Difference between NP-Hard and NP-Complete : NP-hard. NP-Complete. NP-Hard problems. @PeterRaeves All NP-complete problems are NP-hard, by definition: NP-complete = (NP and NP-hard). The inverse is not true: there are problems (such as the Halting Problem) in NP-hard that are not in NP-complete. NP (not solvable in polynomial time) -- that's not what NP means. NP is Non-deterministic-polynomial. All problems in P are also in NP ** NP HARD PROBLEM: NP COMPLETE PROBLEM : Description : NP-Hard problems (say X) can be solved if and only if there is a NP-Complete problem (say Y) can be reducible into X in polynomial time**. NP-Complete problems can be solved by deterministic algorithm in polynomial time. Solution : To solve this problem, it must be a NP problem NP-Complete Algorithms. The next set is very similar to the previous set. Taking a look at the diagram, all of these all belong to , but are among the hardest in the set. Right now, there are more than 3000 of these problems, and the theoretical computer science community populates the list quickly Reducibility- If we can convert one instance of a problem A into problem B (NP problem) then it means that A is reducible to B. NP-hard-- Now suppose we found that A is reducible to B, then it means that B is at least as hard as A. NP-Complete-- The group of problems which are both in NP and NP-hard are known as NP-Complete problem

NP-Complete means that a problem is both NP and NP-Hard. It means that we can verify a solution quickly (NP), but its at least as hard as the hardest problem in NP (NP-Hard). I don't really know what it means for it to be non-deterministic. Non-determinism is an alternative definition of NP The complexity class of problems of this form is called NP, an abbreviation for nondeterministic polynomial time. A problem is said to be NP-hard if everything in NP can be transformed in polynomial time into it even though it may not be in NP. Conversely, a problem is NP-complete if it is both in NP and NP-hard * Some people say the Degree Constrained Minimum Spanning Tree problem (DCMST) is NP-hard for a reason, other people say DCMST is NP-complete for a different reason*. As is explained in the other answers, the word NP-complete means that a problem belongs to NP and is NP-hard

The group of problems what are both NP and NP hard is called as NP Complete Problems. The intersection of NP and NP hard set of problems are called as NP complete. If you have any doubts, please write in the comment section below **NP-complete**. A problem that is **NP-complete** can be solved in polynomial time iff all other **NP-complete** problems can also be solved in polynomial time **NP-hard**. If an **NP-hard** problem can be solved in polynomial time then all **NP-complete** problems can also be solved in polynomial tim In computational complexity theory, NP-hardness (non-deterministic polynomial-time hardness) is the defining property of a class of problems that are informally at least as hard as the hardest problems in NP . A simple example of an NP-hard problem is the subset sum problem

$\begingroup$ @Cybernetic this is what I found on Wikipedia, so it seems like NP-Hard is more universal, since we reduce NP-Complete G to NP-Hard H. ``` Another definition is to require that there be a polynomial-time reduction from an NP-complete problem G to H.[1]:91 As any problem L in NP reduces in polynomial time to G, L reduces in turn to H in polynomial time so this new definition. NP -complete is deterministic+NP Hard If sat tendsto L then sat is NP hard and NP complete as well while L is NP hard but if for that : if in future someone writes Nondetermistic algorithm then it will become NP complete * NP-Complete *.* NP-Complete *is a complexity class which represents the set of all problems X in NP for which it is possible to reduce any other NP problem Y to X in polynomial time.. Intuitively this means that we can solve Y quickly if we know how to solve X quickly. Precisely, Y is reducible to X, if there is a polynomial time algorithm f to transform instances y of Y to instances x = f(y) of.

