Question: Obtain Even Parity hamming code for 1010. Prove that hamming code is an error detecting and correcting code.

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- For a 4-bit code there are 3 parity bits
**p1, p2 and p3**at location**1, 2 and 4**resp. - So, the code will be:
**“p1 p2 n1 p3 n2 n3 n4”**where,**n1, n2, n3, n4**are bits of the code and**p1,p2 and p3**are parity bits to be calculated - Therefore, the code for even parity is calculated as below:

- Therefore the even parity hamming code is:
**1011010.** - Consider that the same calculated code is sent , but it is received with an error in one particular bit, say bit7. S the code word received is:
**1011011**. - Hence, to identify the error and correct it the receiver follows the procedure as below:
- Parity check for bits
**4,5,6,7 : 1011 = 1 (odd parity) = c1** - Parity check for bits
**2,3,6,7: 0111 = 1 (odd parity) = c2** - Parity check for bits
**1,3,5,7: 1101 = 1(odd parity) = c3** - Therefore the receiver now arranges the error correcting bits as
**c1 c2 c3 = 111**and gets the error bit as bit7 and complements it. - Therefore the correct code is:
**1011010**

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