I want to find end to end distance of this skeleton image

filename = 'https://www.mathworks.com/matlabcentral/answers/uploaded_files/1045155/insect.jpg';

k = imread(filename);
k = rgb2gray(k);
%imshow(insect);
%applyig threshold to convert the image into binary so that backgroung can
%be separated
BI = k < 160;
%imshow(BI)
BI = bwareaopen(BI,40);
%imshow(BI)
%finding the skeleton of image
k = bwskel(BI);
%imshow(k);
[a b]=size(k);
output=zeros(a,b);
for i=2:a-1
    for j=2:b-1
        ws=[k(i-1,j-1),k(i-1,j),k(i-1,j+1),k(i,j-1),k(i,j)...
            ,k(i,j+1),k(i+1,j-1),k(i+1,j),k(i+1,j+1)];
        kg=sum(ws);
        if(k(i,j)==1 && kg==2)
            output(i,j)=1;
        end
    end
end
figure;
output=imdilate(output,strel('diamond',5));
imshow(output+k);

i have this code with which i highlighted the end points but dont know how to go further

[Walter Roberson]

It is not obvious to me which are the endpoints, unless you mean the endpoints of every one of the (sometimes very short) line segments ?

[Adarsh Tripathi]

Thanks for your reply @walter Roberson Actually I attached the image before the code named insect in which end points are clearly specified i don't know how this below image is visible. And also just after sometime of posting i solved it myself but again thanks for your early reply.

Answer 1

[Image Analyst]

What exactly is the end to end distance? Do you mean the Feret diameter bwferet?

Do you mean the sum of all white pixels to the neighbor pixels? Have you tried pdist2, and extract all pairs just 1 or sqrt(2) pixels apart, sum them, and divide by 2? Something like (untested):

[y, x] = find(k); % Find coordinates.
xy = [x(:), y(:)]; % Combine into matrix.
distances = pdist2(xy, xy); % Find distance of every point to every other point.
mask = distances < mean([sqrt(2), sqrt(3)]); % Find pixels that are adjacent or on nearest diagonal ONLY.
sumOfDistances = sum(distances(mask)) / 2 % Divide by 2 since we get the sum for each endpoint so it's counted twice.

[Adarsh Tripathi]

Thanks for your reply @image analyst Actually I attached the image before the code named insect in which end points are clearly specified i don't know how this below image is visible. And also just after sometime of posting i solved it myself but again thanks for your early reply.

[Image Analyst]

What was your solution? Your original code is not it.

[Adarsh Tripathi]

ds = imageDatastore("frames/");

for l = 1:1198

img = readimage(ds,l);

%k = imread(img);

k = rgb2gray(img);

%imshow(insect);

%applyig threshold to convert the image into binary so that backgroung can

%be separated

BI = k < 160;

%imshow(BI)

BI = bwareaopen(BI,40);

%imshow(BI)

%finding the skeleton of image

k = bwskel(BI);

%imshow(k);

[a, b]=size(k);

output=zeros(a,b);

for i=2:a-1

for j=2:b-1

ws=[k(i-1,j-1),k(i-1,j),k(i-1,j+1),k(i,j-1),k(i,j)...

,k(i,j+1),k(i+1,j-1),k(i+1,j),k(i+1,j+1)];

kg=sum(ws);

if(k(i,j)==1 && kg==2)

output(i,j) = 1;

[m,n] = find(output)

end

end

end

distance = sqrt((n(1)-n(2))^2+(((m(1)-m(2))^2)))

end

this is my solution but it still creating problems and not applicable on all images

[Image Analyst]

That's so obviously not right. Not sure why you originally thought so. Just look at what you're doing. You're scanning the image with a 3x3 window and counting the number of pixels in each window if the center pixel is 1 and the count is 2, add 1 to the output image there. What about if there are 3 pixels in the window? Like along a diagonal would be okay. What about 3 where the two outer pixels are touching each other? That's probably not okay and should not be counted. (Maybe that would never occur with a skeleton -- I'd have to double check) So I hope you now realize why I said to identify the pattern that are okay and only count those. (Or I know I told someone that yesterday, maybe it was not you). There will be 8 patterns with a single pixel in the perimeter, and 8*5=40 patterns where there are two non-touching pixels in the perimeter. So there are 48 patterns possible that you would consider and 208 that you'd ignore (even if they occur, which they might not).

And by the way, your distance =sqrt() is completely wrong. It looks just at the first two pixels, not all of them.

[Adarsh Tripathi]

thanks for your reply @Image Analystit is really helpfull so now let me state my real problem i am doing a science project in which i have 36000 images of a insect extracted from 10 minutes video in which the insect is changing its postion and i have to find end to end distance of the insect for every image after making skeleton i am completely stuck in it from one week the above code i posted in that i thought one way to specify the end pixels is that they have only one white nieghbour and therefore nieghbouring pixel count will 2 and by this condition i can specify the coordinates of them and then find distance, as you know its completely wrong and not working the problem i am facing is that i am getting sometime only one pair of x,y for some image and sometime i am getting 4 or 5 five pairs for some image while for every time i should get only two pair for every image.that condition which i apllied for m(2) and n(2) in distance formula is for this that getting four or five pairs.i am attachimg some images also.i am new to matlab please help me ill be grate full to you.

[Image Analyst]

What is the end-to-end distance? Do you just want to find all endpoints (like tips of the worm-like thing) and find the distance between the two that are farthest apart? Hopefully this is what you want because this is really easy with bwferet, or bwmorph and pdist2

Or do you want to find the length of the blob? But the blob overlaps, The width of the worm doesn't change so a reasonable metric might be to just track the total area of the blob across frame numbers. If you want it to be super accurate, you can find crossing points and add in an extra bit of area for those locations.

[Adarsh Tripathi]

please see the attached image i want to find this

Do you just want to find all endpoints (like tips of the worm-like thing) and find the distance between the two that are farthest apart? Hopefully this is what you want because this is really easy with bwferet, or bwmorph and pdist2

yes thats what i want but end points should be two only ?

please guid me how i can do this because i am new to matlab so i dont know even easy things properly.

[Walter Roberson]

I wonder if you could describe why that distance is significant? I would have expected it to be more significant to trace along the curve, to measure the length of the creature, instead of calculating something about how much it is curled up at the moment?

[Adarsh Tripathi]

@walter Roberson That is significant because changing of that distance will give the activity of the worm with time.