remove prototype pathops code

experimental/Intersection/Extrema.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 "DataTypes.h"
-#include "Extrema.h"
-
-static int validUnitDivide(double numer, double denom, double* ratio)
-{
-    if (numer < 0) {
-        numer = -numer;
-        denom = -denom;
-    }
-    if (denom == 0 || numer == 0 || numer >= denom)
-        return 0;
-    double r = numer / denom;
-    if (r == 0) { // catch underflow if numer <<<< denom
-        return 0;
-    }
-    *ratio = r;
-    return 1;
-}
-
-/** From Numerical Recipes in C.
-
-    Q = -1/2 (B + sign(B) sqrt[B*B - 4*A*C])
-    x1 = Q / A
-    x2 = C / Q
-*/
-static int findUnitQuadRoots(double A, double B, double C, double roots[2])
-{
-    if (A == 0)
-        return validUnitDivide(-C, B, roots);
-
-    double* r = roots;
-
-    double R = B*B - 4*A*C;
-    if (R < 0) {  // complex roots
-        return 0;
-    }
-    R = sqrt(R);
-
-    double Q = (B < 0) ? -(B-R)/2 : -(B+R)/2;
-    r += validUnitDivide(Q, A, r);
-    r += validUnitDivide(C, Q, r);
-    if (r - roots == 2 && AlmostEqualUlps(roots[0], roots[1])) { // nearly-equal?
-        r -= 1; // skip the double root
-    }
-    return (int)(r - roots);
-}
-
-/** Cubic'(t) = At^2 + Bt + C, where
-    A = 3(-a + 3(b - c) + d)
-    B = 6(a - 2b + c)
-    C = 3(b - a)
-    Solve for t, keeping only those that fit between 0 < t < 1
-*/
-int findExtrema(double a, double b, double c, double d, double tValues[2])
-{
-    // we divide A,B,C by 3 to simplify
-    double A = d - a + 3*(b - c);
-    double B = 2*(a - b - b + c);
-    double C = b - a;
-
-    return findUnitQuadRoots(A, B, C, tValues);
-}
-
-/** Quad'(t) = At + B, where
-    A = 2(a - 2b + c)
-    B = 2(b - a)
-    Solve for t, only if it fits between 0 < t < 1
-*/
-int findExtrema(double a, double b, double c, double tValue[1])
-{
-    /*  At + B == 0
-        t = -B / A
-    */
-    return validUnitDivide(a - b, a - b - b + c, tValue);
-}