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Both the payload and $2\text{D}$ (even) parity bits are shown below. Some of these bit(s) have been flipped. Indicate the row and column of the flipped bit(s) in format (col, row); e.g., top left bit is $(0,0)$.
\[\begin{array}{l}
1000 \; 0001 \; 1111 \; 1010 \text { | } 0 \\
1011 \; 0110 \; 0010 \; 1111 \text { | } 0 \\
0110 \; 1110 \; 0111 \; 0010 \text { | } 1 \\
1101 \; 1010 \; 1000 \; 1101 \text { | } 0 \\
1100 \; 0100 \; 0011 \; 0011 \text { | } 0 \\
\overline{0110 \;0111\; 0101\; 1001 \text { | } 1} \\
\end{array}\]
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A).So here (4,3) it should be 1 and at (5,10) it should be 1 then it will give you the correct 2D even parity code

and  there is another positions instead of correcting the above two positions, B). correction can be done at (5,3) it should be 1 and at (4,10) it should be 1.

[ NOTE : choose one either A or B  ] 

by mistakenly commented copied into an answer by me ...sorry

 

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