I was given an example in class. It goes as follows:

Let the alphabet be $\{a, b\}$. $L$ is the string with one $b$ and at least two $a$'s Find a regular expression that generates this language.

This was my professor's answer: $r = aaa^*ba^*+abaa^*+baaa^*$

My question with this solution is:

Wouldn't this regular expression FORCE $aa$, $aba$, and $baa$ to be together, respectively? Like, there doesn't seem to be a constraint that $a$ must follow $a$ in the first term, or that $a$ must precede $b$ which must precede $a$ in the second, etc.

That's why this was my answer: $r = a^*aa^*aa^*ba^* + a^*aa^*ba^*aa^* + a^*ba^*aa^*aa^*$

It still forces there to be one $a$ and at least two $a$'s, but it doesn't make them be right next to each other.

Which solution is correct? I hope I explained myself correctly


3 Answers 3


Both answers are correct.

First note that the following regular expression works:


The first term catches every word of $L$ that has a pair of $a$s anywhere after the $b$, the second catches every word that has the $b$ between two $a$s, and the third catches every word that has two $a$s anywhere after the $b$. Of course, many words are caught by more than one term, but that’s perfectly all right. It doesn’t matter that I specify that the two $a$s are together: in any string of two or more $a$s, there are two that are adjacent, and I can always choose them to be the first two or the last two.

Your answer is simply an inefficient version of this one. Your $a^*aa^*aa^*ba^*$ is exactly equivalent to $a^*aaba^*$, both generating every string that has at least two $a$s before the $b$. Similarly, your $a^*aa^*ba^*aa^*$ is equivalent to $a^*abaa^*$, both generating every string that has at least one $a$ on each side of the $b$. And your third term is similarly equivalent to mine.

This can be simplified if we observe that either the $b$ comes first, in which case we must have $baaa^*$, or it comes last, in which case we must have $a^*aab$, or it comes between two $a$s, in which case we must have $a^*abaa^*$. Thus, the expression


will work just fine.

Your professor simplified it a little differently. First, $aaa^*ba^*$ catches every word of $L$ that has at least two $a$s before the $b$. If a word of $L$ does not have two $a$s before the $b$, then either it starts $aba$, in which case it’s caught by $abaa^*$, or it starts $baa$, in which case it’s caught by $baaa^*$.

  • $\begingroup$ Sorry, I think the formatting of the answer caused her original answer to not show all the * and instead italicized the b. Is her answer that you see now also incorrect? @Brian M. Scott $\endgroup$ Oct 9, 2015 at 17:37
  • $\begingroup$ @FrostyStraw: Okay, give me a few minutes to rewrite my answer. $\endgroup$ Oct 9, 2015 at 17:39
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    $\begingroup$ Both his and the prof's answers are correct; if you leave the $a^*$ on both sides of the first and third terms, you can get rid of the $a^*$ from the second term. $\endgroup$ Oct 9, 2015 at 17:42

Note that $a^*aa^*$ is equivalent to $aa^*$. Using that, your expressions reduce to:

$$r = aaa^*ba^* + aa^*baa^* + a^*baaa^*$$

This expression does generate that language. But it can be reduced further. Note that aaba would match both the first and second terms of the expression. In fact, you can reduce the expression to just this:

$$r = aaa^*ba^* + aba + a^*baaa^*$$

Your prof's expression was:

$$r = aaa^*ba^* + abaa^* + a^*baaa^*$$

which can likewise be reduced to

$$r = aaa^*ba^* + aba + a^*baaa^*$$


Since the alphabet is so small we can enumerate all possibilities the only thing which could be close to b is a, and there must be at least 2 of them. Those number of permutations of 2 among 3: $\frac{3!}{2!} = 3$ : aab,aba,baa, then we need to fill out the rest with potential a's to both left and right.

So yet another expression would be $a^*((aab) + (aba) + (baa))a^*$


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