# [Haskell-cafe] least fixed points above something

Dan Doel dan.doel at gmail.com
Fri Mar 20 20:07:47 EDT 2009

```On Friday 20 March 2009 5:23:37 am Ryan Ingram wrote:
> On Fri, Mar 20, 2009 at 1:01 AM, Dan Doel <dan.doel at gmail.com> wrote:
> > However, to answer Luke's wonder, I don't think fixAbove always finds
> > fixed points, even when its preconditions are met. Consider:
> >
> >  f []     = []
> >  f (x:xs) = x:x:xs
> >
> >  twos = 2:twos
>
>
> > fixAbove f x = x `lub` fixAbove f (f x)
>
> Probably doesn't work (or give good performance) with the current
> implementation of "lub" :)
>
> But if "lub" did work properly, it should give the correct answer for
> fixAbove f (2:undefined).

This looked good to me at first, too (assuming it works), and it handles my
first example. But alas, we can defeat it, too:

f (x:y:zs) = x:y:x:y:zs
f zs       = zs

Now:

f (2:_|_) = _|_
f _|_     = _|_

fix f = _|_

fixAbove f (2:_|_) = 2:_|_ `lub` _|_ `lub` _|_ ...
= 2:_|_

Which is not a fixed point of f. But there are actually multiple choices of
fixed points above 2:_|_

f (2:[]) = 2:[]
forall n. f (cycle [2,n]) = cycle [2,n]

I suppose the important distinction here is that we can't arrive at the fixed
point simply by iterating our function on our initial value (possibly
infinitely many times). I suspect this example won't be doable without an
oracle of some kind.

Ah well.
-- Dan
```