# hw7 - be corrected by the table(it goes to 01011 p38 2.18...

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MATH 3360 HOMEWORK SOLUTION 7 (Some computational problems are omitted.) p16, 1.7 u + 0 = u and 0 + u = u . In the former the original message is u and the error is 0 and in the latter the original message is 0 and the error is u . In both case we receive u and therefore cannot recognize the error u if it happens. p37, 2.5 Code words are 00000 , 10110 , 01011 , 11101. Here is a list of cosets and corresponding leading-elements (note that in some of the cosets the leading-element is not unique.) coset leading - element (00000 , 10110 , 01011 , 11101) 00000 (10000 , 00110 , 11011 , 01101) 10000 (01000 , 11110 , 00011 , 10101) 01000 (00100 , 10010 , 01111 , 11001) 00100 (00010 , 10100 , 01001 , 11111) 00010 (00001 , 10111 , 01010 , 11100) 00001 (11000 , 01110 , 10011 , 00101) 11000 (10001 , 00111 , 11010 , 01100) 10001 11111 and 01011 are decoded as 11101 and 01011 respectively. The answers to the next two questions are not unique. For example, in the table above, 11000 has two errors and gets corrected to 00000, and 00011, if it is originally 00000 with two errors, then it cannot
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Unformatted text preview: be corrected by the table (it goes to 01011). p38, 2.18 Suppose the parity matrix has two identical columns i and j , then it is easy to see that a string of “000 ... 010 .. 0000 ... 010 ... 000” with only 1’s at position i and j is a code word. It has weight 2 so the linear code C cannot correct single errors. Suppose the matrix has a zero column i , then the string of “000 ... 010 .... 000” with only 1 at position i is a code word and it has weight 1. Conversely suppose that C cannot correct single errors, then the minimal weight is either 1 or 2. Suppose 000 ... 010 .. 0000 ... 010 ... 000 is a code word, then the corresponding columns where 1 appear are identical (they must be both 0 or both 1 for the same row). Suppose 00000 ... 010 .... 0000 is a code word with only one 1, then similarly the i-th column of H is zero. 1...
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