Hough Transform Accumulator to Cartesian

Question

I'm studying a course on vision systems and one of the questions posed was;

For the accumulator shown;

1. Determine the most likely r,θ combination representing the straight line of the greatest strength in the original image.

From my understanding of the accumulator this would be r = 60, θ = 150 as the 41 votes is the highest number of votes in this cluster of large votes. Am I correct with this combination?

2. And hence calculate the equation of this line in the form y = mx + c

I'm not sure of the conversion steps required to convert the r = 60, θ = 150 to y = mx + c with the information given since r = 60, θ = 150 denotes 1 point on the line.

3. State the resolution of your answer and give your reasoning

I assume the resolution is got to do with some of the steps in the auscultation and not the actual resolution of the original image since that's irrelevant to the edges detected in the image.

Any guidance on the above 3 points would be greatly appreciated!

Answer 1

1. Yes, this is correct.

2. This is asking you what the slope and intercept are of the line given r and theta. r and theta are not one point on the line, they are one point of the accumulator. r and theta describe a line using the line equation in polar coordinates: . This is the cool thing about the hough transform, every line in one space, (i.e. image space) can be described by a point in another space (r, theta). This could be done with m and b from the line equation , but as we all know, m is undefined for vertical lines. This is the reason the polar line equation is used. It is important to note that the line described by the HT r and theta refers to a line from the origin extending to the actual line in the image. This means your image line y = mx + b equation will need to be orthogonal to the polar equation. The wiki article on the HT describes this well and shows examples. I would recommend drawing a diagram of your r and theta extending to a line like this:

   

   Then use trig to get two points on the red line. Two points are enough to give you m and b from the line equation.

3. I'm not entirely sure what "resolution" refers to in this context. But it does seem like your line estimator will have some precision loss since r is every 20 mm and theta is every 15 degrees. Perhaps it is asking what degree of error you could get given an accumulator of this resolution.