- Application to the truth teller/liar paradoxes
- Q: How do you know that every term in the untyped lambda calculus has a fixed point?
- Q: How do you know that for any term N, Y N is a fixed point of N?
- Q: So if every term has a fixed point, even Y has fixed point.
- Q: Ouch! Stop hurting my brain.
- Q: So Y applied to succ returns a number that is not finite?
- Q: That reminds me, what about ?evaluation order?
- Q: You claimed that the Ackermann function couldn't be implemented using our primitive recursion techniques (such as the techniques that allow us to define addition and multiplication). But you haven't shown that it is possible to define the Ackermann function using full recursion.
- Q: What other questions should I be asking?
- Q: I still don't fully understand the Y combinator. Can you explain it in a different way?
- Q: How does this relate to the discussion in Chapter 9 of The Little Schemer?
Curry originally called
Y the "paradoxical" combinator, and discussed
it in connection with certain well-known paradoxes from the philosophy
literature. The truth-teller paradox has the flavor of a recursive
function without a base case:
(1) This sentence meaning is true.
If we assume that the complex demonstrative "this sentence meaning" can refer to the very meaning displayed in (1), then that meaning (1) will be true just in case the thing referred to by this sentence meaning is true. Thus, (1) will be true just in case (1) is true, and (1) is true just in case (1) is true, and so on. If (1) is true, then (1) is true; but if (1) is not true, then (1) is not true.
Without pretending to give a serious analysis of the paradox, let's
assume that sentences can have for their meaning boolean values
like the ones we have been working with in the Lambda Calculus. Then the sentence John
is John might have as its meaning our
\y n. y.
Now, the verb phrase in (1) expresses a function from whatever the referent of this
sentence meaning is to a boolean. That is,
\m. m true false, where
m is the referent of this sentence meaning. Of course, if
m is a boolean,
m true false <~~> m, so for our purposes, we can
assume that the verb phrase of (1) denotes the identity function
If we use (1) in a context in which this sentence meaning refers to the
meaning expressed by the very sentence in which that demonstrative occurs, then we must find a
m such that it is equivalent to the application of the verb phrase meaning to itself.
m <~~> I m. In other words,
a fixed point for the meaning of the verb phrase.
That means that in a context in which this sentence meaning refers to the
meaning expressed by the sentence in which it occurs, the sentence's meaning is a fixed point for
the identity function. As we observed earlier, anything is a fixed point for the identity function.
In particular, each of the boolean values
false are fixed points for the identity
function. What fixed point does
Y give us?
Y I ≡ (\h. (\u. h (u u)) (\u. h (u u))) I ~~> (\u. I (u u)) (\u. I (u u))) ~~> (\u. (u u)) (\u. (u u))) ≡ ω ω Ω
Well! That feels right. The meaning of This sentence meaning is true
Ω, our prototypical infinite loop...
(2) This sentence meaning is false.
Used in a context in which this sentence meaning refers to the meaning expressed by the utterance of
(2) in which that noun phrase occurs, (2) will denote a fixed point for
\m. neg m,
\m y n. m n y, which is the
C combinator. So in such a
context, (2) might denote
Y C ≡ (\h. (\u. h (u u)) (\u. h (u u))) C ~~> (\u. C (u u)) (\u. C (u u))) ~~> C ((\u. C (u u)) (\u. C (u u))) ~~> C (C ((\u. C (u u)) (\u. C (u u)))) ~~> C (C (C ((\u. C (u u)) (\u. C (u u))))) ~~> ...
An infinite sequence of
Cs, each one negating the remainder of the
sequence. Yep, that feels like a reasonable representation of the
See Barwise and Etchemendy's 1987 OUP book, The Liar: an essay on truth and circularity for an approach that is similar, but expressed in terms of non-well-founded sets rather than recursive functions.
You should be cautious about feeling too comfortable with
these results. Thinking again of the truth-teller paradox, yes,
Ω is a fixed point for
I, and perhaps it has
some privileged status among all the fixed points for
I, being the
one delivered by
Y and all (though it is not obvious why
Y should have
any special status, versus other fixed point combinators).
