How do I use fix, and how does it work?

Question

I was a bit confused by the documentation for fix (although I think I understand what it's supposed to do now), so I looked at the source code. That left me more confused:

fix :: (a -> a) -> a
fix f = let x = f x in x

How exactly does this return a fixed point?

I decided to try it out at the command line:

Prelude Data.Function> fix id
...

And it hangs there. Now to be fair, this is on my old macbook which is kind of slow. However, this function can't be too computationally expensive since anything passed in to id gives that same thing back (not to mention that it's eating up no CPU time). What am I doing wrong?

Answer 1

You are doing nothing wrong. fix id is an infinite loop.

When we say that fix returns the least fixed point of a function, we mean that in the domain theory sense. So fix (\x -> 2*x-1) is not going to return 1, because although 1 is a fixed point of that function, it is not the least one in the domain ordering.

I can't describe the domain ordering in a mere paragraph or two, so I will refer you to the domain theory link above. It is an excellent tutorial, easy to read, and quite enlightening. I highly recommend it.

For the view from 10,000 feet, fix is a higher-order function which encodes the idea of recursion. If you have the expression:

let x = 1:x in x

Which results in the infinite list [1,1..], you could say the same thing using fix:

fix (\x -> 1:x)

(Or simply fix (1:)), which says find me a fixed point of the (1:) function, IOW a value x such that x = 1:x... just like we defined above. As you can see from the definition, fix is nothing more than this idea -- recursion encapsulated into a function.

It is a truly general concept of recursion as well -- you can write any recursive function this way, including functions that use polymorphic recursion. So for example the typical fibonacci function:

fib n = if n < 2 then n else fib (n-1) + fib (n-2)

Can be written using fix this way:

fib = fix (\f -> \n -> if n < 2 then n else f (n-1) + f (n-2))

Exercise: expand the definition of fix to show that these two definitions of fib are equivalent.

But for a full understanding, read about domain theory. It's really cool stuff.

Answer 2

I don't claim to understand this at all, but if this helps anyone...then yippee.

Consider the definition of fix. fix f = let x = f x in x. The mind-boggling part is that x is defined as f x. But think about it for a minute.

x = f x

Since x = f x, then we can substitute the value of x on the right hand side of that, right? So therefore...

x = f . f $ x -- or x = f (f x)
x = f . f . f $ x -- or x = f (f (f x))
x = f . f . f . f . f . f . f . f . f . f . f $ x -- etc.

So the trick is, in order to terminate, f has to generate some sort of structure, so that a later f can pattern match that structure and terminate the recursion, without actually caring about the full "value" of its parameter (?)

Unless, of course, you want to do something like create an infinite list, as luqui illustrated.

TomMD's factorial explanation is good. Fix's type signature is (a -> a) -> a. The type signature for (\recurse d -> if d > 0 then d * (recurse (d-1)) else 1) is (b -> b) -> b -> b, in other words, (b -> b) -> (b -> b). So we can say that a = (b -> b). That way, fix takes our function, which is a -> a, or really, (b -> b) -> (b -> b), and will return a result of type a, in other words, b -> b, in other words, another function!

Wait, I thought it was supposed to return a fixed point...not a function. Well it does, sort of (since functions are data). You can imagine that it gave us the definitive function for finding a factorial. We gave it a function that dind't know how to recurse (hence one of the parameters to it is a function used to recurse), and fix taught it how to recurse.

Remember how I said that f has to generate some sort of structure so that a later f can pattern match and terminate? Well that's not exactly right, I guess. TomMD illustrated how we can expand x to apply the function and step towards the base case. For his function, he used an if/then, and that is what causes termination. After repeated replacements, the in part of the whole definition of fix eventually stops being defined in terms of x and that is when it is computable and complete.

Answer 3

You need a way for the fixpoint to terminate. Expanding your example it's obvious it won't finish:

fix id
--> let x = id x in x
--> id x
--> id (id x)
--> id (id (id x))
--> ...

Here is a real example of me using fix (note I don't use fix often and was probably tired / not worrying about readable code when I wrote this):

(fix (\f h -> if (pred h) then f (mutate h) else h)) q

WTF, you say! Well, yes, but there are a few really useful points here. First of all, your first fix argument should usually be a function which is the 'recurse' case and the second argument is the data on which to act. Here is the same code as a named function:

getQ h
      | pred h = getQ (mutate h)
      | otherwise = h

If you're still confused then perhaps factorial will be an easier example:

fix (\recurse d -> if d > 0 then d * (recurse (d-1)) else 1) 5 -->* 120

Notice the evaluation:

fix (\recurse d -> if d > 0 then d * (recurse (d-1)) else 1) 3 -->
let x = (\recurse d -> if d > 0 then d * (recurse (d-1)) else 1) x in x 3 -->
let x = ... in (\recurse d -> if d > 0 then d * (recurse (d-1)) else 1) x 3 -->
let x = ... in (\d -> if d > 0 then d * (x (d-1)) else 1) 3

Oh, did you just see that? That x became a function inside our then branch.

let x = ... in if 3 > 0 then 3 * (x (3 - 1)) else 1) -->
let x = ... in 3 * x 2 -->
let x = ... in 3 * (\recurse d -> if d > 0 then d * (recurse (d-1)) else 1) x 2 -->

In the above you need to remember x = f x, hence the two arguments of x 2 at the end instead of just 2.

let x = ... in 3 * (\d -> if d > 0 then d * (x (d-1)) else 1) 2 -->

And I'll stop here!

Answer 4

How I understand it is, it finds a value for the function, such that it outputs the same thing you give it. The catch is, it will always choose undefined (or an infinite loop, in haskell, undefined and infinite loops are the same) or whatever has the most undefineds in it. For example, with id,

λ <*Main Data.Function>: id undefined
*** Exception: Prelude.undefined

As you can see, undefined is a fixed point, so fix will pick that. If you instead do (\x->1:x).

λ <*Main Data.Function>: undefined
*** Exception: Prelude.undefined
λ <*Main Data.Function>: (\x->1:x) undefined
[1*** Exception: Prelude.undefined

So fix can't pick undefined. To make it a bit more connected to infinite loops.

λ <*Main Data.Function>: let y=y in y
^CInterrupted.
λ <*Main Data.Function>: (\x->1:x) (let y=y in y)
[1^CInterrupted.

Again, a slight difference. So what is the fixed point? Let us try repeat 1.

λ <*Main Data.Function>: repeat 1
[1,1,1,1,1,1, and so on
λ <*Main Data.Function>: (\x->1:x) $ repeat 1
[1,1,1,1,1,1, and so on

It is the same! Since this is the only fixed point, fix must settle on it. Sorry fix, no infinite loops or undefined for you.