opencv how to get middle line of a contour - c++

Question

I have an image from which I get the contour of using findContours. This products something that looks like the following: (showing the "inner and outer contour").
Is there a way for me to get the "midpoint" of these two contours? ie some kind of polyline that would fit exactly in between the two lines seen in the image, such that the distance at any point on the resultant time is the same from it to the top contour as is from it to the bottom contour?
More complicated example would be something as follows:
Please note, that it doesnt matter too much what happens at intersections, so long as nothing traces back on itself, so the result of the more complicated example would need multiple lines.

Answer 1

There is a way to get the "midpoint" of the two contours, but I don't think there is an existing OpenCV solution.
You may use the following stages:
Convert image to Grayscale, and apply binary threshold.
You may use cvtColor(... COLOR_BGR2GRAY) and threshold(...) OpenCV functions.
Fill the pixels outsize the area between lines with white color.
You may use floodFill OpenCV function.
Apply "distance transform" to the binary image.
You may use distanceTransform OpenCV function.
Use CV_DIST_L2 for euclidean distance.
Apply Dijkstra's algorithm for finding the shortest paths between most left and most right nodes.
Representing "distance transform" result (image) as weighted graph and applying Dijkstra's algorithm is the most challenging stage.
I implemented the solution in MATLAB.
The MATLAB implemented is used as a "proof of concept".
I know you were expecting C++ implementation, but it requires a lot of work.
The MATLAB implementation uses im2graph function, I downloaded from here.
Here is the MATLAB implementation:
origI = imread('two_contours.png'); % Read input image
I = rgb2gray(origI); % Convert RGB to Grayscale.
BW = imbinarize(I); % Convert from Grayscale to binary image.
% Fill pixels outsize the area between lines.
BW2 = imfill(BW, ([1, size(I,2)/2; size(I,1), size(I,2)/2]));
% Apply "distance transform" (find compute euclidean distance from closest white pixel)
D = bwdist(BW2);
% Mark all pixels outsize the area between lines with zero.
D(BW2 == 1) = 0;
figure;imshow(D, []);impixelinfo % Display D matrix as image
[M, N] = size(D);
% Find starting point and end point - assume we need to find a path from left side to right side.
x0 = 1;
[~, y0] = max(D(:, x0));
x1 = N;
[~, y1] = max(D(:, x1));
% https://www.mathworks.com/matlabcentral/fileexchange/46088-dijkstra-lowest-cost-for-images
StartNode = y0;
EndNode = M*N - (M-y1-1);
conn = 8;%4 or 8 - connected neighborhood for linking pixels
% Use 100 - D, because graphshortestpath searches for minimum weight (and we are looking for maximum weight path).
CostMat = 100 - D;
G = im2graph(CostMat, conn);
%Find "shortest" path from StartNode to EndNode
[dist, path, pred] = graphshortestpath(G, StartNode, EndNode);
% Mark white path in image J image
J = origI;R = J(:,:,1);G = J(:,:,2);B = J(:,:,3);
R(path) = 255;G(path) = 255;B(path) = 255;
J = cat(3, R, G, B);
figure;imshow(J);impixelinfo % Display J image
Result:
D - Result of distance transform:
J - Original image with "path" marked with white color:
Update:
For the new example you can define three paths.
The solution becomes more complicated.
The example is not generalized to solve all the cases.
There must be a simpler solution, I just can't think of one.
tmpI = imread('three_contours.png'); % Read input image
origI = permute(tmpI, [2, 1, 3]); % Transpose image
I = rgb2gray(origI); % Convert RGB to Grayscale.
BW = imbinarize(I); % Convert from Grayscale to binary image.
% Fill pixels outsize the area between lines.
%BW2 = imfill(BW, ([1, size(I,2)/2; size(I,1), size(I,2)/2]));
BW2 = imfill(BW, ([1, 1; size(I,1), size(I,2); size(I,2)/2, 1]));
% Apply "distance transform" (find compute euclidean distance from closest white pixel)
D = bwdist(BW2);
% Mark all pixels outsize the area between lines with zero.
D(BW2 == 1) = 0;
figure;imshow(D, []);impixelinfo % Display D matrix as image
[M, N] = size(D);
% Find starting point and end point - assume we need to find a path from left side to right side.
x0 = 1;
[~, y0a] = max(D(1:M/2, x0));
% Y coordinate of second point
[~, y0b] = max(D(M/2:M, x0));
y0b = y0b + M/2;
x1 = N;
[~, y1] = max(D(:, x1));
% https://www.mathworks.com/matlabcentral/fileexchange/46088-dijkstra-lowest-cost-for-images
StartNodeA = y0a;
StartNodeB = y0b;
EndNode = M*N - (M-y1-1);
conn = 8;%4 or 8 - connected neighborhood for linking pixels
% Use 100 - D, because graphshortestpath searches for minimum weight (and we are looking for maximum weight path).
D(D==0) = -10000; % Increase the "cost" where D is zero
CostMat = 1000 - D;
G = im2graph(CostMat, conn);
%Find "shortest" path from StartNode to EndNode
[dist, pathA, pred] = graphshortestpath(G, StartNodeA, EndNode);
[dist, pathB, pred] = graphshortestpath(G, StartNodeB, EndNode);
[dist, pathC, pred] = graphshortestpath(G, StartNodeA, StartNodeB);
% Mark white path in image J image
J = origI;R = J(:,:,1);G = J(:,:,2);B = J(:,:,3);
R(pathA) = 255;
G(pathB) = 255;
B(pathC) = 255;
J = cat(3, R, G, B);
J = permute(J, [2, 1, 3]); % Transpose image
figure;imshow(J);impixelinfo % Display J image
Three lines: