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NP-hard - Wikipedia, the free encyclopedia It is also easy to see that the halting problem is not in NP since all problems in NP are decidable and the halting problem is not. This intuition is supported by the fact that if we can find an algorithm A that solves one of these problems H in polynomial time, we can construct a polynomial time algorithm for any problem L in NP by first performing the reduction from L to H and then running the algorithm A. If it is also the case that L is in NP, then L is called NP-complete. en.wikipedia.org /wiki/NP-hard   (507 words)

NP-easy - Wikipedia, the free encyclopedia In other words, a problem X is NP-easy if and only if there exists some problem Y in NP such that X is polynomial-time Turing reducible to Y. This means that given an oracle for Y, there exists an algorithm that solves X in polynomial time (possibly by repeatedly using that oracle). In complexity theory, the complexity class NP-easy is the set of function problems that are solvable in polynomial time by a deterministic Turing machine with an oracle for some decision problem in NP. An example of an NP-easy problem is the problem of sorting a list of strings. en.wikipedia.org /wiki/NP-easy   (266 words)

Luboš Motl's reference frame: NP-hard landscape problems You know that it is hard to prove that some apparently difficult problems are NP (non-polynomial), which is the source of the open questions about the so-called NP-completeness, so I am curious how the proof roughly looks like. I think NP is not "non-polynomial" but rather "nondeterministic-polynomial" -- a problem in NP is one which can be solved in polynomial time by a computer which is allowed to branch nondeterministically. The NP = P? problem is one of the seven milinium math problems and there is no excuse not knowing what it is. motls.blogspot.com /2006/01/np-hard-landscape-problems.html   (676 words)

np.vs.p A problem X is NP-hard if it is hard enough that every problem in NP can be solved in polynomial time _if_ you only could solve problem X in polynomial time. This is what is called a "NP-complete" problem: a NP problem such that, if it is in P, then NP = P. If a (not necessarily NP) problem has this same property then it is called "NP-hard". This means that all NP-complete problems are roughly equally hard to solve. www.math.niu.edu /~rusin/known-math/97/np.vs.p   (982 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. 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 is the set of decision problems solvable in polynomial time on a nondeterministic Turing machine. www-static.cc.gatech.edu /grads/z/Howard.Zhou/micellaneous/gre_cs_sub/np_complete.htm   (210 words)

NP- NP-complete problems are hard not because they are in class NP, but because they are the hardest problems in class NP. This is generalized from the computer-science terms NP-hard and NP-complete; NP-complete problems all seem to be very hard, but so far no one has found a proof that they are. NP is the set of Nondeterministic-Polynomial problems, those that can be completed by a nondeterministic Turing machine in an amount of time that is a polynomial function of the size of the input; a solution for one NP-complete problem would solve all the others. www.catb.org /~esr/jargon/html/N/NP-.html   (146 words)

lect08.txt 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). Thus, once we know of a hard problem P (one that is NP-complete), then any problem we can reduce P to must also be hard (NP-complete) "circuit-satisfiability" is a decision problem that can be reduced to by any problem in NP. NP is the set of all decision problems with short proofs for "yes" answers. www.cs.ucdavis.edu /~krovetz/courses/ecs122a_Su02/lect08.txt   (1482 words)

Mathematical Programming Glossary - N In that sense, NP-hard problems are at least as hard as NP-complete problems. 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. Formally, a problem is NP if there exists an algorithm with polynomial time complexity that can certify a solution. carbon.cudenver.edu /~hgreenbe/glossary/N.html   (2146 words)

NP-Complete Problems (2") NP-Complete problems: the set of problems that are both NP and NP-Hard. (2') NP-hard problems: the set of all problems L such that any NP problems can be reduced to L in polynomial times. NP problems thus deal with decision problems as opposed to optimization problems. dcm.cl.uh.edu /yue/courses/csci5432/npcomplete.asp   (742 words)

Stas Busygin's NP-Completeness Page 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. 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. 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. www.busygin.dp.ua /npc.html   (896 words)

