Hi Casey, I was already aware of the translation thing, but didn't want to complicate.
Lot's of ways to skin a cat. I wrote a Lispy solution, then had the feeling I could improve on it w/Haskell. Picking the right tool takes practice. Thanks, Michael --- On Sun, 11/8/09, Casey Hawthorne <cas...@istar.ca> wrote: From: Casey Hawthorne <cas...@istar.ca> Subject: Re: [Haskell-cafe] Area from [(x,y)] using foldl To: haskell-cafe@haskell.org Date: Sunday, November 8, 2009, 5:44 PM Sorry, I forgot to add that if the polygon is very far from the origin, you may have overflow or increased round off error; it is better to translate the polygon back to the origin, before doing the area calculation. How about these BETTER type signatures. -- Area of a Polygon import Data.List type X = Double type Y = Double type Area = Double poly1 = [(0,1),(5,0),(3,4)]::[(X,Y)] areaPoly :: [(X,Y)] -> Area areaPolyCalc :: (Area,(X,Y)) -> (X,Y) -> (Area,(X,Y)) <Spoiler Alert! -- Functions Below!> areaPoly (pt:pts) = 0.5 * (fst (foldl' areaPolyCalc (0,pt) pts)) areaPolyCalc (sum,(x,y)) (xNext,yNext) = (sum + (x * yNext - xNext * y),(xNext,yNext)) -- Regards, Casey _______________________________________________ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
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