Problem Set Answers: Error Detection and Correction

1. Each word has an odd number of 1 bits so this code is using odd parity.
2. 1. 1010 1100
   2. 0101 0101
   3. 1000 0100
   4. 0111 0010
3. 1. 010 1010
   2. 010 1101
   3. 111 1111
4. 1. error in D4
   2. no error
   3. error in C1
5. 1. 1001 1001
   2. 0101 0101
   3. 1110 0001
6. 1. 1 error in C2
   2. 2 errors
   3. 1 error in D2


7. If there is an error in a check bit, then the syndrome will have one 1 bit. If there is an error in a data bit, then the syndrome will have more than one 1 bit.
8. 1. 1010 0011 1001
   2. 0111 1001 1110
   3. 1010 0110 1100
9. 1. error in d2
   2. error in c4
   3. no error

10. 1.      m + r + 1 <= 2r
            10+ r + 1 <= 2r
            11 + r <= 2r
            which implies that r should be 4

    2. The code word for 10 0110 0110 is arranged as follows:

       
       	data bits     1    0   0   1   1   0       0   1   1       0
       	assignments  D10  D9  D8  D7  D6  D5  C8  D4  D3  D2  C4  D1  C2  C1
           positions     14  13  12  11  10   9   8   7   6   5   4   3   2   1
       
            check bit 1 checks D1,D2,D4,D5,D7,D9 so C1 is 0
            check bit 2 checks D1,D3,D4,D6,D7,D10, so C2 is 0
            check bit 4 checks D2,D3,D4,D8,D9,D10, so C4 is 1
            check bit 8 checks D5,D6,D7,D8,D9,D10 so C8 must be 1
       
            The code word is as follows
       
       	bits           1   0   0   1   1   0   1   0   1   1   1   0   0   0
       	assignments  D10  D9  D8  D7  D6  D5  C8  D4  D3  D2  C4  D1  C2  C1
           positions     14  13  12  11  10   9   8   7   6   5   4   3   2   1

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