proto_odd in ...
By the definitions of
Y2, we know that
even <~~> proto_even even odd, and
odd <~~> proto_odd even odd. Hence the bound variables
proto_...functions have the values we want.
evenwill be bound to (something convertible with) the underlined portion of
oddwill be bound to (something convertible with) the underlined portion of
(More challenging.) Using our derivation of
Yfrom this week's notes as a model, construct a pair
Y2that 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
hby evolving it into
h's fixed point. We suggested the thought exercise, how might you instead evolve
Tand then use
T T Tas
h's fixed point. Try solving this problem first. It may help give you the insights you need to define a
Y2. Here are some hints.
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 ...