OpenCV/C - Improve the speed of a time consuming loop? - c++

I've recently implemented an image warping method using OpenCV 2.31(C++). However, this
method is quite time consuming... After some investigations and improvement
I succeed to reduce the processing time from 400ms to about 120ms which is quite nice.
I achieve this result by unrolling the loop (It reduces the time from 400ms to 330ms)
then I enabled optimization flags on my VC++ compiler 2008 express edition (Enabled O2 flag) - this last fix improved the processing to around 120ms.
However, since I have some other processing to implement around this warp, I'd like to reduce down this processing time even more to 20ms - lower than this value will be better of course, but I don't know if it's possible!!!...
One more thing, I'd like to do this using freely available libraries.
All suggestions are more than welcomed.
Bellow, you'll find the method I'm speaking about.
thanks for your help
Ariel B.
cv::Mat Warp::pieceWiseWarp(const cv::Mat &Isource, const cv::Mat &s, TYPE_CONVERSION type)
{
cv::Mat Idest(roi.height,roi.width,Isource.type(),cv::Scalar::all(0));
float xi, xj, xk, yi, yj, yk, x, y;
float X2X1,Y2Y1,X2X,Y2Y,XX1,YY1,X2X1_Y2Y1,a1, a2, a3, a4,b1,b2,c1,c2;
int x1, y1, x2, y2;
char k;
int nc = roi.width;
int nr = roi.height;
int channels = Isource.channels();
int N = nr * nc;
float *alphaPtr = alpha.ptr<float>(0);
float *betaPtr = beta.ptr<float>(0);
char *triMaskPtr = triMask.ptr<char>(0);
uchar *IdestPtr = Idest.data;
for(int i = 0; i < N; i++, IdestPtr += channels - 1)
if((k = triMaskPtr[i]) != -1)// the pixel do belong to delaunay
{
cv::Vec3b t = trianglesMap.row(k);
xi = s.col(1).at<float>(t[0]); yi = s.col(0).at<float>(t[0]);
xj = s.col(1).at<float>(t[1]); yj = s.col(0).at<float>(t[1]);
xk = s.col(1).at<float>(t[2]); yk = s.col(0).at<float>(t[2]);
x = xi + alphaPtr[i]*(xj - xi) + betaPtr[i]*(xk - xi);
y = yi + alphaPtr[i]*(yj - yi) + betaPtr[i]*(yk - yi);
//...some bounds checking here...
x2 = ceil(x); x1 = floor(x);
y2 = ceil(y); y1 = floor(y);
//2. use bilinear interpolation on the pixel location - see wiki for formula...
//...3. copy the resulting intensity (GL) to the destination (i,j)
X2X1 = (x2 - x1);
Y2Y1 = (y2 - y1);
X2X = (x2 - x);
Y2Y = (y2 - y);
XX1 = (x - x1);
YY1 = (y - y1);
X2X1_Y2Y1 = X2X1*Y2Y1;
a1 = (X2X*Y2Y)/(X2X1_Y2Y1);
a2 = (XX1*Y2Y)/(X2X1_Y2Y1);
a3 = (X2X*YY1)/(X2X1_Y2Y1);
a4 = (XX1*YY1)/(X2X1_Y2Y1);
b1 = (X2X/X2X1);
b2 = (XX1/X2X1);
c1 = (Y2Y/Y2Y1);
c2 = (YY1/Y2Y1);
for(int c = 0; c < channels; c++)// Consider implementing this bilinear interpolation elsewhere in another function
{
if(x1 != x2 && y1 != y2)
IdestPtr[i + c] = Isource.at<cv::Vec3b>(y1,x1)[c]*a1
+ Isource.at<cv::Vec3b>(y2,x1)[c]*a2
+ Isource.at<cv::Vec3b>(y1,x2)[c]*a3
+ Isource.at<cv::Vec3b>(y2,x2)[c]*a4;
if(x1 == x2 && y1 == y2)
IdestPtr[i + c] = Isource.at<cv::Vec3b>(y1,x1)[c];
if(x1 != x2 && y1 == y2)
IdestPtr[i + c] = Isource.at<cv::Vec3b>(y1,x1)[c]*b1 + Isource.at<cv::Vec3b>(y1,x2)[c]*b2;
if(x1 == x2 && y1 != y2)
IdestPtr[i + c] = Isource.at<cv::Vec3b>(y1,x1)[c]*c1 + Isource.at<cv::Vec3b>(y2,x1)[c]*c2;
}
}
if(type == CONVERT_TO_CV_32FC3)
Idest.convertTo(Idest,CV_32FC3);
if(type == NORMALIZE_TO_1)
Idest.convertTo(Idest,CV_32FC3,1/255.);
return Idest;
}

I would suggest:
1.Change division by a common factor to multiplication.
i.e. from a = a1/d; b = b1/d to d_1 = 1/d; a = a1*d_1; b = b1*d_1
2.Eliminate the four if test to a single bilinear interpolation.
