All Courses
Explore Courses UPSC Exam UPSC Test Series Free Notes EduRev Infinity Get the App Login Join for Free
Sign in Open App
Computer Science Engineering (CSE) Exam  >  Computer Science Engineering (CSE) Questions  >  Let S be an NP-complete problem. Q and R are ... Start Learning for Free
SaveJoin the Discussion on the App
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?
FREE This question is part of
Download PDF Attempt Test
Download PDF Attempt this Test
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.
Free Test
FREE
Start Free Test
Start Free Test
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. 
< Previous Question Next Question >
Explore Courses for Computer Science Engineering (CSE) exam

Similar Computer Science Engineering (CSE) Doubts

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.

Top Courses for Computer Science Engineering (CSE)View all

Top Courses for Computer Science Engineering (CSE)

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?
Question Description
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? for Computer Science Engineering (CSE) 2024 is part of Computer Science Engineering (CSE) preparation. The Question and answers have been prepared according to the Computer Science Engineering (CSE) exam syllabus. Information about 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? covers all topics & solutions for Computer Science Engineering (CSE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for 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?.
Solutions for 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? in English & in Hindi are available as part of our courses for Computer Science Engineering (CSE). Download more important topics, notes, lectures and mock test series for Computer Science Engineering (CSE) Exam by signing up for free.
Here you can find the meaning of 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? defined & explained in the simplest way possible. Besides giving the explanation of 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?, a detailed solution for 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? has been provided alongside types of 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? theory, EduRev gives you an ample number of questions to practice 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? tests, examples and also practice Computer Science Engineering (CSE) tests.
Explore Courses for Computer Science Engineering (CSE) exam

Top Courses for Computer Science Engineering (CSE)

Explore Courses

Suggested Free Tests

Signup for Free!
Signup to see your scores go up within 7 days! Learn & Practice with 1000+ FREE Notes, Videos & Tests.
10M+ students study on EduRev
Get App
© EduRev
Education Revolution
Signup to see your scores go up within 7 days!
Access 1000+ FREE Docs, Videos and Tests
Takes less than 10 seconds to signup