**Paste:**#107273**Author(s):**1HaskellADay**Language:**Haskell**Channel:**-**Created:**2014-07-10 16:41:14 UTC**Revisions:**- 2014-07-13 17:25:54 UTC #107436 (diff): No title (1HaskellADay)
- 2014-07-13 17:16:35 UTC #107435 (diff): No title (1HaskellADay)
- 2014-07-13 17:09:25 UTC #107433 (diff): No title (1HaskellADay)
- 2014-07-13 17:08:43 UTC #107432 (diff): No title (1HaskellADay)
- 2014-07-13 17:04:35 UTC #107431 (diff): No title (1HaskellADay)
- 2014-07-10 17:55:48 UTC #107285 (diff): No title (1HaskellADay)
- 2014-07-10 16:41:14 UTC #107273: Brady Bunch (1HaskellADay)

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 154 155 156 157 158 159 160 161 162 163 164 165 166 167 168 169 170 171 172 173 174 175 176 177 178 179 180 181 182 183 184 185 186 187 188 189 190 191 192 193 | import Data.Set (Set) import qualified Data.Set as Set import Data.List import Control.List -- http://lpaste.net/107211 import Data.Peano -- http://lpaste.net/107204 import Control.Monad import Control.Monad.State.Lazy import Control.Arrow {-- Okay! SO! You get to see two solutions to this problem: 1) 'hella'-inefficent (generate(-all-possible-solutions)-then-test) 2) just wrong (whoopsie! I thought permutations would do it. It didn't) before, at the bottom of this file, you get to selectBunch 3) 'hella'-fast (0.2 seconds instead of ~4 seconds for 1) which uses a guarded state-like structure to select only unique members for each set. The thing was 3) 'hella'-fast was also 'hella'-easy to write, once I just simply followed the data-flow and worked with it. It's neat how easy Haskell is to code in. kthxbai! --} -- a(n inefficient) solution to http://lpaste.net/107181 data Buddy = Alexandra | Beth | Carla | Da5id | Elie | Franny | Graeme | Henry | Isaac deriving (Eq, Ord, Enum, Show, Read) buddies :: Set Buddy buddies = Set.fromList [Alexandra .. Isaac] -- given takeout from Control.List, bunch takes out an element -- from buddies for each member of each clique. Easy! rawbunch :: (Eq a, Ord a, Enum a) => Set a -> [Int] -> [[[a]]] rawbunch buddies cliques = let lbuds = Set.toList buddies pclix = map (fromInt . toInteger) cliques in mapM (popClique alexandra [] lbuds) pclix >>= return . map fst >>= \cliques -> guard (uniqueness cliques) >> return cliques where alexandra = toEnum 0 -- from a base set of buddies, populate a clique, yielding it and -- the reduced base pool popClique :: (Eq a, Ord a, Enum a) => a -> [a] -> [a] -> Peano -> [([a], [a])] popClique _ accum base Z = return (accum, base) popClique first accum base (S n) = takeout base >>= \(bud, rest) -> guard (bud >= first) >> popClique bud (bud : accum) rest n uniqueness :: Eq a => [[a]] -> Bool uniqueness [] = True uniqueness (clique : cliques) = u' clique (join cliques) && uniqueness cliques where u' [] _ = True u' (bud : buddies) cliques = notElem bud cliques && u' buddies cliques bunch :: (Eq a, Ord a, Enum a) => Set a -> [Int] -> Set (Set (Set a)) bunch buds cliques = buncher buds cliques rawbunch buncher :: (Eq a, Ord a, Enum a) => Set a -> [Int] -> (Set a -> [Int] -> [[[a]]]) -> Set (Set (Set a)) buncher buds cliques fn = sfl $ msfl (msfl id) $ fn buds cliques where sfl = Set.fromList msfl f = map (Set.fromList . f) brady :: [Set (Set Buddy)] -- geddit? the 'brady bunch'? geddit? ;) brady = Set.toList $ bunch buddies [2,3,4] -- length brady ~> 1260 because 9c2 * 7*3 * 4c4 == 1260 -- ... but it takes forEVAH to return because of my eh-algorithm :( -- where forEVAH is defined to be 'around 4 seconds' {-- Discussion, re: my eh-algorithm. What I did was get all possible solutions and then mapped through the solution set, eliminating the non-unique ones. A MUCH better approach would have been to eliminate the choices made from the pool when a clique is formed so that new cliques only choose members who are not in the old cliques. That would have been a MUCH better approach. --} -- after the fact pensées: {-- So, the grouping into cliques is just really subdividing permutations of the original group. I don't know why I made all this fuss, actually. So, a much simpler, and faster, solution is: --} permutedBunch :: (Eq a, Ord a, Enum a) => Set a -> [Int] -> [[[a]]] permutedBunch set cliques = map (grouping cliques) (permute $ Set.toList set) grouping :: [Int] -> [a] -> [[a]] grouping [] [] = [] grouping (splitter:ss) seed = (second (grouping ss) >>> uncurry (:)) (splitAt splitter seed) -- in hed : grouping ss rest {-- Shoot! It's not better, and it's not faster! Because permute [1,2,3] gives six solutions, but 3c3 gives one! permute isn't the answer! Darn! So I do have to have a stateful state in order to formulate this problem correctly! Good experiment, though. Nice try. Sigh! --} {-- So, here's a stetch of what I need. I need a state such that the state is updated for each sublist so that following lists don't select the same elements, and I need to impose ordering so that I don't have the duplicates: [[1,2], ...] and [[2,1], ...] Is this doable? --} selectBunch :: (Eq a, Ord a, Enum a) => Set a -> [Int] -> [[[a]]] selectBunch set cliques = let pianos = map (fromInt . toInteger) cliques in s' pianos (Set.toList set) [] where s' [] _ accum = return accum s' (p:ps) rest accum = selectClique p rest [] (toEnum 0) >>= \(clique, rem) -> s' ps rem (clique:accum) selectClique :: (Ord a) => Peano -> [a] -> [a] -> a -> [([a], [a])] selectClique Z pool clique _ = return (clique, pool) selectClique (S n) pool clique min = takeout pool >>= \(h, rest) -> guard (h >= min) >> selectClique n rest (h:clique) h {-- Yes, that was doable, surprisingly easily so, as you see above, and the return, seeing that we've inlined the guard, is very fast now: *Main Data.Time.Clock> getCurrentTime >>= \t1 -> putStrLn (show (length (selectBunch buddies [2,3,4]))) >> getCurrentTime >>= return . diffUTCTime t1 gives: 1260 -0.22408s (0.2 seconds) as opposed to: *Main Data.Time.Clock> getCurrentTime >>= \t1 -> putStrLn (show (length (bunch buddies [2,3,4]))) >> getCurrentTime >>= return . diffUTCTime t1 1260 -3.693009s (3.7 seconds) Woot! Victory! The type of selectClique is: selectClique :: (Ord a) => Peano -> [a] -> [a] -> a -> [([a], [a])] as expected --} |

51:9: Warning: Use liftM

Found:

mapM (popClique alexandra [] lbuds) pclix >>= return . map fst

Why not:

liftM (map fst) (mapM (popClique alexandra [] lbuds) pclix)