1. Fitting Curve Models to Edges
Most contours can be well described by combining several “round” elements (e.g., circular arcs) and “straight” elements (straight lines).
First we will study a method for fitting connected straight lines to edge data.
Then we will discuss how to fit a circular arc to a given set of points.
Later we will apply a very general and sophisticated method (the Hough transform) for fitting straight lines to edge points.
March 6, 2018
Computer Vision Lecture 11: Image Segmentation I

2. Fitting Curve Models to Edges
Assume that you found edge data that seem to belong to one contour.
The shape of these data is somewhat non-linear, but you would like to use multiple connected straight lines (polylines) to describe the contour.
Then you can use the technique of polyline splitting.
If you have an open contour, start with a straight line connecting the two (estimated) end points.
If you have a closed contour, start with an appropriate polygon (a rectangle usually works well).
March 6, 2018
Computer Vision Lecture 11: Image Segmentation I

3. Fitting Polylines to Edges
Let us look at an example for an open contour:
First, use a single straight line to connect the end points.
Then find the edge point with the greatest distance from this straight line.
March 6, 2018
Computer Vision Lecture 11: Image Segmentation I

4. Fitting Polylines to Edges
Let us look at an example for an open contour:
Then split the straight line in two straight lines that meet at this point.
Now repeat this process with each of the two new lines.
March 6, 2018
Computer Vision Lecture 11: Image Segmentation I

5. Fitting Polylines to Edges
Let us look at an example for an open contour:
Recursively repeat this process until the maximum distance of any point to the polyline falls below a certain threshold.
The result is the fitting polyline.
March 6, 2018
Computer Vision Lecture 11: Image Segmentation I

6. Fitting Circular Arcs to Edges
Now we will discuss how to fit a circular arc to a set of edge points.
Consider three points p1 = (x1, y1), p2 = (x2, y2), and p3 = (x3, y3), which are pairwise different and do not lie on the same straight line.
Then we already know that there is exactly one circle that goes through all of these points.
We will now describe how to find this circle, i.e., determine its center and radius.
March 6, 2018
Computer Vision Lecture 11: Image Segmentation I

7. Fitting Circular Arcs to Edges
The points (x, y) of a circle centered at (x0, y0) and with radius r are given by the following equation:
(x – x0)2 + (y – y0)2 = r2.
To simplify the computation, we will use a coordinate system whose origin is in point p1 = (x1, y1).
In the new coordinate system we have:
x’ = x – x1
y’ = y – y1
(x’ – x’0)2 + (y’ – y’0)2 = r2.
March 6, 2018
Computer Vision Lecture 11: Image Segmentation I

8. Fitting Circular Arcs to Edges
Since the equation (x’ – x’0)2 + (y’ – y’0)2 = r2 is true for all points (x’, y’) on the circle, it must also apply to p1, p2, and p3.
Substituting (x’, y’) with p1, p2, and p3 gives us:
x’02 + y’02 - r2 = 0
x’22 - 2x’2x’0 + x’02 + y’22 – 2y’2y’0 + y’02 - r2 = 0
x’32 - 2x’3x’0 + x’02 + y’32 – 2y’3y’0 + y’02 - r2 = 0
Subtract the first equation from the other two:
2x’2x’0 + 2y’2y’0 = x’22 + y’22
2x’3x’0 + 2y’3y’0 = x’32 + y’32
March 6, 2018
Computer Vision Lecture 11: Image Segmentation I

9. Fitting Circular Arcs to Edges
This gives us two linear equations with two unknown variables x’0 and y’0.
From these we can easily determine the center (x’0, y’0) of the circle.
The radius of the circle can then be determined by:
r2 = x’02 + y’02.
We simply add (x1, y1) to (x’0, y’0) to get the center (x0, y0) of the circle in the original coordinate system.
Then we have completely determined the circle going through points p1, p2, and p3.
March 6, 2018
Computer Vision Lecture 11: Image Segmentation I

