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 lenghth for each character = no of bits * frequency of occurence: 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.