ISRO CS 2025 Q.37: solution and answer

Official answer: DSource: ISRO's final answer key of 15 May 2026

1. Minimum-distance decoding picks the valid codeword nearest to the received word, where distance is the number of bit positions in which two words differ (Hamming distance).

2. Received word $01000$ against A $=00000$: only the second bit differs, so $d=1$.

3. Received word $01000$ against B $=01110$: the third and fourth bits differ, so $d=2$.

4. Received word $01000$ against C $=11000$: only the first bit differs, so $d=1$.

5. The nearest codewords are A and C, both at distance $1$. There is no unique closest codeword, so the decoder has no basis to prefer one.

6. This agrees with the code's distances: $d(A,B)=3$, $d(A,C)=2$, $d(B,C)=3$, so $d_{\min }=2$. A code corrects $t$ errors only if $d_{\min }\ge 2t+1$; with $d_{\min }=2$ it can detect a single-bit error but cannot correct it.

7. The receiver can tell that $01000$ is not a valid codeword but cannot decide between A and C. Options (a) and (c) each choose one of two equally near codewords, and (b) is farther away at distance $2$, so the answer is (d).