Am Freitag 26 Februar 2010 21:34:28 schrieb Ketil Malde: > Daniel Fischer <daniel.is.fisc...@web.de> skrev: > > Am Freitag 26 Februar 2010 16:50:42 schrieb Ketil Malde: > >>> solutions = [[x1,x2,x3,x4,x5,x6,x7,x8,x9,x10] > >>> > >>> | x1 <- [0..9] > > > > First digit can't be 0, so make it [1 .. 9]. > > Since you use the fact that the last digit must be the 0, pull all > > others from [1 .. 9]. > > Originally, I pulled from alternating odds (x1 <- [1,3..9] etc) and > evens, since this is fairly easy to deduce... I reverted this since the > point was to use brute force.
Yes, but did you forget x10 or did you think that one was too obvious? > > >>> solve :: [Int] -> [[Int]] > > > > Not on a 32-bit system. Word would suffice there, but you don't know > > that in advance, so it'd be Int64 or Integer > > Hm? The Ints are just individual digits here. > Yup. I didn't realise that you don't call val for the 10-digit number(s). If you also did x10 <- [0 .. 9] and checked val [x1, x2, ..., x10] `mod` 10 == 0, it would overflow, that's what I was thinking of. > > I would make the length of the prefix a parameter of solve. > > I thought about generating a list with solutions for increasing lenghts, > so that e.g. 'solve [] !! 10' would solve this particular problem. > That's nice, but I think it'd be ugly with a DFS, much nicer with a BFS, like Rafael did. > > solve prefix = > > case length prefix of > > 10 -> return prefix > > l -> do > > x <- [0 .. 9] > > ... > > > > over the if-then-else. > > Yes, much nicer. Thanks for the feedback! > > -k _______________________________________________ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
