Previously I've posted about the lambda calculus and Church numbers. We'd shown how we can encode numbers as functions using the Church encoding, but we'd not really shown how we could do anything with those numbers.
But before we get into it, let's clear up some stuff about brackets...
Just as it's easier to write $\lambda nfx.$ than $\lambda n.\lambda f.\lambdax.$ because we make the assumption that each application of the function returns a new function, so there is a way of writing out function application without having to use lots of parentheses.1
Where we would normally write
$$ f(x) $$
with parentheses, we could instead write
$$ f\ x $$
under the assumption that each argument associates to the one on the left. So if we had
$$ ((f(x)\ y)\ z) $$
we can write it as
$$ f\ x\ y\ z $$
and something like
$$ (g(x)\ f(x)) $$
$$ g\ x\ (f\ x) $$
As we still need the parentheses to make sure that the $f$ and $x$ get bundled together. We'll need this convention as we go on as things are going to get a little more parenthesis-heavy.
OK, let's get back to the arithmetic.
Say we have the number three:
$$ three \ \equiv \ \lambda f \lambda x.\ f\ (f\ (f x)) $$
(the function $f$ applied to $x$ three times)
And we wanted to get to the number four:
$$ four \ \equiv \ \lambda f \lambda x.\ f\ (f (f\ (f x))) $$
(the function $f$ applied to $x$ four times)
How do we get from $three$ to $four$? Well, the difference is that we just need to apply $f$ one more time.
$$ four \ \equiv \ f\ three $$
We can encode the idea of applying $f$ one more time into a lambda function. We could call it $add-one$ or $increment$ but lets go with $succ$ for 'successor'.
$$ succ \ \equiv \ \lambda n. \lambda f. \lambda x.\ f\ (n\ f\ x) $$
The $n$ is the number we're adding one to - we need to bind in the values of $f$ and $x$ in to the function because they'll need to have $n$ applied to them before we can apply $f$ in the one extra time.
Another way to think of this is that the general signature for a number is $\lambda f. \lambda x.$, and that when we apply $succ$ to a number, we need to get back another number - something else with the signature of $\lambda f. \lambda\ x.$
So the signature of $succ$ - and consequently any unary operation on a number - is $\lambda n.\lambda f.\lambda x$, where $n$ is the number being changed.
In Clojure that looks like:
(fn [n] (fn [f] (fn [x] (f ((n f) x))))))
Yeah, it's a bit verbose in comparison to the lambda calculus version.2 All those parentheses, while great for being explicit about which functions get applied to what, makes it a bit tough on the eyes.
What about Haskell?
\n f x -> f (n f x)
Bit more like the original, eh? Haskell has currying and left-association baked in to its syntax so its lambda expressions look almost exactly the same as the lambda calculus ones. You can see why it's so popular.3
Let's see if we can define addition.
First off, $addition$ is an operation that takes two arguments, two numbers. So we know it needs to look something like:
$$ \lambda m. \lambda n. \lambda f. \lambda x. $$
Where $m$ and $n$ are the numbers being added together. Now all we need to do is work out what comes after the dot.
We could define it in terms of $succ$ - all we need to do is apply $succ$ $m$ many times to $n$:
$$ \lambda m.\lambda n.\lambda f.\lambda x.\ m\ succ\ n\ f\ x $$
And this works,4 but we could probably write something both more intuitive and simpler.
What do we want as the result of $add$? We want a function that applies $f$ to $x$ $n$ many times, and the applies $f$ to the result of that $m$ many times.
$$ add\ (\lambda fx.\ f\ (f\ x))\ (\lambda fx.\ f\ (f\ (f\ x))) = \lambda fx.\ f\ (f\ (f\ (f\ (f\ x)))) $$
We can just write that out with the variables we've been given - first apply $f$ to $x$, $n$ many times.
$$ n\ f\ x $$
and then apply $f$ to that result $m$ many times
$$ m\ f\ (n\ f\ x) $$
$$ add\ \equiv\ \lambda n.\lambda m.\lambda f.\lambda x.\ m\ f\ (n\ f\ x) $$
The order of $n$ and $m$ doesn't matter as they're just the order in which the number of $f$s are applied.5
We've used the word 'times' a lot here when talking about the application of $f$ onto $x$s in the above. But now we'll have to deal with real multiplication.
