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For a cnf-formula ∅ with m variables and cclauses O(cm) , show that you can construct in polynomial time an NFA with states that accept all nonsatisfying assignments, represented as Boolean strings of length m. Conclude that ±Êâ‰²Ó±Ê implies that NFAs cannot be minimized in polynomial time.

Short Answer

Expert verified

It has been concluded that P≠NP implies that NFAs cannot be minimized in polynomial time.

Step by step solution

01

Step-1: Nondeterministic Finite Automaton

The acronym NFA stands for Nondeterministic Finite Automaton. If there are multiple possible transitions from one state on the same input symbol, a Finite Automata FA is said to be non-deterministic.

02

Step-2: CNF formula

Build an NFA Non input ϕthat chooses one of the cclauses nondeterministically and reads input length m, accepting if clause is not satisfied and rejecting otherwise.

N accepts all inputs with a length greater than or equal to m. Nalso hasOcm states, indicating that it can work in polynomial time.

Because the clause is not satisfied if Naccepts a, Nis a nonsatisfying assignment .

As a result,N accepts all of's unsatisfactory assignments.

Let's say that minimising NFAs takes polynomial time. Construct an NFA that accepts unsatisfactory assignments. Also, if and only if is not satisfiable, N accepts any binary string.

In the case of new N' put the new minimization algorithm to the test. If N' includes precisely one state and accepts all binary strings, reject Ï• ; otherwise, accept Ï• .

As a result, 3SAT has a polynomial time algorithm, and so P = NP.

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Most popular questions from this chapter

Let UNARY−SSUM be the subset sum problem in which all numbers are represented in unary. Why does the NP− completeness proof for SUBSET−SUM fail to show UNARY−SSUM is NP− complete? Show that UNARY−SSUM∈P

Show that ALLDFAisinP.

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.

You are given a box and a collection of cards as indicated in the following figure. Because of the pegs in the box and the notches in the cards, each card will fit in the box in either of two ways. Each card contains two columns of holes, some of which may not be punched out. The puzzle is solved by placing all the cards in the box so as to completely cover the bottom of the box (i.e., every hole position is blocked by at least one card that has no hole there). It represents a card and this collection of cards has a solution}. Show that PUZZLE is NP-complete.

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

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