Connecting points on opposite end of mask

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kunal
kunal on 19 Jun 2025 at 13:43
Commented: Image Analyst on 24 Jun 2025 at 2:55
Given an mask image i have certain points as seen in the image, I want to connect the point to the point at the opposite side of it given the list of points/array of point.Well the output would look something like.
  9 Comments
kunal
kunal on 23 Jun 2025 at 4:59
The red points are the corner points where there is a drastic change in the angles and i am later using using connected point to divide this mask area in different 4 point segments that will be similar to rectangle.
Image Analyst
Image Analyst on 24 Jun 2025 at 2:55
I don't think this addressed any of the numbered questions. I don't see how that polygon would be divided into 4 rectangles. Anyway, if you go to the corners, the tips of the thick lines will be pointy arrowheads, not rectangles. But let's say you want to split that thick white line/polygon into 4 or more smaller rectangles. Why? And then what are you going to do with those rectangular masks?
I'm just trying to help but want to make sure you don't go on a wild goose chase for some solution you think will work but might not work.

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Accepted Answer

Matt J
Matt J on 20 Jun 2025 at 14:20
Edited: Matt J on 23 Jun 2025 at 11:33
load('coping_data (1).mat')
img=logical(img);
n=height(points);
f=@(z,n) circshift(z,-n);
mod1=@(z) mod(z-1,n) + 1;
x=points(:,1)'; dx=diff([x,x(1)]);
y=points(:,2)'; dy=diff([y,y(1)]);
i=find(dy<-5 & f(dx,1)<-10 & f(dy,2)>+5 & abs(f(dx,1))<50 ,1);
I=mod1( i+2:i+n/2+1 );
J=flip( mod1( I(end)+(1:n/2) ) );
X=[x(I);x(J)]; Y=[y(I);y(J)];
imshow(img,[])
line(X,Y,'Color','r','LineWidth',3)

More Answers (1)

Image Analyst
Image Analyst on 20 Jun 2025 at 17:54
If you can get the inner boundary and outer boundary as separate arrays, then it would be easy to find the boundary point closest to the red points by using pdist2. If you can't figure out pdist2 then I can do it if you give me the inside boundary and outside boundary.

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