Co-NP - Revision history

en>Lim Wei Quan: Please don't rephrase what you don't understand! It became incorrect!

2014-02-23T12:36:40Z

Please don't rephrase what you don't understand! It became incorrect!

← Older revision		Revision as of 14:36, 23 February 2014
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	~~{{lowercase}}~~
	~~{{unsolved\|computer science\|Is '''NP''' {{=}} '''co-NP''' ?}}~~
	In [[computational complexity theory]], '''co-NP''' is a [[complexity class]]. A [[decision problem]] <math>{\mathcal{X}}</math> is a member of '''co-NP''' if and only if its [[complement (complexity)\|complement]] <math>\overline{\mathcal{X}}</math> is in the complexity class '''[[NP (complexity)\|NP]]'''. In simple terms, '''co-NP''' is the class of problems for which efficiently verifiable proofs of ''no'' instances, sometimes called counterexamples, exist. Equivalently, '''co-NP''' is the set of decision problems where the "no" instances can be accepted in polynomial time by a [[non-deterministic Turing machine]].

	An example of an [[NP-complete\|'''NP'''-complete]] problem is the [[subset sum problem]]: given a finite set of integers, is there a non-empty subset that sums to zero? To give a proof of a "yes" instance, one must specify a non-empty subset that does sum to zero. The complementary problem is in '''co-NP''' and asks: "given a finite set of integers, does every non-empty subset have a non-zero sum?" This problem is not obviously seen to be in '''NP'''.

	~~==Relationship to other classes==~~		There are hundreds of thousands of companies online who 're ready to the money, although they are not committed to delivering the quality are effective. So, if you come across such an enterprise and provide them your money to acquire dissertation help, then completely surely lose your profit.<br><br><br><br>In several words, we are say that the SEO copywriter is alleged to produce copy that appeals to search engines and human readers simultaneously. At least, this is what should grow to be. This kind of web copy just what we can without danger call good SEO freelance writing.<br><br>As mentioned above, programs that enable you to generate your own house business prospects do become different models. For the best result, get the one which searches range websites anyone. These websites will be a collection of classified websites, online marketplaces, message boards, and significantly!<br><br>How to a vendor offering high-quality content Writing Services? This is one of the typically asked questions asked by most within the companies. Well, going through this article will help you with that it. All you need would be read directly on!<br><br>If you realize your writing is a commodity, you're able develop a qualified [https://joinup.ec.europa.eu/people/34546/ writing] career, because you'll realize you should take steps to differentiate YOUR writing from regarding other creators. Once you've differentiated your writing, it's no longer an asset.<br><br>Here's a tip; present you with a "sample" testimonial for your buyers to use. They can also add their own words, merely tell you they're willing to go in what you've composed. Writing a sample testimonial helps to ensure that you'll acquire a testimonial fast - or else you may be waiting for weeks, because individuals are entertained.<br><br>Use extremely name in the title of the blog. Avoid a made-up name, this pointless. You're trying to brand yourself as a writer, so take every opportunity to obtain your name known.<br><br>So just do it and along with this hint. You don't always have anything get rid of. The website cost is cheap, perform design it yourself, and can keep it in check on individual personal. Eventually, hybrid cars grow for the point a person simply need a staff. Wouldn't that be huge? And it's the method to make fast salary.
	~~[[P (complexity)\|'''P''']], the class~~ of ~~polynomial time solvable problems, is a subset~~ of ~~both~~ '~~''NP''' and '''co-NP'''. '''P''' is thought~~ to ~~be a strict subset in both cases (and demonstrably cannot be strict in one case but~~ not the ~~other)~~. ~~'''NP'''~~ and ~~'''co-NP''' are also thought~~ to ~~be unequal~~.<~~ref~~>
	~~{{cite book \| first = John E. \| last = Hopcroft \| title = Introduction~~ to ~~Automata Theory, Languages,~~ and ~~Computation (2nd Edition) \| publisher = Addison-Wesley \| location = Boston \| year = 2000 \| isbn = 0-201-44124-1 }} Chap. 11~~.~~</ref> If so~~, ~~then no '''NP'''-complete problem can~~ be ~~in '''co-NP''' and no [[co-NP-complete\|'''co-NP'''-complete]] problem can be in '''NP'''~~.

	This can ~~be shown as follows~~. ~~Suppose there exists an '''NP'''-complete problem~~ <~~math~~>~~{\mathcal{X}}~~<~~/math~~> that ~~is in '''co-NP'''~~. ~~Since all problems in '''NP''' can~~ be ~~reduced to~~ <~~math~~>~~{\mathcal{X}}~~<~~/math~~>~~, it follows that for every problem in '''NP''' we can construct~~ a ~~[[non~~-~~deterministic Turing machine]] that decides its complement in polynomial time, i.e~~., ~~'''NP''' &sube; '''co-NP'''. From~~ this it ~~follows that the set of complements of the problems in '''NP'''~~ is a ~~subset of the set of complements of the problems in~~ '~~''co-NP''', i~~.e., '''~~co-NP~~'~~'' &sube~~; ~~'''NP'''~~. ~~Thus '''co-NP''' = '''NP'~~''. ~~The proof~~ that no '~~''co~~-~~NP'''-complete problem can~~ be ~~in '''NP'''~~, ~~if '''NP''' ≠ '''co-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 (since otherwise '''NP''' = '''co-NP''')~~.