NP-Complete problems are problems that live in both the NP and NP-Hard classes. This means that NP-Complete problems can be verified in polynomial time and that any NP problem can be reduced to.. De nition 22.3 (NP-complete) A problem is NP-complete if it is in NP and is NP-hard. We can see that NP-complete problems are the hardest problem in NP. The reason is that if A is in NP, and B is a NP-complete problem, then A can be reduced to B. Therefore, if any NP-complete problem has a polynomial time algorithm, then P = NP Polynomial Time Algorithms are the algorithms that takes polynomial time, that is the problem for which we know the algorithm which takes some particular polynomial time while, Exponential Tim NP-Complete - A Rough Guide. This is a rough guide to the meaning of NP-Complete. It is not intended to be an exact definition, NP-Hard: as hard as any NP-problem, or maybe harder. Anyway, I hope this quick and dirty introduction has helped you. NP-Hard and NP-Complete Problems. Let's classify the decision problems - problems that have the Yes or No answers. 2.1. P and NP. The problem belongs to class if it can be solved in polynomial time

** P vs NPSatisfiabilityReductionNP-Hard vs NP-CompleteP=NPPATREON : https://www**.patreon.com/bePatron?u=20475192CORRECTION: Ignore Spelling MistakesCourses on U.. NP-complete problems have no known p-time solution, considered intractable. Tractability Difference between tractability and intractability can be slight Can find shortest path in graph in O(m + nlgn) time, but finding longest simple path is NP-complete hard to it (e.g., NP-complete problem) NP Hard and NP Complete ProblemsWatch More Videos at: https://www.tutorialspoint.com/videotutorials/index.htmLecture By: Mr. Arnab Chakraborty, Tutorials Poi.. NP-complete is just the name given to the intersection of NP and NP-hard, that is, problems in NP that all other problems in NP can be reduced to. I've never heard the term NP-incomplete before, and I don't think it's in common usage

Students believe that every problem assigned to them is NP-complete in diﬃculty level, as they have to ﬁnd the solutions. Teaching Assistants, on the other hand, ﬁnd that their job is only as hard as NP, as they only have to verify the student's answers. When some students confound the TAs, even veriﬁcation becomes hard NP-Hard and NP-Complete Problems An algorithm A is of polynomial complexity is there exist a polynomial p( ) such that the computing time of A is O(p(n)). Definition: P is a set of all decision problems solvable by a deterministic algorithm in polynomial time NP Certification algorithm intuition. ・Certifier views things from managerial viewpoint. ・Certifier doesn't determine whether s ∈ X on its own; rather, it checks a proposed proof t that s ∈ X. Def. Algorithm C(s, t) is a certifier for problem X if for every string s, s ∈ X iff there exists a string t such that C(s, t) = yes. Def. NP is the set of problems for which there exists a. NP: is the set of decision problems that can be verified in polynomial time. NP-Hard: L is NP-hard if for all L' ϵ NP, L' ≤p L. Thus if we can solve L in polynomial time, we can solve all NP problems in polynomial time. NP-Complete L is NP-complete if. L ϵ NP and ; L is NP-hard

- Chapter-8 NP hard and NP Complete problems 8.1 Basic Concepts The computing times of algorithms fall into two groups. Group1- consists of problems whose solutions are bounded by the polynomial of small degree. Example - Binary search o (log n) , sorting o(n log n),.
- 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). NP-complete problems are the hardest problems in NP set. A decision problem L is NP-complete if
- All NP-complete problems are NP-hard, but all NP-hard problems are not NP-complete. An NP-complete problem has the property that any problem in NP can be polynomially reduced to (partial order ) it. Any instance of P 1 can be polynomially reduced/transformed to an instance of P 2
- A problem is NP Complete if problem is both: 1)NP Hard and 2)N
- istic Turing machine in polynomial time, the problem belongs to the complexity class P. All problems in this class have a solution whose time requirement is a polynom on the input size n. i.e. f(n) is of form a k n k +a k−1 n k.
- istic. A decision problem is NP-hard when there exists a polynomial-time many-one reduction of any NP problem to the current NP hard problem.. Basically, to prove a problem NP hard we need to reduce it to a problem which is already labelled NP hard

