- istic machine in polynomial time then. NP is a subset of P. P=NP (from 1 and 3) NPC is a subset of P (2 and 3) But why is P a subset of NPC, if P = NP? This question has been asked before and the answers just pointed to the definitions of NP, P and NP-C
- istic machine can be used as a deter
- P is a subset of NP, so P is contained in NP. The statement is that NP is larger (or at least equal) to P, but P is completely contained within NP. - Gail Apr 14 '10 at 17:3
- The class of NP-complete problems is coincides with the class P. The proof of this theorem has been derived in the second half of the year 2000, and it is contained in the book Positionality principle for notation and calculation the function
- Therefore P is a proper subset of NP

NP-complete problems are special problems in class NP. I.e. they are a subset of class NP. An problem p is NP-complete, if 1. p ∈ NP (you can solve it in polynomial time by a non-deterministic Turing machine) and 2 The P versus NP problem is a major unsolved problem in computer science.It asks whether every problem whose solution can be quickly verified can also be solved quickly. It is one of the seven Millennium Prize Problems selected by the Clay Mathematics Institute, each of which carries a US$1,000,000 prize for the first correct solution.. The informal term quickly, used above, means the existence. If P = NP, then the world would be a profoundly different place than we usually assume it to be. There would be no special value in creative leaps, no fundamental gap between solving a problem and recognizing the solution once it's found ** P is a subset of RP, which is a subset of NP**.Similarly, P is a subset of co-RP which is a subset of co-NP.It is not known whether these inclusions are strict. However, if the commonly believed conjecture P = BPP is true, then RP, co-RP, and P collapse (are all equal). Assuming in addition that P ≠ NP, this then implies that RP is strictly contained in NP

**P** problems are **subset** **of** **NP** problems. **NP** problems are superset of **P** problems. It is not known whether P=NP. However, many problems are known in **NP** with the property that if they belong to **P**, then it can be proved that P=NP. If P≠NP, there are problems in **NP** that are neither in **P** nor in **NP**-Complete. All **P** problems are deterministic in nature ** NP Problems**. NP is the set of all the decision problems that are solvable by non - deterministic algorithms in polynomial time. Since deterministic algorithms are just the special case of non - deterministic ones, so we can conclude that P is the subset of NP. Relation between P and NP. NP Hard Proble

Complexity class NP Let A be a p-time algorithm and k a constant: NP = {L {0, 1}* : a certificate y, |y| = O(|x|k), and an algorithm A s.t. A(x, y) = 1} SUBSET-SUM NP P is a proper subset of NP - arXiv Vanity P is a proper subset of Np One of these is the deceptively simple assertion that P = NP. In fact, in a 2002 poll, 61 mathematicians and computer scientists said that they thought P probably didn't equal NP, to only nine who thought it did — and of those nine, several told the pollster that they took the position just to be contrary

P is subset of NP (any problem that can be solved by deterministic machine in polynomial time can also be solved by non-deterministic machine in polynomial time). Informally, NP is set of decision problems which can be solved by a polynomial time via a Lucky Algorithm, a magical algorithm that always makes a right guess among the given set of choices (Source Ref 1 ) It is then possible to determine that the only deterministic optimization of a NP-complete problem that could prove P = NP would be one that examines no more than a polynomial number of input sets for a given problem The purpose of this article is to examine and limit the conditions in which the P complexity class could be equivalent to the NP complexity class. Proof is provided by demonstrating that as the number of clauses in a NP-complete problem approaches infinity, the number of input sets processed per computation performed also approaches infinity when solved by a polynomial time solution The class NP NP is the set of languages for which there exists an e cient certi er. P is the set of languages for which there exists an e cient certi er thatignores the certi cate. That's the di erence: A problem is in P if we can decided them in polynomial time. It is in NP if we can decide them in polynomial time, if we are given the right certi cate Now, P vs NP actually asks if a problem whose solution can be quickly checked to be correct, then is there always a fast way to solve it. Thus writing in mathematically terms: is NP a subset of P or not? Now coming back to NP complete: these are the really hard problems of the NP problems

