[Haskell-beginners] Understanding recursion in Haskell.
Sean Lee
sean.ysl at gmail.com
Wed Mar 18 04:13:30 EDT 2009
This function drops the first n elements from xs.
> If n <= 0 || null xs
> then xs
if n is less than or equal to 0, we are done dropping the elements
that we intended to drop.
If xs is null, the list is empty and we have no more elements to drop.
So, either way, we can just return xs and stop the recursion.
> else myDrop (n - 1) (tail xs)
Otherwise, we drop the first element from xs and pass that to myDrop.
Of course, when we call myDrop, we give n-1 instead of n because we
dropped the first element from xs.
I think it will be easier if we take an example.
Let's say we want to evaluate myDrop 3 [1,2,3,4,5], expecting the
output to be [4,5].
3 is greater than 0 and [1,2,3,4,5] is not empty.
So, it evaluates to
myDrop (3 - 1) (tail [1,2,3,4,5])
2 is greater than 0 and [2,3,4,5] is not empty
So, it evaluates to
myDrop (2 - 1) (tail [2,3,4,5])
1 is greater than 0 and [3,4,5] is not empty
So, it evaluates to
myDrop (1 -1) (tail [3,4,5])
0 is now equal to 0,
so it evaluates to
[4,5].
Sean
On Wed, Mar 18, 2009 at 6:36 PM, <The_Polymorph at rocketmail.com> wrote:
>
> Hi Adrian,
>
> Thanks for the explanations. Could we perhaps examine one recursive example in detail, i.e., step-by-step, as I'm still confused? Maybe the following program from chapter 2 of
> http://book.realworldhaskell.org?
>
> myDrop n xs = if n <= 0 || null xs
> then xs
> else myDrop (n - 1) (tail xs)
>
>
> Danke,
>
> Caitlin
>
>
> --- On Wed, 3/18/09, Adrian Neumann <aneumann at inf.fu-berlin.de> wrote:
>
> From: Adrian Neumann <aneumann at inf.fu-berlin.de>
> Subject: Re: [Haskell-beginners] Understanding recursion in Haskell.
> To: beginners at haskell.org
> Date: Wednesday, March 18, 2009, 12:05 AM
>
>
> Am 18.03.2009 um 06:28 schrieb Caitlin:
>
>>
>> Hi.
>>
>> As a Haskell
> beginner, I was wondering if someoneone could explain how the following programs function (pardon the pun)?
>>
>
>
> This function takes some type which has an ordering defined, i.e. you can compare its elements to one another
>
>> maximum' :: (Ord a) => [a] -> a
>
> it doesn't work for an empty list
>
>> maximum' [] = error "maximum of empty list"
>
> the maximum of a one element list is the lone element. this is the base case which will be eventually reached by the recursion
>
>> maximum' [x] = x
>
> should the list have more than one element
>
>> maximum' (x:xs)
>
> compare the first element to the maximum of the other elements. if it's greater, it's the maximum
>
>> | x > maxTail = x
>
> otherwise the maximum of the other elements is the maximum of the whole list
>
>> | otherwise = maxTail
>
> how to compute the maximum of the other elements? just use
> this function again. after a while we will only have one element left and reach the base case above.
>
>> where maxTail = maximum' xs
>>
>>
>
> This function takes a number and a list of some type a
>
>> take' :: (Num i, Ord i) => i -> [a] -> [a]
>
> first, ignore the list and check whether n is <= 0. in this case return an empty list. this is the base case, that's eventually reached by the recursion
>
>> take' n _
>> | n <= 0 = []
>
> otherwise, check if the list is empty. this is another base case.
>
>> take' _ [] = []
>
> if neither n<=0 or the list empty, take the first element, x, and put it on front of the prefix of length (n-1) of the other elements. use take' again, to get that prefix. after a while either n is 0 or there are no more elements in the list and we reach the base case
>
>> take' n (x:xs) = x : take' (n-1) xs
>>
>
>>
>
> Take two lists
>
>>
>> zip' :: [a] -> [b] -> [(a,b)]
>
> if either one of them is empty, stop
>
>> zip' _ [] = []
>> zip' [] _ = []
>
> otherwise prepend a tuple, build from the two first elements to the zipped list of the other elements. after a while one of the lists should become empty and the base case is reached.
>
>> zip' (x:xs) (y:ys) = (x,y):zip' xs ys
>>
>>
>>
>> quicksort :: (Ord a) => [a] -> [a]
>
> empty list -> nothing to do
>
>> quicksort [] = []
>> quicksort (x:xs) =
>
> otherwise take the first element of the list and use it to split the list in two halves. one with all the elements that are smaller or equal than x, the other one with all those that are bigger. now sort them and put x in the middle. that should give us a sorted whole. how to sort them? just use quicksort again! after some splitting the lists will become empty
> and the recursion stops.
>
>> let smallerSorted = quicksort [a | a <- xs, a <= x]
>> biggerSorted = quicksort [a | a <- xs, a > x]
>> in smallerSorted ++ [x] ++ biggerSorted
>
>
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>
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--
Sean Lee
PhD Student
Programming Language and Systems Research Group
School of Computer Science and Engineering
University of New South Wales
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