How do i correctly fit a line and force it through a particular point in an image? - matlab

Question

I am to fit planes through various points in an image, but I am having issues with forcing the line through a particular point in the image. This happens particularly when the line is 90 degrees.
My code is as follows:
I = [3 3 3 3 3 2 2
3 3 3 3 2 2 2
3 3 3 3 2 2 2
3 3 1 2 2 2 2
1 1 1 2 2 2 2
1 1 1 1 1 2 2
1 1 1 1 1 1 1];
% force the line through point p
p = [3,3];
% points to fit plane through
edgeA = [3,3.5; 3,4; 2.5,4; 2,4; 1.5,4];
edgeB = [3.5,3; 4,3; 4.5,3; 5,3];
% fit a plane through p and edgeA
xws = [p(2), edgeA(:,2)']';
yws = [p(1), edgeA(:,1)']';
Cws = [xws ones(size(xws))];
dws = yws;
Aeqws = [p(2) 1];
beqws = [p(1)];
planefitA = lsqlin(Cws ,dws,[],[],Aeqws, beqws);
% fit a plane through p and edgeB
xwn = [p(2), edgeB(:,2)']';
ywn = [p(1), edgeB(:,1)']';
Cwn = [xwn ones(size(xwn))];
dwn = ywn;
Aeqwn = [p(2) 1];
beqwn = [p(1)];
planefitB = lsqlin(Cwn ,dwn,[],[],Aeqwn, beqwn);
%%%%% plot the fitted planes:
xAxis = linspace(0, size(I, 2), 12);
%obtain linear curve
fA = planefitA(1)*xAxis + planefitA(2);
fB = planefitB(1)*xAxis + planefitB(2);
%plot the fitted curve
RI = imref2d(size(I),[0 size(I, 2)],[0 size(I, 1)]);
figure, imshow(I, RI, [], 'InitialMagnification','fit')
grid on;
hold on;
plot(xAxis,fA, 'Color', 'b', 'linewidth', 2);
plot(xAxis,fB, 'Color', 'r', 'linewidth', 2);
All the points in edgeB fall on a 90 degrees line. However, the function ends up fitting a wrong line through those points. I know this because using
planefitB = polyfit([p(2), edgeB(:,2)'], [p(1), edgeB(:,1)'], 1);
works for this particular line but the problem is that i have these process repeated so many times at different locations in my image, hence i do not know how to suggest polyfit when the line would be 90 degrees.
Please, any ideas/suggestions on how i could make this work? Many thanks.

Answer 1

This amounts to the least squares solution of only the angle of the line. The offset is fixed by the fact that it has to go through (3,3). The easiest way to express this is by offsetting your data points by the known crossing. That is, subtract (3,3) from your data points, and fit the best m for y=mx, the b being fixed to 0.
For the non-vertical case, you can use a classic least-squares formulation, but don't augment the constant 1 into the Vandermonde matrix:
slope = (edgeA(:,2) - p(2)) \ (edgeA(:,1) - p(1));
This gives exactly the same answer as your constrained lsq solution.
Now for the vertical line: A non-vertical line can be expressed in the standard functional form of y=mx, where the least squares formulation implicitly assumes an independent and a dependent variable. A vertical line doesn't follow that, so the only general choice is a "Total Least Squares" formulation, where errors in both variables are considered, rather than just the residuals in the dependent (y) variable.
The simplest way to write this is to choose a and b to minimize ax - by, in the least squares sense. [x_k -y_k]*[a b].' should be as close to a zero vector as possible. This is the vector closest to the null space of the [x -y] matrix, which can be computed with the svd. Swapping columns and fudging signs lets us just use svd directly:
[u s v] = svd(bsxfun(#minus, edgeA, p));
The last column of v is the closest to the null space, so mapping back to your x/y definitions, (edgeA-p)*v(:,2) is the line, so the y multiplier is in the top position, and the x in the lower, with a sign flip. To convert to y=mx form, just divide:
slope = -v(2,2)/v(1,2);
Note that this answer will be quite a bit different than the normal least squares answer, since you are treating the residuals differently. Also, the final step of computing "slope" won't work in the vertical case for the reasons we've already discussed (it produces Inf), so you are probably better off leaving the line as a normalized 2-vector, which won't have any corner cases.