I'm not sure whether that would help you. You could have a try.

Related

Fsolve equivalent in C++

I am trying to replicate Matlab's Fsolve as my project is in C++ solving an implicit RK4 scheme. I am using the NLopt library using the NLOPT_LD_MMA algorithm. I have run the required section in matlab and it is considerably faster. I was wondering whether anyone had any ideas of a better Fsolve equivalent in C++? Another reason is that I would like f1 and f2 to both tend to zero and it seems suboptimal to calculate the L2 norm to include both of them as NLopt seems to only allow a scalar return value from the objective function. Does anyone have any ideas of an alternative library or perhaps using a different algorithm/constraints to more closely replicate the default fsolve.
Would it be better (faster) perhaps to call the python scipy.minimise.fsolve from C++?
double implicitRK4(double time, double V, double dt, double I, double O, double C, double R){
const int number_of_parameters = 2;
double lb[number_of_parameters];
double ub[number_of_parameters];
lb[0] = -999; // k1 lb
lb[1] = -999;// k2 lb
ub[0] = 999; // k1 ub
ub[1] = 999; // k2 ub
double k [number_of_parameters];
k[0] = 0.01;
k[1] = 0.01;
kOptData addData(time,V,dt,I,O,C,R);
nlopt_opt opt; //NLOPT_LN_MMA NLOPT_LN_COBYLA
opt = nlopt_create(NLOPT_LD_MMA, number_of_parameters);
nlopt_set_lower_bounds(opt, lb);
nlopt_set_upper_bounds(opt, ub);
nlopt_result nlopt_remove_inequality_constraints(nlopt_opt opt);
// nlopt_result nlopt_remove_equality_constraints(nlopt_opt opt);
nlopt_set_min_objective(opt,solveKs,&addData);
double minf;
if (nlopt_optimize(opt, k, &minf) < 0) {
printf("nlopt failed!\n");
}
else {
printf("found minimum at f(%g,%g,%g) = %0.10g\n", k[0],k[1],minf);
}
nlopt_destroy(opt);
return V + (1/2)*dt*k[0] + (1/2)*dt*k[1];```
double solveKs(unsigned n, const double *x, double *grad, void *my_func_data){
kOptData *unpackdata = (kOptData*) my_func_data;
double t1,y1,t2,y2;
double f1,f2;
t1 = unpackdata->time + ((1/2)-(1/6)*sqrt(3));
y1 = unpackdata->V + (1/4)*unpackdata->dt*x[0] + ((1/4)-(1/6)*sqrt(3))*unpackdata->dt*x[1];
t2 = unpackdata->time + ((1/2)+(1/6)*sqrt(3));
y2 = unpackdata->V + ((1/4)+(1/6)*sqrt(3))*unpackdata->dt*x[0] + (1/4)*unpackdata->dt*x[1];
f1 = x[0] - stateDeriv_implicit(t1,y1,unpackdata->dt,unpackdata->I,unpackdata->O,unpackdata->C,unpackdata->R);
f2 = x[1] - stateDeriv_implicit(t2,y2,unpackdata->dt,unpackdata->I,unpackdata->O,unpackdata->C,unpackdata->R);
return sqrt(pow(f1,2) + pow(f2,2));
My matlab version below seems to be a lot simpler but I would prefer the whole code in c++!
k1 = 0.01;
k2 = 0.01;
x0 = [k1,k2];
fun = #(x)solveKs(x,t,z,h,I,OCV1,Cap,Rct,static);
options = optimoptions('fsolve','Display','none');
k = fsolve(fun,x0,options);
% Calculate the next state vector from the previous one using RungeKutta
% update equation
znext = z + (1/2)*h*k(1) + (1/2)*h*k(2);``
function [F] = solveKs(x,t,z,h,I,O,C,R,static)
t1 = t + ((1/2)-(1/6)*sqrt(3));
y1 = z + (1/4)*h*x(1) + ((1/4)-(1/6)*sqrt(3))*h *x(2);
t2 = t + ((1/2)+(1/6)*sqrt(3));
y2 = z + ((1/4)+(1/6)*sqrt(3))*h*x(1) + (1/4)*h*x(2);
F(1) = x(1) - stateDeriv_implicit(t1,y1,h,I,O,C,R,static);
F(2) = x(2) - stateDeriv_implicit(t2,y2,h,I,O,C,R,static);
end

nFinding Xcrossing point using Data Points

I was going through Algorithm which identify crossing point using data points on x axis. I can understand calculation but could not understand purpose of this calculation. As per my understanding, it determine each time new y point and slope and subtract them.
