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The grammar is

$$ S\to aSb\ |\ bSa\ |\ SS\ |\ \epsilon. $$

I think this generates the set of strings with equal numbers of $a$'s and $b$'s based on examples I've done. Is this correct?

Martin Berger
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redundant6939
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  • What have you tried? Have you tried to prove your guess correct? Have you seen the use of proof by induction for this sort of problem? Have you given that a try? Did you get stuck somewhere? Also, have you looked at related questions here, e.g., http://cs.stackexchange.com/q/10605/755 – D.W. Oct 25 '13 at 00:36

2 Answers2

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You conjecture that your grammar generates the language $L$ of words having as many $a$s as $b$s. In order to prove your conjecture, you should do two things:

  1. Prove that every word generated by the grammar belongs to $L$.
  2. Prove that every word in $L$ can be generated by the grammar.

In this case, the first part is easy to prove by structural induction. The second part is more difficult, so here is a hint. Given any non-empty $w \in L$, show that it can be decomposed as either $w = xayb$ or $w = xbya$ for some $x,y \in L$. Given this lemma, you can use induction on $|w|$.

Here is a hint for the lemma. Let $d_k(w) = |\{i \in [k] : w_i = a\}| - |\{i \in [k] : w_i = b\}|$, and let $d(w) = d_0(w),d_1(w),\ldots,d_{|w|}(w)$. For example, $d(aabb) = 0,1,2,1,0$ and $d(baba) = 0,-1,0,-1,0$. A prefix $w_1\ldots w_k \in L$ corresponds to a position $k$ such that $d_k(w) = 0$.

Yuval Filmus
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2

This doesn't seem particularly rigorous as far as arguments go, but it's early here. Besides, this might help with intuition.

To see that the grammar generates words in the language, consider that every rule which adds either $a$ or $b$ adds an equal number of the two. Therefore, any string the grammar generates must have equal numbers of the two symbols.

To see that any word in the language can be generated by the grammar, consider any word $w$. We can definitely write $w$ as one of the following:

  1. $w = axa$
  2. $w = axb$
  3. $w = bxa$
  4. $w = bxb$

In cases 2 and 3, we can have applied the productions $S \rightarrow aSb \mid bSa$; we have that $w$ can be generated by our grammar if $x$ can. We also have that $|x| < |w|$. In cases 1 and 4, we must have used the rule $S \rightarrow SS$ to get $w$; let us now write $w = xy$ where $x$ and $y$ are also strings in the language (exercise: show this is always possible in the cases considered). We have $|x|, |y| < |w|$.

Both strings of length 2 in the language can be generated by the grammar (we could choose the empty string as the base case instead, might make more sense). By recursively applying the analysis in the preceding paragraph, we can find a way to generate any string in the language (since all words are even length, child words are shorter than parent words, and length 2 is the base case).

Patrick87
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