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+19 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

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
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

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

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

R1d5 R1d1 R1do will also do the work.

Three corrupted columns and one corrupted row will be handled by this
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)
it will be (r1,d0),(r4,d2),,(r4,d5) right?
(r4,d5) should be 1. Isn`t the answer should be option d?
0 votes


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



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)

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