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# CSCI 576 Assignment 3 solution

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## Question 1: DCT Coding (15 points)

In this question you will try to understand the working of DCT in the context of JPEG. Below is
an 8×8 luminance block of pixel values and its corresponding DCT coefficients.
188 180 155 149 179 116 86 96
168 179 168 174 180 111 86 95
150 166 175 189 165 101 88 97
163 165 179 184 135 90 91 96
170 180 178 144 102 87 91 98
175 174 141 104 85 83 88 96
153 134 105 82 83 87 92 96
117 104 86 80 86 90 92 103

• Using the 2D DCT formula, compute the 64 DCT values. Assume that you quantize your
DCT coefficients uniformly with Q=100. What does your table look like after
quantization? (2 points)
• In the JPEG pipeline, the quantized DCT values are then further scanned in a zigzag
order. Show the resulting zigzag scan when Q=100, (1 point).

• For this zigzag AC sequence, write down the intermediary notation (2 points)
• Assuming these are luminance values, write down the resulting JPEG bit stream. For the
bit stream you may consult standard JPEG VLC and VLI code tables. You will need to
refer to the code tables from the ITU-T JPEG standard which also uploaded with your
assignment. (2 points)
• What compression ratio do you get for this luminance block ?

## Question 2: Image Dithering (25 points)

Let’s say that we have an original image of size mxn with 8 bits per pixel. A certain sub section
of this image matrix is represented in a 12×8 sub image below and the values are normalized. In
the normalized representation shown below, assume that 0 corresponds to white and 9
corresponds to black.

Answer the following questions You may want to code/script a process to generate the final
outputs, but only final outputs are expected. Also, we have asked you to plot this 12×8 image
block as a gray color image, and its processed outputs as binary black/white images. For reasons
of visible clarity (because a 12×8 image block is very small) you may want to show a zoomed or
magnified picture.

• Plot the image as an 8-bit gray scale map, that is – create a 12×8 image to show the
original gray image block. (2 points)
• If you threshold the above 12×8 image block such that all values below 4.5 were 0 and
above 4.5 were 9, how does your output image B/W block look? Plot an image (3 points)
• We can create a better binary image output by using a dithering matrix. Compute the
binary output of a dithering operation on the gray level 12×8 image using the dithering
matrix D given below. Assume that the sub image block’s top left coordinates indexes
start as [0,0]. Show a graphical binary image plot of the dithered output. (5 points)
D =

5 2 7
1 0 3
6 8 4

• What if the sub image block’s top left coordinate indexes start with [1,1]. Show a
graphical binary image plot of the dithered output. (5 points)

## PROGRAMMING DCT vs DWT – (160 points)

This assignment will help you gain an understanding of issues that relate to image compression,
by comparing and contrasting the frequency space representations using the Discrete Cosine
Transform and the Discrete Wavelet Transform. All input RGB files will be of size 512×512 and
in the same rgb format used in the previous assignments.

You will read an RGB file and convert
the file to an 8×8 block based DCT representation (as used in the JPEG implementation) as well
as a DWT representation (as used in the JPEG2000 implementation). Each will contain 262144
frequency coefficients wither DCT or DWT. Depending on the second parameter n you will
decode both the representations using only n coefficients and display them side to side to
compare your results. Your algorithm, whether encoding or decoding, should work on each
channel independently. Display the two outputs side by side.

## Input to your program will be 2 parameters where:

• The first parameter is the name of the input image file. (All images will be given in the
rgb format and have a size 512×512 for ease in encoding/decoding)
• The second parameter n is an integral number that defines the number of coefficients to
use for decoding. The value of n will be specific – (262144, 65536, 16384,4096, 1024,
256, 64, 16, 4, 1)

The interpretation of this parameter for decoding is different for both
the DCT and DWT cases so as to use the same number of coefficients. Please see the
implementation section for an explanation
Typical invocations to your program would look like
MyExe Image.rgb 262144
Here you are making use of all the coefficients to decode because the total number of
coefficients for each channel are going to be 512*512= 262144. Hence the output for each DCT
and DWT should be exactly the same as there is no loss and should be the original image.
MyExe Image.rgb 65536

Here you are making use of 65536 (one fourth of the total number) of coefficients for decoding.
While the number of coefficients are the same for both DCT and DWT decoding, the exact
coefficient indexes are different. Refer to the implementation section for this.
MyExe Image.rgb 16384

Here you are making use of 16384 (1/16th of the total number) of coefficients for decoding.
While the number of coefficients are the same for both DCT and DWT decoding, the exact
coefficient indexes are different. Refer to the implementation section for this.
Implementation
Your implementation should read the input file and convert (for each channel separately) a DCT
representation and a DWT representation.

## Encoding:

For the DCT conversion, break up the image into 8×8 contiguous blocks of 64 pixels each and
then perform a DCT for each block for each channel. For the DWT conversion, convert each row
(for each channel) into low pass and high pass coefficients followed by the same for each column
applied to the output of the row processing. Recurse through the process as explained in class
through rows first then the columns next at each recursive iteration, each time operating on the
low pass section.

## Decoding:

Based on the input parameter of the number of coefficients to use, you need to appropriately
decode by zeroing out the unrequested coefficients (just setting the coefficients to zero) and then
perform an IDCT or an IDWT. The exact coefficients to zero out are different for both the DCT
and DWT cases and explained next.

For a DCT, you want to select the first m coefficients in a zig zag order for each 8×8 block
such that m = ceil(n/4096) where n is the number of coefficients given as input. 4096 is the
number of 8×8 blocks in a 512×512 image. Thus, m represents the first few coefficients to use for
each 8×8 block during decoding. So for the second test run above (n=65536), you will use m =
ceil (65536/4096) = 16.

Each block will be decoded using the first 16 coefficients in zigzag
order. The remaining coefficients in each 8×8 block can be set to zero prior to decoding. For the
third test run above (n=16384), you will use m = ceil (16384/4096) = 4. If the ratio of n/4096
becomes fractional, then m=1 because of the ceil function, ie you are ensuring using at least one
coefficient of every DCT 8×8 block.

For all your tests, the parameter n for decoding will correspond to the number of coefficients in
the Low Pass outputs for both rows and columns at any level in a DWT. In a DWT, as shown
below, there are 262144 coeffients after the hierarchy is processed. You will appropriately use n
= 262144, or 65536, and so on as per the input. For DWT decoding step set all the remaining
coefficients to zero and proceed with an IDWT.
262144 coeffs 65536 coeffs 16384 coeffs …. 4096 coeffs ….

### What should you submit?

• Your source code, and your project file or makefile. Please do not include any binaries or
data sets. Zip all the files together and name it as
FirstName_LastName_Assignment3.zip. This will help us handle submissions better.
We will compile your program and execute our tests accordingly.
• Along with the program, also submit an electronic document (word, pdf, pagemaker etc)
for the written part and any other extra credit explanations.