# Recent questions tagged p-np-npc-nph 1
I am having difficulty in understanding np and np-hard topic in algorithms. If someone can provide some good source to learn about it in easy manner it would be a real help. Thank you!
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We have constructed a polynomial time reduction from problem A to problem B. Which of the following is not a possible scenario? We know of polynomial time algorithms for both A and B. We only know of exponential time algorithms for both A and B. We only know an ... a polynomial time algorithm for B. We only know an exponential time algorithm for B, but we have a polynomial time algorithm for A.
1 vote
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Consider the complexity class $CO-NP$ as the set of languages $L$ such that $\overline{L} \in NP$, and the following two statements: $S_1: \: P \subseteq CO-NP$ $S_2: \: \text{ If } NP \neq CO-NP, \text{ then } P \neq NP$ Which of the following is/are correct? Only $S_1$ Only $S_2$ Both $S_1$ and $S_2$ Neither $S_1$ nor $S_2$
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According to this article, A problem X can be proved to be NP-complete if an already existing NP-complete problem (say Y) can be polynomial-time reduced to current problem X. The problem also needs to be NP. Now my question is: Do we also need to prove that ... does this reduction have to be only one way to prove a problem is NP-complete? PS: I have already asked this question in cs.stackexchange
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I know that all NP problems can be reduced to Boolean Satisfiability SAT problem. But my question is whether SAT problem can be reduced to other NP complete problems like travelling salesperson problem TSP, 0/1 knapsack problem. In short are the reductions in NPC problems two ways?
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$1)$ If the complement of NP-Complete problem is in NP, then can we also say , for this case complement of NP problem is in NP-Complete ? $2)$ If the complement of NP-Complete problem is in Co-NP, then can we also say, for this case complement of Co-NP problem is in NP-Complete?
1 vote
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What will be the answer to this question? L’ is the complement of language L belongs to NP does not always imply that L’ belongs to NP L’ belongs to P both a and b
1 vote
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If there is an $\rm NP$-complete language $L$ whose complement is in $\rm NP$, then the complement of any language in $\rm NP$ is in $\rm NP$ $\rm P$ Both (a) and (b) None of these
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A formula is said to be a $3$-CF-formula if it is a conjunction (i.e., an AND) of clauses, and each clause has at most $3$ literals. Analogously, a formula is said to be a $3$-DF-formula if it is a disjunction (i.e., an OR) of clauses of at most $3$ literals each. Define the ... $\text{3-DF-SAT}$ is NP-complete Neither $\text{3-CF-SAT}$ nor $\text{3-DF-SAT}$ are in P
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All the places where I have read the Ham-Cycle problem, the graph used is directed. Is the problem of finding Ham-Cycle on an undirected graph also NP-Complete or not?
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All $P$, $NP$ and $NPC$ problems are turing decidable problems. $CFLs$ are in $NP$ area and $CFL's$ are not closed under intersection and complementation. So does it mean that $CFL's$ are undecidable under intersection and complementation. If $CFL$ is undecidable on intersection and complementation then how $NP$ problems can be turing decidable?
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Are NP-Hard problems Semi decidable or Decidable or not even semidecidable? I know NP class is decidable as there is polynomial time NTM.But in the following figure in case there are some NP hard problems that are lying outside of NP.Problems which are NP-hard but not NPC.Can these also be solved in polynomial time by NTM?
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I know that NP-complete problems are the hardest NP problems and every NP problem can be reduced NP-Complete problems in polynomial time. Now, it is said that all NP problems can be solved in Non-deterministic polynomial time, so is it true that ALL NP-COMPLETE PROBLEMS CAN BE SOLVED IN NON-DETERMINISTIC POLYNOMIAL TIME ?
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We have constructed a polynomial time reduction from problem $A$ to problem $B$. Which of the following is a valid inference? If the best algorithm for $B$ takes exponential time, then there is no polynomial time algorithm for $A$ If the best algorithm for $A$ takes ... If we don't know whether there is a polynomial time algorithm for $B$, then there cannot be a polynomial time algorithm for $A$.
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Can Anyone please post the diagram which describes the relation between P, NP, NPSPACE, NP-Hard, EXPONENTIAL TIME etc.
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I know this questions is already asked but no answers are yet given IS P NP STILL PRESENT IN GATE 2018 SYLLABUS? The Official Syllabus EXCLUDES those topics and according to https://gatecse.in/gate-cse-2016-syllabus/ , the P NP is NOT THERE. But still I want someone else to verify it. And please dont say, "it is better to read" because although knowledge is good but time is precious :-)
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did P and NP is still in the syllabus? @ COMPUTABILITY AND COMPLEXITY{TOC}
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$G$ respresents an undirected graph and a cycle refers to a simple cycle (no repeated edges or vertices). Define the following two languages. $SCYCLE=\{(G,k)\mid G \text{ contains a cycle of length at most k}\}$ ... $SCYCLE \leq_{p} LCYCLE$ (i.e, there is a polynomial time many-to-one reduction from $SCYCLE$ to $LCYCLE$).
1 vote
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1) I know that turing decidable means recursive language. But does is also means its decidable? So basically i want to know if REC imples decidability and RE implies undecidability or not. I got confused with word decidable in "turing decidable" 2) http://www. ... ? If they have same expressive power why can't we use DTM in NP decision problems? Thanks for being patient and reading doubt.
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what i have tried to get out the conslusion for P problem is that P problem These are set of those yes/no problems which can solved using polynomic time complexity algroithms. For example if are asked to Compare or sort then numbers.Then using loop we can solve and thus we can also ... fact can be verified.hence this Np problem is now P problem Am i correct with these two terms ????????????????
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Consider the following language $\mathsf{PRIMES} = \Biggl\{ \underbrace{111 \dots 11}_{p \: \text{ times} } : \: p \: \text{ is prime } \Biggl\}$ Then, which of the following is TRUE? $\mathsf{PRIMES}$ is regular $\mathsf{PRIMES}$ is undecidable $\mathsf{PRIMES}$ is decidable in polynomial time $\mathsf{PRIMES}$ is context free but not regular $\mathsf{PRIMES}$ is NP-complete and P $\neq$ NP
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Assume $P \neq NP$. Which of the following is not TRUE? 2-SAT in NP 2-SAT in coNP 3-SAT is polynmial-time reducible to 2-SAT 4-SAT is polynmial-time reducible to 3-SAT 2-SAT in P
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Let L1 = Φ and L2 = Σ*. Then which of the following is correct? L1 is NP-complete, L2 is not NP-complete L1 is not NP-complete, L2 is NP-complete Both L1 and L2 are NP-complete Neither L1 nor L2 is NP-complete I know the answer is d my query is in solution they are saying that L1 and L2 are P problems but L1 and L2 are undecidable how can they be P?
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A multivariate polynomial in $n$ variables with integer coefficients has a binary root if it is possible to assign each variable either 0 or 1, so that the polynomial evaluates to 0. For example, the multivariate polynomial $-2x_1^3 -x_1x_2+2$ has the binary root $(x_1=1, x_2=0)$. Then ... in polynomial time is NP-hard, but not in NP is in NP, but not in P and not NP-hard is both in NP and NP-hard
Consider the following statements: Checking if a given $undirected$ graph has a cycle is in $\mathsf{P}$ Checking if a given $undirected$ graph has a cycle is in $\mathsf{NP}$ Checking if a given $directed$ graph has a cycle is in $\mathsf{P}$ Checking if a given $directed$ ... TRUE? Choose from the following options. Only i and ii Only ii and iv Only ii, iii, and iv Only i, ii and iv All of them