I want to insert Image also but I do not have 10 reputation point.
Please let me know if I need to provide info.
//This function determine the crossing point when the graph is intersecting x axis.
XPoint findXingPoint(Curve & curve, double rfix, double vcc, int type, int pull_up)
{
//curve class contain x and y data point
// rfix is fixed value which is generally 50
// vcc also fix value
// type contain 0 and 1
//pull up to identify graph
XPoint p = { 0.0, 0.0 };
double r_fix = rfix;
double m = -1 / r_fix;
double c = vcc / r_fix;
if(type)
c=0;
double r, s, X1, X2, Y1, Y2, Y3, Y4;
r = (m * curve[0].first) + c;
// r is a kind of y value which determine from x and y point
s = abs(r);
for (Curve::iterator i = curve.begin(); i != curve.end(); i++) {
curve_point p = (*i);
double xcurve = p.first;
double ycurve = p.second;
double yloadline = m * xcurve + c;
double B = ycurve - yloadline;
if (B >= 0 && r >= B) {
r = B;
X1 = xcurve;
Y1 = yloadline;
Y2 = ycurve;
}
if (B <= 0 && r >= abs(B)) {
s = abs(B);
X2 = xcurve;
Y4 = yloadline;
Y3 = ycurve;
}
}
#could not understand purpose of B calculation
if (s == 0)
X1 = X2;
if (r == 0)
X2 = X1;
if (X1 != X2) {
double m1, m2, c1, c2;
m1 = (Y3 - Y2) / (X2 - X1);
m2 = (Y4 - Y1) / (X2 - X1);
c1 = Y3 - (m1 * X2);
c2 = Y4 - (m2 * X2);
// CASE m1==m2 should be handled.
p.x = (c2 - c1) / (m1 - m2);
p.y = (m2 * p.x) + c2;
} else {
p.x = X1;
p.y = Y1;
}
#not able to also understand calculation
if (verbosityValue >= 1)
loginfo<<"found crossing point # " << p.x << " " << p.y << endl;
return p;
}
Output:
first
found crossing point # 7.84541e-08 -1.96135e-09 with type 0
found crossing point # 0.528564 0.0182859 with type 1
second
found crossing point # 0.654357 -0.0163589 with type 0
found crossing point # 1.25827 4.31937e-05 with type 1
This appears to be a straightforward implementation. For a given point x,y on the curve, find the slope, draw a line through the slope, find where that line would cross the x axis, and then find the new y value at that point. For well-behaved functions, the new y value is a lot closer to 0 than the initial y value. You iterate until the approximation is good enough.
If you look at the graph, you see that the middle part is quite straight. Hence, the slope of the line is a locally good approximation of the curve, and your new y value is a lot closer to zero (At least 10x times, probably 100x, looking at the graph.0. If you start on the the steeper slopes on either side, you'll need one extra step. Your second x point will be on the middle part.

distance from given point to given ellipse

I have an ellipse, defined by Center Point, radiusX and radiusY, and I have a Point. I want to find the point on the ellipse that is closest to the given point. In the illustration below, that would be S1.
