/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Q53P Let A⊆1* be any unary language... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Let A⊆1* be any unary language. Show that if A is NP-complete, then P = NP. (Hint: Consider a polynomial time reduction f from SATto A. For a formula ϕ, let ϕ0100be the reduced formula where variables x1, x2, x3, and x4 inϕ are set to the values 0, 1, 0, and 0, respectively. What happens when you apply f to all of these exponentially many reduced formulas?)

Short Answer

Expert verified

There are at most n.p(|ϕ|)≤|ϕ|.p(|ϕ|)nodes on any f(ϕv)≤p(|ϕ|)such path, and there are at most nodes on any for all v∈V.

Step by step solution

01

NP Class

A nondeterministic turning machine assigns a problem to the NP (nondeterministic polynomial time) class if it can be solved in polynomial time.


02

Explanation

Let A be the unary NP complete language that satisfies the condition ϕ∈SAT⇔1f(ϕ)∈A. The formula for calculating 1f(ϕ)the amount of time required is p|ϕ|. Let T=V,Ebe a satisfiable call tree for an input formula. For example v∈V, consider ϕbe the V formula with an argument.

Consider two nodes v,v'neither of which is an ancestor of the other. At the time of the v' call, it is known whether p|ϕ|is satisfiable, therefore the hashmap is defined for f(ϕv) . As a result f(ϕv)=f(ϕv') . There are at most n nodes on any such path., and there are at most n.p(|ϕ|)≤|ϕ|.p(|ϕ|) nodes on any f(ϕv)≤p(|ϕ|) for all v∈V.

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 91Ó°ÊÓ!

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

A cut in an undirected graph is a separation of the vertices V into two disjoint subsets S and T . The size of a cut is the number of edges that have one endpoint in S and the other in T . Let MAX-CUT=<G,K>|Ghasacutofsizekormore.

Show that MAX-CUT is NP-complete. You may assume the result of Problem 7.26. (Hint: Show that≠SAT⩽PMAXCUT. The variable gadget for variable x is a collection of 3c nodes labeled with x and another nodes labeled with x . The clause gadget is a triangle of three edges connecting three nodes labeled with the literals appearing in the clause. Do not use the same node in more than one clause gadget. Prove that this reduction works.)

Let ? be a 3cnf-formula. An ≠-assignment to the variables of ? is one where each clause contains two literals with unequal truth values. In other words, an ≠ -assignment satisfies ? without assigning three true literals in any clause.

a. Show that the negation of any ≠ -assignment to ? is also an ≠ -assignment.

b. Let ≠ SAT be the collection of 3cnf-formulas that have an ≠ -assignment. Show that we obtain a polynomial time reduction from 3SAT to ≠ SAT by replacing each clause ci

(y1∨y2∨y3)$$

with the two clauses

(y1∨y2∨zi)and(zi¯∨y3∨b)

Where ziis a new variable for each clause,ci and b is a single additional new variable.

c. Conclude that ≠SAT isNP-complete.

A 2cnf-formula is an AND of clauses, where each clause is an OR of at most two literals. Let . Show that2SAT={ϕ|ϕ is a satisfiable 2CNF formula}. Show that 2SAT∈P.

This problem investigates resolution, a method for proving the unsatisfiability of cnf-formulas. Let ϕ=C1∧C2∧…∧Cmbe a formula in cnf, where the Ciare its clauses. Let C=CiCi is a clause of ϕ. In a resolution step, we take two clauses Caand Cbin C, which both have some variable occurring positively in one of the clauses and negatively in the other. Thus, Ca=x∨y1∨y2∨…∨ykand Cb=x¯∨z1∨z2∨…∨zl, where the and are literals. We form the new clause y1∨y2∨…∨yk∨z1∨z2∨…∨zland remove repeated literals. Add this new clause to C. Repeat the resolution steps until no additional clauses can be obtained. If the empty clause ( ) is in C, then declare ϕunsatisfiable. Say that resolution is sound if it never declares satisfiable formulas to be unsatisfiable. Say that resolution is complete if all unsatisfiable formulas are declared to be unsatisfiable.

a. Show that resolution is sound and complete.

b. Use part (a) to show that 2SAT∈P.

Modify the algorithm for context-free language recognition in the proof of Theorem 7.16 to give a polynomial time algorithm that produces a parse tree for a string, given the string and a CFG, if that grammar generates the string.

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.