# By writing the parity code (even) and threefold repetition code for all possible four-bit straight binary numbers, prove that the Hamming distance in the two cases is at least 2 in the case of the parity code and 3 in the case of the repetition code.

+1 vote
248 views

+1 vote
by
selected by (user.guest)

Solution

The generation of codes is shown in Table 2.10. An examination of the parity code numbers reveals that the number of bit disagreements between any pair of code words is not less than 2. It is either 2 or 4. It is 4, for example, between 00000 and 10111, 00000 and 11011, 00000 and 11101, 00000 and 11110 and 00000 and 01111. In the case of the threefold repetition code, it is either 3, 6, 9 or 12 and therefore not less than 3 under any circumstances.

## Related questions

+1 vote
1 answer 344 views
0 answers 1.1k views
0 answers 400 views
1 answer 2.8k views
+1 vote
1 answer 260 views
0 answers 384 views
+1 vote
0 answers 290 views
0 answers 396 views
1 answer 562 views
+1 vote
1 answer 911 views
0 answers 411 views
0 answers 345 views
0 answers 246 views
0 answers 225 views
+1 vote
1 answer 232 views
1 answer 348 views
+1 vote
1 answer 3.0k views
+1 vote
1 answer 1.1k views
3 answers 3.7k views
1 answer 1.1k views
+1 vote
1 answer 581 views
1 answer 8.2k views
0 answers 377 views
+1 vote
0 answers 423 views
1 answer 5.5k views
1 answer 3.0k views
1 answer 2.5k views
0 answers 336 views
+1 vote
1 answer 651 views
0 answers 221 views
1 answer 680 views
+1 vote
1 answer 3.7k views
+1 vote
2 answers 4.3k views
0 answers 974 views
+1 vote
1 answer 390 views
–1 vote
1 answer 1.8k views
+1 vote
1 answer 366 views
+1 vote
1 answer 1.0k views
+1 vote
1 answer 1.6k views
+1 vote
0 answers 169 views
1 answer 1.3k views
+1 vote
1 answer 639 views
0 answers 234 views
+1 vote
1 answer 433 views
+1 vote
1 answer 596 views
+1 vote
1 answer 5.6k views