Lecture 4: Hamming Distance

If we are trying to make sure that the correct message is entually recieved when there could be errors in its transmission, it would be nice if we could figure out, from some advance estimate of the number of errors that will occur in transmission, like being able to estimate just how mischevious your little brother is. Since we are transmitting only 0s and 1s, the only way that we can have an error is the change a 0 into a 1, a 1 into a 0 or an erasure which turns either a 0 or a 1 into a smudge. But now lets return to our example about just repeating the message of either 0 or 1 a bunch of times and focus on errors. How many errors of changing a 1 to a 0 (or vice versa with the opposite message) would be nessecary in order to make the majority vote turn out differently, so that the message would be recieved wrong?

Let's ask a slightly different question. What is the smallest number of errors must be made not just to turn the code word sent into a different codeword, in other words, another string of bits that actually would be sent using this code? This number of errors is called the Hamming distance of the code. What is the Hamming distance of the repeating the message of 0 or 1 three times code? What about the repeating n times code? What about the parity check code on four bits? The parity check code on n bits?

Hints and Solutioins:

How many errors of changing a 1 to a 0 (or vice versa with the opposite message) would be nessecary in order to make the majority vote turn out differently, so that the message would be recieved wrong? We would need to change a majority (more than half) of the bits to be recieved wrong.

What is the Hamming distance of the repeating the message of 0 or 1 three times code? What about the repeating n times code? We would need to turn a codeword (eitiher all 1s or all 0s) into another sensible codeword that wasn't the one that we started with, so we would need to change every bit so that we go from the all 1s codeword to the all 0s codeword or vice versa, so the Hamming distance is n for the repeat n times code.

What about the parity check code on four bits? The parity check code on n bits? We said before than any even number of errors gets us back to a (different) codeword in the parity check code, so if we want a different codeword with the fewest possible errors, we need the smallest even number more than 0, which is 2.

So how is this question, of what the Hamming distance is differrent from the first one of how many bits needed to be changed in order to get a different result in the majority vote interpretation of the repeating code? If you get nonsense is that different from somehow getting a different sensible codeword? How? In general, we are saving ourselves from getting a different message. With a codeword that doesn't make sense, we at least don't do the wrong thing, we know that we got nonsense instead of a contradictory message from the one that was sent.

This work was made possible through a grant from the National Science Foundation.