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The problem 3-SAT and 2-SAT are
- a)both in P
- b)both NP complete
- c)NP-complete and in P respectively
- d)undecidable and NP-complete respectively
Correct answer is option 'C'. Can you explain this answer?
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Verified Answer
The problem 3-SAT and 2-SAT area)both in Pb)both NP completec)NP-comp...
The Boolean satisfiability problem (SAT) is a decision problem, whose instance is a Boolean expression written using only AND, OR, NOT, variables, and parentheses. The problem is: given the expression, is there some assignment of TRUE and FALSE values to the variables that will make the entire expression true? A formula of propositional logic is said to be satisfiable if logical values can be assigned to its variables in a way that makes the formula true.
3-SAT and 2-SAT are special cases of k-satisfiability (k-SAT) or simply satisfiability (SAT), when each clause contains exactly k = 3 and k = 2 literals respectively.
2-SAT is P while 3-SAT is NP Complete.
Most Upvoted Answer
The problem 3-SAT and 2-SAT area)both in Pb)both NP completec)NP-comp...
Explanation:
The problem 3-SAT and 2-SAT belongs to the category of Boolean Satisfiability Problems (SAT), which are decision problems in computer science.
3-SAT:
The 3-SAT problem is the problem of determining whether there exists an assignment of truth values to variables in a given Boolean formula such that the formula evaluates to true. In 3-SAT, the given Boolean formula is in conjunctive normal form (CNF) and consists of clauses, each containing exactly three literals connected with OR operators, and the clauses are connected with AND operators.
2-SAT:
The 2-SAT problem is a special case of the SAT problem where the given Boolean formula is in CNF and consists of clauses, each containing exactly two literals connected with OR operators, and the clauses are connected with AND operators.
P vs NP:
P refers to the class of decision problems that can be solved in polynomial time, while NP refers to the class of decision problems for which a solution can be verified in polynomial time. The question of whether P = NP is one of the most important open problems in computer science.
Answer:
The problem 3-SAT is NP-complete, which means it belongs to the class of decision problems for which a solution can be verified in polynomial time. It is also NP-complete, which means it is one of the hardest problems in NP and is believed not to have a polynomial-time algorithm.
On the other hand, the problem 2-SAT is in P, which means it can be solved in polynomial time. This is because 2-SAT can be solved using algorithms such as the strongly connected components algorithm or the implication graph algorithm, both of which have polynomial time complexity.
Therefore, the correct answer is option 'C' - 3-SAT is NP-complete and 2-SAT is in P.
The problem 3-SAT and 2-SAT belongs to the category of Boolean Satisfiability Problems (SAT), which are decision problems in computer science.
3-SAT:
The 3-SAT problem is the problem of determining whether there exists an assignment of truth values to variables in a given Boolean formula such that the formula evaluates to true. In 3-SAT, the given Boolean formula is in conjunctive normal form (CNF) and consists of clauses, each containing exactly three literals connected with OR operators, and the clauses are connected with AND operators.
2-SAT:
The 2-SAT problem is a special case of the SAT problem where the given Boolean formula is in CNF and consists of clauses, each containing exactly two literals connected with OR operators, and the clauses are connected with AND operators.
P vs NP:
P refers to the class of decision problems that can be solved in polynomial time, while NP refers to the class of decision problems for which a solution can be verified in polynomial time. The question of whether P = NP is one of the most important open problems in computer science.
Answer:
The problem 3-SAT is NP-complete, which means it belongs to the class of decision problems for which a solution can be verified in polynomial time. It is also NP-complete, which means it is one of the hardest problems in NP and is believed not to have a polynomial-time algorithm.
On the other hand, the problem 2-SAT is in P, which means it can be solved in polynomial time. This is because 2-SAT can be solved using algorithms such as the strongly connected components algorithm or the implication graph algorithm, both of which have polynomial time complexity.
Therefore, the correct answer is option 'C' - 3-SAT is NP-complete and 2-SAT is in P.
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The problem 3-SAT and 2-SAT area)both in Pb)both NP completec)NP-complete and in P respectivelyd)undecidable and NP-complete respectivelyCorrect answer is option 'C'. Can you explain this answer?
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