10. Fitting Circular Arcs to Edges
In practice we usually have more than three edge points through which we want to fit a circular arc.
One possible way to do the fitting in such a case is to take the two end points of those edges as points p1 and p2.
Then we successively take every other edge point as point p3 and fit a circular arc to each generated set of points p1, p2, and p3.
In each case, we measure the deviation of the arc from all edge points.
The arc with minimum deviation is chosen as the curve model for these edges.
March 6, 2018
Computer Vision Lecture 11: Image Segmentation I

11. Computer Vision Lecture 11: Image Segmentation I
Hough Transform
The Hough transform is a very general technique for feature detection.
In the present context, we will use it for the detection of straight lines as contour descriptors in edge point arrays.
We could use other variants of the Hough transform to detect circular and other shapes.
We could even use it outside of computer vision, for example in data mining applications.
So understanding the Hough transform may benefit you in many situations.
March 6, 2018
Computer Vision Lecture 11: Image Segmentation I

12. Computer Vision Lecture 11: Image Segmentation I
Hough Transform
The Hough transform is a voting mechanism.
In general, each point in the input space votes for several combinations of parameters in the output space.
Those combinations of parameters that receive the most votes are declared the winners.
We will use the Hough transform to fit a straight line to edge position data.
To keep the description simple and consistent, let us assume that the input image is continuous and described by an x-y coordinate system.
March 6, 2018
Computer Vision Lecture 11: Image Segmentation I

13. Computer Vision Lecture 11: Image Segmentation I
Hough Transform
A straight line can be described by the equation:
y = mx + c
The variables x and y are the parameters of our input space, and m and c are the parameters of the output space.
For a given value (x, y) indicating the position of an edge in the input, we can determine the possible values of m and c by rewriting the above equation:
c = -xm + y
You see that this represents a straight line in m-c space, which is our output space.
March 6, 2018
Computer Vision Lecture 11: Image Segmentation I

14. Computer Vision Lecture 11: Image Segmentation I
Hough Transform
Example: Each of the three points A, B, and C on a straight line in input space are transformed into straight lines in output space. x y
input space m c
output space C C
winner parameters B B A A
The parameters of their crossing point (which would be the winners) are the parameters of the straight line in input space.
March 6, 2018
Computer Vision Lecture 11: Image Segmentation I

15. Computer Vision Lecture 11: Image Segmentation I
Hough Transform
Hough Transform Algorithm:
Quantize input and output spaces appropriately.
Assume that each cell in the parameter (output) space is an accumulator (counter). Initialize all cells to zero.
For each point (x, y) in the image (input) space, increment by one each of the accumulators that satisfy the equation.
Maxima in the accumulator array correspond to the parameters of model instances.
March 6, 2018
Computer Vision Lecture 11: Image Segmentation I

16. Computer Vision Lecture 11: Image Segmentation I
Hough Transform
The Hough transform does not require preprocessing of edge information such as ordering, noise removal, or filling of gaps.
It simply provides an estimate of how to best fit a straight line (or other curve model) to the available edge data.
If there are multiple straight lines in the image, the Hough transform will result in multiple peaks. You can search for these peaks to find the parameters for all the corresponding straight lines.
March 6, 2018
Computer Vision Lecture 11: Image Segmentation I

17. Improved Hough Transform
Here is some practical advice for doing the Hough transform (e.g., for some future assignment).
The m-c space described on the previous slides is simple but not very practical. It cannot represent vertical lines, and the closer the orientation of a line gets to being vertical, the greater is the change in m required to turn the line significantly.
We are going to discuss an alternative output space that requires a bit more computation but avoids the problems of the m-c space.
March 6, 2018
Computer Vision Lecture 11: Image Segmentation I

18. Improved Hough Transform
As we said before, it is problematic to use m (slope) and c (intercept) as an output space.
Instead, it is a good idea to use the orientation  and length d of the normal of a straight line to describe it.
The normal n of a straight line l is perpendicular to l and connects l with the origin of the coordinate system.
The range of  is from 0 to 360, and the range of d is from 0 to the length of the image diagonal.
Note that we can skip the  interval from 180 to 270, because it would require a negative d.
Let us assume that the image is 450×450 units large.
March 6, 2018
Computer Vision Lecture 11: Image Segmentation I

