Computer Science Engineering (CSE) Exam > Computer Science Engineering (CSE) Questions > Let S be an NP-complete problem. Q and R are ...
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Let S be an NP-complete problem. Q and R are other two problems not known to be NP. Q is polynomial time reducible to S and S is polynomial time reducible to R. Which of the following statements is true?
- a)R is NP-complete
- b)R is NP-hard
- c)Q is NP-complete
- d)Q is NP-hard
Correct answer is option 'B'. Can you explain this answer?
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Most Upvoted Answer
Let S be an NP-complete problem. Q and R are other two problems not kn...
Overview:
In this scenario, we are given that problem S is NP-complete, and problems Q and R are not known to be NP. It is also known that Q is polynomial time reducible to S, and S is polynomial time reducible to R. We need to determine which of the following statements is true.
Explanation:
To understand the answer, let's break down the given information and the options provided:
1. S is NP-complete:
This means that S is one of the hardest problems in the class of NP problems. Any problem in NP can be reduced to S in polynomial time.
2. Q is polynomial time reducible to S:
This means that there exists a polynomial time reduction from problem Q to problem S. In other words, we can transform instances of Q into instances of S in polynomial time.
3. S is polynomial time reducible to R:
This means that there exists a polynomial time reduction from problem S to problem R. In other words, we can transform instances of S into instances of R in polynomial time.
Now, let's analyze the given options:
a) R is NP-complete:
We cannot conclude that R is NP-complete because it is not known whether R is in NP or not. The reduction from S to R only tells us that R is at least as hard as S.
b) R is NP-hard:
This statement is true. Since S is NP-complete and there exists a polynomial time reduction from S to R, we can conclude that R is NP-hard. NP-hardness means that a problem is at least as hard as the hardest problems in NP, even if it is not in NP itself.
c) Q is NP-complete:
We cannot conclude that Q is NP-complete because it is not known whether Q is in NP or not. The reduction from Q to S only tells us that S is at least as hard as Q.
d) Q is NP-hard:
We cannot conclude that Q is NP-hard because it is not known whether Q is in NP or not. The reduction from Q to S only tells us that S is at least as hard as Q.
Conclusion:
Based on the given information and the definitions of NP-complete and NP-hard, the correct statement is option 'B' - R is NP-hard.
In this scenario, we are given that problem S is NP-complete, and problems Q and R are not known to be NP. It is also known that Q is polynomial time reducible to S, and S is polynomial time reducible to R. We need to determine which of the following statements is true.
Explanation:
To understand the answer, let's break down the given information and the options provided:
1. S is NP-complete:
This means that S is one of the hardest problems in the class of NP problems. Any problem in NP can be reduced to S in polynomial time.
2. Q is polynomial time reducible to S:
This means that there exists a polynomial time reduction from problem Q to problem S. In other words, we can transform instances of Q into instances of S in polynomial time.
3. S is polynomial time reducible to R:
This means that there exists a polynomial time reduction from problem S to problem R. In other words, we can transform instances of S into instances of R in polynomial time.
Now, let's analyze the given options:
a) R is NP-complete:
We cannot conclude that R is NP-complete because it is not known whether R is in NP or not. The reduction from S to R only tells us that R is at least as hard as S.
b) R is NP-hard:
This statement is true. Since S is NP-complete and there exists a polynomial time reduction from S to R, we can conclude that R is NP-hard. NP-hardness means that a problem is at least as hard as the hardest problems in NP, even if it is not in NP itself.
c) Q is NP-complete:
We cannot conclude that Q is NP-complete because it is not known whether Q is in NP or not. The reduction from Q to S only tells us that S is at least as hard as Q.
d) Q is NP-hard:
We cannot conclude that Q is NP-hard because it is not known whether Q is in NP or not. The reduction from Q to S only tells us that S is at least as hard as Q.
Conclusion:
Based on the given information and the definitions of NP-complete and NP-hard, the correct statement is option 'B' - R is NP-hard.
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Community Answer
Let S be an NP-complete problem. Q and R are other two problems not kn...
- If the NP-Hard problem is reducible to another problem in polynomial time, then that problem is also NP-hard which means every NP problem can be reduced to this problem in polynomial time.
- If Q is reducible to S in polynomial time, Q could be NP because all NP problems can be reduced to S. Since Q could be Np therefore Q could be P also as P is a subset of NP. Also, Q could be NPC because every NPC problem can be reduced to another NPC problem in polynomial time.
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Let S be an NP-complete problem. Q and R are other two problems not known to be NP. Q is polynomial time reducible to S and S is polynomial time reducible to R. Which of the following statements is true?a)R is NP-completeb)R is NP-hardc)Q is NP-completed)Q is NP-hardCorrect answer is option 'B'. Can you explain this answer?
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