Now I already have code, but there is a logical error somewhere in it, and I seem to be unable to find it. I broke the problem down to the following code example:
#include <vector>
#include <opencv2/core/core.hpp>
#include <opencv2/highgui/highgui.hpp>
#include <math.h>
using namespace std;
void dostuff();
int main()
{
dostuff();
return 0;
}
typedef std::vector<cv::Point> vectorOfCvPoints;
void dostuff()
{
const double ellipseCenterX = 250;
const double ellipseCenterY = 250;
const double ellipseRadiusX = 150;
const double ellipseRadiusY = 100;
vectorOfCvPoints datapoints;
for (int i = 0; i < 360; i+=5)
{
double angle = i / 180.0 * CV_PI;
double x = ellipseRadiusX * cos(angle);
double y = ellipseRadiusY * sin(angle);
x *= 1.4;
y *= 1.4;
x += ellipseCenterX;
y += ellipseCenterY;
datapoints.push_back(cv::Point(x,y));
}
cv::Mat drawing = cv::Mat::zeros( 500, 500, CV_8UC1 );
for (int i = 0; i < datapoints.size(); i++)
{
const cv::Point & curPoint = datapoints[i];
const double curPointX = curPoint.x;
const double curPointY = curPoint.y * -1; //transform from image coordinates to geometric coordinates
double angleToEllipseCenter = atan2(curPointY - ellipseCenterY * -1, curPointX - ellipseCenterX); //ellipseCenterY * -1 for transformation to geometric coords (from image coords)
double nearestEllipseX = ellipseCenterX + ellipseRadiusX * cos(angleToEllipseCenter);
double nearestEllipseY = ellipseCenterY * -1 + ellipseRadiusY * sin(angleToEllipseCenter); //ellipseCenterY * -1 for transformation to geometric coords (from image coords)
cv::Point center(ellipseCenterX, ellipseCenterY);
cv::Size axes(ellipseRadiusX, ellipseRadiusY);
cv::ellipse(drawing, center, axes, 0, 0, 360, cv::Scalar(255));
cv::line(drawing, curPoint, cv::Point(nearestEllipseX,nearestEllipseY*-1), cv::Scalar(180));
}
cv::namedWindow( "ellipse", CV_WINDOW_AUTOSIZE );
cv::imshow( "ellipse", drawing );
cv::waitKey(0);
}
It produces the following image:
You can see that it actually finds "near" points on the ellipse, but it are not the "nearest" points. What I intentionally want is this: (excuse my poor drawing)
would you extent the lines in the last image, they would cross the center of the ellipse, but this is not the case for the lines in the previous image.
I hope you get the picture. Can anyone tell me what I am doing wrong?
Consider a bounding circle around the given point (c, d), which passes through the nearest point on the ellipse. From the diagram it is clear that the closest point is such that a line drawn from it to the given point must be perpendicular to the shared tangent of the ellipse and circle. Any other points would be outside the circle and so must be further away from the given point.
So the point you are looking for is not the intersection between the line and the ellipse, but the point (x, y) in the diagram.
Gradient of tangent:
Gradient of line:
Condition for perpedicular lines - product of gradients = -1:
When rearranged and substituted into the equation of your ellipse...
...this will give two nasty quartic (4th-degree polynomial) equations in terms of either x or y. AFAIK there are no general analytical (exact algebraic) methods to solve them. You could try an iterative method - look up the Newton-Raphson iterative root-finding algorithm.
Take a look at this very good paper on the subject:
http://www.spaceroots.org/documents/distance/distance-to-ellipse.pdf
Sorry for the incomplete answer - I totally blame the laws of mathematics and nature...
EDIT: oops, i seem to have a and b the wrong way round in the diagram xD
There is a relatively simple numerical method with better convergence than Newtons Method. I have a blog post about why it works http://wet-robots.ghost.io/simple-method-for-distance-to-ellipse/
This implementation works without any trig functions:
def solve(semi_major, semi_minor, p):
px = abs(p[0])
py = abs(p[1])
tx = 0.707
ty = 0.707
a = semi_major
b = semi_minor
for x in range(0, 3):
x = a * tx
y = b * ty
ex = (a*a - b*b) * tx**3 / a
ey = (b*b - a*a) * ty**3 / b
rx = x - ex
ry = y - ey
qx = px - ex
qy = py - ey
r = math.hypot(ry, rx)
q = math.hypot(qy, qx)
tx = min(1, max(0, (qx * r / q + ex) / a))
ty = min(1, max(0, (qy * r / q + ey) / b))
t = math.hypot(ty, tx)
tx /= t
ty /= t
return (math.copysign(a * tx, p[0]), math.copysign(b * ty, p[1]))
Credit to Adrian Stephens for the Trig-Free Optimization.
Here is the code translated to C# implemented from this paper to solve for the ellipse:
http://www.geometrictools.com/Documentation/DistancePointEllipseEllipsoid.pdf
Note that this code is untested - if you find any errors let me know.
//Pseudocode for robustly computing the closest ellipse point and distance to a query point. It
//is required that e0 >= e1 > 0, y0 >= 0, and y1 >= 0.
//e0,e1 = ellipse dimension 0 and 1, where 0 is greater and both are positive.
