programming - 2008-Dec-24

• Dec_24_2008
  
  Alternate formulation is something like... Need to verify "Average" and
  simplify (if possible).
  

  Let
     r = radix
     n = number of bits
  
  And
     i - log_2 (r)
     j = the number of bytes needed to make a character
  
  Then
     Best = ceil(i)/jn
     Average = (ceil(i) + 2 - 2^(1 + ceil(i))/r))/jn
     Worst = floor(i)/jn
  

  
  Link for this entry: http://home.ccil.org/~remlaps/weblog_2008/programming_index.html#ThesisFormula24Dec08
  

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• Dec_22_2008
  
  All representations of all numbers stored in a 3D matrix.
  X axis = base
  Y axis = numeral
  Z axis = digits (index 0 = ones place)
  
  

  scp44@tux64-07:~/programming/cpp$ cat dnum.cpp
  #include "NumSpace.h"
  #include 
  #include 
  
  using namespace std;
  
  int main(int argc, char *argv[])
  {
  
           int MB=atoi(argv[1]);
           int MN=atoi(argv[2]);
  
           NumSpace abc(MB,MN);
           abc.Print();
           abc.AllReps(47);
           for (int lcv=1;lcv<5;lcv++)
                   cout << abc.GetDigit(13,2,lcv);
           cout <
  #include 
  #include "NumSpace.h"
  
  int NumSpace::DigitMatrix[MAXBASE][MAXNUM][MAXLEN];
  
  NumSpace::NumSpace(int X, int Y)
  {
           // log_2(N) = log(N)/log(2)
      int Z=(int) ceil(log(Y+1) / log(2));
      Xs = X;
      Ys = Y;
      Zs = Z;
  
     for (int Dig=0; Dig<=Z; Dig++)
     {
             // std::cout << "Digit: " << Dig << std::endl;
           for (int Base=2; Base<=Xs; Base++)
           {
                   int DigVal=0, DigInd=1;
                   if ( Dig >0 ) DigVal = 1;
                   int length=pow(Base,Dig);
                   int Top=Dig+1;
                   if ( Dig > 0 ) Top--;
  
                   for (int Num=0; Num = Base ) DigVal = 0;
                           }
                   }
          }
     }
  };
  
  void NumSpace::Print()
  {
  for (int Base=2; Base<=Xs; Base++)
  {
           std::cout << std::endl << "Base " << Base << std::endl;
  
           for (int Num=2; Num<=Ys; Num++)
           {
                   std::cout << Num << ": ";
                   for (int Dig=Zs; Dig>=0; Dig--)
                           std::cout << DigitMatrix[Base][Num][Dig] << ", ";
                   std::cout << std::endl;
           }
  }
  };
  
  void NumSpace::AllReps(int N)
  {
           std::cout << "All reps for " << N << std::endl;
           for (int B = 2; B < Xs; B++ )
           {
                   std::cout << "Base " << B << ": ";
                   for (int Dig=Zs; Dig>=0; Dig--)
                           std::cout << DigitMatrix[B][N][Dig] << ", ";
                   std::cout << std::endl;
           }
  }
  
  int NumSpace::GetDigit(int N, int B, int D)
  {
           int nd=log(N+1)/log(B);
           return DigitMatrix[B][N][nd-D+1];
  }
  

  
  Link for this entry: http://home.ccil.org/~remlaps/weblog_2008/programming_index.html#NumberSpace22Dec08
  

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• Dec_13_2008
  
  Read later. From December's Journal of the ACM...
  
  "Linear-time disk-based implicit graph search" - Richard E. Korf
  http://doi.acm.org/10.1145/1455248.1455250
  
  Link for this entry: http://home.ccil.org/~remlaps/weblog_2008/programming_index.html#DiskGraphSearch12Dec08
  

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