Prove that NP is closed under karp reduction?












5












$begingroup$


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









$endgroup$








  • 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
















5












$begingroup$


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









$endgroup$








  • 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














5












5








5


0



$begingroup$


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









$endgroup$




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













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Apr 6 at 19:02









Ankit BahlAnkit Bahl

965




965








  • 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




    $begingroup$
    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










1 Answer
1






active

oldest

votes


















7












$begingroup$

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:



$A (i)$




  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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    1 Answer
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    active

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    1 Answer
    1






    active

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    active

    oldest

    votes









    7












    $begingroup$

    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:



    $A (i)$




    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











    $endgroup$


















      7












      $begingroup$

      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:



      $A (i)$




      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











      $endgroup$
















        7












        7








        7





        $begingroup$

        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:



        $A (i)$




        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











        $endgroup$



        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:



        $A (i)$




        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














        share|cite|improve this answer



        share|cite|improve this answer








        edited Apr 9 at 12:18

























        answered Apr 6 at 20:05









        Ankit BahlAnkit Bahl

        965




        965






























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