Am Donnerstag 24 Dezember 2009 03:30:54 schrieb Patrick LeBoutillier:
How do you do it with drinkPerms? The constant element is in the middle (with Nations it seems easier since it's the first one).
insertAt :: Int -> a -> [a] -> [a] insertAt k x xs = case splitAt k xs of (front,back) -> front ++ x:back drinkPerms = map (insertAt 2 Milk) $ permutations [Coffee, Tea, Water, Beer] Another possibility to have x inserted at a fixed position in all permutations of xs is to use do (fs,bs) <- picks k xs pf <- permutations fs pb <- permutations bs return (pf ++ x:pb) which has the advantage that the permutations of the front are shared (if the back is longer than the front, it might be better to swap lines 2 and 3 to share the permutations of the back) and avoids the many splits. picks :: Int -> [a] -> [([a],[a])] picks k xs | k == 0 = [([],xs)] | k == l = [(xs,[])] | k > l = [] | otherwise = pickHelper l k xs where l = length xs pickHelper s t yys@(y:ys) | s == t = [(yys,[])] | otherwise = [(y:zs,ws) | (zs,ws) <- pickHelper (s-1) (t-1) ys] ++ [(zs,y:ws) | (zs,ws) <- pickHelper (s-1) t ys] It's by far not as nice as keeping the first element fixed, but you can easily keep more than one position fixed. And when you have a couple more items, this approach is enormously faster than filtering.