[Haskell-cafe] type metaphysics

[Joachim Breitner]

Hi,

Am Montag, den 02.02.2009, 14:41 -0800 schrieb Dan Piponi:
> 2009/2/2 Luke Palmer <lrpalmer at 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
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