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 NP-complete - Wikipedia, the free encyclopedia
To prove that an NP problem A is in fact an NP-complete problem it is sufficient to show that an already known NP-complete problem completely reduces to an A - see also 'tagging'
In complexity theory, the NP-complete problems are the most difficult problems in NP ("non-deterministic polynomial time") in the sense that they are the ones most likely not to be in P.
The easiest way to prove that some new problem is NP-complete is first to prove that it is in NP, and then to reduce some known NP-complete problem to it. /wiki/NP-complete   (1750 words)

 Complexity classes P and NP - Wikipedia, the free encyclopedia
Similarly, NP is the set of languages expressible in existential second-order logic — that is, second-order logic restricted to exclude universal quantification over relations, functions, and subsets.
This is in fact a common approach to attack problems in NP not known to be in P (see RP, BPP).
Furthermore, the result P= NP would imply many other startling results that are currently believed to be false, such as NP = co-NP and P= PH. /wiki/Complexity_classes_P_and_NP   (2857 words)

 Stas Busygin's NP-Completeness Page
Hence complete problems are hardest in their own classes and as they exist we may choose any of them to advance solving techniques for the entire class.
The concept of complete problems for a class is generalized to hard problems for the class by inclusion of all other problems, whose polynomial time algorithm gives polynomial time solvability for the class.
And it is also in NP as when someone gives us another list asserting it is a sorted version of the first one, we can easily look through it to check the order and compare elements to verify are they the same as initial. /npc.html   (896 words)

That is, an NP-complete problem p is one such that, roughly, if A_p solves p, then every q in NP can be solved by a variation on A_p that runs essentially as fast as A_p.
This completes the computation of f; and it is a polynomial-time computation (in fact, linear time), in the size of the graph.
all other problems q in NP can be "reduced in polynomial time" to solving p: there is a function f computable in polynomial-time such that for all inputs i to p, the correct answer to q(i) is yes iff the correct answer to p(f(i)) is yes. /class/spring2003/cmsc351/notes/complexity.html   (1866 words)

NP is the set of all decision problems with short proofs for "yes" answers.
Because of the formal definition of NP (see your book), it is also clear that there is no problem in P but not in NP.
It is hard to believe that all problems in NP could be solved in polynomial time, so it is strongly believed that NPC is not in P (implying P!= NP). /~krovetz/courses/ecs122a_Su02/lect08.txt   (1482 words)

 Clay Mathematics Institute
Specifically, an NP problem is said to be NP-complete if the existence of a polynomial time solution for that problem implies that all NP problems have a polynomial time solution.
However, if any NP-complete problem turns out to be of type P —- to have a polynomial time solution — - than NP must equal P. We therefore expect all NP-complete problems to be non-P, but no one can yet prove this.
NP is not the same as non-P. A problem is NP if you can check whether a proposed solution actually is a solution in polynomial time. /Popular_Lectures/Minesweeper   (2085 words)

 NP-Complete Problems
A compendium of NP optimization problems: a nice collection of NP Complete optimization problems.
(2") NP-Complete problems: the set of problems that are both NP and NP-Hard.
NP problems thus deal with decision problems as opposed to optimization problems. /yue/courses/csci5432/npcomplete.asp   (742 words)

The notion of NP-completeness, introduced by Cook [Coo71] and independently by Levin [Lev73], has provided a rigorous mathematical definition for measuring intractability of NP problems.
Completeness proofs of these problems are given in Sections 4, 5, and 6.
In particular, they showed that any NP search problem with a p-samplable distribution is reducible to an NP search problem with a uniform distribution under a randomized reduction. /mat/avg/avgnp/node2.html   (987 words)

 NP-Completeness and Origami
NP-complete problems have the property that every problem in NP can be converted into any NP-complete problem, which means that if you could knock off one of these incorrigibles in polynomial time, you could use the same approach to solve all NP problems and make millions of dollars along the way.
The traveling salesman problem and several others are in a special corner of NP, called NP-complete, which means that they are hard in a particular way.
The traveling salesman is in a class of problems, called NP for “nondeterministic polynomial time,” which may or may not be solvable in polynomial time, but whose solutions, once found, can be checked in polynomial time. /periodicals/EandS/articles/LXVII1/NPcompleteness.html   (844 words)

