Is $L = \{ a^ib^j : 0 < j < i < 2j\}$ context free? How can it be shown?

Question

Is $L = \{ a^ib^j : 0 < j < i < 2j\}$ context free? If so, can there be a pushdown automaton described for it? If not, does the pumping lemma apply?

Answer 1

To show a language is context free we can create a context free grammar for it then show that it generates all strings in the language.

Consider the following grammar $G$.

S -> aaaRbb
R -> aRb | aaRb | eps

We must prove that $L(G) = L$.

Proof: The variable $S$ in the grammar starts off with the smallest string in $L$. This string is $aaabb$ with $j=2$ and $i=3$ (since $0 < j = 2 < i = 3 < 2j = 4$).

The repetition happens within the variable $R$. Consider the string $s = a^ib^j \in L$. We want to see whether the string $s' = a^{i+1}b^{j+1}$ is in $L$. The first three conditions are trivial ($0 < j+1 < i+1$). The last condition is quick to check: \begin{align} i+1 &< 2(j+1) \\ i+1 &< 2j+2 \\ i &< 2j+1 \\ \end{align} So $s' \in L$. We can show $s' = a^{i+2}b^{j+1} \in L$ similarly.

The rules R -> aaaRb, R -> aR, and R -> Rb cannot be in our grammar as it can generate strings that are not in the language. As an example, the strings aaaaaabb, aaaabb, and aaabbb can be generated if we add the above rules which are not in $L$.

Thus, $G$ is complete. Hence, $L(G) = L$.

As for creating a pushdown automata, there are many guides on how to construct a PDA from CFG. The following links will be useful (just from a quick google search).

1. Convert CFG to CNF
2. Convert CFG to GNF
3. Construct PDA from CFG

Answer 2

What's the smallest possible j, and which i goes with it (to allow j < i < 2j). So that's the minimum number of a's and b'. Beyond that, you need one or two a's for every additional b.

That should be enough to find a grammar.