Discussion
The number of errors can be corrected using the code words of group code with the generator :
\[ G = \begin{vmatrix} 1 & 1 & 0 \\ 0 & 1 & 1 \end{vmatrix} \]
Answer: D
The code can correct 0 errors.
The code words are all sums (mod 2) of the rows of G: 000, 110, 011 and 110 + 011 = 101.
\[ \begin{aligned} d_{min} &= \text{least non-zero weight} = 2 \\ t &= \lfloor (d_{min} - 1)/2 \rfloor = \lfloor 1/2 \rfloor = 0 \end{aligned} \]
With a minimum distance of 2 the code can detect a single error but cannot correct any, because a received word one bit away from two code words is equally close to both. Correcting one error would need a minimum distance of at least 3, so 1, 2 and 3 are all too high.
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