Note: A trivial example of NP, but (presumably) not NP-complete is finding the bitwise AND of two strings of N boolean bits.
Eppstein's longer, but very good introduction to NP-completeness, with sections like Why should we care?, Examples of problems in different classes, and how to prove a problem is NP-complete.
Informally, a problem is NP-complete if answers can be verified quickly, and a quick algorithm to solve this problem can be used to solve all other NP problems quickly. /dads/HTML/npcomplete.html   (288 words)

 Complexity Zoo - Qwiki
NP, and indeed NP intersect coNP, are not contained in BQP [BBB97].
The class of decision problems L in NP such that, if the answer is "yes," then a proof can be constructed in polynomial time given access only to an oracle for L. Contains NPC.
There exists a problem that is complete for E under polynomial-time Turing reductions but not polynomial-time truth-table reductions [Wat87]. /wiki/Complexity_Zoo   (6419 words)

TSP Given a complete graph G=(V, E), with edge costs, and an integer K, is there a simple cycle that visits all vertices and has total cost /courses/srollins/cs312/web/slides/np.ppt   (52 words)

 NP-Complete Problems
If a polynomial time solution is found for any of the NP-Complete problems, then every NP problem can be solved in polynomial time.
Or, are there problems that are in the class NP that are not in the class P. No one has been able to prove that they are or are not the same class.
Then, it must be possible to map or transform all other NP problems to this problem in a polynomial amount of time. /~karenv/cs101/lectures2/np.complete.html   (168 words)

 P, NP, CO-NP, NP-complete, NP-hard
The complexity class NP-complete is the set of problems that are the hardest problems in NP, in the sense that they are the ones most likely not to be in P. If you can find a way to solve an NP-complete problem quickly, then you can use that algorithm to solve all NP problems quickly.
NP is the set of decision problems solvable in polynomial time on a nondeterministic Turing machine.
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. /grads/z/Howard.Zhou/micellaneous/gre_cs_sub/np_complete.htm   (210 words)

So if we believe that P and NP are unequal, and we prove that some problem is NP-complete, we should believe that it doesn't have a fast algorithm.
But there are problems that are in NP, not known to be in P, and not likely to be NP-complete; for instance the code-breaking example I gave earlier.
A problem is in NP if you can quickly (in polynomial time) test whether a solution is correct (without worrying about how hard it might be to find the solution). /~eppstein/161/960312.html   (3273 words)

 CBofN - Glossary - N
This means that if a fast algorithm exists for an NP-complete problem, then any problem that is in NP can be solved with the same algorithm.
NP-Complete A problem type in which any instance of any other NP class problem can be translated to in polynomial time.
NP Nondeterministic polynomial time problems; a class of computational problems that may or may not be solvable in polynomial time but are expressed in such a way that candidate solutions can be tested for correctness in polynomial time. /books/FLAOH/cbnhtml/glossary-N.html   (581 words)

 CSC-105 2000S : What implications do you see in the existence of NP-complete problems?
If a problem is NP and all other NP problems are polynomial-time reducible to it, the problem is NP-complete.
Thus, finding an efficient algorithm for any NP-complete problem implies that an efficient algorithm can be found for all such problems, since any problem belonging to this class can be recast into any other member of the class.
A problem is called NP (nondeterministic polynomial) if its solution (if one exists) can be guessed and verified in polynomial time; nondeterministic means that no particular rule is followed to make the guess. /~rebelsky/CS105/Questions/question.42.html   (1297 words)

 Mathematical Programming Glossary - N
A problem, p, is NP-complete if it is NP and for any problem in NP, there exists a polynomial time algorithm to reduce it to p.
Among these problems that are not known to be in P (or in ~P), there is a subclass of problems known as NP-complete: those for which either all are solvable in polynomial time, or none are.
Formally, a problem is NP if there exists an algorithm with polynomial time complexity that can certify a solution. /~hgreenbe/glossary/N.html   (2146 words)

 P-versus-NP page
His paper is available at and seems to prove that P is not equal to NP, if "you are willing to believe that the property of a complexity class to be closed or openend to the nondeterministic extension operator it's an invariant of complexity theory".
His proof is not yet complete, but he states that what remains to be done is technical work along the lines of traditional combinatorics.
It shows that OWMF is in NP, but cannot be solved in polynomial time, since there is no faster algorithm other then exhaustive search for it. /~gwoegi/P-versus-NP.htm   (2412 words)

