Homework 4 Solution

0000 1 1111 0000 1 1111 0 0000 1 0 1111 11 0000 00 11

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Unformatted text preview: 000 00 111 11 00 111 11111 00000 1111111 11 11 1 1 11 11 11 0000000 00 00 0 0 00 00 00 111 000 11 00 111 000 11 1 00 0 111 000 11111 00000 111 000 1 1 1 11 11 1 1 1 11 11 0 0 0 00 00 0 0 0 00 00 1 2 3 4 5 6 7 8 9 10 1 0 1111 0000 1 0 1 0 1 1111 0 0 0000 1 1111 0000 1 1111 0 0000 1 0 1111 11 0000 00 11 1111 00 0000 11 00 111 000 1 0 1 11 00 11 00 11 00 1 111 000 00 11 00 1 11 00 11 00 0 11 1 111111 00 0 000000 11 00 1111 0000 111 000 11 00 1 11 0 00 11111 111111 00000 000000 11 00 11 1 00 0 1 0 1 11 0 00 1 0 111 1 111 11 000 0 000 00 11 00 1 0 1 11 0 00 11 00 111 000 10 000 00 11 00 111 11 1 00 111 0001 11 111 11 0 00 000 00 1 0 000 1 00 0 1 11 011 0 111 1 000 0 1 0 1 111 0 000 1 1 00 0 11 111 11 11111 0 000 00 00000 1 1 0 111 1 000 0 1 0 1 111 1 0 000 0 1 0 11 111 11 000 00 111 000 1 1 0 1 0 1 11 0 00 11 00 1 0 111 0 000 6 111 000 11 00 9 1 0 78 111 0001 111 000 11 00 1 0 0 111 000 111 000 11 00 1 0 01 111 1 000 0 11 00 111 000 111 000 11 00 1 0 1 0 11 00 111 1 000 0 11 00 1 0 11 00 111 000 3 11 00 45 111 0001 11 00 0 111 000 11 00 111 000 11 00 11 1 00 0 11 1 00 0 1 2 We thus have the following prefix-free code: symbol relative frequency codeword a 1 00000 b 2 00001 c 3 0001 d 4 1100 e 5 1101 f 6 001 g 7 100 h 8 101 i 9 111 j 10 01 (b) From (a), the expected length of a codeword is 1 (1(5) 55 + 2(5) + 3(4) + 4(4) + 5(4) + 6(3) + 7(3) + 8(3) + 9(3) + 10(2)) = For a message with 1000 symbols, we would therefore expect to use a total of (1000)(173/55) ≈ 3145.45 0’s and 1’s. 5 173 55 ≈ 3.14545....
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