NP-complete Definition: The complexity class of decision problems for which answers can be checked for correctness, given a certificate, by an algorithm whose run time is polynomial in the size of the input (that is, it is NP) and no other NP problem is more than a polynomial factor harder. 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. Note: A trivial example of NP, but (presumably) not NP-complete is finding the bitwise AND of two strings of N boolean bits. www.nist.gov /dads/HTML/npcomplete.html   (288 words)

The Theory of NP-Completeness Because NP is such a large class of problems, most NP-hard problems you encounter will in fact be complete, and the issue can always be settled by giving a (usually simple) verification strategy for the problem. Through a complicated proof, it has been established that satisfiability is at least as hard as any problem in NP. For a variety of social and technical reasons, it is well accepted that satisfiability is a hard problem, one for which no worst-case polynomial-time algorithm exists. www2.toki.or.id /book/AlgDesignManual/BOOK/BOOK3/NODE111.HTM   (438 words)

The Geomblog: Minimum Weight Triangulation is NP-hard MWT is still not known to be in NP, because it is not known whether the number of bits needed to express edge lengths in an optimal solution is polynomial in the input size. They claim that their result can be extended to the case when edge weights are "rounded", and thus show NP-Completeness for this case. If it all checks out, it's a major result, resolving the first problem on the CG Open Problems list, as well as whittling down to two the number of problems from Garey and Johnson's original list of geomblog.blogspot.com /2006/01/minimum-weight-triangulation-is-np.html   (489 words)

Brian's Digest: NP Problems Whether your problem is in NP is a different question: you have to verify whether a nondeterministic "guess" could be verified in polynomial time. It is quite possible that the appropriate mapping from a particular NP complete problem to the general solution procedure would map some of the practical cases onto the intractable 'special cases'. Whether your problem is in NP is a ->different question: you have to verify whether a nondeterministic ->"guess" could be verified in polynomial time. www.worms.ms.unimelb.edu.au /digest/np_problems96.html   (4667 words)

P - NP NP is said to be a class of intractable problems since we cannot find the certificate Y efficiently (in polynomial time). NP  : A decision problem Ø is in NP if and only if there exists a polynomial time algorithm A such that, X is a "Yes" instance of Ø if and only if there exists some Y (Y is polynomial in the size of the input X, Y k Note: if A belongs to NP, then problem B also has to be in NP, else using the above reduction we can solve A in polynomial time. lass.cs.umass.edu /~purukulk/theory/p-np.htm   (413 words)

NPC-2.txt A problem A is NP-Hard if EVERY problem in NP has a reduction to A. A problem A is NP-Complete if it is NP-Hard AND it belongs in NP. The hard part is show that EVERY other problem in NP reduces to CSAT. Recap: NP is the class of decision problems whose YES instances have polynomial-time checkable proofs. www.cs.ucsb.edu /~suri/cs130b/NPC-2.txt   (584 words)

G12FCO -- Formal Computation, lecture 13 Clearly, an enumeration problem is at least as hard as the corresponding decision problem. If, against expectations, it turns out that NP = P, then we will know that there is a polynomial-time solution to your problem, and it is extremely likely that the proof will show us some way to construct it. If NP /= P, then there is no efficient way to solve your problem, except by having the luck to guess a solution. www.maths.nott.ac.uk /personal/anw/G12FCO/lect13.html   (1556 words)

P-versus-NP page His paper is available at http://arxiv.org/abs/cs.CC/0411033 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". 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. The title of the paper is "Proving that P is not equal to NP and that P is not equal to the intersection of NP and co-NP" www.win.tue.nl /~gwoegi/P-versus-NP.htm   (2412 words)

NP In the meantime, some pages at this point that are interesting: The Art of Proving Hardness Steven Skiena's hints and another interesting page on NP P and we know that the reduction to Q is in polynomial time, then for any nondeterministic decision algorithm A in NP we can obtain a deterministic Z in P. Hence, if SAT is in P, then P=NP. We have already seen that satisfiability is in NP www.ecst.csuchico.edu /~amk/foo/csci356/notes/ch11/NP5.html   (1373 words)

lecture23.txt Indeed, optimization versions of NP-hard problems have widely different behavior with respect to approximation even though the decision versions all seem to be "as hard" as each other. Today, we will insist on satisfying the polynomial-time requirement (wish) for NP-hard problems but will relax the ``correctness'' requirement so as to be able deal with NP-hard problems in `real life'. Speaking of `real life', although last week we restricted our discussions to decision problems as the theory of NP, P, and NP-completeness is stated in terms of decision problems, in actuality many of the decision problems we talked about show up as `optimization problems'. theory.lcs.mit.edu /classes/6.046/spring04/lectures/lecture23.txt   (1094 words)

