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Chapter 4: 12P (page 211)

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

Short Answer

Expert verified

A is decidable.

Step by step solution

01

Decidable languages

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

02

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.

鈥淥n inputM

  1. Construct a DFA, Dwhich accepts every string containing an odd number of 1s.
  2. Construct a DFA, B such thatLB=LM鈭LD.
  3. Test whether LB=蠒,using the EDFAdecider T from the Theorem 4.4.
  4. If Taccepts, accept; if T rejects, reject.鈥

Therefore, the language A=M|M鈥is鈥鈥a鈥鈥DFA鈥鈥that鈥鈥doesn't鈥鈥accept鈥鈥any鈥鈥string鈥鈥contianing鈥鈥an鈥鈥odd鈥number鈥of鈥鈥1s is decidable, and has been proven.

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Most popular questions from this chapter

Let ALLDFA={A|A鈥鈥is鈥a鈥DFA andLA=鈭*}. Show thatALLDFAis decidable.

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 C be a language. Prove that C is Turing-recognizable if a decidable language D exists such that C={x|鈭yx,y鈭D}.

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.

A useless state in a pushdown automaton is never entered on any input string. Consider the problem of determining whether a pushdown automaton has any useless states. Formulate this problem as a language and show that it is decidable
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