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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?













1












$begingroup$


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 $B in NP$ $implies$ $A in NP$










share|cite|improve this question







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.







$endgroup$







  • 2




    $begingroup$
    Try using the definitions.
    $endgroup$
    – Yuval Filmus
    2 hours ago










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















1












$begingroup$


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 $B in NP$ $implies$ $A in NP$










share|cite|improve this question







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.







$endgroup$







  • 2




    $begingroup$
    Try using the definitions.
    $endgroup$
    – Yuval Filmus
    2 hours ago










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













1












1








1





$begingroup$


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 $B in NP$ $implies$ $A in NP$










share|cite|improve this question







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






share|cite|improve this question







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







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




share|cite|improve this question






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 2 hours ago









Ankit BahlAnkit Bahl

262




262




New contributor




Ankit Bahl is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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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.







  • 2




    $begingroup$
    Try using the definitions.
    $endgroup$
    – Yuval Filmus
    2 hours ago










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












  • 2




    $begingroup$
    Try using the definitions.
    $endgroup$
    – Yuval Filmus
    2 hours ago










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







2




2




$begingroup$
Try using the definitions.
$endgroup$
– Yuval Filmus
2 hours ago




$begingroup$
Try using the definitions.
$endgroup$
– Yuval Filmus
2 hours ago












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




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










1 Answer
1






active

oldest

votes


















2












$begingroup$

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








New contributor




Ankit Bahl is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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    2












    $begingroup$

    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








    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.






    $endgroup$

















      2












      $begingroup$

      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








      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.






      $endgroup$















        2












        2








        2





        $begingroup$

        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








        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:



        $A (i)$



        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








        New contributor




        Ankit Bahl is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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        share|cite|improve this answer



        share|cite|improve this answer






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        answered 1 hour ago









        Ankit BahlAnkit Bahl

        262




        262




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        Ankit Bahl is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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