Midterm
CMSC 426
October 18, 2005
1. 20 points: Background Subtraction: Suppose you have a video camera set up in
front of a concrete wall, taking 30 images per second, and you want to detect
when people walk by. The wall is always a uniform grey with an intensity of 128.
However, there is a light snow. Each snow flake appears in a single pixel and
turns it to an intensity of 255. About one pixel in 100 has a snow flake in it.
There is no noise affecting any of these measurements.
a. Suppose two people walk by. Person A has a uniform intensity of 250.
Person B has a uniform intensity of 123. Each person is about 100 pixels
big. Which of these outcomes are possible, depending on the parameters
used: 1) A and B are both identified as foreground; 2) A is identified as
foreground but B is not; 3) B is identified as foreground, but A is not.
250 and 123 are equally distant from the most similar training examples of 255 and 128.
However, there are 99 times as many examples with the value of 128. This means that
123 will always be regarded as more likely to be background than 250. This is true
regardless of the choice of sigma. Consequently, we can set the threshold to make 250
and 123 both appear to be background, both appear to be foreground, or we can make 123
appear to be background while 250 appears to be foreground. But the reverse is not
possible.
b. Explain how you might improve performance by using a new background
subtraction algorithm that uses a Markov model of pixel intensity. Be as
explicit as possible, for full credit.
If we use a 3x3 window for classification, we will see that background is generally either
all 128, or 8 values of 128 with one flake of snow (255). Any other event is very rare in
the background. This will make it much easier to tell the difference between a flake of
snow and a region that is all 253 (or even all 255) because it comes from a person.
2. 15 points: Provide a 2D filter that will compute the second derivative of the
image in the x direction.
Convolution is associative, so we can form this filter by convolving together two first
derivative filters. This would be:
-.5 0 .5 * -.5 0 .5 = -.25 0 .5 0 -.25
It is important to remember that when convolving filters together, we must have zeros at
the boundary. This is because convolution with 0 -.5 0 .5 0 has the same effect as
convolving with -.5 0 .5, whereas convolving with -.5 -.5 0 .5 .5 has a very different
effect.
