Haskell Quiz/Geodesic Dome Faces/Solution Jkramar
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< Haskell Quiz  Geodesic Dome Faces(Difference between revisions)
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Latest revision as of 15:33, 18 November 2008
This problem seems to be strongly IObound, so actually computing the geodesic faces is not too timesensitive. Hence there is time for computing the faces from the vertices by trying each triple of vertices to check if it's a face. (In the problem statement, the data for each polyhedron includes a list of the faces, not just the vertices.)
Originally this program represented vectors with ordinary tuples, and in some places that was definitely more understandable. But of course it's more fun this way.
import Prelude hiding (sum, foldr1, foldr, maximum, mapM_) 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 = (sum.).liftA2 (*) 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 = fmap$foldr1 (.+).(<*>t).fmap (.*)  chooses xs!!n gives the combinations of n elements from xs chooses :: [a] > [[[a]]] chooses = foldr consider$[[]]:repeat [] where consider x cs = zipWith (flip (++)) cs$map (x:)<$>[]:cs  inefficient function to generate all the positivelyoriented faces of the  trianglefaced polyhedron with vertices at vs faces :: (Real a) => [V3 a] > [Tri a] faces vs = filter isFace $ orient <$> tris where orient t@(V a b c) = if (signum$det t)==1 then t else V a c b tris = (\[a,b,c]>V a b c) <$> chooses vs!!3 farth (V a b c) = maximum.(dot ((a.b) `cross` (a.c))<$>) isFace t = farth t (toList t) == farth t vs  each triangle side is broken into n pieces, unlike in the problem statement,  where they use n+1 for some reason shatter :: (Fractional a) => Int > Tri a > [Tri a] shatter n t = mmul t'<$>coords where basis = V (V 1 0 0) (V 0 1 0) (V 0 0 1) fwd = [(V k j (n1kj).+) <$> basisk<[0..n1], j<[0..n1k]] bwd = [(V k j (n+1kj).) <$> basisk<[1..n], j<[1..nk]] t' = ((1/fromIntegral n).*)<$>t coords = (fmap fromIntegral<$>)<$>fwd++bwd geode :: (RealFloat a) => [V3 a] > Int > [Tri a] geode vs n = fmap normize<$>(shatter n=<<faces vs) where normize v = (1/sqrt (dot v v)).*v cyc :: V3 a > [V3 a] cyc (V a b c) = [V a b c, V b c a, V c a b] tetrahed :: (Floating a) => [V3 a] tetrahed = [V (x*sqrt 1.5) (sqrt 2/3) (1/3)x<[1,1]]++ [V 0 (2*sqrt 2/3) (1/3), V 0 0 1] octahed :: (Num a) => [V3 a] octahed = cyc =<< [V x 0 0x<[1,1]] icosahed :: (Floating a) => [V3 a] icosahed = cyc =<< [V x (y*(1+sqrt 5/2)) 0x<[1, 1], y<[1, 1]]  this is the test the Ruby Quiz people used main :: IO () main = mapM_ print (geode octahed 51::[Tri Double])
Here is a reimplementation using the vector and matrix types from the hmatrix package:
import Numeric.LinearAlgebra import Data.List import Control.Applicative ((<$>)) cross :: Vector Double > Vector Double > Vector Double cross v w = fromList$map (det.fromColumns.(:[v,w]))$toColumns$ident 3  chooses xs!!n gives the combinations of n elements from xs chooses :: [a] > [[[a]]] chooses = foldr consider$[[]]:repeat [] where consider x cs = zipWith (flip (++)) cs$map (x:)<$>[]:cs cyc :: [a] > [[a]] cyc = map (take 3).take 3.tails.cycle  inefficient function to generate all the positivelyoriented faces of the  trianglefaced polyhedron with vertices at vs faces :: [Vector Double] > [[Vector Double]] faces vs = filter isFace $ orient <$> chooses vs!!3 where orient t@[a,b,c] = if (signum$det$fromColumns t)==1 then t else [a,c,b] farth = (maximum.).map.dot.sum.map (\(v:w:_)>cross v w).cyc isFace ps = farth ps ps == farth ps vs  each triangle side is broken into n pieces, unlike in the problem statement,  where they use n+1 for some reason shatter :: Int > [Vector Double] > [[Vector Double]] shatter n' t = map (fromColumns t*/n<>)<$>fwd++bwd where n = fromIntegral n'; basis = toColumns$ident 3 fwd = [(fromList [k, j, n1kj]+) <$> basisk<[0..n1], j<[0..n1k]] bwd = [(fromList [k, j, n+1kj]) <$> basisk<[1..n], j<[1..nk]] geode :: [Vector Double] > Int > [[Vector Double]] geode vs n = map proj<$>(shatter n=<<faces vs) where proj v = v*/sqrt (v<.>v) tetrahed :: [Vector Double] tetrahed = map fromList$[[x*sqrt 1.5, sqrt 2/3, 1/3]x<[1, 1]]++ [[0, 2*sqrt 2/3, 1/3], [0, 0, 1]] octahed :: [Vector Double] octahed = map fromList$cyc =<< [[x, 0, 0]x<[1, 1]] icosahed :: [Vector Double] icosahed = map fromList$cyc =<< [[x, y*(1+sqrt 5/2), 0]x<[1, 1], y<[1, 1]]  this is the test the Ruby Quiz people used main :: IO () main = mapM_ print$geode octahed 51