No. Not all NP-hard problems are NP-complete. The halting problem is NP-hard, for instance. Perhaps more pertinently, there are NP-hard problems that are decidable, e.g. circuit minimization, that are not known to be in $\Sigma^P_1$ and so cannot be reduced to an NP-complete problem in polynomial-time Since Minesweeper Consistency has been shown to be in **NP** **and** **is** **NP** **Hard**, by de nition, it is **NP** **Complete**. Sources: A very approachable paper and summary on both the Minesweeper consistency. In layman's terms, a problem is said to be NP-hard if discovery of a fast algorithm to solve the problem means that there will be fast algorithms to solve all NP-hard problems. For a more technical explanation, keep reading. Computer scientists ch.. P is the set of decision problems solvable in time polynomial in the size of the input, where time is typically measured in terms of the number of basic mathematical operations performed. An example would be basic multiplication of two numbers. Ev.. More NP-Complete and NP-hard Problems Traveling Salesperson Path Subset Sum Partition . 2 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

np-complete vs np-hard: Comparison between np-complete and np-hard based on user comments from StackOverflow. If there exists a np-hard problem that is not in np to the best of my knowledge no such problem has been proved to fall in this category at this moment of time such problem is harder than np-complete problems. As i understand it. NP-complete problems are the problems that are both NP-hard, and in NP. Proving that a problem is NP is usually trivial, but proving that a problem is NP-hard is not. Boolean satisfiability (SAT) is widely believed to be NP-hard, and thus the usual way of proving that a problem is NP-complete is to prove that there's a polynomial time transformation of the problem to SAT NP-hard NP-complete NP P Figure 22.1: Relationships between P,NP,NP-complete and NP-hard (When P6=NP) an NP-complete problem can be reduced to it; if we can do that, we know all problems in NP can be reduced to it by the de nition of NP-hard. Which NP-complete problem, then, should be reduced to it NP-HARD AND NP-COMPLETE fNON DETERMINISTIC ALGORITHMS. Algorithms with following property are termed deterministic algorithm. The result of... fThe classes NP- hard and NPcomplete. An algorithm A is of polynomial complexity if there exist a polynomial P () such... fCommonlyb belived relationship. Therefore your problem is not NP-complete, nor NP-hard, as flow problems can be solved in polynomial time. Share. Improve this answer. Follow edited Oct 26 '20 at 10:46. answered Oct 26 '20 at 9:25. Kuifje Kuifje. 6,036 1 1 gold badge 7 7 silver badges 26 26 bronze badge

Question: Explain What We Mean By NP, NP-hard, NP- Complete And P What Is The Difference Between NP-hard And NP- Complete? This problem has been solved! See the answer. Show transcribed image text. Expert Answer 100% (1 rating) P:Polynomial time solving P is a set of problems which can be solved in polynomial time hard as every problem in NP. NP X Y1 Y2 Y3 P Y4. NP-completeness and P=NP Theorem If X is NP-complete, then X is solvable in polynomial time if and only if P = NP. You can prove a problem is NP-complete by reducing a known NP-complete problem to it. We know the following problems are NP-complete: Vertex Cover Independent Se P, NP, NP Hard, NP Complete. Focus of this lecture = get the definitions across. This will be a part of midterm2. Only some small part of the next lecture will be part of the midter

