Question

A discrete memoryless source is described by the alphabet x={x1​,x2​,…,xx1​, and the corresponding probability vector p={0.2,0.12,0.06,0.15,0.07,0.1,0.13,0.17}. Design a Huffman code for this source; find L, the average codeword length for the Huffman code; and determine the efficiency of the code.

Answer 1

1. Huffman Code:

`\(x_1\) - 111 \(x_2\) - 110 \(x_3\) - 1111 \(x_4\) - 10 \(x_5\) - 1101 \(x_6\) - 01 \(x_7\) - 00 \(x_8\) - 1110`

2. Average Codeword Length (L):
`\(L = 3 \, \text{bits}\)`

3. Efficiency of the Code:
`\(Efficiency = (H(x))/(L) \approx 0.958\)`

Step-by-step explanation:

To design a Huffman code, we start by constructing a binary tree where each leaf node represents a symbol from the alphabet \(x\) with the probabilities given. The basic idea is to assign shorter codewords to more probable symbols. After constructing the tree, the codewords are assigned by traversing the tree, with left branches representing '0' and right branches representing '1'. The resulting Huffman code for the given source is provided in the first part.

To find the average codeword length L, we use the formula:

`\[ L = \sum_(i=1)^(n) p_i \cdot l_i \]`

`where \(p_i\) is the probability of symbol \(x_i\) and \(l_i\) is the length of the codeword for \(x_i\). Substituting the values, we get \(L = 3 \, \text{bits}\).`

The efficiency of the code Efficiency is calculated as the ratio of the entropy of the source H(x) to the average codeword length L. In this case,
`\(H(x) = 0.721\)`, so
`\(Efficiency \approx (0.721)/(3) \approx 0.958\)`. The efficiency value close to 1 indicates an effective Huffman code for the given source.