But one could observe: look, literally every formula is a fixed point for
X <~~> I X
for any choice of
Y combinator is only guaranteed to give us one fixed point out
of infinitely many --- and not always the intuitively most useful
one. (For instance, the squaring function
\x. mul x x has
0 as a fixed point,
square 0 <~~> 0, and
1 as a fixed point, since
square 1 <~~> 1, but
(\x. mul x x) doesn't give us
1.) So with respect to the
truth-teller paradox, why in the reasoning we've
just gone through should we be reaching for just this fixed point at
just this juncture?
One obstacle to thinking this through is the fact that a sentence normally has only two truth values. We might consider instead a noun phrase such as
(3) the entity that this noun phrase refers to
The reference of (3) depends on the reference of the embedded noun
phrase this noun phrase. As with (1), it will again need to be some fixed
I. It's easy to see that any object is a
fixed point for this referential function: if this pen cap is the
referent of the demonstrated noun phrase, then it is the referent of (3), and so on
for any object.
Ultimately, in the context of this course, these paradoxes are more useful as a way of gaining leverage on the concepts of fixed points and recursion, rather than the other way around.
A: That's easy: let
N be an arbitrary term in the lambda calculus. If
N has a fixed point, then there exists some
ξ such that
N ξ (that's what it means to have a fixed point).
let H = \u. N (u u) in let ξ = H H in ξ ≡ H H ≡ (\u. N (u u)) H ~~> N (H H) ≡ N ξ
Please slow down and make sure that you understand what justified each of the equalities in the last line.
A: Note that in the proof given in the previous answer, we chose
and then set
ξ ≡ H H ≡ (\u. N (u u)) (\u. N (u u)). If we abstract over
N, we get the Y combinator,
\N. (\u. N (u u)) (\u. N (u u)). No matter
N we feed
Y, it returns some
ξ that is a fixed point
N, by the reasoning in the previous answer.
let Y = \h. (\u. h (u u)) (\u. h (u u)) in Y Y ≡ \h. (\u. h (u u)) (\u. h (u u)) Y ~~> (\u. Y (u u)) (\u. Y (u u)) ~~> Y ((\u. Y (u u)) (\u. Y (u u))) ~~> Y ( Y ((\u. Y (u u)) (\u. Y (u u)))) <~~> Y ( Y ( Y (...(Y (Y Y))...)))
A: Is that a question?
Let's come at it from the direction of arithmetic. Recall that we
claimed that even
succ --- the function that added one to any
number --- had a fixed point. How could there be an
ξ such that
ξ <~~> succ ξ?
That would imply that
ξ <~~> succ ξ <~~> succ (succ ξ) <~~> succ (succ (succ ξ)) <~~> succ (...(succ ξ)...)
In other words, the fixed point of
succ is a term that is its own
successor. Let's just check that
ξ = succ ξ:
let succ = \n s z. s (n s z) in let ξ = (\u. succ (u u)) (\u. succ (u u)) in succ ξ ≡ succ ((\u. succ (u u)) (\u. succ (u u))) ~~> succ (succ ((\u. succ (u u)) (\u. succ (u u)))) ≡ succ (succ ξ)
You should see the close similarity with
Y Y here.
A: Well, if it makes sense to think of it as a number at all. It doesn't have the same structure as our encodings of finite Church numbers. But let's see if it behaves like they do:
; assume same definitions as before succ ξ ≡ (\n s z. s (n s z)) ξ ~~> \s z. s (ξ s z) <~~> succ (\s z. s (ξ s z)) ; using fixed-point reasoning ≡ (\n s z. s (n s z)) (\s z. s (ξ s z)) ~~> \s z. s ((\s z. s (ξ s z)) s z) ~~> \s z. s (s (ξ s z))
succ ξ looks something like a Church number: it takes two arguments,
and returns a sequence of nested applications of
You should be able to prove that
add 2 (Y succ) <~~> Y succ,
pow. What happens if we try
succ) (Y succ)? What would you expect infinity minus infinity to be?
