# Haskell Quiz/Geodesic Dome Faces/Solution Jkramar

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

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-- matrix multiplication |
-- matrix multiplication |
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mmul :: (Num a) => V3 (V3 a) -> V3 (V3 a) -> V3 (V3 a) |
mmul :: (Num a) => V3 (V3 a) -> V3 (V3 a) -> V3 (V3 a) |
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− | mmul t s = foldr1 (.+) <$> (flip (.*)<$>t<*>) <$> s |
+ | mmul t = fmap$foldr1 (.+).(<*>t).fmap (.*) |

normize :: (Floating a) => V3 a -> V3 a |
normize :: (Floating a) => V3 a -> V3 a |

## Revision as of 19:20, 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. (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, 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 = (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$foldr1 (.+).(<*>t).fmap (.*) normize :: (Floating a) => V3 a -> V3 a normize v = (1/sqrt (dot v v)).*v -- chooses xs!!n gives the combinations of n elements from xs chooses :: [a] -> [[[a]]] chooses = foldr consider$[[]]:repeat [] where consider x cs = zipWith (flip (++)) cs$map (x:)<$>[]:cs tris :: [V3 a] -> [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 -- inefficient function to generate all the positively-oriented faces of the -- triangle-faced polyhedron with vertices at vs faces :: (Real a) => [V3 a] -> [Tri a] faces vs = filter isFace $ orient <$> tris vs where farth (V a b c) = maximum.(dot ((a.-b) `cross` (a.-c))<$>) isFace t@(V a b c) = farth t [a,b,c] == 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 :: (Integral a, Fractional b) => a -> Tri b -> [Tri b] shatter n t = mmul t'<$>coords where basis = V (V 1 0 0) (V 0 1 0) (V 0 0 1) fwd = [(V k j (n-1-k-j).+) <$> 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 :: (Integral a, RealFloat b) => [V3 b] -> a -> [Tri b] geode vs n = 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]] main :: IO () main = mapM_ print (geode octahed (51::Int)::[Tri Double])