//y0,y1 = initial point on ellipse axis (center of ellipse is 0,0)
//x0,x1 = intersection point
double GetRoot ( double r0 , double z0 , double z1 , double g )
{
double n0 = r0*z0;
double s0 = z1 - 1;
double s1 = ( g < 0 ? 0 : Math.Sqrt(n0*n0+z1*z1) - 1 ) ;
double s = 0;
for ( int i = 0; i < maxIter; ++i ){
s = ( s0 + s1 ) / 2 ;
if ( s == s0 || s == s1 ) {break; }
double ratio0 = n0 /( s + r0 );
double ratio1 = z1 /( s + 1 );
g = ratio0*ratio0 + ratio1*ratio1 - 1 ;
if (g > 0) {s0 = s;} else if (g < 0) {s1 = s ;} else {break ;}
}
return s;
}
double DistancePointEllipse( double e0 , double e1 , double y0 , double y1 , out double x0 , out double x1)
{
double distance;
if ( y1 > 0){
if ( y0 > 0){
double z0 = y0 / e0;
double z1 = y1 / e1;
double g = z0*z0+z1*z1 - 1;
if ( g != 0){
double r0 = (e0/e1)*(e0/e1);
double sbar = GetRoot(r0 , z0 , z1 , g);
x0 = r0 * y0 /( sbar + r0 );
x1 = y1 /( sbar + 1 );
distance = Math.Sqrt( (x0-y0)*(x0-y0) + (x1-y1)*(x1-y1) );
}else{
x0 = y0;
x1 = y1;
distance = 0;
}
}
else // y0 == 0
x0 = 0 ; x1 = e1 ; distance = Math.Abs( y1 - e1 );
}else{ // y1 == 0
double numer0 = e0*y0 , denom0 = e0*e0 - e1*e1;
if ( numer0 < denom0 ){
double xde0 = numer0/denom0;
x0 = e0*xde0 ; x1 = e1*Math.Sqrt(1 - xde0*xde0 );
distance = Math.Sqrt( (x0-y0)*(x0-y0) + x1*x1 );
}else{
x0 = e0;
x1 = 0;
distance = Math.Abs( y0 - e0 );
}
}
return distance;
}
The following python code implements the equations described at "Distance from a Point to an Ellipse" and uses newton's method to find the roots and from that the closest point on the ellipse to the point.
Unfortunately, as can be seen from the example, it seems to only be accurate outside the ellipse. Within the ellipse weird things happen.
from math import sin, cos, atan2, pi, fabs
def ellipe_tan_dot(rx, ry, px, py, theta):
'''Dot product of the equation of the line formed by the point
with another point on the ellipse's boundary and the tangent of the ellipse
at that point on the boundary.
'''
return ((rx ** 2 - ry ** 2) * cos(theta) * sin(theta) -
px * rx * sin(theta) + py * ry * cos(theta))
def ellipe_tan_dot_derivative(rx, ry, px, py, theta):
'''The derivative of ellipe_tan_dot.
'''
return ((rx ** 2 - ry ** 2) * (cos(theta) ** 2 - sin(theta) ** 2) -
px * rx * cos(theta) - py * ry * sin(theta))
def estimate_distance(x, y, rx, ry, x0=0, y0=0, angle=0, error=1e-5):
'''Given a point (x, y), and an ellipse with major - minor axis (rx, ry),
its center at (x0, y0), and with a counter clockwise rotation of
`angle` degrees, will return the distance between the ellipse and the
closest point on the ellipses boundary.
'''
x -= x0
y -= y0
if angle:
# rotate the points onto an ellipse whose rx, and ry lay on the x, y
# axis
angle = -pi / 180. * angle
x, y = x * cos(angle) - y * sin(angle), x * sin(angle) + y * cos(angle)
theta = atan2(rx * y, ry * x)
while fabs(ellipe_tan_dot(rx, ry, x, y, theta)) > error:
theta -= ellipe_tan_dot(
rx, ry, x, y, theta) / \
ellipe_tan_dot_derivative(rx, ry, x, y, theta)
px, py = rx * cos(theta), ry * sin(theta)
return ((x - px) ** 2 + (y - py) ** 2) ** .5
Here's an example:
rx, ry = 12, 35 # major, minor ellipse axis
x0 = y0 = 50 # center point of the ellipse
angle = 45 # ellipse's rotation counter clockwise
sx, sy = s = 100, 100 # size of the canvas background
dist = np.zeros(s)
for x in range(sx):
for y in range(sy):
dist[x, y] = estimate_distance(x, y, rx, ry, x0, y0, angle)
plt.imshow(dist.T, extent=(0, sx, 0, sy), origin="lower")
plt.colorbar()
ax = plt.gca()
ellipse = Ellipse(xy=(x0, y0), width=2 * rx, height=2 * ry, angle=angle,
edgecolor='r', fc='None', linestyle='dashed')
ax.add_patch(ellipse)
plt.show()
Which generates an ellipse and the distance from the boundary of the ellipse as a heat map. As can be seen, at the boundary the distance is zero (deep blue).
