Daniel Fischer ha scritto:

Am Dienstag, 6. Mai 2008 22:40 schrieb patrik osgnach:

Brent Yorgey ha scritto:

On Tue, May 6, 2008 at 8:20 AM, patrik osgnach <patrik.osgnach@gmail.com>

wrote:

Hi. I'm learning haskell but i'm stuck on a generic tree folding
exercise. i must write a function of this type
treefoldr::(Eq a,Show a)=>(a->b->c)->c->(c->b->b)->b->Tree a->c
Tree has type
data (Eq a,Show a)=>Tree a=Void | Node a [Tree a] deriving (Eq,Show)
as an example treefoldr (:) [] (++) [] (Node '+' [Node '*' [Node 'x' [],
Node 'y' []], Node 'z' []])
must return "+∗xyz"
any help?
(sorry for my bad english)

Having a (Tree a) parameter, where Tree is defined as an algebraic data
type, also immediately suggests that you should do some pattern-matching
to break treefoldr down into cases:

treefoldr f y g z Void = ?
treefoldr f y g z (Node x t) = ?

-Brent

so far i have tried
treefoldr f x g y Void = x

Yes, nothing else could be done.

treefoldr f x g y (Node a []) = f a y

Not bad. But actually there's no need to treat nodes with and without children differently.
Let's see:

treefoldr f x g y (Node v ts)

should have type c, and it should use v. We have
f :: a -> b -> c
x :: c
g :: c -> b -> b
y :: b
v :: a.

The only thing which produces a value of type c using a value of type a is f, so we must have

treefoldr f x g y (Node v ts) = f v someExpressionUsing'ts'

where

someExpressionUsing'ts' :: b.

The only thing we have which produces a value of type b is g, so
someExpressionUsing'ts' must ultimately be g something somethingElse.
Now take a look at the code and type of foldr, that might give you the idea.

Cheers,
Daniel

treefoldr f x g y (Node a (t:ts)) = treefoldr f x g (g (treefoldr f x g
y t) y) (Node a ts)
but it is incorrect. i can't figure out how to build the recursive call
thanks for the answer
Patrik

thanks for the tip.
so, if i have understood correctly i have to wirite something like:
treefoldr f x g y (Node a ts) = f a (g (treefoldr f x g y (head ts)) (g (treefoldr f x g y (head (tail ts)) (g ...
it looks like a list foldr so...

treefoldr f x g y Void = x

treefoldr f x g y (Node a ts) = f a (foldr (g) y (map (treefoldr f x g y) ts))
it seems to work. i'm not yet sure it is correct but is better than nothing
thanks to you all. now i will try to write a treefoldl