UGC NET Computer Science July 2018 - II Question 25

Question 25

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25. A text is made up of the characters A, B, C, D, E each occurring with the probability 0.08, 0.40, 0.25, 0.15 and 0.12 respectively. The optimal coding technique will have the average length of :

Options:

1. (1) 2.4
2. (2) 1.87
3. (3) 3.0
4. (4) 2.15

Answer : (4) 2.15

Explanation Question 25

Given:
Probability for occurrence of each character in text.
A → 0.08
B → 0.40
C → 0.25
D → 0.15
E → 0.12

Now, we apply the Huffman optimal coding technique.
First we have sort the character by the probability of the occurrence in text.
A → 0.08
E → 0.12
D → 0.15
C → 0.25
B → 0.40

We will draw a Huffman tree:

Tree using Optimal Huffman Coding Technique

Now Huffman coding for character:
A = 1110
B = 0
C = 10
D = 110
E = 1111

Now, length for each character = no of bits * frequency of occurrence:
Length of A = 4 * 0.08 = 0.32
Length of B = 1 * 0.4 = 0.4
Length of C = 2 * 0.25 = 0.5
Length of D = 3 * 0.15 = 0.45
Length of E = 4 * 0.12 = 0.48

Now add these length to find average length:
Average length = 0.32 + 0.4 + 0.5 + 0.45 + 0.48
= 2.15

So, option (4) is correct.

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