On Mon, Feb 02, 2009 at 02:41:36PM -0800, Dan Piponi wrote: > 2009/2/2 Luke Palmer <lrpal...@gmail.com>: > > > But Nat ~> Bool is computably uncountable, meaning there is no injective > > (surjective?) > > function Nat ~> (Nat ~> Bool), by the diagonal argument above. > > Given that the Haskell functions Nat -> Bool are computably > uncountable, you'd expect that for any Haskell function (Nat -> Bool) > -> Nat there'd always be two elements that get mapped to the same > value. > > 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. Regards, Reid _______________________________________________ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe