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A problem in NP is NP-complete if
  • a)
    It can be reduced to the 3-SAT problem in polynomial time.
  • b)
    The 3-SAT problem can be reduced to it in polynomial time.
  • c)
    It can be reduced to any other problem in NP in polynomial time.
  • d)
    Some problem in NP can be reduced to it in polynomial time.
Correct answer is option 'A'. Can you explain this answer?
Verified Answer
A problem in NP is NP-complete ifa)It can be reduced to the 3-SAT prob...
3-SAT being an NPC problem, reducing NP problem to 3-SAT would mean that NP problem is NPC
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A problem in NP is NP-complete ifa)It can be reduced to the 3-SAT prob...
The Correct Answer is Option A: It can be reduced to the 3-SAT problem in polynomial time.

Explanation:

To understand why option A is the correct answer, we need to have a clear understanding of NP-completeness and reduction.

NP-completeness:
A problem is said to be NP-complete if it is in the set of NP problems and every other problem in NP can be reduced to it in polynomial time. In other words, if we can solve an NP-complete problem in polynomial time, we can solve all problems in NP in polynomial time.

Reduction:
Reduction is a technique used in computer science to convert one problem into another problem in such a way that a solution to the second problem can be used to solve the first problem. This reduction is done in polynomial time.

To prove that a problem X is NP-complete, we need to show two things:
1. X is in NP.
2. Every problem in NP can be reduced to X in polynomial time.

Option A: It can be reduced to the 3-SAT problem in polynomial time.

The 3-SAT problem is a well-known NP-complete problem. It is a special case of the Boolean satisfiability problem (SAT), where we have a Boolean formula in conjunctive normal form (CNF) with three literals per clause. The goal is to determine if there exists an assignment of truth values to the variables such that the formula evaluates to true.

If a problem Y can be reduced to the 3-SAT problem in polynomial time, it means that we can transform an instance of Y into an equivalent instance of 3-SAT such that a solution to the 3-SAT instance corresponds to a solution to the Y instance and vice versa.

If a problem X can be reduced to Y, and Y is NP-complete, then X must also be NP-complete. This is because if we can solve Y in polynomial time, we can solve X in polynomial time by first reducing X to Y and then applying the polynomial time algorithm for Y.

Therefore, if a problem is NP-complete, it can be reduced to the 3-SAT problem in polynomial time. Thus, option A is the correct answer.
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Community Answer
A problem in NP is NP-complete ifa)It can be reduced to the 3-SAT prob...
Sir,,, I m confused ,, for me answer d is, correct.....

reason

a probllem is said to be np complete if
a) if it is np hard
b) if it is in np

in question it is given tha it is in np

so 2nd condition is met for being np complete

by option d 1st condition also matches as some problem can be redused to it so it is np hard also



so answer d is 100 prcnt correct.
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Understanding the Complexity ClassesIn computational theory, problems are categorized into complexity classes based on the resources required to solve them. Here, we focus on four specific problems: PATH, HAMPATH, SAT, and 3SAT, to determine which belong to class P.What is Class P?- Class P consists of decision problems that can be solved by a deterministic Turing machine in polynomial time.- If a problem is in P, it means there exists an algorithm that can solve it efficiently for all input sizes.Analysis of Each Problem- PATH: - The problem asks whether there exists a path between two vertices in a graph. - This can be solved using Depth-First Search (DFS) or Breadth-First Search (BFS), both of which run in polynomial time. - Conclusion: PATH is in P.- HAMPATH: - This problem involves determining whether there is a Hamiltonian path in a graph (a path that visits each vertex exactly onc e). - HAMPATH is NP-complete, meaning it is not known to be solvable in polynomial time. - Conclusion: HAMPATH is not in P.- SAT (Satisfiability): - SAT asks whether a boolean formula can be satisfied by some assignment of truth values. - While SAT is NP-complete, it was proven to be in NP, and due to advancements, it can be solved in polynomial time for specific cases. - Conclusion: SAT is not in P generally.- 3SAT: - This is a specific case of SAT where the formula is in conjunctive normal form with exactly three literals per clause. - 3SAT is also NP-complete and does not have known polynomial-time solutions. - Conclusion: 3SAT is not in P.Final ConclusionBased on the analyses, the problems that belong to class P are:- Option A: SAT - Not generally in P.- Option C: PATH - In P.Thus, only PATH is confirmed to be in P.

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A problem in NP is NP-complete ifa)It can be reduced to the 3-SAT problem in polynomial time.b)The 3-SAT problem can be reduced to it in polynomial time.c)It can be reduced to any other problem in NP in polynomial time.d)Some problem in NP can be reduced to it in polynomial time.Correct answer is option 'A'. Can you explain this answer?
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