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Fill out the table described in the polynomial time algorithm for context-free language recognition from Theorem7.16forstringw=babaandCFGG:

SRTRTR|aTTR|b

Short Answer

Expert verified

The foregoing four-step computation and answer demonstrate that the polynomial algorithm time has been based on correct theory.

Step by step solution

01

Introduction

The term "time complexity" in computer science refers to the computational complexity that quantifies the length of time required to execute an algorithm.

02

Step – 2: Calculation of polynomial Time algorithm:

i) Solve the Diagonal Algorithm cells i.e.1,1,2,2,3,3,4,4

1,1->b=T,2,2->a=R,3,3->b=T,4,4->a=R

ii) solve1,2

1,1o2,2->TR=R,T2,3:2,2o3,3->RT=S3,4:3,3o4,4->TR=R,T

iii) Solution1,3::

1,2o3,3->RToTT=S2,4:2,2o3,4->RR,RT

or

2,3o3,4->SoR=S

iv) Algorithm Solution1,4::

1,1o2,4=ToS=

or

1,2o3,4=R,ToR,T=RR,RT,TR,TT=S,R,T

or

1,3o4,4=SoR=

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Most popular questions from this chapter

A 2cnf-formula is an AND of clauses, where each clause is an OR of at most two literals. Let . Show that2SAT={ϕ|ϕisasatisfiable2CNFformula}. Show that 2SATP.

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

(y1y2y3)$$

with the two clauses

(y1y2zi)and(zi¯y3b)

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

c. Conclude that SATisNP-complete.

A permutation on the set {1,...,k} is a one-to-one, onto function on this set. When P is a permutation,pt means the composition of p with itself t times. Let

PERMPOWER={hp,q,ti|p=qtwherepandqarepermutationson{1,...,k}andtisabinaryinteger}.

Show thatPERMPOWERP . (Note that the most obvious algorithm doesn’t run within polynomial time.

Is the following formula satisfiable?

(xy)(xy)(xy)(xy).

Let UNARYSSUM be the subset sum problem in which all numbers are represented in unary. Why does the NP completeness proof for SUBSETSUM fail to show UNARYSSUM is NP complete? Show that UNARYSSUMP

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