Is it possible to get the boundary size of an object in an image same after rotation of the image by a particular angle?

Question

NAVYA MOHAN 2021-1-19

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此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/720649-is-it-possible-to-get-the-boundary-size-of-an-object-in-an-image-same-after-rotation-of-the-image-by

回答： Walter Roberson 2021-1-21

采纳的回答： Walter Roberson

square.bmp

在 MATLAB Online 中打开

I have a sample image which contains a square. I rotated the image through a range of angle 30 to 50?

When I take the boundary size of the object in the rotated image (for these angle) I am getting values lesser than the boundary size of the object in the original sample image.

I am attaching a sample code for the same:

======================================================

Irot_new =imread('N:\work\tex\paper\square.bmp');
Irot_new = rgb2gray(Irot_new);
Irot_new = imbinarize(Irot_new);
B = bwboundaries(Irot_new);
boundary = B{1}; 
 x = boundary(:,2);
 y = boundary(:,1);
 boundary_size = size(x)
  
Irot_new = imrotate(Irot_new,-40,'nearest','crop');
Irot_new = imbinarize(Irot_new);
figure
imshow(Irot_new);
Irot_new = bwboundaries(Irot_new);
boundary1 = Irot_new{1}; 
 x = boundary1(:,2);
 y = boundary1(:,1);
 boundary_size = size(x)
%======================================================
%I tried even without using the imrotate with the following code , but still the boundary size differ:
%======================================================
I= imread('N:\work\tex\paper\square.bmp');
I_rgb = rgb2gray(I);
image= imbinarize(I_rgb);
%image padding
[Rows, Cols] = size(image); 
Diagonal = sqrt(Rows^2 + Cols^2); 
RowPad = ceil(Diagonal - Rows) + 2;
ColPad = ceil(Diagonal - Cols) + 2;
imagepad = zeros(Rows+RowPad, Cols+ColPad);
imagepad(ceil(RowPad/2):(ceil(RowPad/2)+Rows-1),ceil(ColPad/2):(ceil(ColPad/2)+Cols-1)) = image;
t=30
for i=1:1:4
	rads=degtorad(t);
	%midpoints
	midx=ceil((size(imagepad,1)+1)/2);
	midy=ceil((size(imagepad,2)+1)/2);
	imagerot=zeros(size(imagepad));
	for i=1:size(imagerot,1)
		for j=1:size(imagerot,2)
			 x= (i-midx)*cos(rads)+(j-midy)*sin(rads);
			 y=-(i-midx)*sin(rads)+(j-midy)*cos(rads);
			 x=round(x)+midx;
			 y=round(y)+midy;
			 if (x>=1 && y>=1 && x<=size(imagepad,2) && y<=size(imagepad,1))
				  imagerot(i,j)=imagepad(x,y); % k degrees rotated image         
			 end
		end
	end
	 B = bwboundaries(imagerot);
	 boundary = B{1}; 
	 x = boundary(:,2);
	 y = boundary(:,1);
	 boundary_size = size(x)
	 figure,imagesc(imagerot);
	 colormap(gray(256));
	 t = t+10;
end

=========================================================================

Can you please suggest why the boundary size varies while its determination? Any solution to keep the boundary size constant.

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Answer 1

Walter Roberson 2021-1-21

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此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/720649-is-it-possible-to-get-the-boundary-size-of-an-object-in-an-image-same-after-rotation-of-the-image-by#answer_603358

在 MATLAB Online 中打开

Img =imread('square.bmp');
Imgg = rgb2gray(Img);
angles = 0:-10:-90;
nang = length(angles);
fignums = [1, 1, 1, 1, 2, 3, 1, 1, 1, 1];
bs = zeros(1, nang);
for angidx = 1 : length(angles)
    ang = angles(angidx);
    Irot_new = imrotate(Imgg,ang,'nearest','crop');
    Irot_bin = imbinarize(Irot_new);
    figure(fignums(angidx))
    imshow(Irot_bin);
    axis equal
    title(sprintf('rotated %d', ang))
    drawnow;
    Irot_new = bwboundaries(Irot_bin);
    boundary1 = Irot_new{1}; 
     x = boundary1(:,2);
     y = boundary1(:,1);
     boundary_size = size(x);
     bs(angidx) = boundary_size(1);
     fprintf('angle %d boundary length %d\n', ang, bs(angidx));
end
disp([angles; bs])

angle 0 boundary length 703

angle -10 boundary length 694

angle -20 boundary length 662

angle -30 boundary length 610

angle -40 boundary length 539

angle -50 boundary length 540

angle -60 boundary length 610

angle -70 boundary length 661

angle -80 boundary length 693

angle -90 boundary length 703

0 -10 -20 -30 -40 -50 -60 -70 -80 -90

703 694 662 610 539 540 610 661 693 703

So 40 and 50 are the same to within round-off error. They differ because your image is not exactly centered.

Your changes are not as regular as you were describing. If you look at the changes, you will see that the first difference is roughly 10, the second is roughly 30, the third is roughly 50, the fourth is roughly 70. Thus there is a distinct pattern, and the 40 and 50 are not unusual.

As to why the change gets faster as you increase the angle: that is just how trig works.

You could probably express the coordinates by formula.

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Answer 2

Image Analyst 2021-1-19

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此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/720649-is-it-possible-to-get-the-boundary-size-of-an-object-in-an-image-same-after-rotation-of-the-image-by#answer_601304

The number of elements that describe a boundary will of course vary. Just think about it a little and you'll realize why. If you want the length of the boundary, you'll have to use sqrt() and the Pythagorean theorem to get the Euclidean distance. Even then, the total boundary length may vary by angle somewhat due to quantization of the coordinates into discrete integer rows and columns.

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NAVYA MOHAN 2021-1-19

编辑：NAVYA MOHAN 2021-1-19

Ok @Image Analyst. But why is that it gives same count for boundary coordinates if the angle of rotation is near to the multiples of 90.

I am getting the boundary length almost same for 0,10,20,30 and then at 70,80,90.. but it varies for 40,50 and 60? For these angles(40,50,60) they look almost same without any variation in size!!

Could you please guide on this?

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