# Encoding Booleans, Tuples, Lists, and Numbers

The Lambda Calculus can represent any computable function?

We need to do some work to show how to represent some of the functions we've become acquainted with.

## Booleans

We'll start with the if ... then ... else ... construction we saw last week:

if M then N else L


For a boolean-valued expression M, the displayed expression should evaluate to whatever N does if M evaluates to 'true, and to whatever L does if M evaluates to 'false. In order to encode or simulate such an if clause in the Lambda Calculus, we need to settle on a way to represent 'true and 'false.

We could simply add the constants 'true and 'false to the language, since it does make sense to extend the Lambda Calculus by adding constants. However, that would also require complicating the interpretation of the language; at the very least, we would need more than just beta-reduction as our engine of computation. In the spirit of computational minimalism, let's see how far we can get with the "pure" lambda calculus, without any special constants.

Let's get rid of the parts of the if statement that are just syntactic window-dressing. That is, let's get rid of the if, the then, and the else:

M  N  L


Recall that our convention is that values associate left to right, so this series of terms would be evaluated as:

((M N) L)


If this expression is supposed to have the meaning of if M then N else L, then we need to find a value for 'true such that when it is substituted in place of M, the expression evaluates to N. The function we're looking for should take two arguments: first N, then L, and it should throw away L and return N.

We've already seen such a function. We called it K: (\x y. x). Let's test:

((K N) L) == (((\x y. x) N) L) ~~> ((\y. N) L) ~~> N


Sucess! In the same spirit, 'false could be K I, which reduces to (\y x. x) (also written as (\y. (\x x))):

(((K I) N) L) == ((((\x y. x) (\x x)) N) L)
~~> (((\y. (\x x)) N) L)
~~> ((\x x) L)
~~> L


So we have seen our first major encoding in the Lambda Calculus: "true" is represented by K, and "false" is represented by K I. We'll be building up a lot of representations in the weeks to come, and they will all maintain the discipline that if a expression is meant to be interpreted as a truth value (i.e. as a Boolean), it will be equivalent to (convertible with) K or K I.

In class, we explained how "and" could be thought of as the function, here written in Kapulet syntax:

lambda p q. if p then q else false


or:

lambda p q. if p then q else p


Given that we know how to express if ... then ... in terms of our encoded Booleans, we can represent this easily in the Lambda Calculus as either:

\p q. p q (\y x. x)


or:

\p q. p q p


Typically there will be more than one way to encode a given function into the Lambda Calculus, even holding fixed all your other decisions about how to encode other functions. In this case, we can use either representation because we're assuming that the p only gets bound to arguments that are either \x y. x or \y x. x. We don't make any promises about what will happen if our encodings are abused, that is, supplied with arguments they weren't designed to accept. If p is bound to \x y. x, then the result of both of the above displayed expressions will be whatever q is bound to. If on the other hand, p is bound to \y x. x, then the two displayed expressions will return the same result, namely that function.

Despite this, the two displayed expressions are two different lambda terms, and are not convertible. It's only within the frame of our assumptions about the restricted arguments we're thinking of supplying them that they behave exactly alike.

You can try out these expressions in the lambda evaluator:

let true = \x y. x in
let false = \y x. x in
let and = \p q. p q p in
and true false


will reduce or "normalize" to \y x. x, or false, just as you'd predict. The let true = stuff isn't officially part of the Lambda Calculus language. It's just a convenient shorthand that lets you use the lambda evaluator more easily. Behind the scenes, the displayed expression above gets translated to:

(\true.
(\false.
(\and. and true false)))
(\x y. x) ; this gets bound to variable "true"
(\y x. x) ; this gets bound to variable "false"
(\p q. p q p) ; this gets bound to variable "and"


We expect you'll agree that the first is easier to write and to read.

In Scheme (Racket), all of these notions can be defined like this:

(define true  (lambda (x) (lambda (y) x)))
(define false (lambda (y) (lambda (x) x)))
(define lambda-and (lambda (p) (lambda (q) ((p q) p))))


and then used like this:

((lambda-and true) false)


which will evaluate to a function, that happens to be the same function false is bound to. Most Scheme interpreters like Racket will helpfully display the function with the name, if any, that was first defined to be bound to it. So we'll see the result as:

#<procedure:false>


The funny calling pattern, where we write ((lambda-and true) false) instead of just (lambda-and true false), is because that's how you have to write curried functions in Scheme. Similarly for why we have (lambda (x) (lambda (y) x)) instead of just (lambda (x y) x).