P and NP Clearly we have P NP . So far, nobody has been able to nd a problem that is in NP, but not in P. So we do not know if P ( NP or P = NP . This is one of the millenium problems, and worth a million dollars. Is it harder to nd solutions than to check them? If P = NP then \creativity can be automated NP problems have their own significance in programming, but the discussion becomes quite hot when we deal with differences between NP, P , NP-Complete and NP-hard. P and NP- Many of us know the difference between them. P- Polynomial time solving. Problems which can be solved in polynomial time, which take time like O(n), O(n2), O(n3)

All problems in P can be solved with polynomial time algorithms, whereas all problems in NP - P are intractable. It is not known whether P = NP. However, many problems are known in NP with the property that if they belong to P, then it can be proved that P = NP. If P ≠ NP, there are problems in NP that are neither in P nor in NP-Complete * It's clear that P is a subset of NP*. The open question is whether or not NP problems have deterministic polynomial time solutions. It is largely believed that they do not. Here is an outstanding recent article on the latest (and the importance) of the P = NP problem: The Status of the P versus NP problem The Big Deal. Here are some facts: NP consists of thousands of useful problems that need to be solved every day.; Some of these are in P.; For the rest, the fastest known algorithms run in exponential time. Although no one has found polynomial-time algorithms for these problems, no one has proven that no such algorithms exist for them either! In fact, it is quite possible that all problems in. NP-complete is a subset of NP, the set of all decision problems whose solutions can be verified in polynomial time; NP may be equivalently defined as the set of decision problems solved in polynomial time on a machine. A problem p in NP is also in NPC if and only if every other problem in NP is transformed into p in polynomial time P 6=NP - A Deﬁnitive Proof by Contradiction Adnan Masood April 1, 2014 Abstract The P versus NP problem use to be a major unsolved problem in computer science. However, as of today, April 1st 2014, the researcher has proven sans rigor that P6=NP.The question of P = NP asks whether every problem whose solution can be quickly veriﬁed by a computer can also be quickly solved b

Subset Sum is NP-complete The Subset Sum problem is as follows: given n non-negative integers w 1;:::;w n and a target sum W, the question is to decide if there is a subset I ˆf1;:::;ngsuch that P i2I w i = W. This is a very special case of the Knapsack problem: In the Knapsack problem, items also have values v i, and the problem was to maximize P i2I v i subject to P i2I w i W. If we set v i = Subset Sum is in NP: If any problem is in NP, then given a certificate, which is a solution to the problem and an instance of the problem (a set S of integer a 1 a N and an integer K) we will be able to identify (whether the solution is correct or not) certificate in polynomial time.This can be done by checking that the sum of the integers in subset S' is equal to K * NP-complete Reductions 1*. DOUBLEProve that 3SAT P-SAT, i.e., show DOUBLE SAT is NP complete by reduction from 3SAT. The 3-SAT problem consists of a conjunction of clauses over n Boolean variables, where each clause is a disjunction of 3 literals, e.g., ( Clearly, P NP. It is widely believed that P 6= NP. To de ne NP-completeness, we need to introduce the concept of a reduction. Reductions: The class of NP-complete problems consists of a set of decision problems (languages) (a subset of the class NP) that no one knows how to solve e ciently, but if there were

P=NP implies that any problem in P is NP-complete. $\endgroup$ - Columbo May 23 '17 at 17:44 $\begingroup$ Yes, I see, you're right about that. My point is about the difference between P = NP and NL = NP though (which affects the answer to your question). $\endgroup$ - Huck Bennett May 23 '17 at 18:5 P is a subset of NP. The question is whether they're identical. The technical difference has to do with (classical) Turing machines and what they're allowed to do. Basically, a Turing machine is a simple computer; it has an internal 'state' and a (infinite). * Whether NP is also a subset of P (i*.e., P = NP) is an open question. No one knows yet. NP-hard is a class of problems which are hardest of all problems in NP. These are the problems such that if we can solve them fast, then we can solve all problems in NP fast and P would equal NP

Now, **P** vs **NP** actually asks if a problem whose solution can be quickly checked to be correct, then is there always a fast way to solve it. Thus writing in mathematically terms: is **NP** a **subset** **of** **P** or not? Now coming back to **NP** complete: these are the really hard problems of the **NP** problems I know that if P were to be a strict subset of RP then P not equal NP, but I can't find any other relation or reason why P would be different than NP? computer-science computational-complexity. Share. Cite. Follow edited Dec 30 '17 at 18:22. hmakholm left over Monica np.array() : Create Numpy Array from list, tuple or list of lists in Python Create an empty 2D Numpy Array / matrix and append rows or columns in python 6 Ways to check if all values in Numpy Array are zero (in both 1D & 2D arrays) - Pytho

