Oh, that is not a precondition. So the answer of yours are correct. I
am working on permutations. I used it in a wrong way.
On Mon, Mar 15, 2010 at 4:39 PM, Daniel Fischer
Am Montag 15 März 2010 08:37:20 schrieb Magicloud Magiclouds:
Sorry, I did not make it clear, since I did not know how to say this in technical terms. With comprehension, I could get all the possibilities that "draw one elem from each list and put them together". But consider this: for example, there are two types of pet, dog and cat. And there are two persons, him and her. So "how many kinds of matches are there (orderless)?" The answer is two: "him with dog and her with cat, him with cat and her with dog". So f [a, b, c] [d, e, f] [g, h, i] = [ [ (a, d, g), (b, e, h), (c, f, i) ] , [ (a, d, g), (b, e, i), (c, f, h) ] , [ (a, d, h), (b, e, i), (c, f, g) ] , [ (a, d, h), (b, e, g), (c, f, i) ] , [ (a, d, i), (b, e, g), (c, f, h) ] , [ (a, d, i), (b, e, h), (c, f, g) ] ... ]
In both, your verbal example and the pseudo-code example, all the groups have the same number of members (equal to the number of groups, which may or may not be coincidental). Is that a precondition, that all groups have the same number of members? If so, would the desired result for three groups of two members each be
f3 [a,b] [c,d] [e,f] = [ [ (a,c,e), (b,d,f) ] , [ (a,c,f), (b,d,e) ] , [ (a,d,e), (b,c,f) ] , [ (a,d,f), (b,c,e) ] ]
and for two groups of three members each
f2 [a,b,c] [d,e,f] = [ [ (a,d), (b,e), (c,f) ] , [ (a,d), (b,f), (c,e) ] , [ (a,e), (b,d), (c,f) ] , [ (a,e), (b,f) , (c,d) ] , [ (a,f), (b,d), (c,e) ] , [ (a,f), (b,e), (c,d) ] ]
?
In that case, look at Data.List.permutations and the zipN functions.
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