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The question is the following:

Construct a CFG for $L_2 = \{w \in \{0, 1\}^* \mid w = w^R\text{ and the number of 1’s in $w$ is divisible by 3}\}$.

I can construct a CFG for $\{w \in \{0,1\}^* \mid w = w^R\}$ as follows

$$S → 1S1\mid 0S0\mid 0\mid 1\mid \epsilon$$

I don't understand how to make $w$ divisible by 3 in my CFG.

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2 Answers 2

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Here is the ACTUAL answer
$S_0 \rightarrow 0S_00 \ |\ 1S_11\ |\ 0\ |\ \epsilon$
$S_1 \rightarrow 0S_10\ |\ 1S_21\ |\ 1$
$S_2 \rightarrow 0S_20\ |\ 1S_01\ $

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  • $\begingroup$ The last 11 is redundant. $\endgroup$
    – xskxzr
    May 2, 2019 at 14:06
  • $\begingroup$ True, thank you for pointing that out. I am very tired... $\endgroup$
    – austinross
    May 2, 2019 at 15:18
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One way is to use the closure of context-free languages to intersection with a regular language.

In this case, though, it is simple enough to construct the grammar explicitly. The idea is to have three different symbols, $S_0,S_1,S_2$, where $S_b$ generates all palindromes in which the number of 1s is equivalent to $b$ mod 3. The productions for $S_0$ are $$ S_0 \to 0S_00 \mid 1S_11 \mid 0 \mid \epsilon. $$ I'll let you figure out the productions for $S_1,S_2$, as well as which symbol is the starting symbol.

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  • $\begingroup$ Starting symbol will be S0. However I am unable to figure out exact productions of S1,S2. Can you provide a few hints perhaps? $\endgroup$
    – hsnsd
    Oct 16, 2018 at 1:55
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    $\begingroup$ No, you’ll have to work it out on your own. $\endgroup$ Oct 16, 2018 at 1:57
  • $\begingroup$ Is this correct? $$ S_1 \to 1\mid 1S_21 \mid S_01S_0 $$ $$ S_2 \to S_1S_0S_1 $$ @Yuval Filmus $\endgroup$
    – hsnsd
    Oct 16, 2018 at 4:15
  • $\begingroup$ @hsnsd No, note the two $S_0$s in $S_01S_0$ may produce different strings. The productions for $S_1$ and $S_2$ are very similar to that for $S_0$. The idea is to consider separately the cases based on the first and the last characters. $\endgroup$
    – xskxzr
    May 2, 2019 at 9:01

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