Given an ellipse E in parametric form and a point P
the square of the distance between P and E(t) is
The minimum must satisfy
Using the trigonometric identities
and substituting
yields the following quartic equation:
Here's an example C function that solves the quartic directly and computes sin(t) and cos(t) for the nearest point on the ellipse:
void nearest(double a, double b, double x, double y, double *ecos_ret, double *esin_ret) {
double ax = fabs(a*x);
double by = fabs(b*y);
double r = b*b - a*a;
double c, d;
int switched = 0;
if (ax <= by) {
if (by == 0) {
if (r >= 0) { *ecos_ret = 1; *esin_ret = 0; }
else { *ecos_ret = 0; *esin_ret = 1; }
return;
}
c = (ax - r) / by;
d = (ax + r) / by;
} else {
c = (by + r) / ax;
d = (by - r) / ax;
switched = 1;
}
double cc = c*c;
double D0 = 12*(c*d + 1); // *-4
double D1 = 54*(d*d - cc); // *4
double D = D1*D1 + D0*D0*D0; // *16
double St;
if (D < 0) {
double t = sqrt(-D0); // *2
double phi = acos(D1 / (t*t*t));
St = 2*t*cos((1.0/3)*phi); // *2
} else {
double Q = cbrt(D1 + sqrt(D)); // *2
St = Q - D0 / Q; // *2
}
double p = 3*cc; // *-2
double SS = (1.0/3)*(p + St); // *4
double S = sqrt(SS); // *2
double q = 2*cc*c + 4*d; // *2
double l = sqrt(p - SS + q / S) - S - c; // *2
double ll = l*l; // *4
double ll4 = ll + 4; // *4
double esin = (4*l) / ll4;
double ecos = (4 - ll) / ll4;
if (switched) {
double t = esin;
esin = ecos;
ecos = t;
}
*ecos_ret = copysign(ecos, a*x);
*esin_ret = copysign(esin, b*y);
}
Try it online!
You just need to calculate the intersection of the line [P1,P0] to your elipse which is S1.
If the line equeation is:
and the elipse equesion is:
than the values of S1 will be:
Now you just need to calculate the distance between S1 to P1 , the formula (for A,B points) is:
I've solved the distance issue via focal points.
For every point on the ellipse
r1 + r2 = 2*a0
where
r1 - Euclidean distance from the given point to focal point 1
r2 - Euclidean distance from the given point to focal point 2
a0 - semimajor axis length
I can also calculate the r1 and r2 for any given point which gives me another ellipse that this point lies on that is concentric to the given ellipse. So the distance is
d = Abs((r1 + r2) / 2 - a0)
As propposed by user3235832
you shall solve quartic equation to find the normal to the ellipse (https://www.mathpages.com/home/kmath505/kmath505.htm). With good initial value only few iterations are needed (I use it myself). As an initial value I use S1 from your picture.
The fastest method I guess is
http://wwwf.imperial.ac.uk/~rn/distance2ellipse.pdf
Which has been mentioned also by Matt but as he found out the method doesn't work very well inside of ellipse.
The problem is the theta initialization.
I proposed an stable initialization:
Find the intersection of ellipse and horizontal line passing the point.
Find the other intersection using vertical line.
Choose the one that is closer the point.
Calculate the initial angle based on that point.
I got good results with no issue inside and outside:
As you can see in the following image it just iterated about 3 times to reach 1e-8. Close to axis it is 1 iteration.
