For the following letter frequency table, a Huffman Code was constructed as shown below. The associated tree is also shown below. Letter Frequency Huffman Code E 120 1 L D 42 42 110 101 0 100 1 0 180 U 37 100 42 0 107 32 C C 32 1110 0 0 M 24 11111 M K 7 111101 Z 2 111100 (a) Is there another code and a tree possible? What is (are) the difference(s)? (b) Compute the redundancy= Average Codeword Length (Lav) - Entropy (H(x)). The entropy H(x) is calculated from H(x) = -Ei=1 Pilogpi. (c) Check if this redundancy is less than p1 + 0.086, where pi is the probability of the most common symbol. (d) Also, check if the following inequality is satisfied: H(x) ≤ Lav ≤ H(x) + 1. Repeat parts (b), (c), (d) for a different tree you built. What conclusions can you draw?

Database System Concepts
7th Edition
ISBN:9780078022159
Author:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan
Publisher:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan
Chapter1: Introduction
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I need someone to explain this problem step by step. I need a briefe explanation to the side of the calculation to understand the problem. Thanks
For the following letter frequency table, a Huffman Code was constructed as shown
below. The associated tree is also shown below.
Letter
Frequency
Huffman Code
E
120
1
L
42
110
E
0
37
D
42
101
79
1
0
186
U
37
100
0
107
C
32
1110
1
0
M
24
11111
K
7
111101
Z
2
111100
(a) Is there another code and a tree possible? What is (are) the difference(s)?
(b) Compute the redundancy= Average Codeword Length (Lav) - Entropy (H(x)).
The entropy H(x) is calculated from H(x) -Ei=1 Pilogpi-
(c) Check if this redundancy is less than p₁ + 0.086, where pi is the probability of
the most common symbol.
(d) Also, check if the following inequality is satisfied:
H(x) ≤ Lav ≤ H(x) + 1.
Repeat parts (b), (c), (d) for a different tree you built. What conclusions can you
draw?
Transcribed Image Text:For the following letter frequency table, a Huffman Code was constructed as shown below. The associated tree is also shown below. Letter Frequency Huffman Code E 120 1 L 42 110 E 0 37 D 42 101 79 1 0 186 U 37 100 0 107 C 32 1110 1 0 M 24 11111 K 7 111101 Z 2 111100 (a) Is there another code and a tree possible? What is (are) the difference(s)? (b) Compute the redundancy= Average Codeword Length (Lav) - Entropy (H(x)). The entropy H(x) is calculated from H(x) -Ei=1 Pilogpi- (c) Check if this redundancy is less than p₁ + 0.086, where pi is the probability of the most common symbol. (d) Also, check if the following inequality is satisfied: H(x) ≤ Lav ≤ H(x) + 1. Repeat parts (b), (c), (d) for a different tree you built. What conclusions can you draw?
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where did you get 20/51 , 7/51 and so on? does entropy is not equal to the sumation of pilog2pi? or why is divede it by 1/pi . how do you calculate Lav

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