Prove that NP is closed under karp reduction?Space(n) not closed under Karp reductions - what about NTime(n)?Class P is closed under rotation?Prove or disprove that $NL$ is closed under polynomial many-one reductions$mathbfNC_2$ is closed under log-space reductionOn Karp reductionwhen can I know if a class (complexity) is closed under reduction (cook/karp)Check if class $PSPACE$ is closed under polyonomially space reductionIs NPSPACE also closed under polynomial-time reduction and under log-space reduction?Prove PSPACE is closed under complement?Prove PSPACE is closed under union?
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Prove that NP is closed under karp reduction?
Space(n) not closed under Karp reductions - what about NTime(n)?Class P is closed under rotation?Prove or disprove that $NL$ is closed under polynomial many-one reductions$mathbfNC_2$ is closed under log-space reductionOn Karp reductionwhen can I know if a class (complexity) is closed under reduction (cook/karp)Check if class $PSPACE$ is closed under polyonomially space reductionIs NPSPACE also closed under polynomial-time reduction and under log-space reduction?Prove PSPACE is closed under complement?Prove PSPACE is closed under union?
$begingroup$
A complexity class $mathbbC$ is said to be closed under a reduction if:
$A$ reduces to $B$ and $B in mathbbC$ $implies$ $A in mathbbC$
How would you go about proving this if $mathbbC = NP$ and the reduction to be the karp reduction? i.e.
Prove that if $A$ karp reduces to $B$ and $B in NP$ $implies$ $A in NP$
complexity-theory
asked yesterday
Ankit BahlAnkit Bahl
663
New contributor
Ankit Bahl is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
A complexity class $mathbbC$ is said to be closed under a reduction if:
$A$ reduces to $B$ and $B in mathbbC$ $implies$ $A in mathbbC$
How would you go about proving this if $mathbbC = NP$ and the reduction to be the karp reduction? i.e.
Prove that if $A$ karp reduces to $B$ and $B in NP$ $implies$ $A in NP$
complexity-theory
asked yesterday
Ankit BahlAnkit Bahl
663
New contributor
Ankit Bahl is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
3
$begingroup$
Try using the definitions.
$endgroup$
– Yuval Filmus
yesterday
$begingroup$
@YuvalFilmus thanks for the advice, this helped me figure it out!
$endgroup$
– Ankit Bahl
yesterday
add a comment |
$begingroup$
A complexity class $mathbbC$ is said to be closed under a reduction if:
$A$ reduces to $B$ and $B in mathbbC$ $implies$ $A in mathbbC$
How would you go about proving this if $mathbbC = NP$ and the reduction to be the karp reduction? i.e.
Prove that if $A$ karp reduces to $B$ and $B in NP$ $implies$ $A in NP$
complexity-theory
asked yesterday
Ankit BahlAnkit Bahl
663
New contributor
Ankit Bahl is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
A complexity class $mathbbC$ is said to be closed under a reduction if:
$A$ reduces to $B$ and $B in mathbbC$ $implies$ $A in mathbbC$
How would you go about proving this if $mathbbC = NP$ and the reduction to be the karp reduction? i.e.
Prove that if $A$ karp reduces to $B$ and $B in NP$ $implies$ $A in NP$
complexity-theory
complexity-theory
asked yesterday
Ankit BahlAnkit Bahl
663
New contributor
Ankit Bahl is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked yesterday
Ankit BahlAnkit Bahl
663
New contributor
Ankit Bahl is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked yesterday
Ankit BahlAnkit Bahl
663
New contributor
Ankit Bahl is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked yesterday
Ankit BahlAnkit Bahl
663
asked yesterday
Ankit BahlAnkit Bahl
663
663
New contributor
Ankit Bahl is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Ankit Bahl is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Ankit Bahl is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
3
$begingroup$
Try using the definitions.
$endgroup$
– Yuval Filmus
yesterday
$begingroup$
@YuvalFilmus thanks for the advice, this helped me figure it out!
$endgroup$
– Ankit Bahl
yesterday
add a comment |
3
$begingroup$
Try using the definitions.
$endgroup$
– Yuval Filmus
yesterday
$begingroup$
@YuvalFilmus thanks for the advice, this helped me figure it out!
$endgroup$
– Ankit Bahl
yesterday
3
3
$begingroup$
Try using the definitions.
$endgroup$
– Yuval Filmus
yesterday
$begingroup$
Try using the definitions.
$endgroup$
– Yuval Filmus
yesterday
$begingroup$
@YuvalFilmus thanks for the advice, this helped me figure it out!
$endgroup$
– Ankit Bahl
yesterday
$begingroup$
@YuvalFilmus thanks for the advice, this helped me figure it out!
$endgroup$
– Ankit Bahl
yesterday
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
I was able to figure it out. In case anyone was wondering:
$B in NP$ means that there exists a non-deterministic polynomial time algorithm for $B$. Let's call that $b(i)$, where i is the input to $B$.
