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Let π be a problem that belongs to the class NP. Then which one of the following is TRUE?
  • a)
    There is no polynomial time algorithm for πA
  • b)
    If πA can be solved deterministically in polynomial time, then P = NP
  • c)
    If πA is NP-hard, then it is NP-complete
  • d)
    πA may be undecidable
Correct answer is option 'C'. Can you explain this answer?
Verified Answer
LetπAbe a problem that belongs to the class NP. Then which one of t...
it is given that πA ∈ NP
∴ If πA is NP-hard, and since it is given that πA ∈ NP , this means that πA is NP-complete
∴ choice (c) is correct.
Notice that choice (a) is incorrect since as P ∈ NP, some NP problems are actually P and hence polynomial time algorithm can exist for these.
Choice (b) is incorrect since, If πA can be solved deterministicaily in polynomial time, it does not generate that P=NP, unless of-course it is additionally known that πA is NP-complete.
Choice (d) is incorrect since,
All problems belonging to P or NP have to be decidable.
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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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LetπAbe a problem that belongs to the class NP. Then which one of the following is TRUE?a)There is no polynomial time algorithm forπAb)If πA can be solved deterministically in polynomial time, then P = NPc)If πA is NP-hard, then it is NP-completed)πA may be undecidableCorrect answer is option 'C'. Can you explain this answer?
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