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Chapter 7: Q49P (page 275)


Let f:N鈫�Nbe any function where fn=onlogn. Show that TIMEfn contains only the regular languages.

Short Answer

Expert verified

Thus, it is only a regular language. It is accepted by the TIMEfn.

Step by step solution

01

To Regular Language

A language is a collection of strings made up of characters or symbols from a specific alphabet.

A regular language is one that may be stated using regular expressions, finite automata, or state machines, both deterministic and non-deterministic.

02

To Explain and remove the form

Suppose language is A={0k1k|k鈮�0}鈭�TIME(nlogn)

  • If w is not of the form, reject M(w).
  • Rep till all w's have been crossed out.
  • Reject ifparityf0's鈮�parityf1's.
  • Remove every other 0 and 1 from the equation.
  • If all other bits are crossed out, accept.

A language is a collection of strings made up of characters or symbols from a specific alphabet.

Thus, it is only a regular language. It is accepted by the TIMEfn.

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Most popular questions from this chapter

In a directed graph, the indegreeof a node is the number of incoming edges and the outdegreeis the number of outgoing edges. Show that the following problem is NP-complete. Given an undirected graph G and a designated subset C of G鈥檚 nodes, is it possible to convert G to a directed graph by assigning directions to each of its edges so that every node in C has in-degree 0 or outdegree 0, and every other node in G has indegree at least 1?

Let DOUBLE-SAT={}has at least two satisfying assignments}. Show thatDOUBLE-SATisNP- complete

Let SET-SPLITTINGcollection of subsets of S, for some k>0, such that elements of S can be colored red or blue so that no Ci has all its elements colored with the same color}. Show that SET-SPLITTINGis NP-complete.

Let CONNECTED={<G>|Gisaconnectedundirectedgraph}.Analyse the algorithm given on page 185 to show that this language is in .

This problem investigates resolution, a method for proving the unsatisfiability of cnf-formulas. Let 蠒=C1鈭�C2鈭�鈥�鈭�Cmbe a formula in cnf, where the Ciare its clauses. Let C=CiCi鈥�is鈥�a鈥�clause鈥�of鈥�蠒. 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鈭�y1鈭�y2鈭�鈥�鈭�ykand Cb=x炉鈭�z1鈭�z2鈭�鈥�鈭�zl, where the and are literals. We form the new clause y1鈭�y2鈭�鈥�鈭�yk鈭�z1鈭�z2鈭�鈥�鈭�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 蠒unsatisfiable. 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鈭�P.
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