In computational complexity theory, co-NP is a complexity class, the complement (complexity) of the complexity class NP (complexity). In simple terms, it is the class of problems for which efficiently verifiable proofs of ''no'' instances, sometimes called ''counterexamples'', exist. For example, there is an NP-complete problem called the subset-sum problem, which asks if any subset of a finite set of integers sums to zero. Its complement problem is in co-NP and asks if ''every'' subset sums to a ''nonzero'' number. A counterexample would be a subset which does sum to zero, which is easy to verify. P is a subset of both NP and co-NP. That subset is thought to be strict in both cases. NP and co-NP are also thought to be unequal. If so, then no NP-complete problem can be in co-NP and no co-NP-complete problem can be in NP. This can be shown as follows. Assume that there is an NP-complete problem that is in co-NP. Since all problems in NP can be reduced to this problem it follows that for all problems in NP we can construct a non-deterministic Turing machine that decides the complement of the problem in polynomial time, i.e., NP is a subset of 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 is a subset of NP. Since we already knew that NP is a subset of co-NP it follows that they are the same. 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 (since otherwise NP = co-NP). One example is integer factorization, the problem of finding the prime number factors of a number. It is in both NP and co-NP, but is generally suspected to be outside P, outside NP-complete, and outside co-NP-complete. == References == * Complexity Zoo: [http://www.complexityzoo.com/#conp coNP] Complexity classes
"This can be shown as follows. Assume that there is an NP-complete problem that is in co-NP. Since all problems in NP can be reduced to this problem it follows that for all problems in NP we can construct a non-deterministic Turing machine that decides the complement of the problem in polynomial time, i.e., NP is a subset of 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 is a subset of NP. Since we already knew that NP is a subset of co-NP it follows that they are the same. The proof for the fact that no co-NP-complete problem can be in NP is symmetrical." The so called explanation above is unfortunately typical of what one finds on the web: unsubstantiated claims. "Since all problems in NP can be reduced to this problem it follows that...", sure it is easy to make this *claim*, but WHY does it follow that...? Then, more of the same: "From this it follows that the set of complements...", well, WHY does it follow? The explanation is useless, because it merely makes unsubstantiated assertions. well this article must be a bit older. Primes is in P! : Personally I think it pretty unreasonable to expect an article on higher mathematics to explain every step of a proof in excruciating detail. Some ability on the part of the reader is to be expected. And Primes is certainly in P, but Integer Factorization is a different problem, as the linked article explains in detail. ---- ''NP and co-NP are also thought to be unequal.'' Does this mean "not equal", or does it mean "disjunct"? --User:Zenogantner 10:25, 17 Nov 2004 (UTC) : It means "not equal". Think for example of P which is non-empty and a subset of both, so clearly their intersection cannot be empty. -- User:Jan Hidders 17:44, 17 Nov 2004 (UTC)
See other meanings of words starting from letter:
Words begining with Co-NP:
Co-NP
Co-NP
Co-NP-Complete
Co-NP-complete
Co-NP
In computational complexity theory, co-NP is a complexity class, the complement (complexity) of the complexity class NP (complexity). In simple terms, it is the class of problems for which efficiently verifiable proofs of ''no'' instances, sometimes called ''counterexamples'', exist. For example, there is an NP-complete problem called the subset-sum problem, which asks if any subset of a finite set of integers sums to zero. Its complement problem is in co-NP and asks if ''every'' subset sums to a ''nonzero'' number. A counterexample would be a subset which does sum to zero, which is easy to verify. P is a subset of both NP and co-NP. That subset is thought to be strict in both cases. NP and co-NP are also thought to be unequal. If so, then no NP-complete problem can be in co-NP and no co-NP-complete problem can be in NP. This can be shown as follows. Assume that there is an NP-complete problem that is in co-NP. Since all problems in NP can be reduced to this problem it follows that for all problems in NP we can construct a non-deterministic Turing machine that decides the complement of the problem in polynomial time, i.e., NP is a subset of 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 is a subset of NP. Since we already knew that NP is a subset of co-NP it follows that they are the same. 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 (since otherwise NP = co-NP). One example is integer factorization, the problem of finding the prime number factors of a number. It is in both NP and co-NP, but is generally suspected to be outside P, outside NP-complete, and outside co-NP-complete. == References == * Complexity Zoo: [http://www.complexityzoo.com/#conp coNP] Complexity classes
Co-NP
"This can be shown as follows. Assume that there is an NP-complete problem that is in co-NP. Since all problems in NP can be reduced to this problem it follows that for all problems in NP we can construct a non-deterministic Turing machine that decides the complement of the problem in polynomial time, i.e., NP is a subset of 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 is a subset of NP. Since we already knew that NP is a subset of co-NP it follows that they are the same. The proof for the fact that no co-NP-complete problem can be in NP is symmetrical." The so called explanation above is unfortunately typical of what one finds on the web: unsubstantiated claims. "Since all problems in NP can be reduced to this problem it follows that...", sure it is easy to make this *claim*, but WHY does it follow that...? Then, more of the same: "From this it follows that the set of complements...", well, WHY does it follow? The explanation is useless, because it merely makes unsubstantiated assertions. well this article must be a bit older. Primes is in P! : Personally I think it pretty unreasonable to expect an article on higher mathematics to explain every step of a proof in excruciating detail. Some ability on the part of the reader is to be expected. And Primes is certainly in P, but Integer Factorization is a different problem, as the linked article explains in detail. ---- ''NP and co-NP are also thought to be unequal.'' Does this mean "not equal", or does it mean "disjunct"? --User:Zenogantner 10:25, 17 Nov 2004 (UTC) : It means "not equal". Think for example of P which is non-empty and a subset of both, so clearly their intersection cannot be empty. -- User:Jan Hidders 17:44, 17 Nov 2004 (UTC)
See other meanings of words starting from letter:
C
CA | CB | CD | CE | CF | CG | CH | CI | CJ | CK | CL | CM | CN | CO | CP | CR | CS | CT | CU | CW | CX | CY | CZ |Words begining with Co-NP:
Co-NP
Co-NP
Co-NP-Complete
Co-NP-complete
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