---------- Forwarded message ----------
From: L Corbijn
a = [1,1,1,1] b = [0,1,2,3] d = [0,0,0,0]
for i in b: for j in c: if (i+j)<3: d[i+j] += a[i]
My just work implementation in Haskell http://hpaste.org/57452
Another people implementation in Haskell with Monad and it turns out complex and very imperatively. http://hpaste.org/57358
Do you have any cool solution in FP way?
Thanks. -Simon
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Oops forgot to reply to the list, There are several ways to make it nicer. Without assumption on what lists a, b and c contain it can be written as d = [sum $ map (a !!) [i | i <- b, j <- c, i + j < 3, i + j == dIndex] | dIndex <- [0..3]] With the assumption that b and c are both [0..3] this can be 'improved' to d = (take 3 . map sum . tail $ inits a) ++ replicate (4 - 3) 0 This generates the first three values by the complicated expression and then adds the extra zero's by using the replicate expression. This works as the value of the i-th element of d is the sum over the first i elements in a if i < 3 and 0 otherwise. A list of lists with the first i elements is generated with 'tail $ inits a' which is then summed and restricted to length 3. An alternative for this is d = (take 3 . snd $ mapAccumL (\acc ai -> (acc + ai, acc + ai)) 0 a) ++ replicate (4 - 3) 0 Where the summation and tail generating is done in the mapAccumL function. Greetings, Lars P.S. Yes the replicate (4-3) 0 can be replaced by [0], but I wanted to explicitly include the length (4) of the expected list. P.P.S. for those of you wondering what c is, looking at the solutions in hpaste c is probably defined as c = [0..3].