Regex for binary multiple of 3

If I may plug my solution for this code golf question! It's a piece of JavaScript that generates regexes (probably inefficiently, but does the job) for divisibility for each base.

This is what it generates for divisibility by 3 in base 2:

/^((((0+)?1)(10*1)*0)(0(10*1)*0|1)*(0(10*1)*(1(0+)?))|(((0+)?1)(10*1)*(1(0+)?)|(0(0+)?)))$/

Edit: comparing to Asmor's, probably very inefficient :)

Edit 2: Also, this is a duplicate of this question.

I know this is an old question, but an efficient answer is yet to be given and this question pops up first for "binary divisible by 3 regex" on Google.

Based on the DFA proposed by the author, a ridiculously short regex can be generated by simplifying the routes a binary string can take through the DFA.

The simplest one, using only state A, is:

0*

Including state B:

0*(11)*0*

Including state C:

0*(1(01*0)*1)*0*

And include the fact that after going back to state A, the whole process can be started again.

0*((1(01*0)*1)*0*)*

Using some basic regex rules, this simplifies to

(1(01*0)*1|0)*

Have a nice day.