[Haskell-beginners] foldr for Nats
kane96 at gmx.de
kane96 at gmx.de
Fri Feb 19 15:24:15 EST 2010
nobody here, who can tell me how it can work?
-------- Original-Nachricht --------
> Datum: Thu, 04 Feb 2010 21:58:45 +0100
> Von: kane96 at gmx.de
> An: Stephen Tetley <stephen.tetley at gmail.com>, Daniel Fischer <daniel.is.fischer at web.de>
> CC: Beginners at haskell.org
> Betreff: Re: [Haskell-beginners] foldr for Nats
> @Stephan thanks for the code so far.
> @Daniel it should look like foldrNat a b S(S(S(S(Z)))) gives a(a(a(a(b))))
>
>
> The thing is the definition have to be
> -------- Original-Nachricht --------
> > Datum: Mon, 1 Feb 2010 21:56:14 +0000
> > Von: Stephen Tetley <stephen.tetley at gmail.com>
> > An: kane96 at gmx.de
> > CC: Beginners at haskell.org
> > Betreff: Re: [Haskell-beginners] foldr for Nats
>
> > Hello
> >
> > I'm suspecting this isn't homework as you've waited a week so would
> > presumably have missed a deadline.
> >
> > As Daniel Fischer wrote, one view of folds is that they replace the
> > constructors of a data type, code follows...
> >
> >
> > data Nat = Z | S Nat deriving (Eq,Ord,Show)
> >
> >
> > -- Look at the type of foldr...
> >
> > -- *GHCi> :t foldr
> > -- foldr :: (a -> b -> b) -> b -> [a] -> b
> >
> > -- It has 2 'constructor replacements':
> > -- (a -> b -> b) & b
> >
> > -- Replacing Z is easy, we can get some code to compile
> > -- by avoiding the hard bit with a wildcard pattern "_"...
> >
> > foldrNat1 :: unknown -> b -> Nat -> b
> > foldrNat1 _ b Z = b
> >
> > -- What to do about the constructor (S ..) takes a bit more
> > -- thought or at least some experimenting. I'll do the later...
> >
> > -- One thing to try, is to simply translate foldr with as few
> > -- changes as possible:
> >
> > -- foldr :: (a -> b -> b) -> b -> [a] -> b
> > -- foldr _ z [] = z
> > -- foldr f z (x:xs) = f x (foldr f z xs)
> >
> > -- Unfortunately this leads to a problem:
> >
> > foldrNat2 :: (Nat -> b -> b) -> b -> Nat -> b
> > foldrNat2 f b Z = b -- Z case is the same as before
> > foldrNat2 f b (S n) = f undefined (foldrNat2 f b n) -- Arggh! undefined
> >
> > -- undefined is useful for prototyping, but its a real
> > -- problem for running code!
> >
> > -- Actually I had another problem as well...
> > --
> > -- The difference between Nat and [a] is that List 'carries' some data
> > -- therefore (Nat -> b -> b) on Nat is not equivalent to (a -> b -> b)
> > -- on [a].
> >
> > -- So rather than change the type signature first, get rid of the
> > -- undefined and see what happens
> >
> > foldrNat3 f b Z = b
> > foldrNat3 f b (S n) = f (foldrNat3 f b n)
> >
> > -- *GHCi> :t foldrNat3
> > -- > (t -> t) -> t -> Nat -> t
> >
> > -- GHCi likes to call type variables t, but the signature is equal to
> >
> > -- foldrNat3 :: (b -> b) -> b -> Nat -> b
> >
> >
> > -- This looks promising - it typechecks!
> > -- So try a test:
> >
> > fromNat :: Nat -> Int
> > fromNat n = foldrNat3 (+1) 0 n
> >
> > demo1 = fromNat (S (S (S Z))) -- 3 ??
> >
> > -- By experimenting we seem to have a good answer,
> > -- other people might prefer a more rigorous proof though.
>
> --
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