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Chapter 7: Q15P (page 323)

Show that Pis closed under the star operation. (Hint: Use dynamic programming. On input y=y1···ynforyi∈Σ, build a table indicating for each i≤j whether the substring yi···yj∈A∗foranyA∈P.)

Short Answer

Expert verified

The solution is,

p=∪kTIME(nk)

Step by step solution

01

To Indication of P

By using dynamic programming system we can apply input and build a table which is indicate p is closed under the star operation.

02

To Classification of Polynomial time

P seems to be a classification representing languages which can be deduced within polynomial time that used a deterministic solitary tape Turing computer. That is correct

p=∪kTIME(nk)

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Most popular questions from this chapter

Show that NPis closed under union and concatenation.

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

Show that if P=NP , a polynomial time algorithm exists that takes an undirected graph as input and finds a largest clique contained in that graph. (See the note in Problem 7.38.)

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.
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