- NP-complete problem, any of a class of computational problems for which no efficient solution algorithm has been found. Many significant computer-science problems belong to this class—e.g., the traveling salesman problem, satisfiability problems, and graph-covering problems.. So-called easy, or tractable, problems can be solved by computer algorithms that run in polynomial time; i.e., for a.
- 本文介绍了NP，NP难，NP完全问题的区别。如果所有NP问题都可以多项式归约到问题A，那么问题A就是 NP-Hard；如果问题A既是NP-Hard又是NP，那么它就是NP-Complete
- 其中，P, NP, NP-Hard, NP-Complete是不同的复杂性类，用于将所有的算法问题进行分类，以确定当前算法的难度。 多项式时间可解的问题 ：如果对于某个确定的常数k,存在一个能在O(n k )时间内求解出某具体问题的算法，就说该具体问题是一个多项式时间可解问题
- istic polynomial time). The difference between these two can be huge. If a P algorithm has 100 elements, and its time to complete working is proportional to N 3 , then it will solve its problem in about 3 hours
- istic polynomial) är en klass av matematiska problem för vilka effektiva lösningar saknas.Den enda lösningen man funnit på ett godtyckligt NP-fullständigt problem är i princip att gå igenom alla tänkbara lösningar och jämföra dem, vilket är ogörligt för andra än små proble
- What is NP-Hard Problem? Answer: This is a problem to which a NP-complete problem is polynomial time Turing reducible. If a problem X in NP-complete is polynomial time Turing reducible to problem Y, problem Y is NP-hard problem
- An NP-Hard problem is any problem, not necessarily in NP that can be reduced to an NP-complete problem in a reasonable i.e polynomial time. In this sense NP-Hard problems are as least as hard as NP problems but they could be harder

- istic Turing machine with an oracle fo
- Np complete 1. NP-Complete Problems Dr. C.V. Suresh Babu 2. • Problems that What if aCross the Line problem has: o An exponential upper bound o A polynomial lower bound • We have only found exponential algorithms, so it appears to be intractable
- I've found this Venn Diagram to be incredibly helpful in understanding the differences between P, NP, NP-Complete, and NP-Hard. You're right that Traveling Salesman would not be considered NP-Complete. It's NP-Hard, but not in NP
- 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
- P, NP, NP-Complete, NP-Hard July 9, 2016 by arjun Leave a Comment It might be because of the name but many graduate students find it difficult to understand $\textsf{NP}$ problems

Save Save Np Hard and Np Complete Problems For Later. 0% 0% found this document useful, Mark this document as useful. 0% 0% found this document not useful, Mark this document as not useful. Embed. Share. Print. Related titles. Carousel Previous Carousel Next And so NP-complete is a nice answer because this says you're exactly as hard as everything in NP--no harder, no easier. If you draw, in this vague sense, computational difficulty on one axis-- which is not really accurate, but I like to do it anyway-- and you have P is all of these easy problems down here Minesweeper and NP-completeness Minesweeper is NP-complete! My original paper appeared under this title in the Spring 2000 issue of the Mathematical Intelligencer (volume 22 number 2, pages 9--15).. It was discussed by Ian Stewart in the Mathematical Recreations column in the Scientific American, in October 2000, and has been discussed in newspapers in the USA (including the Boston Globe on. A NP-hard problem is a problem that is at least as hard as the hardest problem in NP (hard meaning time taken to solve for the best algorithm) The travelling salesman is an NP-complete problem, which happens to be easy to understand, so it's used in a lot of examples

对NP-Hard问题和NP-Complete问题的一个直观的理解就是指那些很难（很可能是不可能）找到多项式时间算法的问题. 因此一般初学算法的人都会问这样一个问题: NP-Hard. A problem is NP-Hard iff a polynomial-time solution for it would imply a polynomial-time solution for every problem in NP. A problem is NP-Complete iff it is NP-Hard and it is in NP itself. Lists. Lots of folks have made lists of NP-Complete and NP-Hard Problems. Here are a few: Wikipedia; Paul Dunne's Annotated List; A compendium by Viggo. NP-Hardness A language L is called NP-hard iff for every L' ∈ NP, we have L' ≤ P L. A language in L is called NP-complete iff L is NP-hard and L ∈ NP. The class NPC is the set of NP-complete problems. P NP NP-Hard NP Note that NP-Complete problems are also NP-hard. However not all NP-hard problems are NP (or even a decision problem), despite having 'NP' as a prefix. That is the NP in NP-hard does not mean 'non-deterministic polynomial time'. Yes this is confusing but its usage is entrenched and unlikely to change

