Chapter 4: 12P (page 211)

Let $A = {<M> | M is a DFA that doesn't accept any string contianing an odd number of 1 s}$ . Show that A is decidable.

Short Answer

Expert verified

A is decidable.

Step by step solution

Decidable languages

A language is said to be decidable if the inputs of a language are accepted by a Deterministic Finite Automata.

Explanation

A language is decidable if the input string of a language is accepted by a Turing Machine. Consider the Turing machine M that decides the language A

The following TM decides A.

“On input $<M>$

Construct a DFA, Dwhich accepts every string containing an odd number of 1s.
Construct a DFA, B such that $L (B) = L (M) \cap L (D) .$
Test whether $L (B) = ϕ,$ using the $E_{D F A}$ decider T from the Theorem 4.4.
If Taccepts, accept; if T rejects, reject.”

Therefore, the language $A = \{<M> | M is a DFA that doesn't accept any string contianing an odd number of 1 s\}$ is decidable, and has been proven.

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Let $A = {<M> | M is a DFA that doesn't accept any string contianing an odd number of 1 s}$ . Show that A is decidable.

Short Answer

Step by step solution

Decidable languages

Explanation

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Let A={M|M is a DFA that doesn't accept any string contianing an odd number of 1s}. Show that A is decidable.

Short Answer

Step by step solution

Decidable languages

Explanation

One App. One Place for Learning.

Most popular questions from this chapter

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Study anywhere. Anytime. Across all devices.

Let $A = {<M> | M is a DFA that doesn't accept any string contianing an odd number of 1 s}$ . Show that A is decidable.