# Encoding Booleans, Tuples, Lists, and Numbers #
-The lambda calculus can represent any computable function?
+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:
+We'll start with the `if ... then ... else ...` construction we saw last week:
if M then N else L
~~> ((\x x) L)
~~> L
-So have seen our first major encoding in the lambda calculus:
+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
## Tuples ##
-In class, we also showed you how to encode a tuple in the Lambda Calculator. 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.)
+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:
letrec
fold_right (f, z) xs = case xs of
- [] then zs;
+ [] 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 x z. ...` rather than the uncurried `lambda (x, z). ...` that I'm expecting here. We will be shifting to the curried form ourselves momentarily.)
+(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.
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 = \c cs. \f z. f c (cs f z)
+ 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 `c` gets bound to our `c`, and its `cs` 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 be `z`. If we just simply said `z` here, then our `cons` function would always be giving us back a singleton list of the form `\f z. f x z` for some `x`. That's what we want in this case, but not in the general case. What we'd like is to use the second argument we fed to `cons`, here `[]` or `\f z. z`, and have it reduce to `z`, with the hope that the same strategy wuld 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 been given an `f` and a `z`, supplied by the consumer or user of the sequence `[c]` that we're building. So let's just give them to `\f z. z` as its arguments, and it gives us `z`. In other words, we make `SOMETHING` be `cs f z`.
+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:
-Formally:
+ \d ds. \f z. f d (ds f z)
+
+Let's step through it formally. Given our definitions:
cons c []
is:
- (\c cs. \f z. f c (cs f z)) c (\f z. z) ~~>
- (\cs. \f z. f c (cs f z)) (\f z. z) ~~>
+ (\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
is:
- (\c cs. \f z. f c (cs f z)) b (\f z. f c z) ~~>
- (\cs. \f z. f b (cs f z)) (\f z. f c z) ~~>
+ (\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 values over the list elements in the appropriate order.
+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.
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, z) xs
+ map g xs
in Kapulet syntax, turns into this in our lambda evaluator:
let empty = \f z. z in
- let cons = \c cs. \f z. f c (cs f 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 = \c cs. \f z. f c (cs f 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
\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 values over `[g a, g b, g c]`, which is `map g [a, b, c]`.
+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 ##
-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.
+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 <code>f ○ g</code>, or in Kapulet as `f comp g`. This is defined to be:
+
+ \x. f (g x)
+
+The composition of a function `f` with itself, namely:
+
+ \x. f (f x)
+
+is sometimes abbreviated as <code>f<sup>2</sup></code>. Similarly, <code>f ○ f ○ f</code> or `\x. f (f (f x))` is sometimes abbreviated as <code>f<sup>3</sup></code>. <code>f<sup>1</sup></code> is understood to be just `f` itself, and <code>f<sup>0</sup></code> 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:
+
+<code>\f z. f<sup>3</sup> z</code>
+
+(The <code>f<sup><em>n</em></sup></code> 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:
+
+<code>0 ≡ \f z. I z ; or \f z. f<sup>0</sup> z</code>
+<code>1 ≡ \f z. f z ; or \f z. f<sup>1</sup> z</code>
+<code>2 ≡ \f z. f (f z) ; or \f z. f<sup>2</sup> z</code>
+<code>3 ≡ \f z. f (f (f z)) ; or \f z. f<sup>3</sup> z</code>
+<code>...</code>
+
+The encoding for `0` is equivalent to `\f z. z`, which we've 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.
+
-*More coming...*
+We'll look at other encodings of lists and numbers later.