The C++ code is here:
double initialAngle(double a, double b, double x, double y) {
auto abs_x = fabs(x);
auto abs_y = fabs(y);
bool isOutside = false;
if (abs_x > a || abs_y > b) isOutside = true;
double xd, yd;
if (!isOutside) {
xd = sqrt((1.0 - y * y / (b * b)) * (a * a));
if (abs_x > xd)
isOutside = true;
else {
yd = sqrt((1.0 - x * x / (a * a)) * (b * b));
if (abs_y > yd)
isOutside = true;
}
}
double t;
if (isOutside)
t = atan2(a * y, b * x); //The point is outside of ellipse
else {
//The point is inside
if (xd < yd) {
if (x < 0) xd = -xd;
t = atan2(y, xd);
}
else {
if (y < 0) yd = -yd;
t = atan2(yd, x);
}
}
return t;
}
double distanceToElipse(double a, double b, double x, double y, int maxIter = 10, double maxError = 1e-5) {
//std::cout <<"p="<< x << "," << y << std::endl;
auto a2mb2 = a * a - b * b;
double t = initialAngle(a, b, x, y);
auto ct = cos(t);
auto st = sin(t);
int i;
double err;
for (i = 0; i < maxIter; i++) {
auto f = a2mb2 * ct * st - x * a * st + y * b * ct;
auto fp = a2mb2 * (ct * ct - st * st) - x * a * ct - y * b * st;
auto t2 = t - f / fp;
err = fabs(t2 - t);
//std::cout << i + 1 << " " << err << std::endl;
t = t2;
ct = cos(t);
st = sin(t);
if (err < maxError) break;
}
auto dx = a * ct - x;
auto dy = b * st - y;
//std::cout << a * ct << "," << b * st << std::endl;
return sqrt(dx * dx + dy * dy);
}

How to calculate the point of intersection between two lines

I am attempting to calculate the point of intersection between lines for a Optical Flow algorithm using a Hough Transform. However, I am not getting the points that I should be when I use my algorithm for calculating the intersections.
I save the Lines as an instance of a class that I created called ImageLine. Here is the code for my intersection method.
Point ImageLine::intersectionWith(ImageLine other)
{
float A2 = other.Y2() - other.Y1();
float B2 = other.X2() - other.X1();
float C2 = A2*other.X1() + B2*other.Y1();
float A1 = y2 - y1;
float B1 = x2 - x1;
float C1 = A1 * x1 + B1 * y1;
float det = A1*B2 - A2*B1;
if (det == 0)
{
return Point(-1,-1);
}
Point d = Point((B2 * C1 - B1 * C2) / det, -(A1 * C2 - A2 * C1) / det);
return d;
}
Is this method correct, or did I do something wrong? As far as I can tell, it should work, as it does for a single point that I hard-coded through, however, I have not been able to get a good intersection when using real data.
Considering the maths side: if we have two line equations:
y = m1 * x + c1
y = m2 * x + c2
The point of intersection: (X , Y), of two lines described by the following equations:
Y = m1 * X + c1
Y = m2 * X + c2
is the point which satisfies both equation, i.e.:
m1 * X + c1 = m2 * X + c2
(Y - c1) / m1 = (Y - c2) / m2
thus the point of intersection coordinates are:
intersectionX = (c2 - c1) / (m1 - m2)
intersectionY = (m1*c1 - c2*m2) / m1-m2 or intersectionY = m1 * intersectionX + c1
Note: c1, m1 and c2, m2 are calculated by getting any 2 points of a line and putting them in the line equations.
(det == 0) is unlikely to be true when you're using floating-point arithmetic, because it isn't precise.
Something like (fabs(det) < epsilon) is commonly used, for some suitable value of epsilon (say, 1e-6).
If that doesn't fix it, show some actual numbers, along with the expected result and the actual result.
For detailed formula, please go to this page.
But I love code so, here, check this code (I get it from github, so all credit goes to the author of that code):
///Calculate intersection of two lines.
///\return true if found, false if not found or error
bool LineLineIntersect(double x1, double y1, //Line 1 start
double x2, double y2, //Line 1 end
double x3, double y3, //Line 2 start
double x4, double y4, //Line 2 end
double &ixOut, double &iyOut) //Output
{
//http://mathworld.wolfram.com/Line-LineIntersection.html
double detL1 = Det(x1, y1, x2, y2);
double detL2 = Det(x3, y3, x4, y4);
double x1mx2 = x1 - x2;
double x3mx4 = x3 - x4;
double y1my2 = y1 - y2;
double y3my4 = y3 - y4;
double xnom = Det(detL1, x1mx2, detL2, x3mx4);
double ynom = Det(detL1, y1my2, detL2, y3my4);
double denom = Det(x1mx2, y1my2, x3mx4, y3my4);
if(denom == 0.0)//Lines don't seem to cross
{
ixOut = NAN;
iyOut = NAN;
return false;
}
ixOut = xnom / denom;
iyOut = ynom / denom;
if(!isfinite(ixOut) || !isfinite(iyOut)) //Probably a numerical issue
return false;
return true; //All OK
}
Assuming your formulas are correct, try declaring all your intermediate arguments as 'double'. Taking the difference of nearly parallel lines could result in your products being very close to each other, so 'float' may not preserve enough precision.

