UGC-NET | UGC-NET CS 2017 Nov – III | Question 35
A text is made up of the characters a, b, c, d, e each occurring with the probability 0.11, 0.40, 0.16, 0.09 and 0.24 respectively. The optimal Huffman coding technique will have the average length of:
(A)
2.40
(B)
2.16
(C)
2.26
(D)
2.15
Answer: (B)
Explanation:
a = 0.11 b = 0.40 c = 0.16 d = 0.09 e = 0.24 we will draw a huffman tree: now huffman coding for character:
a = 1111
b = 0
c = 110
d = 1111
e = 10
length for each character = no of bits * frequency of occurrence:
a = 4 * 0.11
= 0.44
b = 1 * 0.4
= 0.4
c = 3 * 0.16
= 0.48
d = 4 * 0.09
= 0.36
e = 2 * 0.24
= 0.48
Now add these lenght for average length:
0.44 + 0.4 + 0.48 + 0.36 + 0.48 = 2.16
So, option (B) is correct.
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Last Updated :
20 Mar, 2018
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