$A$ karp reducing to $B implies$ that there exists a function $m$ such that $m$ can take an input $i$ to $A$ and map it to some input $m(i)$ for $B$, and if an instance of $i$ is true for $A$ then $m(i)$ is true for B (and vice versa),
Therefore, an algorithm for $A$ can be made as follows:
$A (i)$
- Take input $i$ and apply $m$ to yield $m(i)$
- Apply $b$ with input $m(i)$
This yields an output for $A$. Since both $m$ and $b$ are non-deterministic polynomial time, this algorithm is non-deterministic polynomial time. Therefore $A$ must be in NP.
answered yesterday
Ankit BahlAnkit Bahl
663
New contributor
Ankit Bahl is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
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1 Answer
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1 Answer
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votes
$begingroup$
I was able to figure it out. In case anyone was wondering:
$B in NP$ means that there exists a non-deterministic polynomial time algorithm for $B$. Let's call that $b(i)$, where i is the input to $B$.
$A$ karp reducing to $B implies$ that there exists a function $m$ such that $m$ can take an input $i$ to $A$ and map it to some input $m(i)$ for $B$, and if an instance of $i$ is true for $A$ then $m(i)$ is true for B (and vice versa),
Therefore, an algorithm for $A$ can be made as follows:
$A (i)$
- Take input $i$ and apply $m$ to yield $m(i)$
- Apply $b$ with input $m(i)$
This yields an output for $A$. Since both $m$ and $b$ are non-deterministic polynomial time, this algorithm is non-deterministic polynomial time. Therefore $A$ must be in NP.
answered yesterday
Ankit BahlAnkit Bahl
663
New contributor
Ankit Bahl is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
I was able to figure it out. In case anyone was wondering:
$B in NP$ means that there exists a non-deterministic polynomial time algorithm for $B$. Let's call that $b(i)$, where i is the input to $B$.
$A$ karp reducing to $B implies$ that there exists a function $m$ such that $m$ can take an input $i$ to $A$ and map it to some input $m(i)$ for $B$, and if an instance of $i$ is true for $A$ then $m(i)$ is true for B (and vice versa),
Therefore, an algorithm for $A$ can be made as follows:
$A (i)$
- Take input $i$ and apply $m$ to yield $m(i)$
- Apply $b$ with input $m(i)$
This yields an output for $A$. Since both $m$ and $b$ are non-deterministic polynomial time, this algorithm is non-deterministic polynomial time. Therefore $A$ must be in NP.
answered yesterday
Ankit BahlAnkit Bahl
663
New contributor
Ankit Bahl is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
I was able to figure it out. In case anyone was wondering:
$B in NP$ means that there exists a non-deterministic polynomial time algorithm for $B$. Let's call that $b(i)$, where i is the input to $B$.
$A$ karp reducing to $B implies$ that there exists a function $m$ such that $m$ can take an input $i$ to $A$ and map it to some input $m(i)$ for $B$, and if an instance of $i$ is true for $A$ then $m(i)$ is true for B (and vice versa),
Therefore, an algorithm for $A$ can be made as follows:
$A (i)$
- Take input $i$ and apply $m$ to yield $m(i)$
- Apply $b$ with input $m(i)$
This yields an output for $A$. Since both $m$ and $b$ are non-deterministic polynomial time, this algorithm is non-deterministic polynomial time. Therefore $A$ must be in NP.
answered yesterday
Ankit BahlAnkit Bahl
663
New contributor
Ankit Bahl is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
I was able to figure it out. In case anyone was wondering:
$B in NP$ means that there exists a non-deterministic polynomial time algorithm for $B$. Let's call that $b(i)$, where i is the input to $B$.
$A$ karp reducing to $B implies$ that there exists a function $m$ such that $m$ can take an input $i$ to $A$ and map it to some input $m(i)$ for $B$, and if an instance of $i$ is true for $A$ then $m(i)$ is true for B (and vice versa),
Therefore, an algorithm for $A$ can be made as follows:
$A (i)$
- Take input $i$ and apply $m$ to yield $m(i)$
- Apply $b$ with input $m(i)$
This yields an output for $A$. Since both $m$ and $b$ are non-deterministic polynomial time, this algorithm is non-deterministic polynomial time. Therefore $A$ must be in NP.
answered yesterday
Ankit BahlAnkit Bahl
663
New contributor
Ankit Bahl is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered yesterday
Ankit BahlAnkit Bahl
663
New contributor
Ankit Bahl is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered yesterday
Ankit BahlAnkit Bahl
663
answered yesterday
Ankit BahlAnkit Bahl
663
663
New contributor
Ankit Bahl is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Ankit Bahl is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Ankit Bahl is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
add a comment |
Ankit Bahl is a new contributor. Be nice, and check out our Code of Conduct.
Ankit Bahl is a new contributor. Be nice, and check out our Code of Conduct.
Ankit Bahl is a new contributor. Be nice, and check out our Code of Conduct.
Ankit Bahl is a new contributor. Be nice, and check out our Code of Conduct.
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3
$begingroup$
Try using the definitions.
$endgroup$
– Yuval Filmus
yesterday
$begingroup$
@YuvalFilmus thanks for the advice, this helped me figure it out!
$endgroup$
– Ankit Bahl
yesterday