Hough Transform by a-b parameterization

Let a and b correspond to u and v respectively. The transformation function is given by F_b(x, y ; a) = y-ax as (3). Since the angle of the line is limited to $[\pi/4,\pi/4)$, the range of the parameter a is [-1,1). We have $q_b(a) = \Delta s/2$, $c_{max}(a) = \max_i (-x_i)$ and $c_{min}(a) = \min_i (-x_i)$ for the line segment defined by the parameter a. Therefore, we obtain c_max(a)-c_min(a)=l_x, where l_x denotes the length of the line segment measured along x-axis. We must use a possible maximum length as l_x because the length of the line segment is unknown. Since the image is square, the maximum length is l_x=2N, which is same as the image width.

Let $\Delta a$ and $\overline{\Delta a}$ denote the sampling parameter of aand its upper bound respectively. By (22), we have

$$\Delta a \leq \frac{2\Delta s}{l_x}\ .$$ (23)

Thus, the upper bound is given as follows.

Upper Bound of the Sampling Interval for a-b parameterization:

$$\overline{\Delta a}= \frac{\Delta s}{N}\ .$$ (24)

Here we have an interesting coincidence. In the Hough Transform by a-d parameterization proposed by Svalbe  [2], the transformation function is given by

$$d=y-\frac{2a'x}{N'-1} \ \ , \ \ \ 0\leq\vert a'\vert\leq \frac{N'-1}{2}\ ,$$ (25)

where a' is an integer and the angle between a line and x-axis is in the range $[-\pi /4, \pi /4]$. N' in the above function is defined for the image region with $N'\times N'$ pixels. If we use N, which is used in this document, we have N'=2N+1, and

$$d=y-\left(\frac{a'}{N}\right)\cdot x\ .$$ (26)

The sampling interval of the image is assumed to be $\Delta s=1$ in the a-d parameterization. (a'/N) in the above equation is in the range [-1,1], which is divided into 2N equal parts. Therefore, we have a=(a'/N), and the upper bound $\overline{\Delta a}$derived through our discussion is the same value as the sampling interval of (a'/N) in a-d parameterization,