Before you try and reach at an answer, step back a little and ask yourself what the result ought to be, and what the Church arithmetic way of describing it would be.
Say we had the numbers two and three. If I was back in primary school I'd say that the reason that multiplying them together made six was because six was 'two lots of three' or 'three lots of two'.
So when I want to put this into Church arithmetic, I feel like saying 'six is the application of three lots of the application of two lots of the application of $f$ onto $x$'. Which is a mouthful, for sure, but looks like
$$ six\ \equiv\ \lambda f.\lambda x.\ ((three\ (two\ f))\ x) $$
or, without the parentheses
$$ six\ \equiv\ \lambda f.\lambda x.\ three\ (two\ f)\ x $$
$two\ f$ is a function that applies $f$ two times to whatever it's next argument is. $three\ (two\ f)$ will apply $two\ f$ to its next argument three times. So it will apply it $3\ \times\ 2$ times - 6 times.
And so now we can move from the particular case to the general case; multiplication is:
$$ mult\ \equiv\ \lambda m.\lambda n.\lambda f.\lambda x.\ m\ (n\ f)\ x $$
"$m$ lots of ($n$ lots of $f$) applied to $x$", which is still a mouthful but
So what could exponentiation be? Well, the first thing we know is that this time, order is going to be important - $2^3$ is not the same as $3^2$.
Next, what does exponentiation mean? I mean, really mean? When we did multiplication we saw us doing 'two lots of (three lots of $f$)'. But now we need to do 'two lots of something' three times. The 'three' part has to apply, not to the number of times we do an $f$, nor the number of times we do '$n$ lots of $f$'. But rather it needs to be the number of times we do $n$ to itself.
So if 'three' is the application of $f$ three times to $x$, we can say that $2^3$ is the application of $two$ three times to $f\ x$.
Even. Bigger. Woah.
Another way to look at it: a Church number is already encoding some of the
behaviour of exponentiation. When we use
x we can
think of the number
n acting as $inc^n$ -
inc done to itself
n many times.
This is more explicit if we try it with something other than increment - say
double, aka 'times two'. Let's do it in Haskell - but please feel free to pick
any language you like.
let timesTwo = \x -> 2 * x let four = \f x -> f(f(f(f x))) four timesTwo 1 -- 16
Four lots of
timesTwo is 16; all we need to do is to use the number two
instead, and apply the result to an
f and an
let two = \f x -> f(f x) four two succ 0 -- 16
So function for exponentiation -
m to the power of
n - is:
$$ exp\ \equiv\ \lambda m.\lambda n.\lambda f.\lambda x.\ n\ m\ f\ x $$
But remember $\eta$-reduction? We can just go directly to:
$$ exp\ \equiv\ \lambda m.\lambda n.\ n\ m $$
This is because you know the function you're left with after you've applied $n$ to $m$ is a number - will take an $f$ and an $x$ - you don't need to explicitly bind them in the outer function just in order to pass them unchanged to the inner one.
But that's just a nicety. The important thing is... we've finished!
This post looked at some simple arithmetic using Church numerals. We saw successor
$$ succ\quad \equiv\quad \lambda n.\lambda f.\lambda x.\ f\ (n\ f\ x) $$
$$ add\quad \equiv\quad \lambda m.\lambda n.\lambda f.\lambda x.\ m\ f\ (n\ f\ x) $$
$$ mult\quad \equiv\quad \lambda m.\lambda n.\lambda f.\lambda x.\ m\ (n\ f)\ x $$
$$ exp\quad \equiv\quad \lambda m.\lambda n.\ m\ n $$
An interesting relationship between the last three: the $f$ moves along to the right as the operation becomes 'bigger'.
Next post we'll be taking a short break from arithmetic to take a look at logic using the lambda calculus.
And I'm speaking as a mad Lisp fan, lover of parens where ever they are. ↩
But still terse compared to the mess we'd get in Python. Or Ruby. Yeah, don't try it in Ruby. Oh, and I guess we could use the short hand anonymous function syntax, but I think that'd look even messier... ↩
For functional programming that is. ↩
Get your pencil and paper out if you want to prove it! ↩