Created
September 19, 2011 18:35
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All combinations
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| import Control.Arrow | |
| combine :: [a] -> [[a]] | |
| combine = sequence . uncurry replicate . (length &&& id) | |
| {- | |
| Test run: | |
| Prelude Control.Arrow> combine ['A', 'C', 'G'] | |
| ["AAA","AAC","AAG","ACA","ACC","ACG","AGA","AGC","AGG","CAA","CAC","CAG","CCA","CCC","CCG","CGA","CGC","CGG","GAA","GAC","GAG","GCA","GCC","GCG","GGA","GGC","GGG"] | |
| -} | |
| {- | |
| The version that I came up with before: | |
| combine xs = sequence $ replicate (length xs) xs | |
| -} |
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This is how the arrow version works:
&&&takes two functions* of types(a -> b)and(a -> c)and composes them into one function of type(a -> (b, c)). In our caselengthhas type[a] -> Intandidhas typea -> a. So(length &&& id)has type[a] -> (Int, [a]). The signature ofreplicateisInt -> [a] -> [[a]]. So we need to uncurryreplicatein order to be able to compose it with(length &&& id). And the final part is easy. The implementation ofsequencefunction for list monad gives its Cartesian product. And there we have it, all combinations of elements of the list.* Actually
&&&works on arrows, and functions are arrows.