[Haskell-beginners] code critique

[Jeff Lasslett]

Also, if I understand correctly, solutions 1 & 2 will only produce one
pair, whereas solution 3 (based on list comprehensions) will produce
all pairs that sum to the total.

On 27 July 2011 07:33, Gary Klindt <gary.klindt at uni-konstanz.de> wrote:
> Hey all!
>
> I compared the different methods proposed with my eyes and the profiler.
>
> sumCheck1 _     []     _  = Nothing
> sumCheck1 total (x:xs) ys = if total' == Nothing
>                           then sumCheck total xs ys
>                           else return (x,(ys!!(fromJust total')))
>                           where total' = elemIndex (total-x) ys
>
> sumCheck2 total (x:xs) ys =
>    let diff = total - x
>    in if elem diff ys
>       then Just (x,diff)
>       else sumCheck total xs ys
>
> sumCheck3 i as bs = not $ null [(x,y) | x <- as, y <- bs, x+y==i ]
>
> It is fascinating how small the code can be with list comprehensions.
> Unfortunately the third solution needs much more memory during run time than
> the other two.
> Increasing the list size leads to heavy growth of memory allocation.
> Probably the evaluation of the cross product in sumCheck3 is not very lazy
> (but why not?).
> The first two solutions scale very well.
>
> Cheers
>
>
>
> On 07/26/2011 08:31 AM, aditya siram wrote:
>>
>> List comprehension seems like the easiest way to do it.
>>
>> First here's how to get the cross product of two lists, I'll be using
>> this below:
>> cp as bs = [(x,y) | x<- as, y<- bs ]
>> -- cp [1,2,3] [4,5,6] =
>> [(1,4),(1,5),(1,6),(2,4),(2,5),(2,6),(3,4),(3,5),(3,6)]
>>
>> Then constrain the cross product to tuples where the sum is the given
>> number:
>> sums i as bs = [(x,y) | x<- as, y<- bs, x + y == i]
>> -- sums 4 [1,2,3] [1,2,3] == [(1,3),(2,2),(3,1)]
>>
>> Then check that the list of constrained sums is not empty:
>> sumCheck i as bs = not (null (sums i as bs))
>>
>> -deech
>>
>> On Mon, Jul 25, 2011 at 10:52 PM, Bryce Verdier<bryceverdier at gmail.com>
>>  wrote:
>>>
>>> Hi all,
>>> I'm new to haskell, and I'm trying to get better with it. Recently I
>>> completed one of the challenges from Programming Praxis and I was
>>> wondering
>>> if people would be willing to spend some time and tell me how I could
>>> improve my code.
>>> Thanks in advance,
>>> Bryce
>>> Here is a link to the programming praxis:
>>> http://programmingpraxis.com/2011/07/19/sum-of-two-integers/
>>> And here is my code:
>>> import Data.List
>>> import Data.Maybe
>>> sumCheck :: Int ->  [Int] ->  [Int] ->  Maybe (Int, Int)
>>> sumCheck _ [] _ = Nothing
>>> sumCheck total (x:xs) ys = if total' == Nothing
>>>                                 then sumCheck total xs ys
>>>                                 else return (x, (ys !! ( fromJust
>>> total')))
>>>                            where total' = (total - x) `elemIndex` ys
>>>
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>>>
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>
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