P is a proper subset of NP . By Jerrald Meek. Abstract. It is then possible to determine that the only deterministic optimization of a NP-complete problem that could prove P = NP would be one that examines no more than a polynomial number of input sets for a given problem

- Then P is a subset of NP just as every recursive set is re However unlike P vs from CSC 438F at University of Toront
- P is a proper subset of NP - CORE Reade
- P NP Theorem P NP Proof. Suppose X 2P. Then there is a polynomial-time algorithm A for X. To show that X 2NP, we need to design an e cient certi er B(I;C). Just take B(I;C) = A(I). Every problem with a polynomial time algorithm is in NP
- We believe but unable to prove at this point that P is a proper subset of NP ie from CS 9601 at HK
- e whether the collection contains a subcollection that adds up to t. Theorem: SUBSET SUM NP. •Proof: The following is a verifier V for SUBSET-SUM. V = on input <<S,t>,c>: 1. Test whether c is a collection of numbers that sum to t. 2
- istic polynomial problem, in computational complexity (a subfield of theoretical computer science and mathematics), the question of whether all so-called NP problems are actually P problems. A P problem is one that can be solved in polynomial time, which means that an algorithm exists for its solution such that the number of.
- In the effort to answer the P vs N P P\ \text{vs}\ NP P vs N P question, computer scientists have found a subset of N P NP N P problems that are at least as difficult to solve as any other N P NP N P problem; meaning that an answer to one of these problems will solve every N P NP N P problem. These sets of problems are known as NP-Complete problems. The majority of research regarding the.

I've seen that P != LINSPACE (by which I mean SPACE(n)), but that we don't know if one is a subset of the other. I assume that means that the proof must not involve showing a problem that's in one.. If P ≠ NP, there are problems in NP that are neither in P nor in NP-Complete. The problem belongs to class P if it's easy to find a solution for the problem. The problem belongs to NP, if it's easy to check a solution that may have been very tedious to find. Previous Page Print Page. Next Page . Advertisement Claim 1. Subset Sum is in NP. Proof. Given a proposed set I, all we have to test if indeed P i2I w i = W. Adding up at most n numbers, each of size W takes O(nlogW) time, linear in the input size. To establish that Subset Sum is NP-complete we will prove that it is at least as hard asSAT. Theorem 1. SAT Subset Sum. Proof. To prove the claim we. Proving P=NP would mean a P solution to an NP problem was theoretically possible, but it wouldn't necessarily help you find one. Even if P!=NP, we might still be able to solve some NP problems in P time using quantum computing, although some people believe this is not true [wikipedia.org]. Quantum computing can speed up NP algorithms [wikipedia.org] but may not be able to get all the way to P

Question: 8) While It Is Known That P Is A Subset Of NP, It Is Unknown Whether P Is A Proper Subset, Or If The Two Sets Are Equal. The Exponential Time Hypothesis States That A Decision Problem In NP, Called 3SAT Cannot Be Solved In Sub-exponential Time On A Deterministic Machine P ≠ NP Proof (Millennium Prize Problem Solved using the Proof of X ± Y = B at System 1) Martins Kolawole Alabi Keywords: input, systems, subset sum problem, algorithm, P ≠ NP, the proof of x ± y = b. I. Subset Sum Problem onsider the subset sum problem, an example of a problem that is easy to verify, but whose answer may be difficult to. Question: 1. P And NP Which Best Describes What Is Known Today About The Relationship Between The Classes P And NP A) P ⊂ NP (proper Subset) B) P ⊆ NP (subset) C) P = NP (equal) D) NP ⊆ P (subset) E) NP ⊂ P (proper Subset) F) P ∩ NP = ∅ (disjoint Sets) 2 Arial Times New Roman Verdana Wingdings Wingdings 3 Globe NP and NP-Completeness Outline Outline Decision and Optimization Problems Complexity Class P Complexity Class NP Relation of P and NP Polynomial-Time Reducibility NP-Hard and NP-Complete NP-Hard and NP-Complete TSP Circuit-SAT Knapsack PTAS Backtracking Slide 16 Branch-and-Bound Slide 18 Summary Reference To avoid this you will have to combine again subsets each other and this will bring you back to a previous level of splitting. But this is in the next part of analysis. The real solution of P vs. NP is as follows: Subset sum problem is a complete NP-problem. Subset sum problem can not be solved without using combinatorics at all

