A text is made up of the characters A, B, C, D, E each occurring with the…

Solution: Construct the optimal (Huffman) code to find the minimal average length.

Probabilities: A = 0.08, B = 0.40, C = 0.25, D = 0.15, E = 0.12.

• Step 1: Combine the two smallest probabilities: 0.08 + 0.12 = 0.20 (pair A and E).

• Step 2: New list: 0.20, 0.15, 0.25, 0.40. Combine the two smallest: 0.15 + 0.20 = 0.35 (D with previous pair).

• Step 3: New list: 0.25, 0.35, 0.40. Combine 0.25 + 0.35 = 0.60.

• Step 4: Combine 0.60 + 0.40 = 1.00 to complete the tree.

From this Huffman tree the codeword lengths (depths) are:

• B (0.40): length 1

• C (0.25): length 2

• D (0.15): length 3

• A (0.08) and E (0.12): length 4

Compute the average length L:

L = 0.40*1 + 0.25*2 + 0.15*3 + 0.12*4 + 0.08*4 = 0.40 + 0.50 + 0.45 + 0.48 + 0.32 = 2.15 bits per symbol.

Note: The source entropy is approximately 2.098 bits, so the Huffman average length (2.15) is close to and slightly above the entropy, as expected for an optimal prefix code.

Final answer: 2.15 bits per symbol.