Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 0: Q18P (page 1)

Let EREX↑={R|Ris a regular expression with exponentiation andL(R)=∅ }. Show that EREX↑∈P.

Short Answer

Expert verified

The above problem is solved by using PSPACE\varsubsetneqEXSPACEi.e., the problem of any EXSPACE-complete cannot be inPSPACE .

Step by step solution

01

Using Regular Expressions with Exponentiation

Consider the statement EREX↑={R|Ris regular expression with exponentiation and LR=∅

From the above statement it can be understood that R is a regular expression and the language, which contains this regular expression consists NULL values.

02

Applying the concept of intractability

EREX↑is said to be intractable because it can be demonstrated in such a manner that is complete for the class EXSPACE.

From the above statements we can conclude that “any EXSPACE-complete problem cannot be in PSPACE and it is much less in P”.

Because PSPACE will be same as EXSPACE which is contradicting the corollary discussed above.

Hence, it can be said that EREX↑∈P

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.

Start your free trial

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

Show that EQTM'mEQTM'

For each part, give a relation that satisfies the condition.

  1. Reflexive and symmetric but not transitive
  2. Reflexive and transitive but not symmetric
  3. Symmetric and transitive but not reflexive

Question: Let S={{M}|MisaTMandLM={M'}.Show that S nor S' neither is Turing recognizable.

Modify the proof of Theorem 3.16 to obtain Corollary 3.19, showing that a language is decidable if some nondeterministic Turing machine decides it. (You may assume the following theorem about trees. If every node in a tree has finitely many children and every branch of the tree has finitely many nodes, the tree itself has finitely many nodes.)

A two-dimensional finite automaton (2DIM-DFA) is defined as follows. The input is an m×nrectangle, for any m,n≥2. The squares along the boundary of the rectangle contain the symbol # and the internal squares contain symbols over the input alphabet ∑. The transition function δ:Q×∑∪#→Q×L,R,U,Dindicates the next state and the new head position (Left, Right, Up, Down). The machine accepts when it enters one of the designated accept states. It rejects if it tries to move off the input rectangle or if it never halts. Two such machines are equivalent if they accept the same rectangles. Consider the problem of determining whether two of these machines are equivalent. Formulate this problem as a language and show that it is undecidable.
See all solutions

Recommended explanations on Computer Science Textbooks

Issues in Computer Science

Read Explanation

Computer Programming

Read Explanation

Algorithms in Computer Science

Read Explanation

Game Design in Computer Science

Read Explanation

Data Representation in Computer Science

Read Explanation

Databases

Read Explanation
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