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.
October 16, 2014
Computer Vision
Lecture 12: Image Segmentation II
1
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.
October 16, 2014
Computer Vision
Lecture 12: Image Segmentation II
2
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.
October 16, 2014
Computer Vision
Lecture 12: Image Segmentation II
3
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.
y
c
C
C
winner
parameters
B
B
A
0
A
x
input space
0
m
output space
The parameters of their crossing point (which would be the
winners) are the parameters of the straight line in input space.
October 16, 2014
Computer Vision
Lecture 12: Image Segmentation II
4
Hough Transform
Hough Transform Algorithm:
1. Quantize input and output spaces appropriately.
2. Assume that each cell in the parameter (output)
space is an accumulator (counter). Initialize all
cells to zero.
3. For each point (x, y) in the image (input) space,
increment by one each of the accumulators that
satisfy the equation.
4. Maxima in the accumulator array correspond to
the parameters of model instances.
October 16, 2014
Computer Vision
Lecture 12: Image Segmentation II
5
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.
October 16, 2014
Computer Vision
Lecture 12: Image Segmentation II
6
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.
October 16, 2014
Computer Vision
Lecture 12: Image Segmentation II
7
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.
October 16, 2014
Computer Vision
Lecture 12: Image Segmentation II
8
Improved Hough Transform
0
0

Column j
450
636
representation
of same line in
output space
d
d
Row i
line to be
described
450
0
0
input space

output space
360
The parameters  and d form the output space for our Hough
transform.
October 16, 2014
Computer Vision
Lecture 12: Image Segmentation II
9
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).
October 16, 2014
Computer Vision
Lecture 12: Image Segmentation II
10
Improved Hough Transform
By varying  from 0 to 360 we can find all lines crossing (i0, j0):
Column j
0
0
Row i
1 2
3
d3
450
But how can we
compute parameter d
for each value of ?
d2
d1
(i0, j0)
450October 16, 2014
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.
Computer Vision
Lecture 12: Image Segmentation II
11
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:
October 16, 2014
Computer Vision
Lecture 12: Image Segmentation II
12
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.
October 16, 2014
Computer Vision
Lecture 12: Image Segmentation II
13
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?
October 16, 2014
Computer Vision
Lecture 12: Image Segmentation II
14
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:
October 16, 2014
Computer Vision
Lecture 12: Image Segmentation II
15
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.
October 16, 2014
Computer Vision
Lecture 12: Image Segmentation II
16
Sample Results
Input Image
October 16, 2014
Computer Vision
Lecture 12: Image Segmentation II
17
Sample Results
Edges in input image (Sobel filter output)
October 16, 2014
Computer Vision
Lecture 12: Image Segmentation II
18
Sample Results
-90
d
0

90
180
Hough-transformed edge image
October 16, 2014
Computer Vision
Lecture 12: Image Segmentation II
19
Sample Results
270
d
0

90
180
Five greatest maxima in Hough-transformed image
October 16, 2014
Computer Vision
Lecture 12: Image Segmentation II
20
Sample Results
Lines in input image corresponding to Hough maxima
October 16, 2014
Computer Vision
Lecture 12: Image Segmentation II
21