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@xxxxxxxxx>
> > 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 "+âxyz"
> >> 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
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'
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.
> 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
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