Simple interepretation problem regarding Polynomial Hierarchy?Why isn't this undecidable problem in NP?Do any decision problems exist outside NP and NP-Hard?P is contained in NP ∩ Co-NP?Constructing a promise problem equivalent to XSAT from subset sumIs a problem in NP if it is decided by some non-deterministic, polynomial time turing machine?Is there a task that is solvable in polynomial time but not verifiable in polynomial time?A clarification on $NP=coNP$?NP-complete problem with a polynomial number of yes-instances?Is This A Solution for does P=NP problem?How does a co-NP problem differ from an NP (its complement) one?
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Simple interepretation problem regarding Polynomial Hierarchy?
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Simple interepretation problem regarding Polynomial Hierarchy?
Why isn't this undecidable problem in NP?Do any decision problems exist outside NP and NP-Hard?P is contained in NP ∩ Co-NP?Constructing a promise problem equivalent to XSAT from subset sumIs a problem in NP if it is decided by some non-deterministic, polynomial time turing machine?Is there a task that is solvable in polynomial time but not verifiable in polynomial time?A clarification on $NP=coNP$?NP-complete problem with a polynomial number of yes-instances?Is This A Solution for does P=NP problem?How does a co-NP problem differ from an NP (its complement) one?
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$begingroup$
So $NP$ stands for problems where we have small verifiable witnesses for $YES$ instances and $coNP$ for small verifiable witnesses for $NO$ instances. How does this work for
$P^NP$
$NP^NP$
$coNP^NP$
and so on?
complexity-theory np
asked 9 hours ago
TurboTurbo
1,2676 silver badges18 bronze badges
$endgroup$
add a comment |
$begingroup$
So $NP$ stands for problems where we have small verifiable witnesses for $YES$ instances and $coNP$ for small verifiable witnesses for $NO$ instances. How does this work for
$P^NP$
$NP^NP$
$coNP^NP$
and so on?
complexity-theory np
asked 9 hours ago
TurboTurbo
1,2676 silver badges18 bronze badges
$endgroup$
add a comment |
$begingroup$
So $NP$ stands for problems where we have small verifiable witnesses for $YES$ instances and $coNP$ for small verifiable witnesses for $NO$ instances. How does this work for
$P^NP$
$NP^NP$
$coNP^NP$
and so on?
complexity-theory np
asked 9 hours ago
TurboTurbo
1,2676 silver badges18 bronze badges
$endgroup$
So $NP$ stands for problems where we have small verifiable witnesses for $YES$ instances and $coNP$ for small verifiable witnesses for $NO$ instances. How does this work for
$P^NP$
$NP^NP$
$coNP^NP$
and so on?
complexity-theory np
complexity-theory np
asked 9 hours ago
TurboTurbo
1,2676 silver badges18 bronze badges
asked 9 hours ago
TurboTurbo
1,2676 silver badges18 bronze badges
asked 9 hours ago
TurboTurbo
1,2676 silver badges18 bronze badges
asked 9 hours ago
TurboTurbo
1,2676 silver badges18 bronze badges
asked 9 hours ago
TurboTurbo
1,2676 silver badges18 bronze badges
1,2676 silver badges18 bronze badges
add a comment |
add a comment |
1 Answer
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$begingroup$
There is a logical interpretation of the various levels of the polynomial hierarchy, which extends the witness characterization of $mathsfNP$ and $mathsfcoNP$.
A language $L$ is in $Sigma_k^P$ if there is a polytime predicate $f$ and a polynomial $ell$ such that
$$
x in L Leftrightarrow exists |y_1| le ell(|x|) ; forall |y_2| le ell(|x|) ; cdots Q |y_k| le ell(|x|) ; f(x,y_1,ldots,y_k).
$$
Here:
$exists |y| le ell(|x|)$ means that there exists a number $y$ whose length is at most $ell(|x|)$ such that ...
$forall |y| le ell(|x|)$ mean that for all $y$ whose length is at most $ell(|x|)$, the following holds ...
$Q$ is $exists$ if $k$ is odd and $forall$ if $k$ is even.
Similarly, $L$ is in $Pi_k^P$ if it can be written in a similar way, only starting with $forall$.
As an example, $mathsfNP^mathsfNP$ is $Sigma_2^P$, and consists of all languages such that
$$
x in L Leftrightarrow exists |y_1| leq ell(|x|) ; forall |y_2| leq ell(|x|) ; f(x,y_1,ldots,y_k).
$$
As another example, $mathsfcoNP^mathsfNP$ is $Pi_2^P$.
Your third example is $mathsfP^mathsfNP$, which is $Delta_2^P$. I'm not sure what is the logical characterization.
answered 7 hours ago
Yuval FilmusYuval Filmus
204k15 gold badges198 silver badges360 bronze badges
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add a comment |
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1 Answer
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1 Answer
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$begingroup$
There is a logical interpretation of the various levels of the polynomial hierarchy, which extends the witness characterization of $mathsfNP$ and $mathsfcoNP$.
A language $L$ is in $Sigma_k^P$ if there is a polytime predicate $f$ and a polynomial $ell$ such that
$$
x in L Leftrightarrow exists |y_1| le ell(|x|) ; forall |y_2| le ell(|x|) ; cdots Q |y_k| le ell(|x|) ; f(x,y_1,ldots,y_k).
$$
Here:
$exists |y| le ell(|x|)$ means that there exists a number $y$ whose length is at most $ell(|x|)$ such that ...
$forall |y| le ell(|x|)$ mean that for all $y$ whose length is at most $ell(|x|)$, the following holds ...
$Q$ is $exists$ if $k$ is odd and $forall$ if $k$ is even.
Similarly, $L$ is in $Pi_k^P$ if it can be written in a similar way, only starting with $forall$.