Stack Exchange network consists of 176 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers.. Visit Stack Exchang The question is whether it is a proper subset, i.e. \(P\neq NP\), or whether they are equal, \(P=NP\). If they're equal, it means that the process of finding the solution for these problems is at most polynomially harder than the process of the verification The distinctive feature of NP problems is that they are difficult to find any solution for, but easy to check a proposed solution. As any problem in P can provide its own witness, i.e. a solution found in polynomial time is verified in polynomial time, it is obvious that P is in NP, i.e. P is a subset of NP

630 CHAPTER 10. SOME NP-COMPLETE PROBLEMS Then an instance of a problem P is solvable iﬀthe corre-sponding string belongs to the language L P. This implies that our problems must have a yes-no an-swer, which is not always the usual formulation of opti-mization problems where what is required is to ﬁnd som The literal explanation is that Subset Product problem is NP-complete by a reduction from strongly NP-complete problem such as exact cover by 3-sets. In such strong reduction, the input integers are bounded by some polynomial function in the number of integers in the resulting instance of Subset Product problem P. ROGRAMMING. The Subset Sum Problem is as follows: Given a set of positive integers and a positive target integer , determine whether there exists a subset of whose elements sum to (Neapolitan and Naimipour). This problem has been shown to be NP-complete by reduction t

P is actually a subset of NP, so every problem with a P solution is an NP problem. But there are also some NP problems that (we suspect) aren't in P. There are also problems harder than NP. Graph coloring is NP-complete, but I'm not sure whether the proof that all planar graphs are four-colorable is itself NP or not P vs. NP is about finding algorithms, or computer programs, to solve particular math problems, and whether or not good algorithms exist to solve these problems

NP-completeness Proofs 1. The first part of an NP-completeness proof is showing the problem is in NP. 2. The second part is giving a reduction from a known NP-complete problem. • Sometimes, we can only show a problem NP-hard = if the problem is in P, then P = NP, but the problem may not be in NP The Subset Sum problem is NP-complete. It is in NP, because a veriﬁer can simply check that the given subset is a subset of A and that its sum is equivalent to the target in polynomial time. It is NP-hard via a reduction from 3DM. View the numbers in base. b =1+ max. i. n. x. i. Then. (ii) L is NP-hard, which we can show by giving a reduction from some known NP-complete problem L0 to L, that is, L0 P L. (That is, there is a polynomial time function that transforms an instance L0 into an equivalent instance of L for the other problem). Some Easy Reductions: Next, let us consider some closely related NP-complete problems subsets of U, and an integer k, does there exist a collection of k of these sets whose union is equal to U? Sample application. m available pieces of software. Set U of n capabilities that we would like our system to have. P P = NP. problem X in NP, X. An Annotated List of Selected NP-complete Problems. The standard textbook on NP-completeness is: . Michael Garey and David Johnson: Computers and Intractability - A Guide to the Theory of NP-completeness; Freeman, 1979.. David Johnson also runs a column in the journal Journal of Algorithms (in the HCL; there is an on-line bibliography of all issues) . On the Web the following sites may be of.

NP is the set of languages that have short proofs.coNP is the set of languages that have short refutations. Note that coNP is not the complement of NP.NP coNP is non-empty. It is easy to see that all languages in P are in NP coNP i.e., P NP coNP. It is conjectured that P NP coNP. i.e., there are problems in NP coNP that are not in P.Following are some problems in NP coNP that are not known to. For this problem we are looking for an ordered subset of 3 horses (r) from the set of 4 best horses (n). We are ignoring the other 11 horses in this race of 15 because they do not apply to our problem. We must calculate P(4,3) in order to find the total number of possible outcomes for the top 3 winners

