Prove that NP is closed under karp reduction?
Multi tool use
$begingroup$
A complexity class $mathbb{C}$ is said to be closed under a reduction if:
$A$ reduces to $B$ and $B in mathbb{C}$ $implies$ $A in mathbb{C}$
How would you go about proving this if $mathbb{C} = NP$ 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 10 hours ago
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 $mathbb{C}$ is said to be closed under a reduction if:
$A$ reduces to $B$ and $B in mathbb{C}$ $implies$ $A in mathbb{C}$
How would you go about proving this if $mathbb{C} = NP$ 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 10 hours ago
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
10 hours ago
$begingroup$
@YuvalFilmus thanks for the advice, this helped me figure it out!
$endgroup$
– Ankit Bahl
9 hours ago
add a comment |
$begingroup$
A complexity class $mathbb{C}$ is said to be closed under a reduction if:
$A$ reduces to $B$ and $B in mathbb{C}$ $implies$ $A in mathbb{C}$
How would you go about proving this if $mathbb{C} = NP$ 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 10 hours ago
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 $mathbb{C}$ is said to be closed under a reduction if:
$A$ reduces to $B$ and $B in mathbb{C}$ $implies$ $A in mathbb{C}$
How would you go about proving this if $mathbb{C} = NP$ 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 10 hours ago
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 10 hours ago
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 10 hours ago
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 10 hours ago
Ankit BahlAnkit Bahl
663
asked 10 hours ago
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
10 hours ago
$begingroup$
@YuvalFilmus thanks for the advice, this helped me figure it out!
$endgroup$
– Ankit Bahl
9 hours ago
add a comment |
3
$begingroup$
Try using the definitions.
$endgroup$
– Yuval Filmus
10 hours ago
$begingroup$
@YuvalFilmus thanks for the advice, this helped me figure it out!
$endgroup$
– Ankit Bahl
9 hours ago
3
3
$begingroup$
Try using the definitions.
$endgroup$
– Yuval Filmus
10 hours ago
$begingroup$
Try using the definitions.
$endgroup$
– Yuval Filmus
10 hours ago
$begingroup$
@YuvalFilmus thanks for the advice, this helped me figure it out!
$endgroup$
– Ankit Bahl
9 hours ago
$begingroup$
@YuvalFilmus thanks for the advice, this helped me figure it out!
$endgroup$
– Ankit Bahl
9 hours ago
add a comment |
1 Answer
1
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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 9 hours ago
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.
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add a comment |
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1 Answer
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1 Answer
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$begingroup$
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 9 hours ago
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 9 hours ago
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 9 hours ago
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 9 hours ago
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 9 hours ago
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 9 hours ago
Ankit BahlAnkit Bahl
663
answered 9 hours ago
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
10 hours ago
$begingroup$
@YuvalFilmus thanks for the advice, this helped me figure it out!
$endgroup$
– Ankit Bahl
9 hours ago