hasufell
37b38115ae
There was also a bug to be fixed when doing this. We missed an important pattern match case which casued wrong results.
74 lines
1.6 KiB
Haskell
74 lines
1.6 KiB
Haskell
{-# OPTIONS_HADDOCK ignore-exports #-}
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module Algebra.Vector where
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import Algebra.VectorTypes
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import Diagrams.TwoD.Types
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-- |Checks whether the Point is in a given dimension.
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inRange :: Coord -- ^ X dimension
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-> Coord -- ^ Y dimension
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-> PT -- ^ Coordinates
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-> Bool -- ^ result
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inRange (xlD, xuD) (ylD, yuD) p = x <= xuD && x >= xlD && y <= yuD && y >= ylD
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where
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(x, y) = unp2 p
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-- |Get the angle between two vectors.
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getAngle :: Vec -> Vec -> Double
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getAngle a b =
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acos .
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flip (/) (vecLength a * vecLength b) .
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scalarProd a $
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b
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-- |Get the length of a vector.
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vecLength :: Vec -> Double
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vecLength v = sqrt (x^2 + y^2)
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where
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(x, y) = unr2 v
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-- |Compute the scalar product of two vectors.
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scalarProd :: Vec -> Vec -> Double
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scalarProd v1 v2 = a1 * b1 + a2 * b2
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where
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(a1, a2) = unr2 v1
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(b1, b2) = unr2 v2
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-- |Construct a vector that points to a point from the origin.
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pt2Vec :: PT -> Vec
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pt2Vec = r2 . unp2
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-- |Give the point which is at the coordinates the vector
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-- points to from the origin.
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vec2Pt :: Vec -> PT
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vec2Pt = p2 . unr2
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-- |Construct a vector between two points.
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vp2 :: PT -- ^ vector origin
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-> PT -- ^ vector points here
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-> Vec
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vp2 a b = (pt2Vec b) - (pt2Vec a)
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-- |Checks if 3 points a,b,c build a counterclock wise triangle by
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-- connecting a-b-c. This is done by computing the determinant and
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-- checking the algebraic sign.
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ccw :: PT -> PT -> PT -> Bool
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ccw a b c =
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(bx - ax) *
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(cy - ay) -
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(by - ay) *
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(cx - ax) >= 0
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where
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(ax, ay) = unp2 a
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(bx, by) = unp2 b
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(cx, cy) = unp2 c
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