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np [2015/04/10 21:59] nikolaj |
np [2015/10/10 20:29] nikolaj |
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| @#55EE55: postulate | @#55EE55: $\exists M,p.\ \forall x.\ x\in L\Leftrightarrow \exists u.\ \left(M(x,u) = 1\land \left|u\right|=p(\left|x\right|)\right) $ | | | @#55EE55: postulate | @#55EE55: $\exists M,p.\ \forall x.\ x\in L\Leftrightarrow \exists u.\ \left(M(x,u) = 1\land \left|u\right|=p(\left|x\right|)\right) $ | | ||
- | ==== Discussion ==== | + | ----- |
+ | === Discussion === | ||
>todo: Requirements [[Polynomial time Turing machine]] | >todo: Requirements [[Polynomial time Turing machine]] | ||
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The machine $M$ still depends on the input $x$ in a computational manner, because for $x\notin L$ it is //not allowed// for there to be //any// misleading certificate resulting in the $M(x,u) = 1$. In fact, for $x\notin L$ the machine doesn't tell you anything (the class is defined by that predicate is called ${\bf{NP}}$). If $u$ also knew if $x\notin L$, then it could just be a yes/no answer and //any// language could be decided this way. Instead, the above definition permits the interpretation $M$ to be the //verifier//. | The machine $M$ still depends on the input $x$ in a computational manner, because for $x\notin L$ it is //not allowed// for there to be //any// misleading certificate resulting in the $M(x,u) = 1$. In fact, for $x\notin L$ the machine doesn't tell you anything (the class is defined by that predicate is called ${\bf{NP}}$). If $u$ also knew if $x\notin L$, then it could just be a yes/no answer and //any// language could be decided this way. Instead, the above definition permits the interpretation $M$ to be the //verifier//. | ||
- | === Examples === | + | == Examples == |
A problem in $\mathrm{\bf{NP}}$ is the Boolean satisfiability problem. Finding out if a circuit, given via $x$, can be satisfied is hard. But as $u$ is allowed to be a string among the right inputs, having $M$ be the machine which tests $u$ on a given circuit which returns a positive answer if it works directly answers the initial question. | A problem in $\mathrm{\bf{NP}}$ is the Boolean satisfiability problem. Finding out if a circuit, given via $x$, can be satisfied is hard. But as $u$ is allowed to be a string among the right inputs, having $M$ be the machine which tests $u$ on a given circuit which returns a positive answer if it works directly answers the initial question. | ||
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$ \mathrm{\bf{NP}} = \bigcup_{k\in\mathbb{N}} \mathrm{\bf{NTIME}}(n^k) $ | $ \mathrm{\bf{NP}} = \bigcup_{k\in\mathbb{N}} \mathrm{\bf{NTIME}}(n^k) $ | ||
- | ==== Parents ==== | + | |
+ | ----- | ||
=== Requirements === | === Requirements === | ||
[[Turing machine as partial function]] | [[Turing machine as partial function]] | ||
=== Subset of === | === Subset of === | ||
[[Bit string]] | [[Bit string]] |