`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 > > ----- fac n = let { f = foldr (*) 1 [1..n] } in f -- View this message in context: http://old.nabble.com/How-to-improve-its-performance---tp27940036p27941343.html Sent from the Haskell - Haskell-Cafe mailing list archive at Nabble.com. _______________________________________________ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
