Consider the binary code that consists of only four valid codewords as given…
2017
Consider the binary code that consists of only four valid codewords as given below:
00000,01011,10101,11110
Let the minimum Hamming distance of the code \(p\) and the maximum number of erroneous bits that can be corrected by the code be \(q\). Then the values of \(p\) and \(q\) are
- A.
\(p=3 \)and\(q=1\) - B.
\(p=3 \)and\(q=2\) - C.
\(p=4\)and\(q=1\) - D.
\(p=4\)and\(q=2\)
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Correct answer: A
Solution: compute the minimum Hamming distance p and the error-correcting capability q.
List the codewords: 00000, 01011, 10101, 11110.
Compute pairwise Hamming distances (number of bit positions that differ):
d(00000, 01011) = 3
d(00000, 10101) = 3
d(00000, 11110) = 4
d(01011, 10101) = 4
d(01011, 11110) = 3
d(10101, 11110) = 3
The smallest of these distances is 3, so the minimum Hamming distance p = 3.
The maximum number of errors that can be corrected is q = floor((p - 1) / 2) = floor((3 - 1) / 2) = 1.
Therefore, p = 3 and q = 1.
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