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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 'B'. Can you explain this answer?
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Verified Answer
A problem in NP is NP-complete ifa)It can be reduced to the 3-SAT prob...
A problem in NP becomes NPC if all NP problems can be reduced to it in polynomial time. This is same as reducing any of the NPC problem to it. 3-SAT being an NPC problem, reducing it to a NP problem would mean that NP problem is NPC.
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Most Upvoted Answer
A problem in NP is NP-complete ifa)It can be reduced to the 3-SAT prob...
Explanation:

Reduction to 3-SAT problem:
- A problem is NP-complete if it can be reduced to the 3-SAT problem in polynomial time.
- This means that an instance of the given problem can be transformed into an equivalent instance of the 3-SAT problem in polynomial time.

Reduction from 3-SAT problem:
- In the case of NP-completeness, the 3-SAT problem can also be reduced to the given problem in polynomial time.
- This implies that any instance of the 3-SAT problem can be transformed into an equivalent instance of the given problem in polynomial time.

Significance of the 3-SAT problem:
- The 3-SAT problem is often used as a benchmark for NP-completeness because it is known to be NP-complete.
- If a problem can be reduced to the 3-SAT problem or the 3-SAT problem can be reduced to it, then it is considered NP-complete.
Therefore, in the context of NP-completeness, a problem is NP-complete if the 3-SAT problem can be reduced to it in polynomial time. This signifies the strong connection between the given problem and the 3-SAT problem, which is a well-known NP-complete problem.
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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 timeb)The 3-SAT problem can be reduced to it in polynomial timec)It can be reduced to any other problem in NP in polynomial timed)some problem in NP can be reduced to it in polynomial timeCorrect answer is option 'B'. Can you explain this answer?
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