Intersection work in progress
Review URL: https://codereview.appspot.com/5576043

git-svn-id: http://skia.googlecode.com/svn/trunk@3087 2bbb7eff-a529-9590-31e7-b0007b416f81
diff --git a/experimental/Intersection/LineCubicIntersection.cpp b/experimental/Intersection/LineCubicIntersection.cpp
new file mode 100644
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--- /dev/null
+++ b/experimental/Intersection/LineCubicIntersection.cpp
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+#include "CubicIntersection.h"
+#include "CubicUtilities.h"
+#include "Intersections.h"
+#include "LineUtilities.h"
+
+/*
+Find the interection 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 )
+ */
+
+class LineCubicIntersections : public Intersections {
+public:
+
+LineCubicIntersections(const Cubic& c, const _Line& l, Intersections& i)
+    : cubic(c)
+    , line(l)
+    , intersections(i) {
+}
+
+bool intersect() {
+    double slope;
+    double axisIntercept;
+    moreHorizontal = implicitLine(line, slope, axisIntercept);
+    double A = cubic[3].x; // d
+    double B = cubic[2].x * 3; // 3*c
+    double C = cubic[1].x * 3; // 3*b
+    double D = cubic[0].x; // a
+    A -= D - C + B;     // A =   -a + 3*b - 3*c + d
+    B += 3 * D - 2 * C; // B =  3*a - 6*b + 3*c
+    C -= 3 * D;         // C = -3*a + 3*b
+    double E = cubic[3].y; // h
+    double F = cubic[2].y * 3; // 3*g
+    double G = cubic[1].y * 3; // 3*f
+    double H = cubic[0].y; // e
+    E -= H - G + F;     // E =   -e + 3*f - 3*g + h
+    F += 3 * H - 2 * G; // F =  3*e - 6*f + 3*g
+    G -= 3 * H;         // G = -3*e + 3*f
+    if (moreHorizontal) {
+        A = A * slope - E;
+        B = B * slope - F;
+        C = C * slope - G;
+        D = D * slope - H + axisIntercept;
+    } else {
+        A = A - E * slope;
+        B = B - F * slope;
+        C = C - G * slope;
+        D = D - H * slope - axisIntercept;
+    }
+    double t[3];
+    int roots = cubicRoots(A, B, C, D, t);
+    for (int x = 0; x < roots; ++x) {
+        intersections.add(t[x], findLineT(t[x]));
+    }
+    return roots > 0;
+}
+
+protected:
+    
+double findLineT(double t) {
+    const double* cPtr;
+    const double* lPtr;
+    if (moreHorizontal) {
+        cPtr = &cubic[0].x;
+        lPtr = &line[0].x;
+    } else {
+        cPtr = &cubic[0].y;
+        lPtr = &line[0].y;
+    }
+    double s = 1 - t;
+    double cubicVal = cPtr[0] * s * s * s + 3 * cPtr[2] * s * s * t
+                + 3 * cPtr[4] * s * t * t + cPtr[6] * t * t * t;
+    return (cubicVal - lPtr[0]) / (lPtr[2] - lPtr[0]);
+}
+
+private:
+
+const Cubic& cubic;
+const _Line& line;
+Intersections& intersections;
+bool moreHorizontal;
+
+};
+ 
+bool intersectStart(const Cubic& cubic, const _Line& line, Intersections& i) {
+    LineCubicIntersections c(cubic, line, i);
+    return c.intersect();
+}