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CS 659 Image Processing
Sample Midterm Exam (Closed Book, 90 minutes)
Covering Lectures 1~6. There are 5 questions. Each is 20 points.
1. (20 points)
(a) Let a grayscale image be the size 100-by-200 with 256 gray levels. How many bits in
memory are required to store this grayscale image without adding the overhead?
(b) Let a color image be the size 100-by-200. How many bits in memory are required to store
this color image without adding the overhead?
(c) If we reduce the image size by a half in both row and column. What fraction in the memory
size can we reduce?
Answer:
(a) 100*200*8=160,000 bits
(b) 100*200*8*3=480,000 bits
(c) or 0.25
2. (20 points) Let an image f of 4-by-4 and a mask g of 2-by-2 as follows. Calculate the
convolution f * g and correlation f ° g. Note the origin is located at lower-left corner.
3. (20 points)
(a) Let an image of 2-by-2 as follows. Perform image negative on this image. What is the
resulting image?
3 33
170 255(b) Let an image of 2-by-2 as follows. Perform bit-plane slicing on this image. What are the eight
images on each bit-plane, B0, B1, B2, B3, B4, B5, B6, and B7, from Least Significant Bit (LSB)
B0 to Most Significant Bit (MSB) B7?
3 33
170 255
Answer:
(a) We use 255 to subtract each pixel.
252 222
85 0
(b) We represent each pixel into 8-bit as:
3=00000011,
33=00100001
170=10101010
255=11111111
Therefore,
4. (20 points)
Let a binary image be f and a template be g as follows. Perform image matching using the
equation: Note that is correlation and
is the complement. What is the
resulting image?
15. (20 points) Explain the image equalization and image specification. Briefly describe both
procedures to achieve their goals.
Answer:
In histogram equalization we are trying to maximize the image contrast by applying a gray level
transform which tries to flatten the resulting histogram. It turns out that the gray level transform
that we are seeking is simply a scaled version of the original image's cumulative histogram. That
is, the graylevel transform T is given by T[i] = (G-1)c(i), where G is the number of gray levels
and c(i) is the normalized cumulative histogram of the original image.
r: the gray levels to be enhanced.
Assume r continuous in [0,1]
s=T(r)
a) T(r) single-valued, monotonically increasing
b) 0<=T(r)<=1
When we want to specify a non-flat resulting histogram, we can use the following steps (called
image specification):
1. Specify the desired histogram g(z)
2. Obtain the transform which would equalize the specified histogram, Tg, and its inverse
Tg-1
3. Get the transform which would histogram equalize the original image, s=T[i]
4. Apply the inverse transform Tg-1
on the equalized image, that is z=Tg-1
[s]