to SUBSET SUM, and so SUBSET SUM is NP-complete. Believe it or not, these reductions become routine, eventually. Goddard 19b: 18. Practice Show that VERTEX COVER is NP-complete. (Recall that the removal of a vertex cover de-stroys every edge, and that the input to VERTEX COVE NP-completeness Reduction of 3-Sat to Subset Sum: n variables x i and m clauses c j For each variable x i, construct numbers t i and f i of n + m digits: The i-th digit of t i and f i is equal to 1 For n + 1 j n + m, the j-th digit of t i is equal to 1 if x i is in clause c j n For n + 1 j n + m, the j-th digit of NP completeness 1. Design and Analysis of Algorithms NP-COMPLETENESS 2. Instructor Prof. Amrinder Arora amrinder@gwu.edu Please copy TA on emails Please feel free to call as well Available for study sessions Science and Engineering Hall GWU Algorithms NP-Completeness 2 LOGISTIC And P is a subset of P, but it is not known if P is a proper subset of P. (that's the whole P=NP/P!=NP question.) Log in to post comments By Michael Ralston (not verified) on 08 Jan 2007 #permalin We use NP to designate the class of all nondeterministically polynomial problems. Clearly, P is a subset of NP A very famous open question in Computer Science: P = NP ? To give the 3rd alternative definition of NP, we introduce an imaginary, non-implementable instruction, which we call choose(). Behavior of choose()

NP does not mean Not-P. It is an unfortunate coincidence that causes confusion. NP comes from Nondeterministic Polynomial, as related to nondeterministic, polynomial-time Turing Machines. P ⊂ NP HAMPATH on a NTM: N 1 = On input G, s, t , where G is a directed graph with nodes s and t: 1. Write a list of m numbers, p 1,...,p m, where m i There are multiple possible questions you might be asking here, but the answer is in general no. Tim Converse's answer covers a more natural interpretation, so I'll cover the literal one. A language is a set of strings. A subset is a set entirely. One of the unanswered questions in computer science is a problem called P vs. NP. In 2000 the Clay Institute offered 1 Million dollars each for the solutions to seven key problems, called th Relationship between P and NP All decision problems NP P NP-Complete Problems • The search for an answer to the P=NP question depends on the notion of NP-complete problems, which was introduced by Stephen Cook in 1971. In an informal sense, a problem is NP-complete if it is provably as difficult to solve as any other problem in NP Theorem 17.1 P is a subset of the intersection of NP and Co NP One question we need to consider is: does P = NP? To answer this question, we'll see that a large class of problems such that solving any one of them will imply P = NP. 17.3.2 More examples Independent set: fhG;ki: G has an independent set of size kg

2. P and NP { recap 1.P and NP: formal de nitions 2.Open problem: whether or not P is a proper subset of NP 3.The size of the input can change the classi cation of P or NP However, even with strong restrictions on the inputs, many NPC problems are still NPC. 5/1 The contents. Highly parallel reduction; P-completeness; The basic P-complete problem; Examples of other P-complete problems. We have seen that NC is subset of P, but similarly to the NP-completeness theory, the problem whether P=NC is open and is likely equally difficult as its famous predecessor P=NP.The situation is very similar and also the techniques to deal with the problem are similar Proving NP-Completeness: • Step 1: Subset-Sum ∈ NP - Argue that, given a certificate, you can verify that the certificate provides a solution in polynomial time • Step 2: Show that some known NP-Complete problem is reducible in poly-time to Subset-Sum (i.e., A ≤. p. Subset-Sum) - What known NP-Complete problem do we choose