As an example, $mathsfNP^mathsfNP$ is $Sigma_2^P$, and consists of all languages such that
$$
x in L Leftrightarrow exists |y_1| leq ell(|x|) ; forall |y_2| leq ell(|x|) ; f(x,y_1,ldots,y_k).
$$
As another example, $mathsfcoNP^mathsfNP$ is $Pi_2^P$.
Your third example is $mathsfP^mathsfNP$, which is $Delta_2^P$. I'm not sure what is the logical characterization.
answered 7 hours ago
Yuval FilmusYuval Filmus
204k15 gold badges198 silver badges360 bronze badges
$endgroup$
add a comment |
$begingroup$
There is a logical interpretation of the various levels of the polynomial hierarchy, which extends the witness characterization of $mathsfNP$ and $mathsfcoNP$.
A language $L$ is in $Sigma_k^P$ if there is a polytime predicate $f$ and a polynomial $ell$ such that
$$
x in L Leftrightarrow exists |y_1| le ell(|x|) ; forall |y_2| le ell(|x|) ; cdots Q |y_k| le ell(|x|) ; f(x,y_1,ldots,y_k).
$$
Here:
$exists |y| le ell(|x|)$ means that there exists a number $y$ whose length is at most $ell(|x|)$ such that ...
$forall |y| le ell(|x|)$ mean that for all $y$ whose length is at most $ell(|x|)$, the following holds ...
$Q$ is $exists$ if $k$ is odd and $forall$ if $k$ is even.
Similarly, $L$ is in $Pi_k^P$ if it can be written in a similar way, only starting with $forall$.
As an example, $mathsfNP^mathsfNP$ is $Sigma_2^P$, and consists of all languages such that
$$
x in L Leftrightarrow exists |y_1| leq ell(|x|) ; forall |y_2| leq ell(|x|) ; f(x,y_1,ldots,y_k).
$$
As another example, $mathsfcoNP^mathsfNP$ is $Pi_2^P$.
Your third example is $mathsfP^mathsfNP$, which is $Delta_2^P$. I'm not sure what is the logical characterization.
answered 7 hours ago
Yuval FilmusYuval Filmus
204k15 gold badges198 silver badges360 bronze badges
$endgroup$
add a comment |
$begingroup$
There is a logical interpretation of the various levels of the polynomial hierarchy, which extends the witness characterization of $mathsfNP$ and $mathsfcoNP$.
A language $L$ is in $Sigma_k^P$ if there is a polytime predicate $f$ and a polynomial $ell$ such that
$$
x in L Leftrightarrow exists |y_1| le ell(|x|) ; forall |y_2| le ell(|x|) ; cdots Q |y_k| le ell(|x|) ; f(x,y_1,ldots,y_k).
$$
Here:
$exists |y| le ell(|x|)$ means that there exists a number $y$ whose length is at most $ell(|x|)$ such that ...
$forall |y| le ell(|x|)$ mean that for all $y$ whose length is at most $ell(|x|)$, the following holds ...
$Q$ is $exists$ if $k$ is odd and $forall$ if $k$ is even.
Similarly, $L$ is in $Pi_k^P$ if it can be written in a similar way, only starting with $forall$.
As an example, $mathsfNP^mathsfNP$ is $Sigma_2^P$, and consists of all languages such that
$$
x in L Leftrightarrow exists |y_1| leq ell(|x|) ; forall |y_2| leq ell(|x|) ; f(x,y_1,ldots,y_k).
$$
As another example, $mathsfcoNP^mathsfNP$ is $Pi_2^P$.
Your third example is $mathsfP^mathsfNP$, which is $Delta_2^P$. I'm not sure what is the logical characterization.
answered 7 hours ago
Yuval FilmusYuval Filmus
204k15 gold badges198 silver badges360 bronze badges
$endgroup$
There is a logical interpretation of the various levels of the polynomial hierarchy, which extends the witness characterization of $mathsfNP$ and $mathsfcoNP$.
A language $L$ is in $Sigma_k^P$ if there is a polytime predicate $f$ and a polynomial $ell$ such that
$$
x in L Leftrightarrow exists |y_1| le ell(|x|) ; forall |y_2| le ell(|x|) ; cdots Q |y_k| le ell(|x|) ; f(x,y_1,ldots,y_k).
$$
Here:
$exists |y| le ell(|x|)$ means that there exists a number $y$ whose length is at most $ell(|x|)$ such that ...
$forall |y| le ell(|x|)$ mean that for all $y$ whose length is at most $ell(|x|)$, the following holds ...
$Q$ is $exists$ if $k$ is odd and $forall$ if $k$ is even.
Similarly, $L$ is in $Pi_k^P$ if it can be written in a similar way, only starting with $forall$.
As an example, $mathsfNP^mathsfNP$ is $Sigma_2^P$, and consists of all languages such that
$$
x in L Leftrightarrow exists |y_1| leq ell(|x|) ; forall |y_2| leq ell(|x|) ; f(x,y_1,ldots,y_k).
$$
As another example, $mathsfcoNP^mathsfNP$ is $Pi_2^P$.
Your third example is $mathsfP^mathsfNP$, which is $Delta_2^P$. I'm not sure what is the logical characterization.
answered 7 hours ago
Yuval FilmusYuval Filmus
204k15 gold badges198 silver badges360 bronze badges
answered 7 hours ago
Yuval FilmusYuval Filmus
204k15 gold badges198 silver badges360 bronze badges
answered 7 hours ago
Yuval FilmusYuval Filmus
204k15 gold badges198 silver badges360 bronze badges
answered 7 hours ago
Yuval FilmusYuval Filmus
204k15 gold badges198 silver badges360 bronze badges
204k15 gold badges198 silver badges360 bronze badges
add a comment |
add a comment |
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