Shannon-Fano and Huffman codes for six messages

Messages $m_1 \dots m_6$ have $p = 0.3, 0.08, 0.1, 0.15, 0.25, 0.12$ (sum = 1). Arranged in decreasing order: $m_1$ 0.3, $m_5$ 0.25, $m_4$ 0.15, $m_6$ 0.12, $m_3$ 0.1, $m_2$ 0.08.

Entropy:

$$H = 2.4224\ \text{bits/message}$$

Shannon–Fano code

Split into equal-probability halves: {0.3, 0.25} = 0.55 vs {0.15, 0.12, 0.1, 0.08} = 0.45; then {0.15, 0.12} = 0.27 vs {0.1, 0.08} = 0.18, and so on.

$$\begin{aligned} \bar L_{SF} &= 0.3(2)+0.25(2)+0.15(3)+0.12(3)+0.1(3)+0.08(3) \\ &= 0.6+0.5+0.45+0.36+0.3+0.24 = 2.45\ \text{bits} \\ \eta_{SF} &= \frac{2.4224}{2.45} = 98.87\% \end{aligned}$$

Huffman code

Reduction (sum of two lowest moved up as high as possible):

$$\begin{aligned} \bar L_{H} &= 0.3(2)+0.25(2)+0.15(3)+0.12(3)+0.1(3)+0.08(3) = 2.45\ \text{bits} \\ \eta_{H} &= \frac{2.4224}{2.45} = 98.87\% \end{aligned}$$

Answer: both codes give $\bar L = 2.45$ bits/message and efficiency 98.87 %.