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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. $1$
  2. $2$
  3. $3$
  4. $4$
asked in Computer Networks by Boss (19.1k points)
edited by | 2.7k views
+1
is this the  approach to solve this?

 i got too many combinations and choose minimum among them

or alternate way ?

6 Answers

+26 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
answered by Veteran (60.8k points)
edited by
+1

@Prashant but it will be (r1,d0),(r4,d2),(r4,d5) 

0
R1d5 R1d1 R1do will also do the work.

Three corrupted columns and one corrupted row will be handled by this
0
Please provide material to read about this topic. I am unabe to understand this question
+9 votes
Answer: C

(r1, d5) should be 1.

(r4, d2) should be 0.

(r4, d0) should be 1.
answered by Boss (34k points)
+3
it will be (r1,d0),(r4,d2),,(r4,d5) right?
0
(r4,d5) should be 1. Isn`t the answer should be option d?
0 votes

OTHER WAY

answered by Active (1.8k points)
0 votes
option c is correct ,toggle these ones

1.r0-d5

2.r0-d0

3.r1-d2
answered by (213 points)
–1 vote
3 option c
answered by Active (3.3k points)
–2 votes
here the parity bits can also be corrupted, so they asked minimun so when we correct (r0,d0) to 0 and (r0,d5) to 1 then the minimun errors i found are 2, so the answer will be B, correct me if i am wrong
answered by Boss (10.9k points)
Answer:

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