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Figure 9: (a) True values of the line width w, of the staircase line f,. (b) True values of the asymmetry a. (c) Line
position correction. Note that (c) is rotated by 180? with respect to (a) and (b).
with scale-normalized quantities w, and v, as above. Thus, the predicted line position /,, can be calculated for all w; and
a € [0, 1]. The result is shown in Figure 8(a). We can see that the bias of the line position is largest for small line widths
and large asymmetries. Furthermore, the bias is symmetrical with respect to a — 0.5. Note that the positions are only
defined within the range of a and w defined by (15). Furthermore, the predicted line width w, and gradient ratio r can be
calculated. They are displayed in Figures 8(b) and (c) along with their contour lines.
As is obvious from Figures 8(b) and (c), for a < 0.5 the bias function is invertible since the contour lines of v; and r
intersect almost perpendicularly everywhere. Hence, the inverted bias function can be calculated by a multi-dimensional
root finding algorithm to obtain the true values of w, and a. They are shown in Figures 9(a) and (b). Furthermore, the
absolute value of the correction c, to be applied to the line position in order to obtain the true line position is displayed
in Figure 9(c). The only remaining question is how the extraction algorithm should decide whether the true asymmetry is
smaller or larger than 0.5. This can simply be done based on the gradient directions at one of the two edge points and on
the direction perpendicular to the line. If the dot product of the two vectors is larger than zero, the bright side of the line is
the right side, and vice versa. If the bright side of the line and the side with the weaker gradient are equal, the asymmetry
must be chosen larger than 0.5, otherwise smaller.
The discussion so far has been concerned with the extraction of line points in 1D. It is not immediately obvious how to
derive a 2D extraction algorithm for lines with different polarity. However, if the gradient image is considered, it can be
seen that such a line manifests itself as a dark line there. An example of this is displayed in Figure 10(a), where two rings
of lines of different and equal polarity are displayed. In the gradient image in Figure 10(b) the line of different polarity
(a) Input image (b) Gradient image (c) Extracted line
Figure 10: (a) An image containing two rings of lines with different and equal polarities, respectively. (b) The gradient
image shows that both types of lines result in dark lines in the gradient image. (c) Extracted staircase lines.
International Archives of Photogrammetry and Remote Sensing. Vol. XXXIII, Part B3. Amsterdam 2000. 147