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Chapter 0: Q20P (page 1)

We generally believe that PATH is not NP-complete. Explain the reason behind this belief. Show that proving PATH is not NP-complete would prove P ≠ NP

Short Answer

Expert verified

If PATH is not NP -complete, thenNP≠P$.

Step by step solution

01

To assume PATH would ne NP-complete

EveryAisNPis polynomial time reducible toB.

G is a directed graph that has a directed path.

PATH is not NP - complete:

Let us assume that PATH would be NP - complete.

From the definition of NP - completeness,

02

To explain it’s true or not

ThusP=NP which we believe that it is not true.

Hence, PATH is not NP - complete.

Assume thatP=NP and then show that PATH is NP - complete.

So assume P=NP.

So, PATH is NP - complete.

Thus, if PATH is not NP -complete, thenP≠NP

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Most popular questions from this chapter

Question:Consider the algorithm MINIMIZE, which takes a DFA as input and outputs DFA .

MINIMIZE = “On input , where M=(Q,Σ,δ,q0,A) is a DFA:

1.Remove all states of G that are unreachable from the start state.

2. Construct the following undirected graph G whose nodes are the states of .

3. Place an edge in G connecting every accept state with every non accept state. Add additional edges as follows.

4. Repeat until no new edges are added to G :

5. For every pair of distinct states q and r of and every a∈Σ :

6. Add the edge (q,r) to G if δq,a,δr,a is an edge of G .

7. For each state q,let[q] be the collection of statesq={r∈Q|noedge joins q and r in G }.

8.Form a new DFA M'=Q',Σ,δ',q'0,A'where

Q'={[q]|q∈Q}(ifq=r,onlyoneofthemisinQ'),δ'(q,a)=[δq,a]foreveryq∈Qanda∈Σ,q00=[q0],andA0={[q]|q∈A}

9. Output ( M')”

a. Show that M and M' are equivalent.

b. Show that M0 is minimal—that is, no DFA with fewer states recognizes the same language. You may use the result of Problem 1.52 without proof.

c. Show that MINIMIZE operates in polynomial time.

Using the solution you gave to Exercise 1.25, give a formal description of the machines T1 and T2depicted in Exercise 1.24

Show that A is decidable iff A≤m0*1*.

Question: Let B be the set of all infinite sequences over {0 , 1}. Show that B is uncountable using a proof by diagonalization.

Let. ∑={0,1}LetC1 be the language of all strings that contain a 1 in their middle third.

Let C2be the language of all strings that contain two 1s in their middle third. So C1={xyz|x,z∈∑*andy∈∑*1∑*,where|x|=|z|≥|y|}and C2={xyz|x,z∈∑*and y∈∑*1∑*1∑*,  where |x|=|z|≥|y|}.

a.Show that C1is a CFL.

b. Show thatC2 is not a CFL
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