[Haskell-cafe] Area from [(x,y)] using foldl
nowgate at yahoo.com
Sun Nov 8 18:07:45 EST 2009
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.
--- On Sun, 11/8/09, Casey Hawthorne <caseyh at istar.ca> wrote:
From: Casey Hawthorne <caseyh at istar.ca>
Subject: Re: [Haskell-cafe] Area from [(x,y)] using foldl
To: haskell-cafe at 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
How about these BETTER type signatures.
-- Area of a Polygon
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))
Haskell-Cafe mailing list
Haskell-Cafe at haskell.org
-------------- next part --------------
An HTML attachment was scrubbed...
More information about the Haskell-Cafe