To answer this question you need to understand the hamming distance.

The hamming is the number of difference bits in two vectors.

For example, suppose you have a single parity bit for each codeword. What's the hamming distance?

Suppose you have one data bit to send and a single parity bit. Let the parity bit be even...that is, make an even number of ones in the code word. The codeword is the data + the parity....

Possible data words:

1 or 0

Possible codewords:

11 or 00

Note, the hamming distance is two.

Try another...suppose you have 2 data bits....

possible data bits:

A	B
0	0
0	1
1	0
1	1

Codewords, with parity will be:

A	B	P= A xor B
0	0	0
0	1	1
1	0	1
1	1	0

Note that a single parity bit gives us a hamming distance Dh = 2...no matter what the size of the data word is.

Suppose I have number of bits in the code word equal to n and the number of bits in the message equal to m so that the number check bits is equal to

k = n - m

Suppose n = 3 and m = 1, now the number of check bits is 2. What is the maximum hamming distance between any two code words?

000 or 111

Suppose m=2 and k = 2 so that n = 4. Now I can transmit

ABp1p2

0000

0101

1010

1111

Hamming distance now is 2. How about a 3 bit code?

• ABCp1p2
• 000 00
• 001 10
• 010 01
• 011 11
• 100 10
• 101 00
• 110 11
• 111 01
• So now the hamming distance is two.