CODEWITHSUNDEEP
matlabimage-processingcomputer-visionlinear-algebra
mash
mash

Reputation: 489

Find enclosing rectangle of image object using second moments

My understanding is the second central moment should give me an object's variance. Which I should be able to use as a measure of distribution of pixels from the center of the object, and so I would think I could find the size of the enclosing rectangle of an object using the second moments. Say in Malab I create an image with a white rectangle of dimensions 100x200 and find the center of mass using the first moments, then I calculate the variance

im = zeros(501,400);
im(200:300,100:300) = ones(101,201);
%Sum the moments in a for loop for clarity
for v = 1:rows  
  for u = 1:cols
    val = im(v,u);
    %zeroth moment is just the total pixel count
    m00 = m00 + val;        
    %First moments. pixel value times position is the m10 and m01
    m01 = m01 + (val * v);
    m10 = m10 + (val * u);
   end
end
%centers of mass
uc = m10/m00;
vc = m01/m00;
%Now I find the second central moments wrt centroid 
for v = 1:rows  
  for u = 1:cols
    val = im(v,u);
    u02 = u02 + (v - vc)^2 * val;
    u20 = u20 + (u - uc)^2 * val;
  end
end
hold on;
plot(uc,vc, 'r+');
text(uc + 5,vc + 5,strcat('X:',num2str(uc)),'Color','green','FontSize',12);
text(uc + 5,vc + 20,strcat('Y:',num2str(vc)),'Color','green','FontSize',12);

enter image description here

My value for u02 is 17255850, the square root of which is 4154. Is there a way I can directly get the approximate size of my rectangle using the variance? Thanks for any help.

EDIT: Experimenting with fitting an ellipse using the covariance matrix and trying to calculate the size using the variance as discussed in the comments and I get the image below. The dimensions of the yellow rectangle were calculated with

  %can we estimate size based on standard deviation?
stdX = sqrt(u20/m00);
stdY = sqrt(u02/m00);
rectangle('Position',[uc-2*stdX vc-2*stdY 4*stdX 4*stdY],'EdgeColor','yellow');

enter image description here

Upvotes: 1

Views: 510

Answers (1)

Cris Luengo
Cris Luengo

Reputation: 60680

After you've computed your normalized second order central moments, you get a matrix like this:

| u20/m00  u11/m00 |
| u11/m00  u02/m00 |

The two eigenvalues of this matrix provide a rotationally-invariant measure of the dimensions of the object.

%Normalized second order central moment tensor
mu = [u02/m00, u11/m00; u11/m00, u20/m00];
%The eigenvalues relate to the size
[V,D] = eig(mu);
l1 = D(1,1);
l2 = D(2,2);
%The eigenvectors relate to the orientation
phi = atan2(V(1,2),V(1,1));

Given the eigenvalues, we can determine the radii of the ellipse with the same normalized second order central moments:

a = 2*sqrt(l1);
b = 2*sqrt(l2);

Note that this is true only for a solid ellipse.

One can consider this the best fit ellipse, for some definition of "best".

We cannot determine the size of an arbitrary shape for these values, but if we know that our shape is a rectangle, then we can determine its sizes by knowing that the moments of inertia of a rectangle in 2D are (see Wikipedia, compare moments of 3D cuboid):

| 1/12 m h^2       0     |
|      0      1/12 m w^2 |

Where m is the mass, which in our case is 1 since we already normalized the moments.

Thus,

w = sqrt(12*l1);
h = sqrt(12*l2);

which gives us w = 100.9950 and h = 200.9975 for the case in the OP.

Upvotes: 1

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