Prove that NP is closed under karp reduction? The 2019 Stack Overflow Developer Survey Results Are InSpace(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?
The 2019 Stack Overflow Developer Survey Results Are InSpace(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 Apr 6 at 19:02
Ankit BahlAnkit Bahl
965
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 Apr 6 at 19:02
Ankit BahlAnkit Bahl
965
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.
$endgroup$
4
$begingroup$
Try using the definitions.
$endgroup$
– Yuval Filmus
Apr 6 at 19:09
$begingroup$
@YuvalFilmus thanks for the advice, this helped me figure it out!
$endgroup$
– Ankit Bahl
Apr 6 at 20:06
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 Apr 6 at 19:02
Ankit BahlAnkit Bahl
965
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 Apr 6 at 19:02
Ankit BahlAnkit Bahl
965
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 Apr 6 at 19:02
Ankit BahlAnkit Bahl
965
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 Apr 6 at 19:02
Ankit BahlAnkit Bahl
965
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 Apr 6 at 19:02
Ankit BahlAnkit Bahl
965
asked Apr 6 at 19:02
Ankit BahlAnkit Bahl
965
965
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.
4
$begingroup$
Try using the definitions.
$endgroup$
– Yuval Filmus
Apr 6 at 19:09
$begingroup$
@YuvalFilmus thanks for the advice, this helped me figure it out!
$endgroup$
– Ankit Bahl
Apr 6 at 20:06
add a comment |
4
$begingroup$
Try using the definitions.
$endgroup$
– Yuval Filmus
Apr 6 at 19:09
$begingroup$
@YuvalFilmus thanks for the advice, this helped me figure it out!
$endgroup$
– Ankit Bahl
Apr 6 at 20:06
4
4
$begingroup$
Try using the definitions.
$endgroup$
– Yuval Filmus
Apr 6 at 19:09
$begingroup$
Try using the definitions.
$endgroup$
– Yuval Filmus
Apr 6 at 19:09
$begingroup$
@YuvalFilmus thanks for the advice, this helped me figure it out!
$endgroup$
– Ankit Bahl
Apr 6 at 20:06
$begingroup$
@YuvalFilmus thanks for the advice, this helped me figure it out!
$endgroup$
– Ankit Bahl
Apr 6 at 20:06
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
I was able to figure it out. In case anyone (mans in ECE 406) 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 same for false case),
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 Apr 6 at 20:05
Ankit BahlAnkit Bahl
965
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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$begingroup$
I was able to figure it out. In case anyone (mans in ECE 406) 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 same for false case),
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 Apr 6 at 20:05
Ankit BahlAnkit Bahl
965
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 (mans in ECE 406) 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 same for false case),
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 Apr 6 at 20:05
Ankit BahlAnkit Bahl
965
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 (mans in ECE 406) 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 same for false case),
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 Apr 6 at 20:05
Ankit BahlAnkit Bahl
965
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 (mans in ECE 406) 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 same for false case),
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 Apr 6 at 20:05
Ankit BahlAnkit Bahl
965
New contributor
Ankit Bahl is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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edited yesterday
answered Apr 6 at 20:05
Ankit BahlAnkit Bahl
965
New contributor
Ankit Bahl is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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answered Apr 6 at 20:05
Ankit BahlAnkit Bahl
965
answered Apr 6 at 20:05
Ankit BahlAnkit Bahl
965
965
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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Ic0k3Ig G,H3CIMOJ7kNO,BkHxUmQz
4
$begingroup$
Try using the definitions.
$endgroup$
– Yuval Filmus
Apr 6 at 19:09
$begingroup$
@YuvalFilmus thanks for the advice, this helped me figure it out!
$endgroup$
– Ankit Bahl
Apr 6 at 20:06