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 <caseyh@istar.ca> wrote:

From: Casey Hawthorne <caseyh@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
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