chromium/third_party/skia/src/pathops/SkDCubicLineIntersection.cpp

/*
 * Copyright 2012 Google Inc.
 *
 * Use of this source code is governed by a BSD-style license that can be
 * found in the LICENSE file.
 */
#include "include/core/SkPath.h"
#include "include/core/SkPoint.h"
#include "include/core/SkTypes.h"
#include "include/private/base/SkDebug.h"
#include "src/pathops/SkIntersections.h"
#include "src/pathops/SkPathOpsCubic.h"
#include "src/pathops/SkPathOpsCurve.h"
#include "src/pathops/SkPathOpsDebug.h"
#include "src/pathops/SkPathOpsLine.h"
#include "src/pathops/SkPathOpsPoint.h"
#include "src/pathops/SkPathOpsTypes.h"

#include <cmath>

/*
Find the intersection of a line and cubic by solving for valid t values.

Analogous to line-quadratic intersection, solve line-cubic intersection by
representing the cubic as:
  x = a(1-t)^3 + 2b(1-t)^2t + c(1-t)t^2 + dt^3
  y = e(1-t)^3 + 2f(1-t)^2t + g(1-t)t^2 + ht^3
and the line as:
  y = i*x + j  (if the line is more horizontal)
or:
  x = i*y + j  (if the line is more vertical)

Then using Mathematica, solve for the values of t where the cubic intersects the
line:

  (in) Resultant[
        a*(1 - t)^3 + 3*b*(1 - t)^2*t + 3*c*(1 - t)*t^2 + d*t^3 - x,
        e*(1 - t)^3 + 3*f*(1 - t)^2*t + 3*g*(1 - t)*t^2 + h*t^3 - i*x - j, x]
  (out) -e     +   j     +
       3 e t   - 3 f t   -
       3 e t^2 + 6 f t^2 - 3 g t^2 +
         e t^3 - 3 f t^3 + 3 g t^3 - h t^3 +
     i ( a     -
       3 a t + 3 b t +
       3 a t^2 - 6 b t^2 + 3 c t^2 -
         a t^3 + 3 b t^3 - 3 c t^3 + d t^3 )

if i goes to infinity, we can rewrite the line in terms of x. Mathematica:

  (in) Resultant[
        a*(1 - t)^3 + 3*b*(1 - t)^2*t + 3*c*(1 - t)*t^2 + d*t^3 - i*y - j,
        e*(1 - t)^3 + 3*f*(1 - t)^2*t + 3*g*(1 - t)*t^2 + h*t^3 - y,       y]
  (out)  a     -   j     -
       3 a t   + 3 b t   +
       3 a t^2 - 6 b t^2 + 3 c t^2 -
         a t^3 + 3 b t^3 - 3 c t^3 + d t^3 -
     i ( e     -
       3 e t   + 3 f t   +
       3 e t^2 - 6 f t^2 + 3 g t^2 -
         e t^3 + 3 f t^3 - 3 g t^3 + h t^3 )

Solving this with Mathematica produces an expression with hundreds of terms;
instead, use Numeric Solutions recipe to solve the cubic.

The near-horizontal case, in terms of:  Ax^3 + Bx^2 + Cx + D == 0
    A =   (-(-e + 3*f - 3*g + h) + i*(-a + 3*b - 3*c + d)     )
    B = 3*(-( e - 2*f +   g    ) + i*( a - 2*b +   c    )     )
    C = 3*(-(-e +   f          ) + i*(-a +   b          )     )
    D =   (-( e                ) + i*( a                ) + j )

The near-vertical case, in terms of:  Ax^3 + Bx^2 + Cx + D == 0
    A =   ( (-a + 3*b - 3*c + d) - i*(-e + 3*f - 3*g + h)     )
    B = 3*( ( a - 2*b +   c    ) - i*( e - 2*f +   g    )     )
    C = 3*( (-a +   b          ) - i*(-e +   f          )     )
    D =   ( ( a                ) - i*( e                ) - j )

For horizontal lines:
(in) Resultant[
      a*(1 - t)^3 + 3*b*(1 - t)^2*t + 3*c*(1 - t)*t^2 + d*t^3 - j,
      e*(1 - t)^3 + 3*f*(1 - t)^2*t + 3*g*(1 - t)*t^2 + h*t^3 - y, y]
(out)  e     -   j     -
     3 e t   + 3 f t   +
     3 e t^2 - 6 f t^2 + 3 g t^2 -
       e t^3 + 3 f t^3 - 3 g t^3 + h t^3
 */

class LineCubicIntersections { … };

int SkIntersections::horizontal(const SkDCubic& cubic, double left, double right, double y,
        bool flipped) { … }

int SkIntersections::vertical(const SkDCubic& cubic, double top, double bottom, double x,
        bool flipped) { … }

int SkIntersections::intersect(const SkDCubic& cubic, const SkDLine& line) { … }

int SkIntersections::intersectRay(const SkDCubic& cubic, const SkDLine& line) { … }

// SkDCubic accessors to Intersection utilities

int SkDCubic::horizontalIntersect(double yIntercept, double roots[3]) const { … }

int SkDCubic::verticalIntersect(double xIntercept, double roots[3]) const { … }