can the CImg library draw thick lines

I have been using the CImg library, and have been pleased with how easy it is to integrate and use. However, I now want to draw thick lines (i.e., more than one pixel thick). It is not clear from the API documentation of the draw_line function (here) how this can be done. A second version of the function (just below the first in the documentation) even takes a texture as input, but again no width. It seems strange that such a comprehensive library would not have this feature. Perhaps it's supposed to be done using some kind of transformation? I know I could do it using a polygon (i.e., a rectangle where I would compute the corners of the polygon using a normal to the line), but I fear that would be significantly slower.
Apparently, it is not possible 'out-of-the-box', but creating your own routine that calls multiple times the 'draw_line()' routine of CImg, with one or two pixels shifts should give you the result you want, without much work.
This function can be used to draw thick lines as polygons.
void draw_line(cimg_library::CImg<uint8_t>& image,
const int x1, const int y1,
const int x2, const int y2,
const uint8_t* const color,
const unsigned int line_width)
{
if (x1 == x2 && y1 == y2) {
return;
}
// Convert line (p1, p2) to polygon (pa, pb, pc, pd)
const double x_diff = std::abs(x1 - x2);
const double y_diff = std::abs(y1 - y2);
const double w_diff = line_width / 2.0;
// Triangle between pa and p1: x_adj^2 + y_adj^2 = w_diff^2
// Triangle between p1 and p2: x_diff^2 + y_diff^2 = length^2
// Similar triangles: y_adj / x_diff = x_adj / y_diff = w_diff / length
// -> y_adj / x_diff = w_diff / sqrt(x_diff^2 + y_diff^2)
const int x_adj = y_diff * w_diff / std::sqrt(std::pow(x_diff, 2) + std::pow(y_diff, 2));
const int y_adj = x_diff * w_diff / std::sqrt(std::pow(x_diff, 2) + std::pow(y_diff, 2));
// Points are listed in clockwise order, starting from top-left
cimg_library::CImg<int> points(4, 2);
points(0, 0) = x1 - x_adj;
points(0, 1) = y1 + y_adj;
points(1, 0) = x1 + x_adj;
points(1, 1) = y1 - y_adj;
points(2, 0) = x2 + x_adj;
points(2, 1) = y2 - y_adj;
points(3, 0) = x2 - x_adj;
points(3, 1) = y2 + y_adj;
image.draw_polygon(points, color);
}
Benchmarks with line_width 20 and 3 colors. First time is using this function, second time is drawing single 1 px wide line using image.draw_line().
1000,1000 to 2000,2000: 216 µs / 123 µs
2000,2000 to 8000,4000: 588 µs / 151 µs
3000,1000 to 3020,1000: 21 µs / 5 µs
Basically this code does the same as #vll's answer, but also handles the case when (x1-x2)/(y1-y2) < 0 (I remove the abs function).
void draw_line(cimg_library::CImg<uint8_t>& image,
const int x1, const int y1,
const int x2, const int y2,
const uint8_t* const color,
const uint8_t line_width,
const double opacity=1.0)
{
if (x1 == x2 && y1 == y2) {
return;
}
// Convert line (p1, p2) to polygon (pa, pb, pc, pd)
const double x_diff = (x1 - x2);
const double y_diff = (y1 - y2);
const double w_diff = line_width / 2.0;
// Triangle between pa and p1: x_adj^2 + y_adj^2 = w_diff^2
// Triangle between p1 and p2: x_diff^2 + y_diff^2 = length^2
// Similar triangles: y_adj / x_diff = x_adj / y_diff = w_diff / length
// -> y_adj / x_diff = w_diff / sqrt(x_diff^2 + y_diff^2)
const int x_adj = y_diff * w_diff / std::sqrt(std::pow(x_diff, 2) + std::pow(y_diff, 2));
const int y_adj = x_diff * w_diff / std::sqrt(std::pow(x_diff, 2) + std::pow(y_diff, 2));
// Points are listed in clockwise order, starting from top-left
cimg_library::CImg<int> points(4, 2);
points(0, 0) = x1 - x_adj;
points(0, 1) = y1 + y_adj;
points(1, 0) = x1 + x_adj;
points(1, 1) = y1 - y_adj;
points(2, 0) = x2 + x_adj;
points(2, 1) = y2 - y_adj;
points(3, 0) = x2 - x_adj;
points(3, 1) = y2 + y_adj;
image.draw_polygon(points, color, opacity);
}