Find regular expression for a binary sequence dividable by 3

Question

I am trying to find a short regular expression that matches to all binary sequences that are dividable by 3.

This is homework. It would be great if I could only get some hints before the final solution :D

Using the DFA I would get the following states

 |0|1
A|A|B
B|C|A
C|B|C

That would give

$$A=0A+1B\\ B=0C+1A\\ C=0B+1C $$

I eliminate the recursion of $1C$ which is $1*$ and $0A$ which is $0*$

$$ A=0*+1B\\ B=0C+1A\\ C=0B+1*$$

I replace C in B

$$B=0(0B+1*)+1A$$

I need to get rid of the $0B$ somehow to be able to place it into A. $0B$ part should be again a recurision.. But I am lost from here on.

Any hints?

Thanks a lot

Hm not sure? But union is commutativ so could be $1*+0B$ so the + can be removed? — Chris, Feb 18 '14 at 15:43
There is a shorter version $(1(010)1|0)+$ which I try to get to — Chris, Feb 18 '14 at 16:21

score 0 · Answer 1 · answered Feb 18 '14 at 18:04

0

From $C=0B+1C$ we have $$ C = 1^*0B $$ Then $B=0C+1A$ becomes $B=1^*0B+1A$, from which he find $$ B = (1^*0)^*1A$$ Now $A=0A+1B+\epsilon$ becomes $A= 0A+(1^*0)^*1A +\epsilon= (0 | (1^*0)^*1)A+\epsilon $ and from this ultimately $$A=(0 | (1^*0)^*1)^*$$ (or with the last $^*$ replaced with $^+$ if you want to avoid $\epsilon$ as match)

answered Feb 18 '14 at 18:04

Hagen von Eitzen

374,180

how did you get from $B=1∗0B+1A$ to $B=(1∗0)∗1A$? – Chris Feb 18 '14 at 18:10

Find regular expression for a binary sequence dividable by 3

1 Answers1