Question
Apply Huffman's algorithm to construct an optimal binary prefix code for the letters and its frequencies in the table given below (Show the complete steps).
Letters A B C D E F G
Frequency 15 25 5 7 10 13 9
Find out an average number of bits required per character. Also, Implement Huffman's coding algorithm and run for the given problem instance. Support your code with suitable comments for better readability
Answer :
Word Count : 599
We are given the letters A, B, C, D, E, F, G with frequencies: A = 15, B = 25, C = 5, D = 7, E = 10, F = 13, G = 9 --- Step 1: Arrange frequencies in ascending order C(5), D(7), G(9), E(10), F(13), A(15), B(25) --- Step 2: Combine two lowest frequencies C(5) + D(7) = 12 → Node1(12) Remaining: Node1(12), G(9), E(10), F(13), A(15), B(25) --- Step 3: Arrange again G(9), E(10), Node1(12), F(13), A(15), B(25) --- Step 4: Combine two lowest G(9) + E(10) = 19 → Node2(19) Remaining: Node1(12), F(13), A(15), Node2(19), B(25) --- Step 5: Arrange again Node1(12), F(13), A(15), Node2(19), B(25) --- Step 6: Combine two lowest Node1(12) + F(13) = 25 → Node3(25) Remaining: A(15), Node2(19), Node3(25), B(25) --- Step 7: Arrange again A(15), Node2(19), Node3(25), B(25) --- Step 8: Combine two lowest A(15) + Node2(19) = 34 ___ _______ _______ _______ ___ _____ ___ _______ ________ ________ __________ __________.
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We are given the letters A, B, C, D, E, F, G with frequencies: A = 15, B = 25, C = 5, D = 7, E = 10, F = 13, G = 9 --- Step 1: Arrange frequencies in ascending order C(5), D(7), G(9), E(10), F(13), A(15), B(25) --- Step 2: Combine two lowest frequencies C(5) + D(7) = 12 → Node1(12) Remaining: Node1(12), G(9), E(10), F(13), A(15), B(25) --- Step 3: Arrange again G(9), E(10), Node1(12), F(13), A(15), B(25) --- Step 4: Combine two lowest G(9) + E(10) = 19 → Node2(19) Remaining: Node1(12), F(13), A(15), Node2(19), B(25) --- Step 5: Arrange again Node1(12), F(13), A(15), Node2(19), B(25) --- Step 6: Combine two lowest Node1(12) + F(13) = 25 → Node3(25) Remaining: A(15), Node2(19), Node3(25), B(25) --- Step 7: Arrange again A(15), Node2(19), Node3(25), B(25) --- Step 8: Combine two lowest A(15) + Node2(19) = 34 ___ _______ _______ _______ ___ _____ ___ _______ ________ ________ __________ __________.
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