[Haskell-cafe] help in tree folding
Brent Yorgey
byorgey at gmail.com
Thu May 8 07:18:39 EDT 2008
On Wed, May 7, 2008 at 6:28 AM, patrik osgnach <patrik.osgnach at gmail.com>
wrote:
> 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 at 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
>
If it typechecks and you have used all the parameters, then it is probably
correct! =) That may sound trite, but it is often true.
-Brent
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