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

michael rice nowgate at yahoo.com
Sun Nov 8 16:42:13 EST 2009

```I see what one problem is, what happens when I end up with (x,y):[]? However, I'm confused about how Haskell is "expecting" and "inferring" upon compilation.

Michael

--- On Sun, 11/8/09, michael rice <nowgate at yahoo.com> wrote:

From: michael rice <nowgate at yahoo.com>
Subject: Re: [Haskell-cafe] Area from [(x,y)] using foldl
Date: Sunday, November 8, 2009, 4:30 PM

That's certainly better than mine, but I'm lost again, with the following. What seemed like a simple improvement doesn't compile.

Michael

===============

This works.

area :: [(Double,Double)] -> Double
area ps = abs \$ (/2) \$ area' (last ps) ps
where area' _ [] = 0
area' (x0,y0) ((x,y):ps) = (x0-x)*(y0+y) + area' (x,y) ps

*Main> let p = [(0.0,0.0),(1.0,0.0),(1.0,1.0),(0.0,1.0),(0.0,0.0)]
*Main> area (last p) p
1.0
*Main>

===============

This doesn't.

area :: [(Double,Double)] -> Double
area p = abs \$ (/2) \$ area' (last p):p

where area' [] = 0
area' ((x0,y0),(x,y):ps) = ((x0-x)*(y0+y)) + area' (x,y):ps

Subject: Re: [Haskell-cafe] Area from [(x,y)] using foldl
To: "michael rice" <nowgate at yahoo.com>
Date: Sunday, November 8, 2009, 3:52 PM

On Sun, Nov 8, 2009 at 9:04 PM, michael
rice <nowgate at yahoo.com> wrote:

Of course! Back to the drawing board.

If I understand the problem correctly, I'm not convinced that foldl is the right approach (nevermind that foldl is almost never what you want, foldl' and foldr being the correct choice almost always). My proposition would be the following :

> area ps = abs . (/2) . sum \$ zipWith (\(x,y) (x',y') -> (x - x') * (y + y')) ps (tail \$ cycle ps)

I think it express the algorithm more clearly.

--
Jedaï

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