How can I detect a triangle in my image?

Please I need help for detect triangles in the image I attached. I don't know how to do it.

Answer 1

[Abolfazl Chaman Motlagh]

Hi. in case your problem is exactly what you attached. here is an easy solution i can provide:

I = imread('image.jpeg'); %read image
I = rgb2gray(I); % transform image to grayscale
C = corner(I); % find corners in image
G = imgradient(I); % create gradient of image
BW = G>max(G(:))*0.1; % make a binary image by thresholding gradient image
CC = bwconncomp(BW); % extract connected components in binary image
ind = sub2ind(size(I),C(:,2),C(:,1)); % transform coordinate of corner points to linear index
for component=1:CC.NumObjects
    % determine which corners belong to which components
    Index_in_componet{component} = ismember(ind,CC.PixelIdxList{component});
    % check if number of corner belong to a components is equal 3
    Is_triangle(component) = sum(Index_in_componet{component}) == 3;
end
triangle = find(Is_triangle); % find the component with 3 corner
% find XY cordinate of Corners in Triangle
[Triangle_points_Y,Triangle_points_X] = ind2sub(size(I),ind(Index_in_componet{triangle}));
% find XY cordinate of All points in selected component
[Row,Col] = ind2sub(size(I),CC.PixelIdxList{triangle});
% Visualization 
J = zeros(size(I));
for i=1:numel(Row)
J(Row(i),Col(i))=1;
end
figure;
subplot(1,2,1);
imshow(I)
hold;
scatter(Triangle_points_X,Triangle_points_Y,50,'b','filled')
subplot(1,2,2);
imshow(J);

The Output :

[Damien Lahire Gomez-Martinho]

Thanks, but if the number of triangles changed (maybe 2 o 3 triangles), I had an error with that code.

[Abolfazl Chaman Motlagh]

yes of course because the codes after triangle,it is asumed that there is 1 triangle in image.

so this will do it :

I = imread('image2.jpeg'); %read image
I = rgb2gray(I); % transform image to grayscale
C = corner(I); % find corners in image
G = imgradient(I); % create gradient of image
BW = G>max(G(:))*0.1; % make a binary image by thresholding gradient image
CC = bwconncomp(BW); % extract connected components in binary image
ind = sub2ind(size(I),C(:,2),C(:,1)); % transform coordinate of corner points to linear index
for component=1:CC.NumObjects
    % determine which corners belong to which components
    Index_in_componet{component} = ismember(ind,CC.PixelIdxList{component});
    % check if number of corner belong to a components is equal 3
    Is_triangle(component) = sum(Index_in_componet{component}) == 3;
end
triangle = find(Is_triangle); % find the component with 3 corner
J = zeros(size(I));
figure;
subplot(1,2,1);
imshow(I)
hold;
for T = triangle
    % find XY cordinate of Corners in Triangle
    [Triangle_points_Y,Triangle_points_X] = ind2sub(size(I),ind(Index_in_componet{T}));
    % find XY cordinate of All points in selected component
    [Row,Col] = ind2sub(size(I),CC.PixelIdxList{T}  );
    % Visualization
    for i=1:numel(Row)
        J(Row(i),Col(i))=1;
    end
    scatter(Triangle_points_X,Triangle_points_Y,50,'b','filled')
end
subplot(1,2,2);
imshow(J);

[Damien Lahire Gomez-Martinho]

Thank you Abolfazl, so helpful info!

Answer 2

[yanqi liu]

yes，sir，may be use the Circularity，such as

clc; clear all; close all;
img = imread('https://www.mathworks.com/matlabcentral/answers/uploaded_files/834120/image.jpeg');
bw = ~im2bw(img);
stats = regionprops(bw, 'BoundingBox', 'Circularity');
cis = cat(1, stats.Circularity);
id =  cis<1 & cis>0.5;
figure; imshow(img, []); 
hold on; rectangle('position', stats(id).BoundingBox, 'EdgeColor', 'r', 'LineWidth', 2)

[Damien Lahire Gomez-Martinho]

Thanks, but if the number of triangles changed (maybe 2 o 3 triangles), I had an error with that code.

[yanqi liu]

yes，sir，may be use

clc; clear all; close all;
img = imread('https://www.mathworks.com/matlabcentral/answers/uploaded_files/834620/image.jpeg');
bw = ~im2bw(img);
stats = regionprops(bw, 'BoundingBox', 'Circularity');
cis = cat(1, stats.Circularity);
id =  find(cis<1 & cis>0.5);
figure; imshow(img, []); 
for i = 1 : length(id)
hold on; rectangle('position', stats(id(i)).BoundingBox, 'EdgeColor', 'r', 'LineWidth', 2)
end

[Damien Lahire Gomez-Martinho]

very helpful sir.

Thank You!!

[Image Analyst]

@yanqi liu, a fun trick that you might want to know about to convert a scalar field of a structure into a vector is to enclose it in brackets:

cis = [stats.Circularity];

I do it this way all the time instead of using cat(1,)

[yanqi liu]

@Image Analyst yes，it is great，thanks

Answer 3

[Image Analyst]

Lots of ways to do it. Here is one:

rgbImage = imread('image.jpeg');
[rows, columns, numberOfColorChannels] = size(rgbImage)
subplot(2, 2, 1);
imshow(rgbImage);
impixelinfo
mask = rgb2gray(rgbImage) < 128;
subplot(2, 2, 2);
imshow(mask);
impixelinfo
[labeledImage, numRegions] = bwlabel(mask);
props = regionprops(mask, 'Area')
allAreas = [props.Area]
index = find(allAreas == 2635)
triangle = ismember(labeledImage, index);
subplot(2, 2, 3);
imshow(triangle);

[Damien Lahire Gomez-Martinho]

Thanks you Sir!

[Image Analyst]

Damien, since there are several ways that you can pick out triangles from the other objects, you need to specify what about the triangle blobs makes them unique. I mean, we could pick out triangles based on intensity, location, shape, size, etc. So you need to specify the range of all those things so that the right features are measured. You could check on all of those features, but you may not have to if you know certain things. You might need to specify only one, like I did when I specified the area. But I just as well could have done it by saying that the top-most blob in the image is the triangle if we know for a fact that the top blob will be a triangle. Or if the triangle is a unique intensity, like a gray level of 193 while all the other blobs have a different gray level, but could otherwise be located randomly, then we could just pick out blobs with a gray level of 193. If there could be any number of triangles, and they could be located anywhere, and they could be any brightness and size then you'd need a more sophisticated algorithm than one that simply looked at one particular feature.