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I am supposed to create CFG for this languague:

$L= \{w : w \in \{a, b\}^*, |w_b| = 3k, k \geq 0 \}$

where $|w_b|$ is count of terminals $b$ in $w$.

For example:

aa - OK, no 'b'

abb - wrong, only 2 'b'

aaabbb - OK, 3 times 'b'

aababbb - wrong, 4 times 'b'

abbbbbaaa - wrong, 5 times 'b'

abababbbaaab - OK, 6 times 'b'

and so on...

I can't come up with any solution. Any advice?

My goal is to design context-free grammar, not automaton or regular expression (i don't know how to design automatons or RE yet).

What about CFG

G = {{S,A}, {a,b}, R, S}

where R rules are:

1] S -> S A b A b A b A S
2] S -> A
3] S -> ε
4] A -> a A
5] A -> ε

Explanation:

rule 2] is for cases when there are no 'b' symbols in w

rule 3] is for case of empty string

rule 4] is for adding 'a' symbols between 'b', e.g. baaaabab, babaab

rule 5] is for cases, when there are multiple 'b' next to each other, e.g. abbbaaa

Is this CFG ok?

Gilles 'SO- stop being evil'
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John
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3 Answers3

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Hint: A string with exactly one $a$ would look like this: $$ (\text{any number of } b's)\,a\,(\text{any number of } b's) $$ and a grammar to generate the language of these strings is $S\rightarrow AaA,\ \ A\rightarrow bA\mid \epsilon$.

Generalize.

Hendrik Jan
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Rick Decker
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Your language is regular. You can create a DFA or NFA for your language, and then convert it mechanically into a regular grammar. As an example, here is a grammar for the language of all even-length words over $\Sigma = \{a\}$, which was generated this way from a DFA with two states:

$$ \begin{align*} &S \to aT \mid \epsilon \\ &T \to aS \end{align*} $$

The two nonterminals $S,T$ correspond to the two states; the starting symbol corresponds to the initial state; the productions $S \to aT$ and $T \to aS$ correspond to the transition function; and the production $S \to \epsilon$ corresponds to $S$ being an accepting state.

Yuval Filmus
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Your CFG looks correct, but it can be made simpler. The production $S \rightarrow SAbAbAbAS$ can be replaced by $S \rightarrow SbAbAbA$ (i.e. remove the first $A$ and last $S$) . Given any string containing $3k$ $b$'s $(k \ge 1)$, the suffix of the string starting with the third $b$ from the right can be generated by the $bAbAbA$ part. For example, if the string is $aababababbab$, then the rightmost $bbab$ can be generated from $bAbAbA$.

Ashwin Ganesan
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