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I am new to automata and learning to make regular expression for languages. But I have been stuck on this one.

Suppose we have a language L, Language of all strings that has exactly 1 triple “b” defined over alphabet set Σ = {a, b}
Now after several tries, I came up with this (a* (ab)* (ba)* )* bbb (a* (ab)* (ba)* )* but then I realize that this is wrong too because the string abbbabababb doesn't fit on this.

Kindly someone point out at my mistake or help me solve it as I have spent almost an hour on this.

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  • $\begingroup$ Perhaps you can convert FM of that into a regular expression. $\endgroup$ – Mr. Sigma. Dec 2 '18 at 11:14
  • $\begingroup$ I don't have or know the FA of this. I am trying to convert it directly into RE. @Mr.Sigma. You got any ideas?? $\endgroup$ – Tom Dec 2 '18 at 11:17
  • $\begingroup$ Not directly relevant, but I find it easier to write $(A^*B^*C^*)^*$ as $(A+B+C)^*$ $\endgroup$ – chi Dec 2 '18 at 11:45
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To be clearer, we use "triple $b$'s" to mean three consecutive $b$'s.

What you would like to figure out first is a detailed description or characterization of a string that has not triple $b$'s and that does not end at an $b$, the part of the string that is before that triple $b$.

That string should start with zero or more $a$'s, possibly followed by one or two $b$'s followed by one or more $a$'s, possibly followed by one or two $b$'s followed by one or more $a$'s, and so on for some rounds. That is, $a^*((b\mid bb)aa^*)^*$.

So, a regular expression could be $a^*((b\mid bb)aa^*)^*bbb(aa^*(b\mid bb))^*a^*$, or written symmetrically, $a^*((b\mid bb)aa^*)^*bbb(a^*a(bb\mid b))^*a^*$.

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  • $\begingroup$ Wow, you explained it really well. And I guess your answer is correct. But the answer posted above you is somewhat different and also seems to be correct. So I guess both of you are right? $\endgroup$ – Tom Dec 2 '18 at 12:00
  • $\begingroup$ this won't generate bbb $\endgroup$ – Mr. Sigma. Dec 2 '18 at 12:00
  • $\begingroup$ Corrected. The symmetry is, in fact, useful in understanding and verification. $\endgroup$ – John L. Dec 2 '18 at 12:10
  • $\begingroup$ Now, both solutions are correct? $\endgroup$ – Tom Dec 2 '18 at 12:21
  • $\begingroup$ Yes. Sigma's answer, $(a^{*} (ba)^*(bba)^*)^*bbb(a^*(ab)^*(abb)^*)^*$, can be rephrased as some rounds of strings that does not end with $b$, followed by $bbb$, followed by some rounds of strings that does not starts with $b$. $\endgroup$ – John L. Dec 2 '18 at 12:25
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It seems you are almost there. You just need to care substrings with $abb$. One possible way is $R.E = (a^{*} (ba)^*(bba)^*)^*bbb(a^*(ab)^*(abb)^*)^*$

Note that I came upon this regular expression from the $FM$ of language you described. So, another way to find $R.E$ is from $FM$ directly.

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