Re: [Haskell-cafe] Area from [(x,y)] using foldl

On Sun, 8 Nov 2009 15:07:45 -0800 (PST), you wrote:

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

Since Haskell is a pure functional language, it adds lazy evaluation as another tool to your modularity toolbox. Lazy evaluation separates control from computation, so, if the computation of sum is going off the rails (and I don't mean Ruby on Rails), one can prematurely terminate the calculation, without evaluating more points.

--- On Sun, 11/8/09, Casey Hawthorne wrote:

From: Casey Hawthorne 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))

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