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Dynamic Huffman Coding

Dynamic Huffman Coding. Computer Networks Assignment. T. Stage 1 (First occurrence of t ) r / 0 t (1) Order: 0, t (1) * r represents the root

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Dynamic Huffman Coding

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  1. Dynamic Huffman Coding Computer Networks Assignment

  2. T • Stage 1 (First occurrence of t ) r / \ 0 t(1) • Order: 0,t(1) * r represents the root * 0 represents the null node *t(1) denotes the occurrence of T with a frequency of 1

  3. TE • Stage 2 (First occurrence of e) r / \ 1 t(1) / \ 0 e(1) • Order: 0,e(1),1,t(1)

  4. TEN • Stage 3 (First occurrence of n ) r / \ 2 t(1) / \ 1 e(1) / \ 0 n(1) • Order: 0,n(1),1,e(1),2,t(1) : Misfit

  5. Reorder: TEN r / \ t(1) 2 / \ 1 e(1) / \ 0 n(1) • Order: 0,n(1),1,e(1),t(1),2

  6. TENN • Stage 4 ( Repetition of n ) r / \ t(1) 3 / \ 2 e(1) / \ 0 n(2) • Order: 0,n(2),2,e(1),t(1),3 : Misfit

  7. Reorder: TENN r / \ n(2) 2 / \ 1 e(1) / \ 0 t(1) • Order: 0,t(1),1,e(1),n(2),2 • t(1),n(2) are swapped

  8. TENNE • Stage 5 (Repetition of e ) r / \ n(2) 3 / \ 1 e(2) / \ 0 t(1) • Order: 0,t(1),1,e(2),n(2),3

  9. TENNES • Stage 6 (First occurrence of s) r / \ n(2) 4 / \ 2 e(2) / \ 1 t(1) / \ 0 s(1) • Order: 0,s(1),1,t(1),2,e(2),n(2),4

  10. TENNESS • Stage 7 (Repetition of s) r / \ n(2) 5 / \ 3 e(2) / \ 2 t(1) / \ 0 s(2) • Order: 0,s(2),2,t(1),3,e(2),n(2),5 : Misfit

  11. Reorder: TENNESS r / \ n(2) 5 / \ 3 e(2) / \ 1 s (2) / \ 0 t(1) • Order : 0,t(1),1,s(2),3,e(2),n(2),5 • s(2) and t(1) are swapped

  12. TENNESSE • Stage 8 (Second repetition of e ) r / \ n(2) 6 / \ 3 e(3) / \ 1 s(2) / \ 0 t(1) • Order : 0,t(1),1,s(2),3,e(3),n(2),6 : Misfit

  13. Reorder: TENNESSE r / \ e(3) 5 / \ 3 n(2) / \ 1 s(2) / \ 0 t(1) • Order : 1,t(1),1,s(2),3,n(2),e(3),5 • N(2) and e(3) are swapped

  14. TENNESSEE • Stage 9 (Second repetition of e ) r 0/ \1 e(4) 5 0/ \1 3 n(2) 0/ \1 1 s(2) 0/ \1 0 t(1) • Order : 1,t(1),1,s(2),3,n(2),e(4),5

  15. ENCODING The letters can be encoded as follows: • e : 0 • n : 11 • s : 101 • t : 1001

  16. Average Code Length Average code length = i=0,n (length*frequency)/ i=0,nfrequency = { 1(4) + 2(2) + 3(2) + 1(4) } / (4+2+2+1) = 18 / 9 = 2

  17. ENTROPY Entropy = -i=1,n (pilog2 pi) = - ( 0.44 * log20.44 + 0.22 * log20.22 + 0.22 * log20.22 + 0.11 * log20.11 ) = - (0.44 * log0.44 + 2(0.22 * log0.22 + 0.11 * log0.11) / log2 = 1.8367

  18. TENNESSE 9 0/ \1 5 e(4) 0/ \1 s(2) 3 0/ \1 t(1) n(2) ENCODING E : 1 S : 00 T : 010 N : 011 Average code length = (1*4 + 2*2 + 2*3 + 3*1) / 9 = 1.89 Ordinary Huffman Coding

  19. SUMMARY The average code length of ordinary Huffman coding seems to be better than the Dynamic version,in this exercise. But, actually the performance of dynamic coding is better. The problem with static coding is that the tree has to be constructed in the transmitter and sent to the receiver. The tree may change because the frequency distribution of the English letters may change in plain text technical paper, piece of code etc. Since the tree in dynamic coding is constructed on the receiver as well, it need not be sent. Considering this, Dynamic coding is better. Also, the average code length will improve if the transmitted text is bigger.

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