[Haskell-cafe] Area from [(x,y)] using foldl
michael rice
nowgate at yahoo.com
Sun Nov 8 15:04:03 EST 2009
Of course! Back to the drawing board.
Thanks,
Michael
--- On Sun, 11/8/09, Eugene Kirpichov <ekirpichov at gmail.com> wrote:
From: Eugene Kirpichov <ekirpichov at gmail.com>
Subject: Re: [Haskell-cafe] Area from [(x,y)] using foldl
To: "michael rice" <nowgate at yahoo.com>
Cc: haskell-cafe at haskell.org
Date: Sunday, November 8, 2009, 2:56 PM
The type of foldl is:(b -> a -> b) -> b -> [a] -> b
What do you expect 'a' and 'b' to be in your algorithm?
2009/11/8 michael rice <nowgate at yahoo.com>
Here's an (Fortran) algorithm for calculating an area, given one dimensional
arrays of Xs and Ys. I wrote a recursive Haskell function that works, and one using
FOLDL that doesn't. Why would Haskell be "expecting" (t, t) out of ((*) (xold-x) (yold+y))?
Michael
====================
AREA = 0.0
XOLD = XVERT(NVERT)
YOLD = YVERT(NVERT)
DO 10 N = 1, NVERT
X = XVERT(N)
Y = YVERT(N)
AREA = AREA + (XOLD - X)*(YOLD + Y)
XOLD = X
YOLD = Y
10
CONTINUE
AREA = 0.5*AREA
====================
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>
====================
area :: [(Double,Double)] -> Double
area p = foldl (\ (xold,yold) (x,y) -> ((*) (xold-x) (yold+y))) 0 ((last p):p)
Prelude> :l area
[1 of 1] Compiling Main ( area.hs, interpreted )
area.hs:29:40:
Occurs check: cannot construct the infinite type: t =
(t, t)
Expected type: (t, t)
Inferred type: t
In the expression: ((*) (xold - x) (yold + y))
In the first argument of `foldl', namely
`(\ (xold, yold) (x, y) -> ((*) (xold - x) (yold + y)))'
Failed, modules loaded: none.
Prelude>
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--
Eugene Kirpichov
Web IR developer, market.yandex.ru
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