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Given the recursively enumerable language (\(L_{RE}\)), the context sensitive language (\(L_{CS}\)), the recursive language (\(L_{REC}\)), the context free language (\(L_{CF}\)) and deterministic context free language (\(L_{DCF}\)). The relationship between these families is given by

This question was previously asked in
UGC NET 2014 Paper 1 Question Paper (28-Dec-2014)
The correct answer is

\(L_{DCF} \subseteq L_{CF} \subseteq L_{CS} \subseteq L_{REC} \subseteq L_{RE}\)

The language families form a chain of inclusions (part of the Chomsky hierarchy). Each class is recognised by a more powerful machine than the one before it, so we place the smallest family first.

Every deterministic PDA is also a PDA, so \(L_{DCF} \subseteq L_{CF}\). The inclusion is proper: \(\{ww^R\}\) is context free but not deterministic context free.

Every context free language (without λ) can be generated by a context sensitive grammar and accepted by a linear bounded automaton, so \(L_{CF} \subseteq L_{CS}\).

Membership in a context sensitive language is decidable (an LBA has only finitely many configurations), so every such language is recursive: \(L_{CS} \subseteq L_{REC}\).

A recursive language is accepted by a Turing machine that always halts, which is in particular a Turing machine, so \(L_{REC} \subseteq L_{RE}\). The halting problem shows the reverse fails.

Chains that start with \(L_{CF} \subseteq L_{DCF}\) or end with \(L_{RE} \subseteq L_{REC}\) reverse a true inclusion, so the only valid chain is \(L_{DCF} \subseteq L_{CF} \subseteq L_{CS} \subseteq L_{REC} \subseteq L_{RE}\).

Hence, the answer is \(L_{DCF} \subseteq L_{CF} \subseteq L_{CS} \subseteq L_{REC} \subseteq L_{RE}\).

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