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1. L2 = E{0,1)

All the words in L2 with equal number of 0's and 1's and every prex of w has at most two more of either 0's or 1's.

2. All words that does not contain “bbb” as substring.

i think this is :( e + b + bb)(a + ab + abb)*

Thanks in advance!

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I'm not sure if you're a native speaker. Hoewever, could you please correct your spelling? Especially the description of 1. is quite unclear to me. –  tohecz Nov 27 '12 at 14:46
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migrated from math.stackexchange.com Nov 27 '12 at 21:59

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1 Answer

Your point 2, is obviously correct.

As for the point 1, we construct the automaton: let's remember how much more 0's we've read than 1's. Then we have transitions:

S0  by 0 to S1
S0  by 1 to S-1
S1  by 0 to S2
S1  by 1 to S0
S-1 by 0 to S0
S-1 by 1 to S-2
S2  by 1 to S1
S-2 by 0 to S-1

S0 is initial, only S0 is final. This way we are sure, that in whatever state we are, the number of $0$'s and $1$'s that were read so far differs by at most $2$ and in the final state the number are the same.

Then the language is $$ \Bigl(0(01)^*1+1(10)^*0\Bigr)^*$$

The large parenthesis corresponds to all paths from S0 to S0, and we can repeat it as many times as needed.

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I hope I understood the question correctly. –  tohecz Nov 27 '12 at 14:59
    
Nice answer. +1 –  Rick Decker Nov 27 '12 at 21:06
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