It's possible to do the next few weeks of assignment without using a Scheme interpreter, however we do recommend you get Scheme installed on your computer, and get started learning Scheme. It will help you understand the ideas better to experiment with it.

There's also a (slow, bare-bones, but perfectly adequate) version of Scheme available for online use at http://tryscheme.sourceforge.net/. (Unfortunately, though, that Scheme implementation will only display the result of ((lambda-and true) false) as #<procedure lambda>. You won't be able to see if it's the function assigned to true or the function assigned to false. You'd have to feed that result some more arguments to see how it behaved.)

You should also be experimenting with this site's lambda evaluator.

## Tuples

In class, we also showed you how to encode a tuple in the Lambda Calculus. We did it with an ordered triple, but the strategy generalizes in a straightforward way. (Some authors just use this strategy to define pairs, then define triples as pairs whose second member is another pair, and so on. Yech. If you keep a firm grip on your wits, that can also be made to work, but it's extremely likely that people who code in that way are going to lose their grip at some point and get themselves in a corner where they'll regret having made that decision about how to encode triples. And they will be forced to add further complexities at later points, that they're probably not anticipating now. The strategy presented here is as elegant as it first looks, and will help you program more hygienically even when your attention lapses.)

Our proposal was to define the triple (a, b, c) as:

\h. h a b c


To extract the first element of this, you'd write:

(\h. h a b c) fst_of_three


where fst_of_three is the function \x y z. x:

(\h. h a b c) (\x y z. x) ~~>
(\x y z. x) a b c ~~>
(\y z. a) b c ~~>
(\z. a) c ~~>
a


Here are the corresponding definitions in Scheme (Racket):

(define make-triple (lambda (a) (lambda (b) (lambda (c)
(lambda (h) (((h a) b) c))))))

(define fst_of_three (lambda (x) (lambda (y) (lambda (z) x))))
(define snd_of_three (lambda (x) (lambda (y) (lambda (z) y))))
(define trd_of_three (lambda (x) (lambda (y) (lambda (z) z))))


Then:

(define p (((make-triple 10) 20) 30))
(p fst_of_three)  ; will evaluate to 10
(p snd_of_three)  ; will evaluate to 20


If you're puzzled by having the triple to the left and the function that "uses" or "consumes" or operates on it to the right, think about why it's being done this way: the triple is a package that will be used in coordination with some function for operating on its elements. We don't know in advance what that function will be. And it's not obvious how to make the triple be some structure that the function can "look inside" and extract the elements from. We're still trying to figure out how to define such structures! But what we can do is make the triple take the function as an argument, and return the result of operating on its elements with that function. In other words, the triple is a higher-order function: a function that expects another function as an argument.

(Consider the similarities between this definition of a triple and a generalized quantifier. This is in fact our first taste of "continuations" in the course, which are a systematic pattern for inverting the naive order of who-is-the-argument? and who-is-the-operator?)

If you really want to, you can disguise what's going on like this:

(define lifted-fst_of_three (lambda (trip) (trip fst_of_three)))


Then you could say:

(lifted-fst_of_three t)


(t fst_of_three)


Of course, the last is still what's happening under the hood.

(Remark: (lifted-f (((make-triple 10) 20) 30)) stands to ((((make-triple 10) 20) 30) f) as ((((make-triple 10) 20) 30) f) stands to (((f 10) 20) 30).)