NP-complete Problems and Physical Reality ﬁnding a Steiner tree is NP-hard, (2) soap bubbles ﬁnd a Steiner tree in polynomial time, (3) soap bubbles are classical objects, and (4) classical physics can be simulated by a Turing machine with polynomial slowdown, it follows that P = NP What We Know 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. Interesting read: Interviews with leading thinker Adjective []. NP-complete (not comparable) (computing theory, of a decision problem) That is both NP (solvable in polynomial time by a non-deterministic Turing machine) and NP-hard (such that any (other) NP problem can be reduced to it in polynomial time).2001, Thomas H Cormen, Charles E Leiserson, Ronald L Rivest, Clifford Stein, Introduction To Algorithms, The MIT Press, 2nd Edition, page 968 NP-hard problems don't have to be decision problems, and they do not have to be in NP. Formally: A problem X is NP-hard, if there is an NP-complete problem Y , such that Y is reducible to X in polynomial time

5 NP -HARD AND NP -COMPLETE PROBLEMS •Group2 -contains problems whose best known algorithms are non polynomial. •Example -Traveling salesperson problem 0(n22n), knapsack problem 0(2n/2) etc. •There are two classes of non polynomial time problems 1) NP- hard Complexity Classes. Definition of NP class Problem: - The set of all decision-based problems came into the division of NP Problems who can't be solved or produced an output within polynomial time but verified in the polynomial time.NP class contains P class as a subset. NP problems being hard to solve. Note: - The term NP does not mean not polynomial NP-complete is a class of problems which are in NP and are NP-hard. NP-complete problems transform to each-other by polynomial-time many-one reductions so if a polynomial-time algorithm exists for any one of them, then polynomial algorithms exist for all of them Modified by Dr. ISSAM ALHADID 15/4/2019 Decision and Optimization Problems. Polynomial Time vs. Non-Polynomial Time. Three General Problem Categories. The Sets P and NP. NP-Complete Problems. NP-Hard Problems. NP-Hard vs. NP-Complete Problems. Most of the problems we have dealt with thus far in the class are optimization problems, in which the goal is to find the maximum or minimum value of a. Or as Nate mentioned, NP-hard problems might be impossible to solve in general. You can tell the difference in difficulty between NP-hard and NP-complete problems because the class NP includes everything easier than its toughest problems--if a problem is not in NP, it is harder than all the problems in NP

Another **NP-complete** problem is to decide if there exist k star-shaped polygons whose union is equal to a given simple polygon, for some parameter k. The optimization problem, i.e., finding the minimum number (least k) of star-shaped polygons whose union is equal to a given simple polygon, is **NP-hard** - A problem X is NP-hard if there is an NP-complete problem Y that can be polynomially Turing reduced to it (i.e., Y P T X) - Note that any polynomial-time algorithm for X would translate into one for Y. Since Y is NP-complete, this would imply that P = NP A problem is NP-hard if every problem in NP can be Cook-reduced to it. 6 An oracle for B is a magical black-box which solves any instance of problem B in one step. A Cook reduction is Indeed, the condition of the second statement is vacuously satisfied by many problems (such as all the NP-hard ones), but that doesn't make them all NP-complete. For a concrete example, the trivial empty language is not NP-complete, because there's no way to run a mapping-reduction to the empty problem from any other problem

- No, its not known to be NP-complete, and it would be very surprising if it were. This is because its decision version is known to be in $\text{NP} \cap \text{co-NP}$
- istic Turing machine
- Status: NP-complete even if all tile positions are known. PSPACE-hard even to approximate the strategy with maximum probability of success when some tiles are covered and unknown. #P-complete to count solutions (personal communication from Michiel de Bondt)

