static inline void
FindInflectionApproximationRange(BezierControlPoints aControlPoints,
Float *aMin, Float *aMax, Float aT,
Float aTolerance)
{
SplitBezier(aControlPoints, nullptr, &aControlPoints, aT);
Point cp21 = aControlPoints.mCP2 - aControlPoints.mCP1;
Point cp41 = aControlPoints.mCP4 - aControlPoints.mCP1;
if (cp21.x == 0 && cp21.y == 0) {
// In this case s3 becomes lim[n->0] (cp41.x * n) / n - (cp41.y * n) / n = cp41.x - cp41.y.
*aMin = aT - CubicRoot(double(aTolerance / (cp41.x - cp41.y)));
*aMax = aT + CubicRoot(aTolerance / (cp41.x - cp41.y));
return;
}
Float s3 = (cp41.x * cp21.y - cp41.y * cp21.x) / hypotf(cp21.x, cp21.y);
if (s3 == 0) {
// This means within the precision we have it can be approximated
// infinitely by a linear segment. Deal with this by specifying the
// approximation range as extending beyond the entire curve.
*aMin = -1.0f;
*aMax = 2.0f;
return;
}
Float tf = CubicRoot(abs(aTolerance / s3));
*aMin = aT - tf * (1 - aT);
*aMax = aT + tf * (1 - aT);
}
static float CubicRoot(float aValue) {
if (aValue < 0.0) {
return -CubicRoot(-aValue);
}
else {
return powf(aValue, 1.0f / 3.0f);
}
}
float SolveBezier(float x, float p0, float p1, float p2, float p3)
{
// check for valid f-curve
// we only take care of monotonic bezier curves, so there has to be exactly 1 real solution
tl_assert(p0 <= x && x <= p3);
tl_assert((p0 <= p1 && p1 <= p3) && (p0 <= p2 && p2 <= p3));
double a, b, c, t;
double x3 = -p0 + 3*p1 - 3*p2 + p3;
double x2 = 3*p0 - 6*p1 + 3*p2;
double x1 = -3*p0 + 3*p1;
double x0 = p0 - x;
if(x3 == 0.0 && x2 == 0.0)
{
// linear
// a*t + b = 0
a = x1;
b = x0;
if(a == 0.0)
return 0.0f;
else
return -b/a;
}
else if(x3 == 0.0)
{
// quadratic
// t*t + b*t +c = 0
b = x1/x2;
c = x0/x2;
if(c == 0.0)
return 0.0f;
double D = b*b - 4*c;
t = (-b + sqrt(D))/2;
if(0.0 <= t && t <= 1.0001f)
return t;
else
return (-b - sqrt(D))/2;
}
else
{
// cubic
// t*t*t + a*t*t + b*t*t + c = 0
a = x2 / x3;
b = x1 / x3;
c = x0 / x3;
// substitute t = y - a/3
double sub = a/3.0;
// depressed form x^3 + px + q = 0
// cardano's method
double p = b/3 - a*a/9;
double q = (2*a*a*a/27 - a*b/3 + c)/2;
double D = q*q + p*p*p;
if(D > 0.0)
{
// only one 'real' solution
double s = sqrt(D);
return CubicRoot(s-q) - CubicRoot(s+q) - sub;
}
else if(D == 0.0)
{
// one single, one double solution or triple solution
double s = CubicRoot(-q);
t = 2*s - sub;
if(0.0 <= t && t <= 1.0001f)
return t;
else
return (-s - sub);
}
else
{
// Casus irreductibilis ... ,_,
double phi = acos(-q / sqrt(-(p*p*p))) / 3;
double s = 2*sqrt(-p);
t = s*cos(phi) - sub;
if(0.0 <= t && t <= 1.0001f)
return t;
t = -s*cos(phi+pi/3) - sub;
if(0.0 <= t && t <= 1.0001f)
return t;
else
return -s*cos(phi-pi/3) - sub;
}
}
}
