Average of two closed curves

Dear everyone,

Is there a way in Matlab to average two closed curves? say I want to take the average of a unit square and a unit circle? How do I do this in Matlab?

[KALYAN ACHARJYA]

It would be better to undestand the problem, if you show the same in mathematical expression?

[KSSV]

Read about mean

Answer 1

[Stephan]

Hi,

try:

% Calculate Parts to 90°
angle1 = linspace(0,pi/4,180)';
% Radius of circle = length of square = 1
r = 1;
% Calculate the mean 
r_mean = ((sqrt(r.*tan(angle1).^2 +r.^2))+r)./2;
% Calculate the coordinates
x = [r_mean.*cos(angle1); r_mean.*cos(angle1)];
y = [r_mean.*sin(angle1); -r_mean.*sin(angle1)];
% Build the vectors to plot
xvals = [x; -x; y; y];
yvals = [y; y; -x; x];
% plot
scatter(xvals,yvals)

a further possibility is working with polyshape objects:

% construct the polyshape object
poly = polyshape(xvals, yvals);
pgon = simplify(convhull(poly));

% construct the square
rect = polyshape([-1 1 1 -1], [-1 -1 1 1]);

% construct the circle
t = 0:0.05:2*pi;
x1 = cos(t);
y1 = sin(t);
circ = polyshape(x1,y1);

% plot them all
hold on
plot(rect)
plot(pgon)
plot(circ)
hold off
xlim([-1.1 1.1])
ylim([-1.1 1.1])

which gives:

Best regards

Stephan

[Jullienne Franz]

Thank you very much for your suggestion Stephan. How do I extend the code to general cases? say I have two set of points, each forming two closed curves and I want to take their average creating a new closed curve.

[Stephan]

In your question it was possible to exploit symmetries and to base the computation on simple vector operations. I think it depends on how your points are awarded. Being able to apply the mean to vectors of different amounts but equal angles is easy. But we do not know much about your points and what you you mean by general case.

[John D'Errico]

+1. Yes. I was going to show how to use polyshape in this, but I see that Stephan has done a good job of it.

[Jullienne Franz]

Thank you Stephan for your suggestions. I will try to incorporate your ideas in my code and hope to resolve my problem.

Answer 2

[John D'Errico]

Given two closed curves, say a circle and a square, I would first formulate the problem as a pair of functions in polar coordinates. Thus, represent these two closed curves in polar form, where we will have r as a function of theta. Once you have those relationships, the problem is absolutely trivial.

So, for a unit curcle, we have

r1 = @(theta) ones(size(theta);

For a comparable square centered at the origin, thus of side length 2, I might get trickier...

r2 = @(theta) min(1./abs(cos(theta)), 1./abs(sin(theta)));
r3 = @(theta) (r1(theta) + r2(theta))/2;

Now, all is, as I said, trivial.

theta = linspace(0,2*pi,1000);
plot(cos(theta).*r1(theta),sin(theta).*r1(theta),'b-')
hold on
plot(cos(theta).*r2(theta),sin(theta).*r2(theta),'g-')
plot(cos(theta).*r3(theta),sin(theta).*r3(theta),'r-')
axis equal
grid on

As I said, trivial.

The problem becomes considerably less trivial if you have some fully general closed curve. For example...

pxy = randn(10,2);
pxy(end+1,:) = pxy(1,:);
plot(pxy(:,1),pxy(:,2),'o-')

You cannot dispute this is a closed curve. Yet interpolation along the curve is now a bit more problematic. Thus conversion to polar coordinates is now a bit less useful. Even a simple figure 8 curve is now less automatic. So the complexity of that closed curve is important. Can it be simply represented as a single valued parametric function in polar form? If so, then all is trivial again. If not, then you need to decide how you wish to interpolate, and what is the meaning of that mean in your complex case.

Otherwise, you are doing something possibly no more meaningful than taking the average of apples and oranges, the result may be just grape juice.

[Jullienne Franz]

Thank you John for your insightful comment. Actually I have two irregular simple closed curves (star shapes) and I only have the coordinates of their boundaries. I get your point that everything is trivial when they can be represented by a single valued parametric function in polar form. So, maybe I could take the centroid of the two shapes and from there calculate the mean distance of the points lying on the same ray.

[John D'Errico]

Yup. That would be entirely adequate, as good as anything else. As long as they contain some point that can serve as an origin, then rotating that ray around the center is no different from what I did.

[Jullienne Franz]

Thank you John for your suggestions. I will try to work on these ideas and hope to resolve my problem.