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Pascal's Formula Recusrion (not triangle)



Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Redesign a module to be genericCritique of Pascal's Triangle ClassCondensing code that generates Pascal's trianglePerceptron algorithmBlocking reads when writes are happening on two flowsMultithreaded Pascal's Triangle CalculatorCalculate Kth Row of Pascal's TriangleCompare 2 unordered, rooted trees for shape-isomorphismFollow-up 1: Compare 2 unordered, rooted trees for shape-isomorphismPascal's Triangle - Java Recursion



.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty margin-bottom:0;








0












$begingroup$


So i am trying to recurse through Pascal's formula which is (n-1, k-1) + (n-1, k) but i am having trouble going through both the sides. i am thinking about it as a box trace ..here is what i have so far. i know my base cases are correct but my actual recurse method is incorrect



static void printSubsets(int []B, int k, int i)
//int n = 0;

if(k==0)
System.out.println(setToString(B));


if(k > B.length - i)
return ;


else if (B[i]!=)
printSubsets(B, k, i+1);



else if (B[i]==1)
printSubsets(B, k-1, i+1);;











share







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Check out our Code of Conduct.







$endgroup$


















    0












    $begingroup$


    So i am trying to recurse through Pascal's formula which is (n-1, k-1) + (n-1, k) but i am having trouble going through both the sides. i am thinking about it as a box trace ..here is what i have so far. i know my base cases are correct but my actual recurse method is incorrect



    static void printSubsets(int []B, int k, int i)
    //int n = 0;

    if(k==0)
    System.out.println(setToString(B));


    if(k > B.length - i)
    return ;


    else if (B[i]!=)
    printSubsets(B, k, i+1);



    else if (B[i]==1)
    printSubsets(B, k-1, i+1);;











    share







    New contributor




    Eternity is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
    Check out our Code of Conduct.







    $endgroup$














      0












      0








      0





      $begingroup$


      So i am trying to recurse through Pascal's formula which is (n-1, k-1) + (n-1, k) but i am having trouble going through both the sides. i am thinking about it as a box trace ..here is what i have so far. i know my base cases are correct but my actual recurse method is incorrect



      static void printSubsets(int []B, int k, int i)
      //int n = 0;

      if(k==0)
      System.out.println(setToString(B));


      if(k > B.length - i)
      return ;


      else if (B[i]!=)
      printSubsets(B, k, i+1);



      else if (B[i]==1)
      printSubsets(B, k-1, i+1);;











      share







      New contributor




      Eternity is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.







      $endgroup$




      So i am trying to recurse through Pascal's formula which is (n-1, k-1) + (n-1, k) but i am having trouble going through both the sides. i am thinking about it as a box trace ..here is what i have so far. i know my base cases are correct but my actual recurse method is incorrect



      static void printSubsets(int []B, int k, int i)
      //int n = 0;

      if(k==0)
      System.out.println(setToString(B));


      if(k > B.length - i)
      return ;


      else if (B[i]!=)
      printSubsets(B, k, i+1);



      else if (B[i]==1)
      printSubsets(B, k-1, i+1);;









      java





      share







      New contributor




      Eternity is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.










      share







      New contributor




      Eternity is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.








      share



      share






      New contributor




      Eternity is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.









      asked 3 mins ago









      EternityEternity

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      New contributor




      Eternity is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.





      New contributor





      Eternity is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.






      Eternity 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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