### Curried and Uncurried functions in the Lambda Calculus

As we've explained before, an uncurried function is one that takes multiple arguments in a single bundle, as a tuple, like this:

f (x, y, z)


Whereas a curried function is one that takes multiple arguments in sequence, like this:

g x y z


That is, g is a function expecting one argument (here x), that evaluates to a second function, that itself expects another argument (here y), and so on. (So g is a higher-order function: the result of applying it to argument x returns another function.) In discussing Kapulet and Scheme and OCaml and Haskell, we sometimes worked with uncurried functions, and other times with curried ones. Now that you've seen how to build and work with tuples in the Lambda Calculus, you can write uncurried functions there too. That is, you can write functions f that will expect arguments of the form \h. h x y z. But in the end, f is going to have to apply that argument to some auxiliary handler function g anyway, where g takes its arguments in curried form. So if you can, in the Lambda Calculus it's easiest to just work with curried functions like g in the first place, rather than uncurried, tuple-expecting arguments like f.

In some cases you can't do this, because you'll be partaking of some general pattern that only makes room for a single argument --- like the "starting value" z in the fold_right function discussed below; yet you're performing some task that really requires you to stuff a couple of values into that position. Tuples are ideal for that purpose. But in for run-of-the-mill functions you're defining in the Lambda Calculus, if multiple arguments need to be passed to a function, and it's up to you whether to pass them in curried or uncurried/tuple style, you should default to the curried style (as in, g x y z). That's the more idiomatic, native style for passing arguments in the Lambda Calculus.

## Lists

There are multiple ways to encode lists, and also multiple ways to encode numbers. We are going to start with what we think are the most natural and elegant encodings. Historically these were the first encodings of numbers but not of lists.

In seminar we discussed two further functions for working with lists or sequences. Reverting to Kapulet syntax, these functions work like this:

fold_right (f, z) [10, 20, 30]


That will evaluate to whatever this does:

f (10, f (20, f (30, z)))


For example, if we let f be (+) and z be 0, then it will be 10 + (20 + (30 + 0)) or 60. Another example, if we let f be (&) and z be [], then fold_right ((&), []) [10, 20, 30] will be 10 & (20 & (30 & [])) or [10, 20, 30], the same sequence we began with.

The other function works like this:

fold_left (f, z) [10, 20, 30]


That will evaluate to whatever this does:

f (f (f (z, 10), 20), 30)


With a commutative operator like (+), it makes no difference whether you say fold_right ((+), z) xs or fold_left ((+), z) xs. But with other operators it will make a difference. We can't say fold_left ((&), []) [10, 20, 30], since that would start by trying to evaluate [] & 10, which would crash. But we could do this:

let
flipped_cons match lambda (xs, x). x & xs  # also expressible as uncurried_flip (&)
in fold_left (flipped_cons, []) [10, 20, 30]


and that would evaluate to flipped_cons (flipped_cons (flipped_cons ([], 10), 20), 30), in other words, to 30 & (20 & (10 & [])), or [30, 20, 10]. So this reverses the sequence we began with.

In class we considered how to express other list operations in terms of these. For example, we saw that we could define length xs as:

fold_right ((lambda (_, z). 1 + z), 0) xs


And we could define map g xs as:

fold_right ((lambda (x, zs). g x & zs), []) xs


Here is a nice website we found that lets you evaluate these things step by step. Just click on the subexpressions to evaluate them.

We also saw in seminar how to define fold_right. We could do this in Kapulet like this:

letrec
fold_right (f, z) xs = case xs of
[] then z;
y & ys then f (y, fold_right (f, z) ys)
end
in fold_right


(In some presentations you may see this expecting f to be a curried function lambda y z. ... rather than the uncurried lambda (y, z). ... that I'm expecting here. We will be shifting to the curried form ourselves momentarily.)

We suggested in class that these functions were very powerful, and could be deployed to do most everything you might want to do with a list. Given how our strategy for encoding booleans turned out, this ought to suggest to you the idea that folding is what we fundamentally do with lists, just as conditional branching is what we fundamentally do with booleans. So we can try to encode lists in terms of lambda expressions that will let us perform folds on them. We could try to do this with either right-folds or left-folds. Either is viable. Some things are more natural if you use right-folds, though, so let's do that.

What should the encoding look like? We don't know what function and what starting value someone might want to fold over our list!

Wait, does that remind you of anything?

Good. I knew it would.