NP-Complete -- The group of problems which are both in NP and NP-hard are known as NP-Complete problem. Now suppose we have a NP-Complete problem R and it is reducible to Q then Q is at least as hard as R and since R is an NP-hard problem. therefore Q will also be at least NP-hard , it may be NP-complete also Since an NP-complete problem, by definition, is a problem which is both NP and NP-Hard, the proof or statement that a problem is NP-Complete consists of two parts: The problem itself is NP-Complete. All other problems in NP class can be polynomial-time reducible to that. (B is poly-time reducible to. co−NP NP−hard NP NP−complete More of what we think the world looks like. 16.4 Reductions (again) and SAT To prove that a problem is NP-hard, we use a reduction argument, exactly like we're trying to prove a lower bound. So we can use a special case of that statement about reductions the you tattooe Lecture Notes CMSC 251 Harder NP−Hard NP NP−Complete P Easy Figure 36: Relationship between P, NP, and NP-complete. problem A in polynomial time. You want to prove that B cannot be solved in polynomial time. Ho

de ne NP-completeness precisely, and give examples of NP-complete problems. We show how to prove that a problem is NP-complete, and give some help for coping with NP-completeness. However, many sources take the term \NP-hard to refer to Cook reducibility. Many important languages are now known to be NP-complete. Before we get to them, le NP-complete and NP-hard problems - NP-complete and NP-hard problems | PowerPoint PPT presentation | free to view . How to solve polluted Water Problems In Agricultural? - Hard water from a bore (well) normally has very excessive floor tension NP -HARD AND NP - COMPLETE PROBLEMS Under this topic we are concerned with distinction between problems that can be solved by a polynomial time algorithm and problems for which no polynomial time algorithm is known. Problems can be divided into two groups: Group1- problems for which solution can be found in polynomial time ie time complexity is bound by a polynomial of small degree NP-complete (complexity) (NPC, Nondeterministic Polynomial time complete) A set or property of computational decision problems which is a subset of NP (i.e. can be solved by a nondeterministic Turing Machine in polynomial time), with the additional property that it is also NP-hard. Thus a solution for one NP-complete problem would solve all problems in.

NP-hard if it is at least as hard as all problems in NP, and NP-complete if it is in NP and it is also NP-hard. An unproven but generally accepted conjecture, which we will assume is true, is that P 6= NP. 2The restriction that the range of the objective function is R might be limiting in some situations, but mos Lecture NP-Completeness Spring 2015 • A problem X is NP-hard if every problem Y ∈ NP reduces to X. - If P =NP,then X/ ∈ P. • A reduction from proble Note: A trivial example of NP, but (presumably) not NP-complete is finding the bitwise AND of two strings of N boolean bits. The problem is NP, since one can quickly (in time Θ(N) ) verify that the answer is correct, but knowing how to AND two bit strings doesn't help one quickly find, say, a Hamiltonian cycle or tour of a graph ** NP-Completeness is probably one of the more enigmatic ideas in the study of algorithms**. NP stands for nondeterministic polynomial time, and is the name for what is called a complexity class to which problems can belong.The important thing about the NP complexity class is that problems within that class can be verified by a polynomial time algorithm

In this post, we will prove that the decision version of the set-covering problem is NP-complete, using a reduction from the vertex covering problem (which is NP-complete). This gives us that we can verify that and that covers in , which proves the problem is in NP. Showing the problem is NP-hard An equivalent de nition is that a problem is NP Complete if it is in NP and is in NP Hard. 1. In order to prove that p is in NP, we need to show that it has a polynomial length certi cate which can be veri ed in polynomial time. We can show that a problem p is NP-Hard by reducin NP-hard, NP-complete definition X is NP-hard problem iff Y reduce to X with polynomial time, for each Y belongs to NP. X is NP-complete problem iff Y reduce to X with polynomial time, for each Y belongs to NP and X is NP. ( this implys that NP-hard. A set or property of computational search problems. A problem is NP-hard if solving it in polynomial time would make it possible to solve all problems in class NP in polynomial time. Some NP-hard problems are also in NP (these are called NP-complete), some are not. If you could reduce an NP problem to an NP-hard problem and then solve it in polynomial time, you could solve all NP problems NP-Hard: A collection of problems that do not have to be in NP, whose solutions are at least as hard as the NP-Complete problems. If a problem is in NP, and it's NP hard, then it is also NP-Complete. In this lecture, we are going to see what it takes to prove that problems belong to these sets