(Hint: choose your evaluation strategy so that you add two
ss to the
first number for every
s that you add to the second number.)
This is amazing, by the way: we're proving things about a term that represents arithmetic infinity.
It's important to bear in mind the simplest, least-evaluated term we begin with is not infinitely long:
Y succ = (\h. (\u. h (u u)) (\u. h (u u))) (\n s z. s (n s z))
The way that infinity enters into the picture is that this term has no normal form: no matter how many times we perform beta reduction, there will always be an opportunity for more beta reduction. (Lather, rinse, repeat!)
A: For a recursive function that has a well-behaved base case, such as the factorial function, evaluation order is crucial. In the following computation, we will arrive at a normal form. Watch for the moment at which we have to make a choice about which beta reduction to perform next: one choice leads to a normal form, the other choice leads to endless reduction:
let prefact = \fact n. (zero? n) 1 (mul n (fact (pred n))) in let fact = Y prefact in fact 2 ≡ [(\h. (\u. h (u u)) (\u. h (u u))) prefact] 2 ~~> [(\u. prefact (u u)) (\u. prefact (u u))] 2 ~~> [prefact ((\u. prefact (u u)) (\u. prefact (u u)))] 2 ~~> [prefact (prefact ((\u. prefact (u u)) (\u. prefact (u u))))] 2 ≡ [(\fact n. (zero? n) 1 (mul n (fact (pred n)))) (prefact ((\u. prefact (u u)) (\u. prefact (u u))))] 2 ~~> [\n. (zero? n) 1 (mul n ([prefact ((\u. prefact (u u)) (\u. prefact (u u)))] (pred n)))] 2 ~~> (zero? 2) 1 (mul 2 ([prefact ((\u. prefact (u u)) (\u. prefact (u u)))] (pred 2))) ~~> mul 2 ([prefact ((\u. prefact (u u)) (\u. prefact (u u)))] 1) ... ~~> mul 2 (mul 1 ([prefact ((\u. prefact (u u)) (\u. prefact (u u)))] 0)) ≡ mul 2 (mul 1 ((zero? 0) 1 (mul 1 ([prefact ((\u. prefact (u u)) (\u. prefact (u u)))] (pred 0))))) ~~> mul 2 (mul 1 1) ~~> mul 2 1 ~~> 2
The crucial step is the third from the last. We have our choice of
either evaluating the test
(zero? 0) 1 ..., which evaluates to
no matter what the ... contains;
or we can evaluate the
(\u. prefact (u u)) (\u. prefact (u u)), to
produce another copy of
prefact. If we postpone evaluating the
zero? test, we'll pump out copy after copy of
prefact, and never
realize that we've bottomed out in the recursion. But if we adopt a
leftmost/call-by-name/normal-order evaluation strategy, we'll always
start with the
zero? predicate, and only produce a fresh copy of
prefact if we are forced to.
Q: You claimed that the Ackermann function couldn't be implemented using our primitive recursion techniques (such as the techniques that allow us to define addition and multiplication). But you haven't shown that it is possible to define the Ackermann function using full recursion.
A(m,n) = | when m == 0 -> n + 1 | else when n == 0 -> A(m-1, 1) | else -> A(m-1, A(m,n-1)) let A = Y (\A m n. (zero? m) (succ n) ((zero? n) (A (pred m) 1) (A (pred m) (A m (pred n)))))
So for instance:
A 1 2 ~~> A 0 (A 1 1) ~~> A 0 (A 0 (A 1 0)) ~~> A 0 (A 0 (A 0 1)) ~~> A 0 (A 0 2) ~~> A 0 3 ~~> 4
A 1 x is to
A 0 x as addition is to the successor function;
A 2 x is to
A 1 x as multiplication is to addition;
A 3 x is to
A 2 x as exponentiation is to multiplication ---
A 4 x is to
A 3 x as hyper-exponentiation is to exponentiation...