19. Improved Hough Transform
Column j
450 0
360
636
output space  d
450
Row i
input space  d
line to be described
representation of same line in output space
The parameters  and d form the output space for our Hough transform.
March 6, 2018
Computer Vision Lecture 11: Image Segmentation I

20. Improved Hough Transform
For any edge point (i0, j0) indicated by our Sobel edge detector, we have to find all parameters  and d for those straight lines that pass through (i0, j0).
We will then increase the counters in our output space located at every (, d) by the edge strength, i.e., the magnitude provided by the Sobel detector.
This way we will find out which parameters (, d) are most likely to indicate the clearest lines in the image.
But first of all, we have to discuss how to find all the parameters (, d) for a given point (i0, j0).
March 6, 2018
Computer Vision Lecture 11: Image Segmentation I

21. Improved Hough Transform
By varying  from 0 to 360 we can find all lines crossing (i0, j0): 2 d2
450
Column j
Row i
But how can we compute parameter d for each value of ?
Idea: Rotate (i0, j0) around origin by - so that it lands on i-axis.
Then the i-coordinate of the rotated point is the value of d. 1 d1 3 d3
(i0, j0)
March 6, 2018
Computer Vision Lecture 11: Image Segmentation I

22. Improved Hough Transform
And how do we rotate a point in two-dimensional space?
The simplest way is to multiply the point vector with a rotation matrix.
We compute the rotated point (iR, jR) as obtained by rotation of point (i0, j0) around the point (0, 0) by the angle  as follows:
March 6, 2018
Computer Vision Lecture 11: Image Segmentation I

23. Improved Hough Transform
We are only interested in the i-coordinate:
iR = i0  cos  - j0  sin 
In our case, we want to rotate by the angle -:
iR = i0  cos(-) - j0  sin(-)
iR = i0  cos  + j0  sin 
Now we can compute parameter d as a function of i0, j0, and :
d(i0, j0; ) = i0  cos  + j0  sin 
By varying  we are now able to determine all parameters (, d) for a given point (i0, j0) and increase the counters in output space accordingly.
March 6, 2018
Computer Vision Lecture 11: Image Segmentation I

24. Improved Hough Transform
We can then define the straight line representing the “winner” in parametric form using a parameter p.
The points (i, j) of the line are then given by the following equation:
By varying p within an appropriate range, we can compute every pixel of our straight line.
But how can we find the vector (i, j) that determines the slope of the line?
March 6, 2018
Computer Vision Lecture 11: Image Segmentation I

25. Improved Hough Transform
The idea here is that the orientation of the straight line is perpendicular to its normal.
Since the orientation of the normal is given by the angle , the orientation of the straight line must be given by  - 90.
Therefore, we get:
March 6, 2018
Computer Vision Lecture 11: Image Segmentation I

26. Improved Hough Transform
Then the complete equation for the straight line with parameters (, d) looks like this:
Just vary parameter p in steps of 1 (to catch every pixel) between approximately -900 to 900 (twice the image size).
Whenever (i, j) is within the image range, then visualize that pixel in the bitmap.
March 6, 2018
Computer Vision Lecture 11: Image Segmentation I

27. Computer Vision Lecture 11: Image Segmentation I
Sample Results
Input Image
March 6, 2018
Computer Vision Lecture 11: Image Segmentation I

28. Sample Results Edges in input image (Sobel filter output)
March 6, 2018
Computer Vision Lecture 11: Image Segmentation I

29. Sample Results d -90 0  90 180 Hough-transformed edge image
March 6, 2018
Computer Vision Lecture 11: Image Segmentation I

30. Sample Results d
270 0 
90
180
Five greatest maxima in Hough-transformed image
March 6, 2018
Computer Vision Lecture 11: Image Segmentation I

31. Sample Results Lines in input image corresponding to Hough maxima
March 6, 2018
Computer Vision Lecture 11: Image Segmentation I