** NP complete problems are problems such that, with some simple steps, any other NP problem can be converted into this problem**. Thus, if you solve any NP-complete problem, all other NP problems come as a 'freebie' (not just the NP-complete ones) \NP-Complete, this will imply that if Lis in P, then P = NP; we will take this as being very strong evidence that L=2P. NP-Completeness Proof: Say that S is (Turing)-NP-Hard and solvable in polynomial time. Consider an arbitrary language L02NP. We have L0!p Sand Sis solvable in polynomial time, so by Theorem 2 in the last set of notes, L02P Tensor rank is NP-complete for any finite field and NP-hard for the rational numbers. Proof. First observe that it is easy to verify the problem is in NP for a finite field, since we have no trouble guessing the vectors ve1)

- 1.12 Showing Other Problems NP Complete Once we have one NP-complete problem we can obtain more using the fol-lowing lemma: Lemma 36.8 If L′ is NP-complete and L′ ≤ P L, then L is NP-hard. If L is also in NP, then L is NP-complete
- NP-Complete는 간단합니다. NP-Hard이면서 NP클래스 안에 있으면 NP-Complete입니다. NP클래스에 속해있으면서, 어떤 기준에 의해 가장 어려운 문제이면 NP-Complete인거죠
- e whether it is also NP-complete or not. = (a) One-Fourth-Path Problem: Given an undirected graph G passes through at least one forth of the vertices in G

Let S be an NP-complete pr oblem and Q and R be two other pr oblems not known t o be in NP. Q is polynomial time reducible to S and S is polynomial-time r educible to R 什么是NP问题概念1:在计算机学科中，存在多项式时间的算法的一类问题，称之为P类问题；而像梵塔问题、推销员旅行问题、（命题表达式）可满足问题这类，至今没有找到多项式时间算法解的一类问题，称之为NP类问题。概念2:多项式时间（Polynomial time）在计算复杂度理论中，指的是一个问题的计算. TSP is as hard as the NP problems, but TSP is not in NP since a solution to TSP can't be checked in polynomial time. A version of the problem that has easy to check solutions is the TS decision problem: given a graph and a distance d , is there a route whose length is less than

En teoría de la complejidad computacional, la clase de complejidad NP-hard (o NP-complejo, o NP-difícil) es el conjunto de los problemas de decisión que contiene los problemas H tales que todo problema L en NP puede ser transformado polinomialmente en H.Esta clase puede ser descrita como aquella que contiene a los problemas de decisión que son como mínimo tan difíciles como un problema.

倾向于接受NP完全问题（NP-Complete或NPC）和NP难题（NP-Hard或NPH）不存在有效算法这一猜想，认为这类问题的大型实例不能用精确算法求解，必须寻求这类问题的有效的近似算法

- An NP problem is an algorithmic problem such that if you have a case of the problem of size , the number of steps needed to check the answer is smaller than the value of some polynomial in .It doesn't mean one can find an answer in the polynomial number of steps, only check it. An NP-complete problem is an NP problem such that if one could find answers to that problem in polynomial number of.
- 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
- Many problems are NP-hard in theory, but practical instance are typically easy to solve or approximate. It's not clear if typical instances of the Bitcoin block problem are hard. Double-spends a relatively rare, as is publishing a transaction in the same block as its parent transaction, so the maximum independent set problem is probably easy in practice today
- This project investigates the potential of computers to solve complex tasks such as games. The paper proves that the complexity of a generalized version of spider solitaire is NP-Complete and uses much of structure of the proof that FreeCell is NP-Hard in the paper Helmert, M. Complexity Results for Standard Benchmark Domains in Planning. Artificial Intelligence 143.2 (2003): 219-62. Print.
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