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If we assume the following: the received bits were encoded using a 2-D even-parity code, what is min #of error is there.

I am a little bit confused , it is clear that the third column has an error and ( second , fourth and sixth ) rows have errors.

For example, If we change the bit location(2,3) the third column and second row will indicate there are no errors and we lift with only two rows (fourth and sixth) with no column. What can I say about this , the minimum # of error in this case?

Can someone help me with this? Thank you

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With this kind of parity only a single bit error can be corrected. So not a single error can be corrected in this example.

Here one column and three rows contain an error. To rows could be corrected by flipping two bits in any arbitrary row (where they intersect with the two rows that contain an error), it wouldn't change their parity bit. The remaining error could be fixed at the intersection of the row and column with the remaining error.

However, since the choice of the first column is arbitrary, the errors cannot be corrected.

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  • \$\begingroup\$ So, the correct answer would be >> Min #of Error is 3 and of course it can't be corrected using this algorithm. \$\endgroup\$ – user34755 Dec 11 '16 at 23:54
  • \$\begingroup\$ Yes, there are three rows with wrong parity. \$\endgroup\$ – Mario Dec 12 '16 at 5:52

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