Why does my implementation of SVG arc conversion not pass QuickCheck?
Asked Answered
S

1

13

I implemented W3s recommended algorithm for converting SVG-path arcs from endpoint-arcs to center-arcs and back in Haskell.

type EndpointArc = ( Double, Double, Double, Double
                   , Bool, Bool, Double, Double, Double )

type CenterArc = ( Double, Double, Double, Double
                 , Double, Double, Double )

endpointToCenter :: EndpointArc -> CenterArc

centerToEndpoint :: CenterArc -> EndpointArc

See full implementation and test-code here.

But I can't get this property to pass:

import Test.QuickCheck
import Data.AEq ((~==))

instance Arbitrary EndpointArc where
    arbitrary = do
        ((x1,y1),(x2,y2)) <- arbitrary `suchThat` (\(u,v) -> u /= v)
        rx                <- arbitrary `suchThat` (>0)
        ry                <- arbitrary `suchThat` (>0)
        phi               <- choose (0,2*pi)
        (fA,fS)           <- arbitrary
        return $ correctRadiiSize (x1, y1, x2, y2, fA, fS, rx, ry, phi)

prop_conversionRetains :: EndpointArc -> Bool
prop_conversionRetains earc =
    let result = centerToEndpoint (endpointToCenter earc)
    in earc ~== result

Sometimes this is due to floating point errors (which seem to exceed ieee754) but sometimes there are NaNs in the result.

(NaN,NaN,NaN,NaN,False,False,1.0314334509082723,2.732814841776921,1.2776112657142984)

Which indicates there is no solution although I think I scale rx,ry as described in F.6.6.2 in W3's document.

import Numeric.Matrix

m :: [[Double]] -> Matrix Double
m = fromList

toTuple :: Matrix Double -> (Double, Double)
toTuple = (\[[x],[y]] -> (x,y)) . toList

primed :: Double -> Double -> Double -> Double -> Double
       -> (Double, Double)
primed x1 y1 x2 y2 phi = toTuple $
    m [[ cos phi, sin phi]
      ,[-sin phi, cos phi]
      ]
    * m [[(x1 - x2)/2]
        ,[(y1 - y2)/2]
        ]

correctRadiiSize :: EndpointArc -> EndpointArc
correctRadiiSize (x1, y1, x2, y2, fA, fS, rx, ry, phi) =
    let (x1',y1') = primed x1 y1 x2 y2 phi
        lambda    = (x1'^2/rx^2) + (y1'^2/ry^2)
        (rx',ry') | lambda <= 1 = (rx, ry)
                  | otherwise   = ((sqrt lambda) * rx, (sqrt lambda) * ry)
    in (x1, y1, x2, y2, fA, fS, rx', ry', phi)
Subspecies answered 16/11, 2014 at 11:26 Comment(0)
S
13

OK, I figured this out myself. The clue was of course in W3s document:

In the case that the radii are scaled up using equation (F.6.6.3), the radicand of (F.6.5.2) is zero and there is exactly one solution for the center of the ellipse.

F.6.5.2 in my code is

(cx',cy') = (sq * rx * y1' / ry, sq * (-ry) * x1' / rx)
                where sq = negateIf (fA == fS) $ sqrt
                         $ ( rx^2 * ry^2 - rx^2 * y1'^2 - ry^2 * x1'^2 )
                         / ( rx^2 * y1'^2 + ry^2 * x1'^2 )

The radicand that it is referring to is

( rx^2 * ry^2 - rx^2 * y1'^2 - ry^2 * x1'^2 )
 / ( rx^2 * y1'^2 + ry^2 * x1'^2 )

But of course, because we are working with floats it's not exactly zero but approximately and sometimes it might be something like -6.99496644301622e-17 which is negative! The square-root of a negative number is a complex number so the calculation returns NaN.

The trick really would be to propagate the fact that rx and ry have been resized to return zero and make sq zero instead of going through the whole calculation unecessarily but the quick fix is just to take the absolute value of the radicand.