Indeed, we'll make our encoded lists consist of higher-order functions that take the f and the starting value z to be folded as arguments. So the list [a, b, c] should look something like this:

\f z. SOMETHING


Now, what should the SOMETHING be? Well, when we supply an f and a z we should get back the right-fold of those over [a, b, c], so the answer should evidently be:

\f z. f a (f b (f c z))


Here we assume f to be a curried function, taking its arguments in the form f c z rather that f (c, z) (that is, f (\h. h c z)), because as we explained at the end of the section on Tuples, the curried form is the idiomatic and native style for passing multiple arguments in the Lambda Calculus.

So if [a, b, c] should be the displayed higher-order function above, what should [c] be? Evidently:

\f z. f c z


Now what should the empty list [] be? Think about it...

Did you arrive at an answer?

I hope it was this one: \f z. z. Because when we fold a function f and a starting value z over the empty list, we just get back whatever the starting value z was.

We saw before what a make-triple function would look like. What about a make-list function, or as we've been calling it, "cons" (& in Kapulet)? Well, we've already seen what the representation of [] and [c] are. In that simplest case, the cons function should take us from the encoding of [], namely \f z. z to the encoding of [c], namely \f z. f c z, as a result. Here is how we can define this:

let cons = \d ds. \f z. f d (ds f z)
in ...


Let's walk through this. We supply this function with our c and [] as arguments. So its d gets bound to our c, and its ds gets bound to our [], namely \f z. z. Then our result is a higher order function of the form \f z. f c SOMETHING. That looks good, what we're after is just this, except the SOMETHING should end up as z. If we just simply substituted z for SOMETHING, then our cons function would always end up giving us back a singleton list of the form \f z. f X z for some X. That is the form we want in the present case, computing cons c [], but not in the general case. What we'd like instead is to do something to the second argument we fed to cons, here [] or \f z. z, and have it reduce to z, with the hope that the same operation would make more complex second arguments reduce to other appropriate values. Well, \f z. z expects an f and a z as arguments, and hey we happen to have an f and a z handy, since we're inside something of the form \f z. f c SOMETHING. The f and z will be supplied by the consumer or user of the sequence [c] that we're building. So let's just give those f and z to \f z. z as its arguments, and we end up with simply z. In other words, where ds is the second argument we fed to cons, we make SOMETHING in \f z. f c SOMETHING be ds f z. That is why we make cons be:

\d ds. \f z. f d (ds f z)


Let's step through it formally. Given our definitions:

cons c []


is:

(\d ds. \f z. f d (ds f z)) c (\f z. z) ~~>
(\ds. \f z. f c (ds f z)) (\f z. z) ~~>
\f z. f c ((\f z. z) f z) ~~>
\f z. f c ((\z. z) z) ~~>
\f z. f c z


which is just the representation of [c] we were after.

Will this work when more complex values are supplied as the second argument to cons? Let's see:

cons b [c]


is:

(\d ds. \f z. f d (ds f z)) b (\f z. f c z) ~~>
(\ds. \f z. f b (ds f z)) (\f z. f c z) ~~>
\f z. f b ((\f z. f c z) f z) ~~>
\f z. f b ((\z. f c z) z) ~~>
\f z. f b (f c z)


which looks right. Persuade yourself that the cons we've defined here, together with the representation \f z. z for the empty list, always give us the correct encoding for complex lists, in terms of a function that takes an f and a z as arguments, and then returns the result of right-folding those arguments over the list elements in the appropriate order.

You may notice that the encoding we're proposing for [] is the same encoding we proposed above for false. There's nothing deep to this. If we had adopted other conventions, we could easily have made it instead be true and [] that had the same encoding.

Now we saw above how to define map in terms of fold_right. In Kapulet syntax, map g xs was:

fold_right ((lambda (x, zs). g x & zs), []) xs


In our Lambda Calculus encoding, fold_right (f, z) xs gets translated to xs f z. That is, the list itself is the operator, just as we saw triples being. So we just need to know how to represent lambda (x, zs). g x & zs, on the one hand, and [] on the other, into the Lambda Calculus, and then we can also express map. Well, in the Lambda Calculus we're working with curried functions, and there's no infix syntax, so we'll replace the first by lambda x zs. cons (g x) zs. But we just defined cons, and the lambda is straightforward. And we also just defined []. So we already have all the pieces to do this. Namely:

map g xs


in Kapulet syntax, turns into this in our lambda evaluator:

let empty = \f z. z in
let cons = \d ds. \f z. f d (ds f z) in
xs (\x zs. cons (g x) zs) empty


