What is the complexity of subset of the power set of a set












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The complexity is $2^n$, if I have a set $B$ and want to find the power set $mathcal{P}(B)$.



But what if I want to find only sets of max size $i$ of $mathcal{P}(B)$? So a subset of $mathcal{P}(B)$ where there only exist sets with sizes less or equal to $i$



Something points in the direction that the complexity is $n^i$ but I have trouble understanding why.






combinatorics discrete-mathematics







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    0












    $begingroup$


    The complexity is $2^n$, if I have a set $B$ and want to find the power set $mathcal{P}(B)$.



    But what if I want to find only sets of max size $i$ of $mathcal{P}(B)$? So a subset of $mathcal{P}(B)$ where there only exist sets with sizes less or equal to $i$



    Something points in the direction that the complexity is $n^i$ but I have trouble understanding why.






    combinatorics discrete-mathematics







    share|cite|improve this question











    $endgroup$















      0












      0








      0





      $begingroup$


      The complexity is $2^n$, if I have a set $B$ and want to find the power set $mathcal{P}(B)$.



      But what if I want to find only sets of max size $i$ of $mathcal{P}(B)$? So a subset of $mathcal{P}(B)$ where there only exist sets with sizes less or equal to $i$



      Something points in the direction that the complexity is $n^i$ but I have trouble understanding why.






      combinatorics discrete-mathematics







      share|cite|improve this question











      $endgroup$




      The complexity is $2^n$, if I have a set $B$ and want to find the power set $mathcal{P}(B)$.



      But what if I want to find only sets of max size $i$ of $mathcal{P}(B)$? So a subset of $mathcal{P}(B)$ where there only exist sets with sizes less or equal to $i$



      Something points in the direction that the complexity is $n^i$ but I have trouble understanding why.






      combinatorics discrete-mathematics




      combinatorics discrete-mathematics






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Jan 3 at 8:37







      Adam Soel

















      asked Jan 3 at 8:22









      Adam SoelAdam Soel

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          Provided that the cardinality of a finite set $mathbf B$ is $n$ the number of subsets of $mathbf B$ with cardinality $j$ equals $binom{n}{j}$, so the number of subset with cardinality that does not exceed $i$ equals:$$sum_{j=0}^ibinom{n}j$$



          There is no closed form of this expression.



          Substituting $i=n$ observe that $sum_{j=0}^nbinom{n}j=2^n$ which is in accordance with your observation that the cardinality of $wp(mathbf B)$ equals $2^n$.






          share|cite|improve this answer









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            $begingroup$

            Provided that the cardinality of a finite set $mathbf B$ is $n$ the number of subsets of $mathbf B$ with cardinality $j$ equals $binom{n}{j}$, so the number of subset with cardinality that does not exceed $i$ equals:$$sum_{j=0}^ibinom{n}j$$



            There is no closed form of this expression.



            Substituting $i=n$ observe that $sum_{j=0}^nbinom{n}j=2^n$ which is in accordance with your observation that the cardinality of $wp(mathbf B)$ equals $2^n$.






            share|cite|improve this answer









            $endgroup$


















              2












              $begingroup$

              Provided that the cardinality of a finite set $mathbf B$ is $n$ the number of subsets of $mathbf B$ with cardinality $j$ equals $binom{n}{j}$, so the number of subset with cardinality that does not exceed $i$ equals:$$sum_{j=0}^ibinom{n}j$$



              There is no closed form of this expression.



              Substituting $i=n$ observe that $sum_{j=0}^nbinom{n}j=2^n$ which is in accordance with your observation that the cardinality of $wp(mathbf B)$ equals $2^n$.






              share|cite|improve this answer









              $endgroup$
















                2












                2








                2





                $begingroup$

                Provided that the cardinality of a finite set $mathbf B$ is $n$ the number of subsets of $mathbf B$ with cardinality $j$ equals $binom{n}{j}$, so the number of subset with cardinality that does not exceed $i$ equals:$$sum_{j=0}^ibinom{n}j$$



                There is no closed form of this expression.



                Substituting $i=n$ observe that $sum_{j=0}^nbinom{n}j=2^n$ which is in accordance with your observation that the cardinality of $wp(mathbf B)$ equals $2^n$.






                share|cite|improve this answer









                $endgroup$



                Provided that the cardinality of a finite set $mathbf B$ is $n$ the number of subsets of $mathbf B$ with cardinality $j$ equals $binom{n}{j}$, so the number of subset with cardinality that does not exceed $i$ equals:$$sum_{j=0}^ibinom{n}j$$



                There is no closed form of this expression.



                Substituting $i=n$ observe that $sum_{j=0}^nbinom{n}j=2^n$ which is in accordance with your observation that the cardinality of $wp(mathbf B)$ equals $2^n$.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Jan 3 at 8:35









                drhabdrhab

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