| A. | 2.15 |
| B. | 3.01 |
| C. | 2.3 |
| D. | 1.78 |
Answer : A Explanation :
Using Hoffman's algorithm, code for a is 1111; b is 0; c is 110; d is 1110; e is 10. Average code length is
4 x.12 + 1 x .4 + 3 x.15 + 4 x.08 + 2 x.25 = 2.15
raj said: (3:04am on Wednesday 5th June 2013)
if code for a is 1111 then it must be multiplied by 15 i.e .12x15 refer link http://www.siggraph.org/education/materials/HyperGraph/video/mpeg/mpegfaq/huffman_tutorial.html
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Option: A Explanation : Explanation will come here. Explanation will come here. Explanation will come here. Explanation will come here. Explanation will come here. |
