Prove that NP is closed under karp reduction?












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
















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







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














3












3








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







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






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 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














  • 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










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








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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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    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








    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.






    $endgroup$


















      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








      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.






      $endgroup$
















        5












        5








        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








        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.
        Check out our Code of Conduct.









        share|cite|improve this answer



        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.









        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.






















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