CS301 Assignment 3 Solution 2021 - VU Answer

CS301 Assignment 3 Solution Spring 2021

VU Answer Provides a Complete CS301 Assignment 3 Solution Spring 2021. Easy to View CS301 Assignment 3 Solution 2021 and Download PDF File.


CS301 UPDATE ASSIGNMENT 3 SOLUTION SPRING 2021 

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According to Announcement Section CS301 Assignment format will Be updated you can follow this Correct solution file.

STEP 1:

Count all the letters including space from the given text message. 


SOLUTION:

Text Message = “virtual university”

Total No. of Letters = 18


STEP 2:

Draw a table with columns name letter, frequency, original bits, encoded bits

SOLUTION:

Letter Name

Frequency

Original Bits

Encoded Bits


ASCII Code

Binary


STEP 3:

Fill the table with letters, frequency and original bits (for original bits get ASCII code of each letter convert the decimal ASCII into 8 bits binary code). 


SOLUTION:

Letters

Frequency

Original bits

Encoding bits

ASCII

Binary

v

2

118

01110110



i

3

105

01101001



r

2

114

01110010



t

2

116

01110100



u

2

117

01110101



a

1

97

01100001



l

1

108

01101100



space

1

32

00100000



n

1

110

01101110



e

1

101

01100101



s

1

115

01110011



y

1

121

01111001




STEP 4:

Draw final Huffman encoding tree with the help of frequency table. (Step by step construction of Huffman encoding tree is not required). 

SOLUTION:

Huffman Encoding Tree:

Recommend Draw your diagram using Data Flow Diagram (DLD) Software

Correct CS301 Assignment 3 Solution 2021



STEP 5:

Get the encoded bits from tree and fill code of each letter in last column of table constructed in step 2. 


SOLUTION:



Letters

Frequency

Original bits

Encoding bits

ASCII

Binary

v

2

118

01110110

011

3*2 = 6

i

3

105

01101001

110

3*3 = 9

r

2

114

01110010

010

3*2 = 6

t

2

116

01110100

000

3*2 = 6

u

2

117

01110101

001

3*2 = 6

a

1

97

01100001

11110

5*1 = 5

l

1

108

01101100

11111

5*1 = 5

space

1

32

00100000

1010

4*1 = 4

n

1

110

01101110

1011

4*1 = 4

e

1

101

01100101

1001

4*1 = 4

s

1

115

01110011

1000

4*1 = 4

y

1

121

01111001

1110

4*1 = 4

Total Encoding Bits = 63


STEP 6:

Calculate the efficiency of Huffman encoding technique. 

(For efficiency use total original bits, total compressed (encoded) bits and find what percentage of memory is saved with the help of Huffman encoding technique).


SOLUTION:


Total Original bits = 18*8 = 144 

Total Encoding bit’s = 63

Percentage of memory bits = 56.25% bit’s memory saved with the help of Huffman Encoding



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