Result of running tools/sanitize_source_files.py (which was added in https://codereview.appspot.com/6465078/)
This CL is part II of IV (I broke down the 1280 files into 4 CLs).
Review URL: https://codereview.appspot.com/6474054
git-svn-id: http://skia.googlecode.com/svn/trunk@5263 2bbb7eff-a529-9590-31e7-b0007b416f81
diff --git a/experimental/Intersection/CubicSubDivide.cpp b/experimental/Intersection/CubicSubDivide.cpp
index 0c6ed42..98250ad 100644
--- a/experimental/Intersection/CubicSubDivide.cpp
+++ b/experimental/Intersection/CubicSubDivide.cpp
@@ -3,21 +3,21 @@
/*
Given a cubic c, t1, and t2, find a small cubic segment.
-
+
The new cubic is defined as points A, B, C, and D, where
s1 = 1 - t1
s2 = 1 - t2
A = c[0]*s1*s1*s1 + 3*c[1]*s1*s1*t1 + 3*c[2]*s1*t1*t1 + c[3]*t1*t1*t1
D = c[0]*s2*s2*s2 + 3*c[1]*s2*s2*t2 + 3*c[2]*s2*t2*t2 + c[3]*t2*t2*t2
-
+
We don't have B or C. So We define two equations to isolate them.
First, compute two reference T values 1/3 and 2/3 from t1 to t2:
-
+
c(at (2*t1 + t2)/3) == E
c(at (t1 + 2*t2)/3) == F
-
+
Next, compute where those values must be if we know the values of B and C:
-
+
_12 = A*2/3 + B*1/3
12_ = A*1/3 + B*2/3
_23 = B*2/3 + C*1/3
@@ -36,9 +36,9 @@
= F
E*27 = A*8 + B*12 + C*6 + D
F*27 = A + B*6 + C*12 + D*8
-
+
Group the known values on one side:
-
+
M = E*27 - A*8 - D = B*12 + C* 6
N = F*27 - A - D*8 = B* 6 + C*12
M*2 - N = B*18
@@ -46,7 +46,7 @@
B = (M*2 - N)/18
C = (N*2 - M)/18
*/
-
+
static double interp_cubic_coords(const double* src, double t)
{
double ab = interp(src[0], src[2], t);
@@ -57,7 +57,7 @@
double abcd = interp(abc, bcd, t);
return abcd;
}
-
+
void sub_divide(const Cubic& src, double t1, double t2, Cubic& dst) {
double ax = dst[0].x = interp_cubic_coords(&src[0].x, t1);
double ay = dst[0].y = interp_cubic_coords(&src[0].y, t1);