Hi, Am Montag, den 02.02.2009, 14:41 -0800 schrieb Dan Piponi: > 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 > -- > Dan > > (PS I think this is hard. But my brain might be misfiring so it might > be trivial.)
> toSame _ = (const True, const True) ;-) Joachim -- Joachim "nomeata" Breitner mail: m...@joachim-breitner.de | ICQ# 74513189 | GPG-Key: 4743206C JID: nome...@joachim-breitner.de | http://www.joachim-breitner.de/ Debian Developer: nome...@debian.org
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