Peterson Davie 2.13

Question

Show that two-dimensional parity provides the receiver enough information to correct any $1$-bit error (assuming the receiver knows only $1$ bit is bad), but not any $2$-bit error.

Answer 1

I would like to explain this answer :) 

In the question above we are talking about parity -- which is one of the methods used for Error Correction and detection 

Now assume data we need to send is 

ROW 1 : 1 1 1

Row 2 : 0 1   1 

Row 3 : 1 0  1 

now what we need to do we need to calculate parity ( assume even parity of 1 ) rowwise and columnwise .

So for rowise ( ROW 1) we will get parity bit as 1 ( we need to make even no of ones by counting   parity and databits ) 

for row 2 : parity bit will be 0 

for row 3 parity bit will be 0 

now again calculate for columwise 

coloum1 : parity bit is 0 

column2 parity bit is 0 

coloumn 3 parity bit is 1 

now sender sends data and even parity bits that it calculate (rowwise and coloumnwise) 

So data send here is : 

Row 1 : 1 1 1 

Row 2 : 0 1  1 

Row 3 : 1 0  1

PRow :  1 0  0   where P row mean rowise parity

Crow :  0   0   1   where C mean colown parity ( calculated above ) 

Now next thing : 

Reciever get data :

Assume he get 

 1   0  1 

0     1   1 

1  0   1 

now he will calcuated parity rowise and columwise 

so for row 1 : parity is 0 

for row 2 : it is 0 

for row 3 it is 0 

now colum wise 

col 1 : 0 

col2 : 1 

col 3 : 1

Now final step : 

if we see parity at sender side  was  ( Rowise parity was ) : 1 0  0 

f we see parity at column  side  was  ( Rowise parity was ) : 0 0  0 

see the first bit is changes from 1 -- 0 so error ( first row ) 

f we see parity at sender side  was  ( columnwise parity was ) : 0 0 1 

f we see parity at column  side  was  ( columwise  parity was ) : 0 1 1

here the third bit doenst match it gt change from 1-- 0 ( second column)

Now the intersection or position of first row and 2 column , we have an error 

And look the under lined 0 ( above ) . It is an erro bit . since we know the postion we flip the bit to 1 

hence this way we can detect 1 bit error 

But sometimes we cant detect also beacuse parity bits might get corrupted with data bits . so that they satisfy even parity criteria .

( you can try for out 2 bits :)) -- Just change any 2 bits and follow same procedure you wont be able to find out position of that on reciever side 

The fact when we cant distinguish or find error even with the help of parity bits . Such errors are called meaningful errors