# Haskell Quiz/Geodesic Dome Faces/Solution Jkramar

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< Haskell Quiz | Geodesic Dome Faces(Difference between revisions)

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farth :: (Real a, Foldable t) => V3 a -> t (V3 a) -> a |
farth :: (Real a, Foldable t) => V3 a -> t (V3 a) -> a |
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− | farth dir = maximum.map (dot dir).toList |
+ | farth dir = maximum.(dot dir<$>).toList |

-- inefficient function to generate all the positively-oriented faces of the |
-- inefficient function to generate all the positively-oriented faces of the |

## Revision as of 08:16, 17 November 2008

This problem seems to be strongly IO-bound, so actually computing the geodesic faces is not too time-sensitive. Hence there is time for computing the faces from the vertices by trying each triple of vertices to check if it's a face.

Originally I wrote this program representing vectors just by ordinary tuples, and in some places that was definitely more understandable. But of course it's more fun this way.

import Prelude hiding (maximum, foldr, foldr1, concat, sequence_, sum) import Control.Monad hiding (sequence_) import Control.Applicative import Data.Foldable data V3 a = V !a !a !a deriving (Read,Show) instance Foldable V3 where foldr f x (V a b c) = f a$f b$f c x instance Functor V3 where fmap f (V a b c) = V (f a) (f b) (f c) instance Applicative V3 where pure a = V a a a V f g h <*> V a b c = V (f a) (g b) (h c) type Tri a = V3 (V3 a) (.+) :: (Num a) => V3 a -> V3 a -> V3 a (.+) = liftA2 (+) (.-) :: (Num a) => V3 a -> V3 a -> V3 a (.-) = liftA2 (-) (.*) :: (Num a) => a -> V3 a -> V3 a (.*) = fmap.(*) dot :: (Num a) => V3 a -> V3 a -> a dot a b = sum$liftA2 (*) a b cross :: (Num a) => V3 a -> V3 a -> V3 a cross (V x y z) (V x' y' z')=V (y*z'-z*y') (z*x'-x*z') (x*y'-y*x') det :: (Num a) => V3 (V3 a) -> a det (V a b c) = (a `cross` b) `dot` c -- matrix multiplication mmul :: (Num a) => V3 (V3 a) -> V3 (V3 a) -> V3 (V3 a) mmul t s = foldr1 (.+) <$> (flip (.*)<$>t<*>) <$> s normize :: (Floating a) => V3 a -> V3 a normize v = (1/sqrt (dot v v)).*v -- chooses xs!!n gives the combinations of n elements from xs chooses :: (Foldable t, Alternative f) => t a -> [f [a]] chooses = foldr consider$pure []:repeat empty where consider x cs = zipWith (flip (<|>)) cs$fmap (x:)<$>empty:cs tris :: (Foldable t, Alternative f) => t (V3 a) -> f (Tri a) tris xs = (\[a,b,c]->V a b c) <$> (chooses xs!!3) orient :: (Num a) => Tri a -> Tri a orient t@(V a b c) = if (==1)$signum$det t then t else V a c b farth :: (Real a, Foldable t) => V3 a -> t (V3 a) -> a farth dir = maximum.(dot dir<$>).toList -- inefficient function to generate all the positively-oriented faces of the -- triangle-faced polyhedron with vertices at vs faces :: (Real a, Foldable t, MonadPlus m) => t (V3 a) -> m (Tri a) faces vs = do t@(V a b c) <- unwrapMonad $ orient <$> tris vs let dir = ((a.-b) `cross` (a.-c)) guard (farth dir [a,b,c]==farth dir vs) >> return t -- each triangle side is broken into n pieces, unlike in the problem statement, -- where they use n+1 for some reason shatter :: (Integral a, Fractional b, MonadPlus m) => a -> Tri b -> m (Tri b) shatter n t = msum$return<$>mmul t'<$>coords where basis = V (V 1 0 0) (V 0 1 0) (V 0 0 1) fwd = [(V k j (n-1-k-j).+) <$> basis|k<-[0..n-1], j<-[0..n-1-k]] bwd = [(V k j (n+1-k-j).-) <$> basis|k<-[1..n], j<-[1..n-k]] t' = ((1/fromIntegral n).*)<$>t coords = (fmap fromIntegral<$>)<$>fwd++bwd geode :: (RealFloat b, Integral a, MonadPlus m, Foldable t) => t (V3 b) -> a -> m (Tri b) geode vs n = return.fmap normize=<<shatter n=<<faces vs cyc :: V3 a -> [V3 a] cyc (V a b c) = [V a b c, V b c a, V c a b] tetrahed :: (Floating a) => [V3 a] tetrahed = [V (x*sqrt 1.5) (-sqrt 2/3) (-1/3)|x<-[-1,1]]++ [V 0 (2*sqrt 2/3) (-1/3), V 0 0 1] octahed :: (Num a) => [V3 a] octahed = cyc =<< [V x 0 0|x<-[-1,1]] icosahed :: (Floating a) => [V3 a] icosahed = cyc =<< [V x (y*(1+sqrt 5/2)) 0|x<-[-1, 1], y<-[-1, 1]] -- this is the test the Ruby Quiz people were doing to see how fast the code is main :: IO () main = sequence_$map print$(geode octahed::Int->[Tri Double]) 51