	~~An example of~~ a ~~problem that is known to belong to both '''NP''' and in '''co~~-~~NP''' is [[integer factorization]]: given positive integers ''m'' and ''n'' determine if ''m'' has a factor less than ''n'' and greater than one~~. ~~Membership in~~ '~~''NP''' is clear; if ''m'' does have such~~ a ~~factor then the factor itself is a certificate~~. ~~Membership in '''co-NP''' is also straightforward: one can~~ just ~~list the prime factors of ''m'', which the verifier can confirm to be valid by multiplication~~ and ~~the [[AKS primality test]]~~.

	~~[[Integer factorization]] is closely related to the [[primality]] problem. Both primality testing and factorization~~ have ~~long been known to be '''NP''' and '''co-NP''' problems~~. The ~~[[AKS primality test]]~~, ~~published in 2002~~, ~~proves that primality testing also lies~~ in ~~'''P'''~~, ~~while factorization may or may not have~~ a ~~polynomial-time algorithm.<ref>~~
	~~Manindra Agrawal, Neeraj Kayal, Nitin Saxena, "[http://www.cse.iitk.ac.in/users/manindra/algebra/primality_v6~~.~~pdf PRIMES is in P]",~~ ''~~Annals of Mathematics'' 160 (2004), no. 2, pp. 781-793~~.~~</ref>~~

	~~== References ==~~
	~~{{Reflist}}~~

	~~== External links ==~~
	* {{CZoo\|coNP\|C#conp}}

	~~{{ComplexityClasses}}~~

	~~{{DEFAULTSORT:Co-Np}}~~
	~~[[Category:Complexity classes]]~~

147.251.201.176: Correctly define the co-[subset sum problem]

2014-01-02T18:55:28Z

Correctly define the co-[subset sum problem]

← Older revision		Revision as of 20:55, 2 January 2014
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			{{unsolved\|computer science\|Is '''NP''' {{=}} '''co-NP''' ?}}
			In [[computational complexity theory]], '''co-NP''' is a [[complexity class]]. A [[decision problem]] <math>{\mathcal{X}}</math> is a member of '''co-NP''' if and only if its [[complement (complexity)\|complement]] <math>\overline{\mathcal{X}}</math> is in the complexity class '''[[NP (complexity)\|NP]]'''. In simple terms, '''co-NP''' is the class of problems for which efficiently verifiable proofs of ''no'' instances, sometimes called counterexamples, exist. Equivalently, '''co-NP''' is the set of decision problems where the "no" instances can be accepted in polynomial time by a [[non-deterministic Turing machine]].

			An example of an [[NP-complete\|'''NP'''-complete]] problem is the [[subset sum problem]]: given a finite set of integers, is there a non-empty subset that sums to zero? To give a proof of a "yes" instance, one must specify a non-empty subset that does sum to zero. The complementary problem is in '''co-NP''' and asks: "given a finite set of integers, does every non-empty subset have a non-zero sum?" This problem is not obviously seen to be in '''NP'''.

			==Relationship to other classes==
			[[P (complexity)\|'''P''']], the class of polynomial time solvable problems, is a subset of both '''NP''' and '''co-NP'''. '''P''' is thought to be a strict subset in both cases (and demonstrably cannot be strict in one case but not the other). '''NP''' and '''co-NP''' are also thought to be unequal.<ref>
			{{cite book \| first = John E. \| last = Hopcroft \| title = Introduction to Automata Theory, Languages, and Computation (2nd Edition) \| publisher = Addison-Wesley \| location = Boston \| year = 2000 \| isbn = 0-201-44124-1 }} Chap. 11.</ref> If so, then no '''NP'''-complete problem can be in '''co-NP''' and no [[co-NP-complete\|'''co-NP'''-complete]] problem can be in '''NP'''.

			This can be shown as follows. Suppose there exists an '''NP'''-complete problem <math>{\mathcal{X}}</math> that is in '''co-NP'''. Since all problems in '''NP''' can be reduced to <math>{\mathcal{X}}</math>, it follows that for every problem in '''NP''' we can construct a [[non-deterministic Turing machine]] that decides its complement in polynomial time, i.e., '''NP''' &sube; '''co-NP'''. From this it follows that the set of complements of the problems in '''NP''' is a subset of the set of complements of the problems in '''co-NP''', i.e., '''co-NP''' &sube; '''NP'''. Thus '''co-NP''' = '''NP'''. The proof that no '''co-NP'''-complete problem can be in '''NP''', if '''NP''' ≠ '''co-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 (since otherwise '''NP''' = '''co-NP''').

			An example of a problem that is known to belong to both '''NP''' and in '''co-NP''' is [[integer factorization]]: given positive integers ''m'' and ''n'' determine if ''m'' has a factor less than ''n'' and greater than one. Membership in '''NP''' is clear; if ''m'' does have such a factor then the factor itself is a certificate. Membership in '''co-NP''' is also straightforward: one can just list the prime factors of ''m'', which the verifier can confirm to be valid by multiplication and the [[AKS primality test]].

			[[Integer factorization]] is closely related to the [[primality]] problem. Both primality testing and factorization have long been known to be '''NP''' and '''co-NP''' problems. The [[AKS primality test]], published in 2002, proves that primality testing also lies in '''P''', while factorization may or may not have a polynomial-time algorithm.<ref>
			Manindra Agrawal, Neeraj Kayal, Nitin Saxena, "[http://www.cse.iitk.ac.in/users/manindra/algebra/primality_v6.pdf PRIMES is in P]", ''Annals of Mathematics'' 160 (2004), no. 2, pp. 781-793.</ref>

			== References ==
			{{Reflist}}

			== External links ==
			* {{CZoo\|coNP\|C#conp}}

			{{ComplexityClasses}}

			{{DEFAULTSORT:Co-Np}}
			[[Category:Complexity classes]]

en>Stephanwehner: Argument was a bit loose

2012-06-08T00:08:23Z

Argument was a bit loose

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