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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$mathbf{NC_2}$ 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?

3

$begingroup$

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

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asked 21 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
  21 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @YuvalFilmus thanks for the advice, this helped me figure it out!
  $endgroup$
  – Ankit Bahl
  20 hours ago
  

  
  
  
  

add a comment | 

3

$begingroup$

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

share|cite|improve this question

asked 21 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
  21 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @YuvalFilmus thanks for the advice, this helped me figure it out!
  $endgroup$
  – Ankit Bahl
  20 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

share|cite|improve this question

asked 21 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.

$endgroup$

share|cite|improve this question

asked 21 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.

share|cite|improve this question

asked 21 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 21 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 21 hours ago

Ankit BahlAnkit Bahl

663

1 Answer
1

active

oldest

votes

5

$begingroup$

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:

$A (i)$

1. Take input $i$ and apply $m$ to yield $m(i)$
   


2. 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.

share|cite|improve this answer

answered 20 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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5

$begingroup$

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:

$A (i)$

1. Take input $i$ and apply $m$ to yield $m(i)$
   


2. 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.

share|cite|improve this answer

answered 20 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 | 

5

$begingroup$

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:

$A (i)$

1. Take input $i$ and apply $m$ to yield $m(i)$
   


2. 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.

share|cite|improve this answer

answered 20 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:

$A (i)$

1. Take input $i$ and apply $m$ to yield $m(i)$
   


2. 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.

share|cite|improve this answer

answered 20 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.

$endgroup$

share|cite|improve this answer

answered 20 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 20 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 20 hours ago

Ankit BahlAnkit Bahl

663