[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
>
> How about
>
> > 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
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