Information sharing

Imagine that Y given to Bob is 00000000 (0).

Then, as we are told it is 1 hamming distance away (number of different bits is 1), so X could be:

10000000 (128), 01000000 (64), 00100000 (32), 00010000 (16), 00001000 (8), 00000100 (4), 00000010 (2) and 00000001 (1).

So now X could be one of 8 different combinations out of the 256 possible combinations in 8 bit. So we are now guessing from a 32th of the original set: 8 combinations / 256 combinations.

Then as per the formula above the question.

$\log_{2}\left ( \frac{1}{\frac{8}{256}} \right )$

$\log_{2}\left ( \frac{1}{0.03125} \right )$

$\log_{2}\left ( 32 \right ) = 5$

Though I got the answer, the route I took may not be correct, of course.

The information content of X is 8. When Alice tells Bob that the Hamming distance of X and Y is equal to 1, all the information that Bob needs in order to know X is the position of the 'wrong' bit in Y. The needed information content is therefore 3, as just 3 bits of information are necessary to locate the 'wrong' bit in the 8-bit word Y. Information given to Bob about X is therefore 8 - 3 = 5.

previous possibilities = 2^{8}
new possibilities = 8 = 2^ {3} ( because, hamming distance is 1)
information = log(previous possibilities/ new possibilities) = 5

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