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 Which of the following statements are TRUE?
1. The problem of determining whether there exists a cycle in an undirected graph is in P.
2. The problem of determining whether there , exists a cycle in an undirected graph is in NP.
3. If a problem A is NP-Complete, there exists a non-deterministic polynomial time algorithm to solve A
  • a)
    1,2 and 3
  • b)
    1 and 3 only
  • c)
    2 only
  • d)
    3 only
Correct answer is option 'A'. Can you explain this answer?
Verified Answer
Which of the following statements are TRUE?1. The problem of determini...
1. By depth first search we can find whether there exits a cycle in an undirected graph in O(n2) times so it is P problem.
2. As so this problem will also be considered as NP problem.
3. This is the definition of NP-complete so it is trivially true.
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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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Which of the following statements are TRUE?1. The problem of determining whether there exists a cycle in an undirected graph is in P.2. The problem of determining whether there , exists a cycle in an undirected graph is in NP.3. If a problem A is NP-Complete, there exists a non-deterministic polynomial time algorithm to solve Aa)1,2 and 3b)1 and 3 onlyc)2 onlyd)3 onlyCorrect answer is option 'A'. Can you explain this answer?
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