2 Feb
2009
2 Feb
'09
5:41 p.m.
2009/2/2 Luke Palmer
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 -- Dan (PS I think this is hard. But my brain might be misfiring so it might be trivial.)