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Chapter 4: 13P (page 211)

Let A={R,S|R and S are regular expressions and LR⊆LS}. Show that A is decidable.

Short Answer

Expert verified

A is decidable.

Step by step solution

01

Explain Decidable Language

A problem is decidable if a Turing machine halts in finite amount of time for every input and gives the answers as “yes” or “no.”A decidable problem has an algorithm to determine the answer for a given input.

02

Show that A is decidable

A language is decidable if a Turing machine accepts the input of a language. Consider that Rand Sare regular expressions.

Consider a Turing machine M that decides the language AOn input string W

  1. Consider that the input has two possibilities.
  2. If the input decodes a pair of regular expressions, then it takes finite time.
  3. If it doesn’t decode a pair of regular expressions, then it stops rejecting.

Therefore, the language given is decidable and has been proved.

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Most popular questions from this chapter

The proof of Lemma 2.41 says that (q,x) is a looping situation for a DPDAP if whenP is started in state q withx∈Γ on the top of the stack, it never pops anything belowx and it never reads an input symbol. Show thatF is decidable, whereF={(P,q,x)|(q,x)isaloopingsituationforP}.

Say that an NFA is ambiguous if it accepts some string along two different computation branches

LetAMBIGNFA={(N)|NisanambiguousNFA}.Show thatAMBIGNFA is decidable. (Suggestion: One elegant way to solve this problem is to construct a suitable DFA and then run EDFAon it.)

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

Let X be the set {1, 2, 3, 4, 5} and Y be the set {6, 7, 8, 9, 10}. We describe the functions

Answer each part and give a reason for each negative answer.

a. Is f one-to-one?

b. Is f onto?

c. Is f a correspondence?

d. Is g one-to-one?

e. Is g onto?

f. Is g a correspondence?

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

Show thatINFINITEPDAis decidable.
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