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 Consider two decision problems Q1, Q2 such that Q1 reduces in polynomial time to 3-SAT and 3-SAT reduces in polynomial time to Q2. Then which one of the following is consistent with the above statement?
  • a)
    Q1 is in NP, Q2 is NP hard
  • b)
    Q2 is in NP, Q1 is NP hard
  • c)
    Both and Q2 are in NP
  • d)
    Both Q1 and Q2 are NP hard
Correct answer is option 'A'. Can you explain this answer?
Verified Answer
Consider two decision problems Q1, Q2 such that Q1 reduces in polynomi...
Given Q1 ≤ 3-SAT and 3-SAT ≤ Q2 3-SAT is NP- complete.
∴ It is NP as well as NP-hard.
  is NP-problem   3-SAT ≤ Qand 3-SAT is NP-hard ⇒ Q2 is NP- hard problem.
So the strongest statements is Q1 is in NP and Q2 is NP-hard.
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Most Upvoted Answer
Consider two decision problems Q1, Q2 such that Q1 reduces in polynomi...
Explanation:
To understand why option A is the correct answer, let's break down the given information and analyze each option.

Given information:
1. Q1 reduces in polynomial time to 3-SAT.
2. 3-SAT reduces in polynomial time to Q2.

Analyzing the options:
a) Q1 is in NP, Q2 is NP-hard:
Since Q1 reduces to 3-SAT in polynomial time, we can conclude that Q1 is in NP (nondeterministic polynomial time). This is because if we have a polynomial-time algorithm to verify the solution to 3-SAT, we can use it to verify the solution to Q1 as well. However, we cannot say anything about Q2 being NP-hard based on the given information.

b) Q2 is in NP, Q1 is NP-hard:
From the given information, we know that 3-SAT reduces to Q2 in polynomial time. If Q2 were in NP, then 3-SAT would also be in NP, as a polynomial-time reduction from an NP-complete problem to another problem implies that the target problem is also NP-complete. Therefore, we cannot conclude that Q2 is in NP based on the given information.

c) Both Q1 and Q2 are in NP:
We cannot conclude that both Q1 and Q2 are in NP based on the given information. While we know that Q1 reduces to 3-SAT in polynomial time, we do not have any information about the reduction of 3-SAT to Q2 in terms of NP.

d) Both Q1 and Q2 are NP-hard:
We cannot conclude that both Q1 and Q2 are NP-hard based on the given information. While we know that 3-SAT reduces to Q2 in polynomial time, we do not have any information about the reduction of Q1 to 3-SAT in terms of NP-hardness.

Conclusion:
Based on the given information, the only consistent statement is option A, where Q1 is in NP and Q2 is NP-hard. This is because Q1 reduces to 3-SAT, which is an NP-complete problem, implying that Q1 is in NP. However, we cannot make any conclusions about Q2 being NP-hard or in NP based on the given information.
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