Try out this:

let empty = \f z. z in
let cons = \d ds. \f z. f d (ds f z) in
let map = \g xs. xs (\x zs. cons (g x) zs) empty in
let abc = \f z. f a (f b (f c z)) in
map g abc


That will evaluate to:

\f z. f (g a) (f (g b) (f (g c) z))


which looks like what we want, a higher-order function that will take an f and a z as arguments and then return the right fold of those arguments over [g a, g b, g c], which is map g [a, b, c].

## Numbers

Armed with the encoding we developed for lists above, Church's method for encoding numbers in the Lambda Calculus is very natural. But this is not the order that these ideas were historically developed.

It's noteworthy that when he was developing the Lambda Calculus, Church had not predicted it would be possible to encode the numbers. It came as a welcome surprise.

The idea here is that a number is something like a list, only without any slot for an interesting, possibly varying head element at each level. Whereas in a list like:

[a, b, c]


we have:

head=a, sublist=(head=b, sublist=(head=c, sublist=[]))


with the number 3 we instead only have:

head=(), subnumber=(head=(), subnumber=(head=(), subnumber=zero))


where () is the zero-length tuple, the one and only member of its type. That is, the head here is a type of thing that cannot interestingly vary. Or we could just eliminate the head, and have:

subnumber=(subnumber=(subnumber=zero))


probably more familiar to you as:

succ (succ (succ zero))


In other words, where a list is inductively built up out of some interesting new datum and its "sublist" or tail, a number is inductively built up out of no interesting new datum and its "subnumber" or predecessor. In this light, a number can be seen as a kind of stunted or undernourished list.

Now, we had a strategy for encoding the list [a, b, c] as:

\f z. f a (f b (f c z))


So perhaps we can apply the same kind of idea to the number 3, and encode it as:

\f z. f   (f   (f   z))


In fact, there's a way of looking at this that makes it look incredibly natural. You may have encountered the idea of a composition of two functions f and g, written as f ○ g, or in Kapulet as f comp g. This is defined to be:

\x. f (g x)


For example, the operation that maps a number n to n2+1 is the composition of the successor function and the squaring function (first we square, then we take the successor).

The composition of a function f with itself, namely:

\x. f (f x)


is sometimes abbreviated as f2. Similarly, f ○ f ○ f or \x. f (f (f x)) is sometimes abbreviated as f3. f1 is understood to be just f itself, and f0 is understood to be the identity function.

So in proposing to encode the number 3 as:

\f z. f   (f   (f   z))


we are proposing to encode it as:

\f z. f3 z

(The fn isn't some bit of new Lambda Calculus syntax. It's just our metalanguage abbreviation for the same expressions we had before. It does help focus our attention on what we're essentially doing here.)

And indeed this is the Church encoding of the numbers:

0 ≡ \f z. z   ; <~~> \f z. I z, or \f z. f0 z
1 ≡ \f z. f z                 ; or \f z. f1 z
2 ≡ \f z. f (f z)             ; or \f z. f2 z
3 ≡ \f z. f (f (f z))         ; or \f z. f3 z
...

The encoding for 0 is what we also proposed as the encoding for [] and for false. Don't read too much into this.

Given the above, can you figure out how to define the succ function? We already worked through the definition of cons, and this is just a simplification of that, so you should be able to do it. We'll make it a homework.

We saw that using the cons operation (Kapulet's &) and [] as arguments to fold_right over a list give us back that very same list. In the Lambda Calculus, that comes to the following:

[a, b, c] cons []


with the appropriate lambda terms substituted throughout, just reduces to the same lambda term we used to encode [a, b, c].

In the same spirit, what do you expect this will reduce to:

3 succ 0


Since 3 is \f z. f (f (f z)), it should reduce to whatever:

succ (succ (succ 0))


does. And if we define succ properly, we should expect that to give us the same lambda term that we used to encode 3 in the first place.

We'll look at other encodings of lists and numbers later.