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Question

The number of errors can be corrected using the code words of group code with the generator :

\(G=\begin{vmatrix}1 & 1 & 0\\0 & 1 & 1\end{vmatrix}\)

This question was previously asked in
MH SET 2020 Paper 2 Philosophy Question Paper (27-Dec-2020)
The correct answer is

0

A code with minimum distance \(d_{\min}\) can correct \(t = \left\lfloor \dfrac{d_{\min}-1}{2} \right\rfloor\) errors. For a group code, the minimum distance equals the smallest weight of a non-zero code word.

Step 1 — The generator has 2 rows, so the messages are 00, 01, 10, 11. Multiplying each by G gives the code words 000, 011, 110 and 101.

Step 2 — The non-zero code words 011, 110 and 101 each have weight 2, so \(d_{\min} = 2\).

Step 3 — \(t = \left\lfloor \dfrac{2-1}{2} \right\rfloor = 0\). A distance-2 code can detect a single error but cannot correct any, because a received word with one flipped bit is equally close to two different code words.

Correcting one error would need a minimum distance of at least 3, and correcting two or three errors would need distances of 5 or 7, which a 3-bit code with four code words cannot reach.

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