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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=x111yforsomexandy}. Show that A is decidable.

Short Answer

Expert verified

A is decidable.

Step by step solution

01

Explain DFA

DFArefers to deterministic finite automata. In DFA, there is only one path for a specific input from a current state to the next state.

02

Show that A is decidable

Consider the language C=w*|wconsists111asasubstring. Consider that C is a regular language with the regular expression 01*11101* and is recognized by the following DFA, Dc.


Now, consider any regular expression with the alphabet .If CLR=, then R generates a string having 111 as a substring, so RA. Also, if CLR=, then R does not generate any string having 111 as a substring, thus RA. By the Kleen’s Theorem, since L(R) is described by the regular expression R, L(R) must be a regular language. Since C and L(R) are regular languages, CLRis regular since the class of regular languages is closed under intersection. Thus, CLRhas some DFA DCLR.The theorem 4.4 shows that EDFA=B|BisaDFAwithLBis decidable, so there is a Turing machine H that decides the EDFA. Apply TM H to DCLRto determine if CLR=. Putting this all together gives us the following Turing machine T to decide A : T = “On input R

where R is a regular expression:

1. Convert Rinto a DFA using the algorithm in the proof of the Kleene’s Theorem.

2. Construct a DFA, DCLR for language CLR from the DFAs Dc and DR.

3. Run TM H that decides EDFAon input DCLR.

4. If Haccepts, reject. If Hrejects, accept.”

Therefore, the Turing machine H decides A . Hence, it has been shown that A is decidable.

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