On Mon, Feb 2, 2009 at 4:18 PM, Reid Barton <rwbar...@math.harvard.edu>wrote:
> > So here's a programming challenge: write a total function (expecting > > total arguments) toSame :: ((Nat -> Bool) -> Nat) -> (Nat -> Bool,Nat > > -> Bool) that finds a pair that get mapped to the same Nat. > > > > Ie. f a==f b where (a,b) = toSame f > > (Warning: sketchy argument ahead.) Let f :: (Nat -> Bool) -> Nat be a > total function and let g0 = const True. The application f g0 can > only evaluate g0 at finitely many values, so f g0 = f (< k) for any k > larger than all these values. So we can write > > > toSame f = (const True, head [ (< k) | k <- [1..], f (const True) == f (< > k) ]) > > and toSame is total on total inputs. Well done! That's not sketchy at all! There is always such a k (when the result type of f has decidable equality) and it is the "modulus of uniform continuity" of f. This is computable directly, but the implementation you've provided might come up with a smaller one that still works (since you only need to differentiate between const True, not all other streams). I guess I should hold off on conjecturing the impossibility of things... :-) Luke
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