Minimum hamming distance of multiple binary words
-
04-11-2019 - |
Question
Our task is to compute the minimum hamming distance for the following 16-bit words:
0000000000000000
0011001100110011
0101010101010101
0000111111110000
0011111111000000
1100110000000000
1111111111111111
This minimum distance is defined as the minimum of all distances between every combination of two of these bitstrings, so the first approach would be to calculate all of these individual distances. I'm pretty sure this is not the right way to do it, since there are 21
possible combinations.
Then I came across this answer on StackOverflow which says that:
We have a theorem that d_min=weight(sum(all codes)); weight is the number of non zeros in the result string . In your example modulo add all string codes like first column of all and second....... then we get code as [ 0 0 1 1 0 ], weight of this is 2 ( no. of non zeros), i.e the minimum distance of hamming code
but is that really true? If we for example take the three bitstrings
0000
1000
1110
the minimum hamming distance is obviously 1
but the calculation based on that answer would result in 0110
and would suggest that the minimum distance is 2
. Am I right about this and is that answer wrong? Or did I miss something really simple that I'm just not noticing?
I also read this answer which - if I understood it correctly - states that there is no simple way of finding the result.
No correct solution