NP . is not a subset of . P. If . NP . is a subset of . P, then . P=NP. Given the present state of knowledge, it is reasonable to assume that . NP . is not equal to . P . and this is the widely held belief. If . P . is not equal to . NP, then the distinction between . P . and . NP\P . is important. All problems in . P . can be solved in. morsch writes Researcher Vinay Deolalikar from HP Labs claims proof that P != NP.The 100-page paper has apparently not been peer-reviewed yet, so feel free to dig in and find some flaws. However, the attempt seems to be quite genuine, and Deolalikar has published papers in the same field in the past. So this may be the real thing Resolving P ≟ NP has proven extremely difficult. In the past 44 years: Not a single correct proof either way has been found. Many types of proofs have been shown to be insufficiently powerful to determine whether P ≟ NP. A majority of computer scientists believe P ≠ NP, but this isn't a large majority Degree constrained subgraph (solving MSMD and DDDkS problems) O. Amines and others in (Amidi O., Peleg D., P rennes S., San J., Saurabh S., 2008.) describe the two NP-hard problem, as follows: The Minimum Subgraph of Minimum Degree problem (MSMD) and the Dual Degree-Dense k-Subgraph problem (DDDkS). 2-induced-CYCLE: Input:Graph G and denote a set of two nodes of G.Question: Is there a subset. •!Again, clearly in NP: just guess a subset and verify that it sums to t. Polynomial-time reductions •!We've now seen several problems that are in NP but don't seem to be in P: HAMPATH, VC, SAT, SUBSET-SUM •!We shall see: if we could somehow solve one of these problems in P-time, we could solve all of them in P-time. •!How

While it is true that the class **P** is the set of all problems which have an algorithmic solution in polynomial time by a deterministic Turing machine, the class **NP** is the set of all problems which have an algorithmic solution in polynomial time by a non-deterministic Turing machine.. There are problems which can be solved in exponential time which are not in **NP**; **NP** is a proper **subset** **of**. Definition: A language B is NP-complete if: 1. B ∈NP 2. Every A in NP is poly-time reducible to B That is, A ≤ P B When this is true, we say B is NP-hard Last time: We showed 3SAT ≤ P CLIQUE ≤ P IS ≤ P VC ≤ P SUBSET-SUM ≤ P KNAPSACK All of them are in NP, and 3SAT is NP-complete, so all of these problems are NP-complete While P NP, it is still an open question whether NP P. We recognized a special class of problems inside NP, which are called NP-complete problems. They are the fundamental problems to tackle in order to solve P vs NP. We gave three examples of NP-complete problems (proof omitted): SAT, Partition, and 3-Partition. Our goal in this lecture is to. SUBSET-SUM: given a collection S={s 1s n} of integers and a target integer t, does a sub-collection of S sum to t? Show that KNAPSACK is NP-complete using a reduction from SUBSET-SUM. We restrict instances of KNAPSACK to have W=V, i.e., all values equal weights. In this case KNAPSACK is exactly the SUBSET-SUM problem, and hence is NP. ‰ P i2S ai • b P P i2S si • t i2S ci ‚ k P i2S si ‚ t X i2S si = t Suppose we have a Yes answer to the new problem, it means we can ﬁnd such a subset S µ [1;2;¢¢¢;n] that satisﬁes the left part of the deduction. Then this subset S is also a solution to the right part. So we must also have a Yes answer to the original problem

Why Is P A Subset Of NP? B. When Is The Satisfiability Problem In P? C. Why Is The Halting Problem Not In NP? Note: Please Provide Detailed Answers For Questions Above. Thanks. This question hasn't been answered yet Ask an expert. Questions related to the Theory of NP. 1). a rected graphs that contain a vertex cover of size ≤ p. Prove that for every p>0, FIXED ∈ P. Proof. The polynomial algorithm for the problem tests all subsets of size pin a given graph with nvertices. If any of the subsets is a vertex cover, the graph is accepted as a member of FIXED. The algorithm is indeed polynomial since there are O(np. This result leads to the conclusion that class P is a proper subset of class NP. This paper solves a long standing open problem of whether NP-complete problems could be solved in polynomial time on a deterministic Turing machine by showing that the indistinguishable binomial decision tree can be formed in a 3-SAT instance 5 Lecture 24: P=NP? 25 NP-Hardness Proofs • Conjecture: A has some property Y. • Proof by reduction from B to A: - Assume A has Y.Then, we know an M exists. - We already know B does not have property Y. - Show how to build S that solves B using M. • Since we know B does not have Y, but having S would imply B has Y, S cannot exist. Therefore, M cannot exist, and A doe NP-CompletenessofSubset-Sum problem Rahul R. Huilgol 11010156 Simrat Singh Chhabra 11010165 Shubham Luhadia 11010176 September 7, 2013 ProblemStatement IntheSUBSETSUMproblem,wearegivenalistofnnumbersA 1,...,A n and a number T and need to decide whether there exists a subset S ⊆[n] suchthat X i S A i=