39 votes 39 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 $n$? $1$ $2$ $3$ $4$ Computer Networks gateit-2008 computer-networks normal error-detection + – Ishrat Jahan asked Oct 29, 2014 • edited Jun 20, 2018 by Milicevic3306 Ishrat Jahan 15.8k views answer comment Share Follow See all 3 Comments See all 3 3 Comments reply set2018 commented Sep 3, 2017 reply Follow Share is this the approach to solve this? i got too many combinations and choose minimum among them or alternate way ? 2 votes 2 votes JashanArora commented Nov 26, 2019 reply Follow Share Odd parity (error) is in 1 row and 3 columns. So, we need to correct 3 * 1 = 3 bits. Is this approach fine? 0 votes 0 votes u_t_k_a_r_s_h commented 5 days ago reply Follow Share What about r1d5, r1d2 and r1d0? 0 votes 0 votes Please log in or register to add a comment.
3 votes 3 votes There will be two possibilities... CASE 1: d7 d6 d5 d4 d3 d2 d1 d0 r0 0 1 0 1 0 0 1 1 r1 1 1 0 0 1 1 1 0 r2 0 0 0 1 0 1 0 0 r3 0 1 1 0 1 0 1 0 r4 1 1 0 0 0 1 1 0 CASE 2: d7 d6 d5 d4 d3 d2 d1 d0 r0 0 1 0 1 0 0 1 1 r1 1 1 0 0 1 1 1 0 r2 0 0 0 1 0 1 0 0 r3 0 1 1 0 1 0 1 0 r4 1 1 0 0 0 1 1 0 so, in either cases minimum three 3-bits are corrupted. scholaraniket answered Sep 5, 2019 • edited Oct 17, 2019 by scholaraniket scholaraniket comment Share Follow See all 0 reply Please log in or register to add a comment.
2 votes 2 votes It is 2D parity scheme and generally we check for 1D parity scheme, i.e., either row and column, so in 2D parity scheme we need to check both row as well as column. And it is given this is even parity so, whichever column and row is not even parity there is corruption happend So if you look carefully column d5 it is odd parity so there is one corruption similarly at d2 and in last d0. So make it even 2D parity we need to change atleast 3 bits So option C is correct. Arvnd answered Dec 5, 2018 Arvnd comment Share Follow See all 0 reply Please log in or register to add a comment.
1 votes 1 votes Can anyone provide any source to read this topic as I'm not getting the question??? pass_i0n answered Sep 20, 2019 pass_i0n comment Share Follow See all 0 reply Please log in or register to add a comment.
–2 votes –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 Pavan Kumar Munnam answered Aug 26, 2016 Pavan Kumar Munnam comment Share Follow See 1 comment See all 1 1 comment reply HeadShot commented Nov 19, 2018 reply Follow Share Refer : https://youtu.be/vNPBfq-gKHE 0 votes 0 votes Please log in or register to add a comment.