 |
CS301 ASSIGNMENT NO. 2 SPRING 2022 |
CS301 ASSIGNMENT NO. 2 SPRING 2022 || 100% RIGHT SOLUTION || DATA STRUCTURES || BY VuTech
CS301
DATA STRUCTURES
ASSIGNMENT NO. 2
SPRING 2022
LAST DATE: 11-08-2022
SOLUTION:
1: Count all the letters including space from the
given text message.
Answer: Total 52 characters
2: Draw a table with column names (letter, frequency,
original bits, encoded bits)
Answer: Mention in 3rd
Step.
3: Fill the
table with letters, frequency, original bits (for original bits get ASCII
decimal code of each letter, convert the decimal ASCII into 8 bits binary code)
and encoded bits (these can be found from Huffman encoding tree as mentioned in
point 5).
Answer:
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|
Frequency |
Original Bits |
Encoded Bits |
|
|
Space |
10 |
00100000 |
01 |
|
n |
6 |
01101110 |
000 |
|
o |
6 |
01101111 |
001 |
|
a |
5 |
01100001 |
1010 |
|
e |
4 |
01100101 |
1100 |
|
t |
3 |
01110100 |
1111 |
|
i |
3 |
01101001 |
1101 |
|
u |
2 |
01110101 |
11101 |
|
c |
2 |
01100011 |
1000 |
|
y |
2 |
01111001 |
10011 |
|
d |
1 |
01100100 |
100100 |
|
s |
1 |
01110011 |
1011001 |
|
h |
1 |
01101000 |
101110 |
|
g |
1 |
01100111 |
111000 |
|
k |
1 |
01101011 |
101111 |
|
w |
1 |
01110111 |
111001 |
|
f |
1 |
01100110 |
1011000 |
|
r |
1 |
01110010 |
101101 |
|
m |
1 |
01101101 |
100101 |
4: Draw final Huffman encoding tree with the help
of frequency table. (Step by step construction of Huffman encoding tree is not
required, just show the final tree in the solution file).
Answer: You can make Tree by using Shapes in Insert tab of MS Word.
5: Get the encoded bits from tree and fill code of
each letter in last column of table constructed in step 2.
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).
Answer:
A total of 93 bits are required
when using the Huffman coding method. Whereas if we send the original message,
the total number of characters is 52, each bit requires 8 bits. So, it will take 52 * 8 =
416 bits to send the original message. This number of bits is 77% less than the
number of bits in the ASCII encoding form.
7: For the calculation of bandwidth saved for
10,000 messages, use the calculation performed in step 6.
Answer:
For Message
10000 students we can use only 33% of total bandwidth and we can save 77% bandwidth.
For 10,000 messages, only 3300 messages of bandwidth
will be required.
KINDLY,
DON’T COPY PASTE
SUBSCRIBE, SHARE, LIKE AND COMMENTS FOR MORE UPDATES
SEND WHATSAPP OR E-MAIL FOR ANY QUERY
0325-6644800
kamranhameedvu@gmail.com