Question

The fixed point combinator doesn’t always produce the right answer given the definition:

fix f = f (fix f)

The following code does not terminate:

fix (\x->x*x) 0

Of course, fix can’t always produce the right answer, but I was wondering, can this be improved?

Certainly for the above example, one can implement some fix that looks like

fix f x | f x == f (f x)  = f x
        | otherwise       = fix f (f x)

and gives the correct output.

What is the reason the above definition (or something even better, as this one only handle function with 1 parameter) is not used instead?

Answer 1

Fixed point combinator finds the least-defined fixed point of a function, which is ⊥ in your case (non-termination indeed is undefined value).

You can check, that in your case

(\x -> x * x) ⊥ = ⊥

i.e. ⊥ really is fixed point of \x -> x * x.

As for why is fix defined that way: the main point of fix is to allow you use anonymous recursion and for that you do not need more sophisticated definition.