Chapter 4: Q28P (page 212)

Let $C = {{G, x} | G is a CFG x is a substring of some y \in L (G)} .$ Show that $C$ is decidable.

Short Answer

Expert verified

The language $C$ is decidable.

Step by step solution

Context free language.

Context free language is a grammar where its language or string that formed by context free grammar is supports pushdown deterministic automata.

The problem is decidable.

Let, $C = {{G, x} | G is a CFG x is a substring of some y \in L (G)} .$

Showing that, $C$ is decidable.

Let’s the following Turing machine $M$ = On input hG, xi,

Where, $(G)$ is a context free grammar.

(a) Let $R$ be the following RE .

$R = Σ * xΣ *$

(b) Create the context free grammar $H$ such that,

$L (H) = L (G) \cap L (R)$

(d) If it accepts, then reject.

(e) Else, accept.

$M$ is clearly a decider since it halts. Also, $M$ accepts ${G, x}$ .

If f generates some string $Σ * xΣ *$ in that is some string with $x$ as a substring.

Hence, the language $C$ is decidable.

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Let $C = {{G, x} | G is a CFG x is a substring of some y \in L (G)} .$ Show that $C$ is decidable.

Short Answer

Step by step solution

Context free language.

The problem is decidable.

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Let C={{G,x}|GisaCFGxisasubstringofsomey∈L(G)}. Show thatC is decidable.

Short Answer

Step by step solution

Context free language.

The problem is decidable.

One App. One Place for Learning.

Most popular questions from this chapter

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Let $C = {{G, x} | G is a CFG x is a substring of some y \in L (G)} .$ Show that $C$ is decidable.