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 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 <[hidden email]> wrote: From: Adrian Neumann <[hidden email]> Subject: Re: [Haskell-beginners] Understanding recursion in Haskell. To: [hidden email] 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 -----Inline Attachment Follows----- _______________________________________________ Beginners mailing list [hidden email] http://www.haskell.org/mailman/listinfo/beginners
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 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,  <[hidden email]> 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 <[hidden email]> wrote: > > From: Adrian Neumann <[hidden email]> > Subject: Re: [Haskell-beginners] Understanding recursion in Haskell. > To: [hidden email] > 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 > > > -----Inline Attachment Follows----- > > _______________________________________________ > Beginners mailing list > [hidden email] > http://www.haskell.org/mailman/listinfo/beginners> > > > > _______________________________________________ > Beginners mailing list > [hidden email] > http://www.haskell.org/mailman/listinfo/beginners> -- 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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 In reply to this post by Caitlin-2 On 18 Mar 2009, at 08:36, [hidden email] 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) In this example, I'm gonna use the "~>" symbol to mean "reduces to",   that is, if you perform one more evaluation step, you get this.  We're   going to try and evaluate myDrop 2 [1,2,3,4]     myDrop 2 [1,2,3,4]     { Reduce myDrop 2 [1,2,3,4] to its right hand side as defined   above } ~> if 2 <= 0 || null [1,2,3,4] then [1,2,3,4] else myDrop (2-1) (tail   [1,2,3,4])     { Reduce 2 <= 0 to False } ~> if False || null [1,2,3,4] then [1,2,3,4] else myDrop (2-1) (tail   [1,2,3,4])     { Reduce null [1,2,3,4] to False ~> if False || False then [1,2,3,4] else myDrop (2-1) (tail [1,2,3,4])      { Reduce False || False to False } ~> if False then [1,2,3,4] else myDrop (2-1) (tail [1,2,3,4])     { Reduce the if expression to its else branch }} ~> myDrop (2-1) (tail [1,2,3,4])     {Reduce myDrop (2-1) (tail [1,2,3,4]) to its right hand side as   defined above } ~> if n <= 0 || null xs then xs else myDrop (n - 1) (tail xs) where n   = (2-1); xs = tail [1,2,3,4]     { Reduce 2 - 1 to 1 } ~> if n <= 0 || null xs then xs else myDrop (n - 1) (tail xs) where n   = 1; xs = tail [1,2,3,4]     { Reduce 1 <= 0 to False } ~> if False || null xs then xs else myDrop (n - 1) (tail xs) where n =   1; xs = tail [1,2,3,4]     { Remove superfluous definition of n (there's only one use of it   now) } ~> if False || null xs then xs else myDrop (1 - 1) (tail xs) where xs   = tail [1,2,3,4]     { Reduce tail [1,2,3,4] to [2,3,4] } ~> if False || null xs then xs else myDrop (1 - 1) (tail xs) where xs   = [2,3,4]     { Reduce null [2,3,4] to False } ~> if False || False then xs else myDrop (1 - 1) (tail xs) where xs =   [2,3,4]     { Reduce False || False to False } ~> if False then xs else myDrop (1 - 1) (tail xs) where xs = [2,3,4]     { Reduce if expression to its else branch } ~> myDrop (1 - 1) (tail xs) where xs = [2,3,4]     { Remove superfluous definition of xs as there's only one use of   it now } ~> myDrop (1 - 1) (tail [2,3,4])     { Reduce myDrop (1 - 1) (tail [2,3,4]) to its right hand side as   defined above } ~> if n <= 0 || null xs then xs else myDrop (n - 1) (tail xs) where n   = (1 - 1); xs = tail [2,3,4]     { Reduce 1 - 1 to 0 } ~> if n <= 0 || null xs then xs else myDrop (n - 1) (tail xs) where n   = 0; xs = tail [2,3,4]     { Reduce 0 <= 0 to True } ~> if True || null xs then xs else myDrop (n - 1) (tail xs) where n =   0; xs = tail [2,3,4]     { Remove superfluous definition of n, as its only used in one   place now } ~> if True || null xs then xs else myDrop (0 - 1) (tail xs) where xs =   tail [2,3,4]     { Reduce True || anything to True } ~> if True then xs else myDrop (0 - 1) (tail xs) where xs = tail [2,3,4]     { Reduce if expression to its then branch } ~> xs where xs = tail [2,3,4]     { Remove superfluous definition of xs, as it only appears once } ~> tail [2,3,4]     { Reduce tail [2,3,4] to [3,4] } ~> [3,4] Hope that helps Bob
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