 Complexity Theory - NP and NP-completeness
NP is the class of problems with efficient `verifying' algorithms.
NP = { f : CHECK(f) is in P}
f is in NP) but for which f is not in P. /~ped/teachadmin/algor/npcomp.html   (1801 words)

 Computational Complexity: Favorite Theorems: Combinatorial NP-Complete Problems
Karp lists three NP problems whose classification was open at the time.
If Cook made the P versus NP question interesting to logicians, Karp made the question important to everyone else.
If P=NP then NP is closed under complementation and polynomial-bounded existential quantification. /2005/05/favorite-theorems-combinatorial-np.html   (382 words)

A problem in NP is NP-complete if every other problem in NP can be expressed in terms of it by means of a polynomial time algorithm.
NP-complete problems are considered to be the hardest problems, since if any problem in NP is shown to be intractable then all NP-complete problems are intractable.
In this paper, two DNA algorithms are presented for the Road Coloring problem, a problem that is in NP but not known to be NP-complete. /~jamie/.Refs/Courses/CS881/charlotte.html   (2332 words)

It is in both NP and co-NP, but is generally suspected to be outside P, outside NP-complete, and outside co-NP-complete.
The proof for the fact that no co-NP-complete problem can be in NP is symmetrical.
If a problem can be shown to be in both NP and co-NP, that is generally accepted as strong evidence that the problem is probably not NP-complete. /Co-NP.html   (393 words)

 NP-complete - Wikipedia, the free encyclopedia
Although the question of NP = co-NP is an open question it is considered unlikely and therefore it is also unlikely that the two definitions of NP-completeness are equivalent.
In complexity theory, the NP-complete problems are the most difficult problems in NP, in the sense that they are the ones most likely not to be in P.
To prove that a NP problem A is in fact a NP-complete problem we must show that an already known NP-complete problem reduces to A. /wiki/NP-complete   (1439 words)

Given a graph G and a positive integer K, is there a simple path in G of length (number of edges on path) at least K? (Note: You may not assume that the end points of the path are provided as part of the problem instance.) Show that the longest path is NP-Complete.
Note that membership in NP for FVS is not as trivial as for CLIQUE or VC -- this problem is on the practice midterm exam.
Thus the reduction consists of constructing a complete graph with K vertices, which can clearly be done in polynomial time. /~amk/foo/csci356/notes/ch11/HW11.html   (2036 words)

 Properties of NP-Complete Sets
We study several properties of sets that are complete for NP.
Moreover, we prove for every L \in NP - P, that there exists a sparse S \in EXP such that L- S is not \le _p^m-hard for NP.
Hence, removing sparse information in P from a complete set leaves the set complete, while removing sparse information in EXP from a complete set may destroy its completeness. /persagen/DLAbsToc.jsp?resourcePath=/dl/proceedings/&toc=comp/proceedings/ccc/2004/2120/00/2120toc.xml&DOI=10.1109/CCC.2004.1313839   (343 words)

 4.3 An NP-Complete Problem: Tiling.
Existence of w is an NP problem since w can be ``guessed'' non-deterministically and verified in P-time.
So, if some NP problem cannot be solved in P-time then neither can be the u-problem.
So, any P-time algorithm extending a given first line to the whole table of matching tiles from a given set would solve all NP problems by converting them to Tiling as shown. /fac/lnd/toc/z/node19.html   (370 words)

15-451 Algorithms 10/21/99 Complexity Theory* NP and NP-completeness* Reductions HW 4 grading today and tomorrow ============================================================================ Preliminaries ============= Poly-time reduction and equivalence ----------------------------------- Problem A is poly-time REDUCIBLE to problem B if given a poly-time alg for B, we can use to produce a poly-time alg for A (A
Version 2: NP is the set of problems such that YES instances have short proofs that the answer is yes. /afs/   (557 words)

 Citations: The Graph Genus Problem is NP-Complete - Thomassen (ResearchIndex)
By, genus testing is NP complete for general graphs.
The main result of Section 5 shows that the genus problem remains NP complete even if we restrict ourselves to apex graphs.
proved that the problem of determining whether a graph has genus g is NP complete. /context/21871/0   (2525 words)

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