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Chapter 4: Q28P (page 212)

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

Short Answer

Expert verified

The languageC is decidable.

Step by step solution

01

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.

02

The problem is decidable.

Let,C={{G,x}|GisaCFGxisasubstringofsomey鈭�L(G)}.

Showing that,Cis decidable.

Let鈥檚 the following Turing machine M= On input hG, xi,

Where, (G) is a context free grammar.

(a) Let Rbe the following RE .

R=危鈭�虫危鈭�

(b) Create the context free grammar Hsuch that,

L(H)=L(G)鈭�L(R)

(c) SubmitG to the decider for ECFG.

(d) If it accepts, then reject.

(e) Else, accept.

Mis clearly a decider since it halts. Also, Maccepts{G,x} .

If f generates some string危鈭�虫危鈭� in that is some string withx as a substring.

Hence, the languageC is decidable.

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Most popular questions from this chapter

Let S={M|M鈥�is鈥�鈥�a鈥�鈥�DFA鈥�鈥�that鈥�鈥�accepts鈥�鈥�wR鈥�whenever鈥�鈥�it鈥�鈥�accepts鈥�鈥�w} Show that S is decidable.

Let INFINITEPDA={(M)|MisaPDAandL(M)isaninfinitelanguage}.

Show thatINFINITEPDAis decidable.

Let ETM={M|M鈥�is鈥�a鈥�TM鈥�and鈥�LM=蠒}.Show that ETM炉, the complement of ETM, is Turing-recognizable.

Let PREFIX鈭�FREEREX=role="math" localid="1659933266276" {{R}|RisaregularexpressionandL(R)isprefix-free}. Show that PREFIX鈭�FREEREXis decidable. Why does a similar approach fail to show thatPREFIX鈭�FREECFG is decidable?

Let A={R|Ris a regular expression describing a language containing at least one string w that has 111 as a substring i.e.,w=x111y鈥�for鈥�some鈥�x鈥�and鈥�y}. Show that A is decidable.
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