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Chapter 8: Q28P (page 360)

BOTHNFA={(M1,M2)|M1andM2areNFAswhereL(M1)∩L(M2)≠ϕ}.

Show that BOTHNFA is NL-complete.

Short Answer

Expert verified

Its reduction occurs in polynomial time as well. So it's NP Difficult. As a result, BOTHNFAis NL-complete.

Step by step solution

01

Step-1: NP complete

There are two conditions that must be met in order to prove that the problem is NP complete.

  1. The problem is in NP.
  2. The problem is NP-hard.
02

Step-2: BOTHNFA is NL complete

M=" on input (M1, M2), so that they are both NFAs.

  1. Mark the initials state of both NFAs using markers.
  2. Repeat the procedure for 2q1*q2time.
  3. Change the position of the marker in both stages in a nondeterministic manner to simulate symbol reading.
  4. Accept if a string exists in stages 2 and 3, else reject."

This occurs in a logarithmic space. So BOTHNFA∈NL.

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Most popular questions from this chapter

Show that EDFAis NL-complete.

Question:Show that if every NP-hard language is also PSPACE-hard, then PSPACE = NP.

a. Let ADD={⟨x,y,z⟩|x,y,z>0 are binary integers and x+y=z}. Show
that ADD∈L.
b. Let PAL_ADD={⟨x,y⟩|x,y>0are binary integers wherex+y is an integer whose binary representation is a palindrome). (Note that the binary representation of the sum is assumed not to have leading zeros. A palindrome is a string that equals its reverse.) Show that PAL_ADD∈L.

Let CNFH1={(ϕ)|ϕ   is a satisfiable cnf-formula where each clause contains any number of positive literals and at most one negated literal. Furthermore, each negated literal has at most one occurrence in ϕ }. Show that CNFH1 is NL- complete.

For each n, exhibit two regular expressions,R  and  S , of length poly(n), whereL(R)≠L(S), but where the first string on which they differ is exponentially long. In other words,L(R) and  L(S) must be different yet agree on all strings of length up to2nd for some constant ε>0.
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