i want to know how can i find g(12), is there a way beside trial and error way ? i know i can find g(n) that differ 1 bit from 1110 since it satisfy d(g(n), g(n+1))=1 such as 1010,1111,0110,1100 but this clearly not a good way to solve this problem , is there pattern?
There is only one possible choice for $g(12)$, and that is most easily found through exhaustive search. But the numbers are produced in a systematic manner (I found the system by drawing two squares with the corners and their coordinates representing the bit strings, and looking at the path taken through them by the sequence $g$):
The rightmost two bits and the leftmost two bits both follow the pattern $$ 00\to01\to11\to10 $$ After the rightmost bits have completed this pattern once, the leftmost two bits advance once, and then the rightmost bits follow this pattern in reverse. Rinse and repeat.
So while there is only one valid choice for $g(12)$, I guess that the table continues $$ \begin{array}{|c|c|} \hline 12 & 1010\\ 13 & 1011\\ 14 & 1001\\ 15 & 1000\\ \hline \end{array} $$ But no matter whether I'm right or not, after choosing between $g(13) = 1011$ and $g(13) = 1000$, $g(14)$ must be $1001$ and $g(15)$ is the one you didn't choose for $g(13)$.