Let J={w|eitherw=0xfor some,xATM orw=1yfor some yATM¯}. Show that neither JnorJis Turing-recognizable.

Short Answer

Expert verified

Both J and J are not Turing Recognizable.

Step by step solution

01

 Introduction to Turing Recognizable

A language L is said to be Turing Recognizable if and only if there exist any Turing Machine (TM) which recognize it i.e., TM halt and accept strings belong to language L and will reject or not halt on the input strings that doesn’t belong to language L.

02

 Proving neither J nor J are Turing Recognizable

In order to prove J and Jboth are not Turing-Recognizable,

First we will show thatATMmJ.

To prove this, we will construct TM Q

Q=oninputM,w

Write symbol 0 and M,wthen in the tape and make it halt. We have:

localid="1662181827790" M,wATMoutputofQJ

So, we are able to obtain a reduction mapping of ATM to J .

Now, we will show thatATMmJ¯ . For this we will construct a Turing Machine R:
R=oninputM,w

Write symbol 1 and then M,win the tape and make it halt. We have:

M,wATMoutputofRJ

This is similar to:

M,wATMoutputofRJ¯

So, we are able to obtain a reduction mapping of ATM to J.

Sincelocalid="1662181817969" ATMmJmJ,. This signifies that J is not Turing-recognizable as ATM is not Turing-recognizable.

In same way, since. This signifies that J is not Turing recognizable.

Hence both J and Jare not Turing-recognizable.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Give a counter example to show that the following construction fails to prove that the class of context-free languages is closed under star. Let A be a CFL that is generated by the CFG G=(V,,R,S). Add the new ruleSSS and call the resulting grammar. This grammar is supposed to generate A*.

We generally believe that PATH is not NP-complete. Explain the reason behind this belief. Show that proving PATH is not NP-complete would prove P ≠ NP

A queue automaton is like a push-down automaton except that the stack is replaced by a queue. A queue is a tape allowing symbols to be written only on the left-hand end and read only at the right-hand end. Each write operation (we’ll call it a push) adds a symbol to the left-hand end of the queue and each read operation (we’ll call it a pull) reads and removes a symbol at the right-hand end. As with a PDA, the input is placed on a separate read-only input tape, and the head on the input tape can move only from left to right. The input tape contains a cell with a blank symbol following the input, so that the end of the input can be detected. A queue automaton accepts its input by entering a special accept state at any time. Show that a language can be recognized by a deterministic queue automaton iff the language is Turing-recognizable.

Let

A/B={w|wxaAforsomexB}Show that ifAis context free andBis regular, thenA/Bis context free

In the following solitaire game, you are given anm×m board. On each of itsm2 positions lies either a blue stone, a red stone, or nothing at all. You play by removing stones from the board until each column contains only stones of a single color and each row contains at least one stone. You win if you achieve this objective. Winning may or may not be possible, depending upon the initial configuration. LetSOLITAIRE={<G>|G is a winnable game configuration}. Prove that SOLITAIREis NP-complete.

See all solutions

Recommended explanations on Computer Science Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free