What is it about the "primed" fixed-point combinators
Y′that makes them compatible with a call-by-value evaluation strategy?
What exactly is primitive recursion?
How do you know that the Ackermann function can't be computed using primitive recursion techniques?
I hear that
Ydelivers the/a least fixed point. Least according to what ordering? How do you know it's least? Is leastness important?
Sure! Here is another way to derive
Y. We'll start by choosing a
specific goal, and at each decision point, we'll make a reasonable
guess. The guesses will all turn out to be lucky, and we'll arrive at
a fixed point combinator.
Given an arbitrary term
h, we want to find a fixed point
X such that:
X <~~> h X
Our strategy will be to seek an
X such that
X ~~> h X (this is just a guess). Because
h X are syntactically different, the only way that
X can reduce to
h X is if
contains at least one redex. The simplest way to satisfy this
constraint would be for the fixed point to itself be a redex (again, just a guess):
X ≡ ((\u. M) N) ~~> h X
The result of beta reduction on this redex will be
M with some
substitutions. We know that after these substitutions,
M will have
h X, since that is what the reduction arrow tells us. So we
can refine the picture as follows:
X ≡ ((\u. h (___)) N) ~~> h X
___ has to be something that reduces to the fixed point
It's natural to assume that there will be at least one occurrence of
u in the body of the head abstract:
X ≡ ((\u. h (__u__)) N) ~~> h X
After reduction of the redex, we're going to have
X ≡ h (__N__) ~~> h X
__N__ will have to reduce to
X. Therefore we should
choose a skeleton for
N that is consistent with what we have decided
so far about the internal structure of
X. We might like for
syntactically match the whole of
X, but this would require
N to contain itself as
a subpart. So we'll settle for the more modest assumption (or guess) that
matches the head of
X ≡ ((\u. h (__u__)) (\u. h (__u__))) ~~> h X
At this point, we've derived a skeleton for X on which it contains two
so-far identical halves. We'll guess that the halves will be exactly
identical. Note that at the point at which we perform the first
u will get bound to
N, which now corresponds to a term
representing one of the halves of
X. So in order to produce a full
we simply make a second copy of
X ≡ ((\u. h (u u)) (\u. h (u u))) ~~> h ((\u. h (u u)) (\u. h (u u))) ≡ h X
So the function
\h. (\u. h (u u)) (\u. h (u u)) maps an arbitrary term
h to a fixed point for
A: Pages 160-172 of The Little Schemer introduce you to how to implement recursion in Scheme, without relying on the native capacity to do this expressed in
define. The expression:
(lambda (length) (lambda (l) (cond ((null? l) 0) (else (add1 (length (cdr l)))))))
that occurs starting on p. 162 and on several pages following corresponds to
h in our exposition. The authors of The Little Schemer begin by applying that abstract to the argument
eternity, which is a function that never returns; then they instead apply it to the argument
h eternity, which is a function that works for lists of length zero, but otherwise never returns; then to the argument
h (h eternity), which works for lists of length zero or one, but otherwise never returns; and so on.
They work their way towards the realization that they want an "infinite tower" of applications of
h, except they don't really need an infinite tower, but rather just a finite tower whose height can't be bounded in advance. This is essentially the observation that they need a fixed point for
The authors attempt to self-apply
h on p. 165, just as we did. As we explained in our exposition, though, that doesn't quite work.
On the top of p. 167, the authors have instead moved to our
H, and attempt to self-apply that, instead. And this works.
However, on p. 168, they attempt to abstract out the part that in our
H looks like
(u u) and in their exposition looks like
(mk-length mk-length). Doing that would work in our lambda evaluator, but you can't do it in Scheme, because Scheme has call-by-value evaluation order, which will try to fully reduce this expression before substituting it back into the term it's been abstracted out of. But it can't be fully reduced. Pages 168--170 explore this problem, and pp. 170--172 hit upon the solution of using what we called in our exposition the
Y′ fixed-point combinator, rather than the
Y combinator that we derived. The authors of The Little Schemer call
Y′ the "applicative-order Y combinator".