CSCI 3104 Class notes Page 19 The only example of an NP-hard problem that is not inside NP discussed in class was the "Halting Problem," which is the problem of deciding for a given program and input whether or not that program on that input will run into an infinite loop (execute forever). A problem is "NP-hard" if all problems in NP can be transformed into it. This is a classical problem that has been known for a long time to be unsolvable by any program. www.cs.colorado.edu /~karl/3104.fall95/19.html   (527 words)

lesson1.txt A language is in NP if there exists a Nondeterministic Turing Machine with time bounded polynomially in the size of the input, which can decide the language. This can be stated formally -- the verifier model described states that instance I belonging to a language L in NP has a short proof, but yet we conjecture that these proofs cannot be constructed in (guaranteed, deterministic) polynomial time. We conjecture that P not equal to NP. www.cs.ucla.edu /classes/winter04/cs280CO/lesson1.txt   (1210 words)

Re: ID/IDREF makes XML generation NP-hard So finding a valid instance is as hard a > problem as solving the 3SAT problem, and 3SAT is known to be NP-hard. However, the NP completeness result is not true if we do not allow #FIXED attributes. Next by Date: [off-topic] Re: If XML is too hard for a programmer, perhaps he'd b e better off as a crossing guard www.stylusstudio.com /xmldev/200303/post00840.html   (398 words)

WordSearch is NP-hard A boolean circuit is a graph made of wires, logical gates, etc; that produces a boolean output value from boolean input values. PROOF: ------ The WordSearch decision problem is obviously in NP, as any claimed solution can be checked in polynomial time. fvdp.homestead.com /files/potm_ws_proof.html   (1062 words)

Slashdot Tetris Is Hard: NP-Hard We study the offine version because its hardness captures much of the difficulty of playing Tetris; intuitively, it is only easier to play Tetris with complete knowledge of the future, so the difficulty of playing the offine version suggests the diffculty of playing the online version. The Compendium of NP Optimization Problems [nada.kth.se] is a great place to look for real world examples of NP problems. This has no bearing on the NP completeness of the problem. slashdot.org /articles/02/10/24/2251234.shtml?tid=127   (5713 words)

10.5.1 NP-Hardness and NP-Completeness Informally, an NP-hard problem is a problem that is at least as hard as any problem in NP. Similarly, if any problem in NP is not polynomial-time solvable, then no NP-complete problem will be polynomial-time solvable. Thus NP-completeness is at the crux of deciding whether or not NP = P. lcm.csa.iisc.ernet.in /dsa/node228.html   (302 words)

NP-complete from FOLDOC Many (but not all) naturally arising problems in class NP are in fact NP-complete. (NPC, non-deterministic Polynomial time complete) A set or property of computational decision problems which is a subset of NP (i.e. Thus a solution for one NP-complete problem would solve all problems in NP. lgxserver.uniba.it /lei/foldop/foldoc.cgi?NP-complete   (145 words)

Perl Regular Expression Matching is NP-Hard First, we need to show that it is NP-hard; the proofs on this page do that. Second, we need to show that regex matching is in NP. Return to: Universe of Discourse main page perl.plover.com /NPC   (381 words)

Protein Design is NP-hard -- Pierce and Winfree 15 (10): 779 -- Protein Engineering Design and Selection NP: Contains problems for which polynomial-time algorithms are Step 1: that PRODES belongs to the class NP. we now know that protein design is as hard as any problem in peds.oxfordjournals.org /cgi/content/full/15/10/779   (2015 words)

A compendium of NP optimization problems This is a continuously updated catalog of approximability results for NP optimization problems. The compendium is also a part of the book Graph Theory: Vertex Ordering, Network Design: Cuts and Connectivity. www.nada.kth.se /~viggo/wwwcompendium/wwwcompendium.html   (82 words)

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