update pathops to circle sort

src/pathops/SkDCubicToQuads.cpp

 /*
 http://stackoverflow.com/questions/2009160/how-do-i-convert-the-2-control-points-of-a-cubic-curve-to-the-single-control-poi
 */
 
 /*
 Let's call the control points of the cubic Q0..Q3 and the control points of the quadratic P0..P2.
 Then for degree elevation, the equations are:
 
 Q0 = P0
 Q1 = 1/3 P0 + 2/3 P1
 Q2 = 2/3 P1 + 1/3 P2
 Q3 = P2
 In your case you have Q0..Q3 and you're solving for P0..P2. There are two ways to compute P1 from
  the equations above:
 
 P1 = 3/2 Q1 - 1/2 Q0
 P1 = 3/2 Q2 - 1/2 Q3
 If this is a degree-elevated cubic, then both equations will give the same answer for P1. Since
  it's likely not, your best bet is to average them. So,
 
 P1 = -1/4 Q0 + 3/4 Q1 + 3/4 Q2 - 1/4 Q3
 
-
 SkDCubic defined by: P1/2 - anchor points, C1/C2 control points
 |x| is the euclidean norm of x
 mid-point approx of cubic: a quad that shares the same anchors with the cubic and has the
  control point at C = (3·C2 - P2 + 3·C1 - P1)/4
 
 Algorithm
 
 pick an absolute precision (prec)
 Compute the Tdiv as the root of (cubic) equation
 sqrt(3)/18 · |P2 - 3·C2 + 3·C1 - P1|/2 · Tdiv ^ 3 = prec
@@ skipping 144 matching lines @@
         for (int idx = 0; idx < last; ++idx) {
             part = subDivide(inflectT[idx], inflectT[idx + 1]);
             addTs(part, precision, inflectT[idx], inflectT[idx + 1], ts);
         }
         part = subDivide(inflectT[last], 1);
         addTs(part, precision, inflectT[last], 1, ts);
         return;
     }
     addTs(*this, precision, 0, 1, ts);
 }