[Haskell-beginners] code critique

[Gary Klindt]

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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