i got too many combinations and choose minimum among them

or alternate way ?

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34 votes

Data transmitted on a link uses the following $2D$ parity scheme for error detection:

Each sequence of $28$ bits is arranged in a $4\times 7$ matrix (rows $r_0$ through $r_3$, and columns $d_7$ through $d_1$) and is padded with a column $d_0$ and row $r_4$ of parity bits computed using the Even parity scheme. Each bit of column $d_0$ (respectively, row $r_4$) gives the parity of the corresponding row (respectively, column). These $40$ bits are transmitted over the data link.

$$\small \begin{array}{|c|c|c|c|c|c|c|c|c|c|} \hline

&\bf{d_7}&\bf{d_6}&\bf{d_5}&\bf{d_4}&\bf{d_3}&\bf{d_2}&\bf{d_1}&\bf{d_0}\\

\hline

\bf{r_0}&0&1&0&1&0&0&1&\bf{1}\\\hline

\bf{r_1}&1&1&0&0&1&1&1&\bf{0}\\\hline

\bf{r_2}&0&0&0&1&0&1&0&\bf{0}\\\hline

\bf{r_3}&0&1&1&0&1&0&1&\bf{0}\\\hline

\bf{r_4}&\bf{1}&\bf{1}&\bf{0}&\bf{0}&\bf{0}&\bf{1}&\bf{1}&\bf{0}\\

\hline\end{array}$$

The table shows data received by a receiver and has $n$ corrupted bits. What is the minimum possible value of $n$?

- $1$
- $2$
- $3$
- $4$

38 votes

Best answer

$$\small \begin{array}{|c|c|c|c|c|c|c|c|c|c|} \hline

&\bf{d_7}&\bf{d_6}&\bf{d_5}&\bf{d_4}&\bf{d_3}&\bf{d_2}&\bf{d_1}&\bf{d_0}\\

\hline

\bf{r_0}&0&1&0&1&0&0&1&\bf{1}\\\hline

\bf{r_1}&1&1&\boxed0&0&1&1&1&\bf{0}\\\hline

\bf{r_2}&0&0&0&1&0&1&0&\bf{0}\\\hline

\bf{r_3}&0&1&1&0&1&0&1&\bf{0}\\\hline

\bf{r_4}&\bf{1}&\bf{1}&\bf{0}&\bf{0}&\bf{0}&\boxed{\bf{0}}&\bf{1}&\boxed{\bf{1}}\\

\hline\end{array}$$

Here, we need to change minimum $3$ bits, and by doing it we get correct parity column wise and row wise (Correction marked by boxed number)

C is answer

&\bf{d_7}&\bf{d_6}&\bf{d_5}&\bf{d_4}&\bf{d_3}&\bf{d_2}&\bf{d_1}&\bf{d_0}\\

\hline

\bf{r_0}&0&1&0&1&0&0&1&\bf{1}\\\hline

\bf{r_1}&1&1&\boxed0&0&1&1&1&\bf{0}\\\hline

\bf{r_2}&0&0&0&1&0&1&0&\bf{0}\\\hline

\bf{r_3}&0&1&1&0&1&0&1&\bf{0}\\\hline

\bf{r_4}&\bf{1}&\bf{1}&\bf{0}&\bf{0}&\bf{0}&\boxed{\bf{0}}&\bf{1}&\boxed{\bf{1}}\\

\hline\end{array}$$

Here, we need to change minimum $3$ bits, and by doing it we get correct parity column wise and row wise (Correction marked by boxed number)

C is answer

22 votes

Sherrinford question is about error detection not correction so whatever way the table becomes valid is correct the minimum value is gonna stay 3

0

7 votes

Here we have odd parity at row r1 and columns d5,d2 and d0.

Now since only 1 row shows error it can be (r1,d0) or (r1,d2) or (r1,d5)-----> any one of the three possible choices.

suppose it is (r1,d2).

Now still we are left with two errors at d0 and d5 but there is no error in any other row. It means error is at same row but two columns d0 and d5 and hence row parity could not detect it.

example- it could be r2,d0 and r2,d5 or r3,d0 and r3,d5 or any such choice.

so we can have minimum three bit error here.

**Answer- C**