`allPairs list = [(x,y) | x <- list, y <- list] ` is not what `combination` does !
let allPairs list = [(x,y) | x <- list, y <- list] allPairs [1,2,3] [(1,1),(1,2),(1,3),(2,1),(2,2),(2,3),(3,1),(3,2),(3,3)]
combination [1,2,3] [[1,2,3],[2,3],[1,3],[3],[1,2],[2],[1],[]]
Alexander Solla-2 wrote:
On Mar 17, 2010, at 6:14 PM, Daniel Fischer wrote:
I found it surprisingly not-slow (code compiled with -O2, as usual). There are two points where you waste time.
I found one big point where time is wasted: in computing the powerset of a list. He's making 2^n elements, and then iterating through them all and filtering, but only needs n^2 or n `choose` 2 of the (depending on the semantics for his "groups").
The answer is to do something like:
allPairs list = [(x,y) | x <- list, y <- list]
to get it done in n^2 time. _______________________________________________ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
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