(cx',cy') = (sq * rx * y1' / ry, sq * (-ry) * x1' / rx)
                where sq = negateIf (fA == fS) $ sqrt $ abs
                         $ ( rx^2 * ry^2 - rx^2 * y1'^2 - ry^2 * x1'^2 )
                         / ( rx^2 * y1'^2 + ry^2 * x1'^2 )

After that there are some remaining floating point issues. Firstly the error exceeds what is allowed for by ieee754's ~== operator so I made my own approxEq

approxEq (x1a, y1a, x2a, y2a, fAa, fSa, rxa, rya, phia) (x1b, y1b, x2b, y2b, fAb, fSb, rxb, ryb, phib) =
       abs (x1a - x1b  ) < 0.001
    && abs (y1a - y1b  ) < 0.001
    && abs (x2a - x2b  ) < 0.001
    && abs (y2a - y2b  ) < 0.001
    && abs (y2a - y2b  ) < 0.001
    && abs (rxa - rxb  ) < 0.001
    && abs (rya - ryb  ) < 0.001
    && abs (phia - phib) < 0.001
    && fAa == fAb
    && fSa == fSb

prop_conversionRetains :: EndpointArc -> Bool
prop_conversionRetains earc =
    let result = centerToEndpoint (trace ("FIRST:" ++ show (endpointToCenter earc)) (endpointToCenter earc))
    in earc `approxEq` trace ("SECOND:" ++ show result) result

Which starts bringing cases where fA is getting flipped. Spot the magic number:

FIRST:(-5.988957688551294,-39.5430169665332,64.95929681921707,29.661347617532357,5.939852349879405,-1.2436798376040206,3.141592653589793)

SECOND:(4.209851895761209,-73.01839718538467,-16.18776727286379,-6.067636747681732,False,True,64.95929681921707,29.661347617532357,5.939852349879405)

*** Failed! Falsifiable (after 20 tests):
(4.209851895761204,-73.01839718538467,-16.18776781572145,-6.0676366434916655,True,True,64.95929681921707,29.661347617532357,5.939852349879405)

You got it! fA = abs dtheta > pi is in centerToEndpoint so if it's therabouts then it can go either way.

So I took out the fA condition and increased the number of tests in quickcheck

approxEq (x1a, y1a, x2a, y2a, fAa, fSa, rxa, rya, phia) (x1b, y1b, x2b, y2b, fAb, fSb, rxb, ryb, phib) =
       abs (x1a - x1b  ) < 0.001
    && abs (y1a - y1b  ) < 0.001
    && abs (x2a - x2b  ) < 0.001
    && abs (y2a - y2b  ) < 0.001
    && abs (y2a - y2b  ) < 0.001
    && abs (rxa - rxb  ) < 0.001
    && abs (rya - ryb  ) < 0.001
    && abs (phia - phib) < 0.001
    -- && fAa == fAb
    && fSa == fSb

main = quickCheckWith stdArgs {maxSuccess = 50000} prop_conversionRetains

Which shows that the threshold approxEq is still not lax enough.

approxEq (x1a, y1a, x2a, y2a, fAa, fSa, rxa, rya, phia) (x1b, y1b, x2b, y2b, fAb, fSb, rxb, ryb, phib) =
       abs (x1a - x1b  ) < 1
    && abs (y1a - y1b  ) < 1
    && abs (x2a - x2b  ) < 1
    && abs (y2a - y2b  ) < 1
    && abs (y2a - y2b  ) < 1
    && abs (rxa - rxb  ) < 1
    && abs (rya - ryb  ) < 1
    && abs (phia - phib) < 1
    -- && fAa == fAb
    && fSa == fSb

Which I can finally get to pass reliably with a high number of tests. Well its all just to make some funny graphics anyway... I am sure it's accurate enough :)

Subspecies answered 16/11, 2014 at 18:24 Comment(0)

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