dynamic iterative programming to generate combinat

Updated with my own version of the program :

I'm trying to do an iterative dynamic programming to generate n choose k combination.

Say I've 4 vectors of values

Now I use addition as my aggregate function and I want to generate 4 choose 2 combinations of vectors as follows :

v1v2 : 3 3 3
v1v3 : 4 4 4
v1v4 : 5 5 5
v2v3 : 5 5 5
v2v4 : 6 6 6
v3v4 : 7 7 7

A naive way to do this would be to go through every pair and find the result. if the N and k are very large this is very very inefficient. So an alternative way to this would be recursive/iterative dynamic programming. Recursion for very large N and k would take up a lot of memory so the ideal way to this would be iterative dynamic programming which can be done as follows :

Consider the following table row header is N and the column header is k and our objective is to find N choose k :

We can find N choose k combination using dynamic program by following way :

The approach is as follows :

Block[0,1] and Block[0,2] always returns [ ]. {[ ] denotes an empty value as there are no values there}.
Block[1,1] receives [ ], computes {v1} + [ ] (which is v1 itself) , saves it to Block [1,1].
Block[1,2] receives [ ], does {v1} + [ ] & {v2} + [ ], saves it to Block [1,2].
Block[1,3] receives [ ], does {v1} + [ ], {v2} + [ ] + {v3} U [ ], Block [1,3].
Block[2,4] receives :
- [{v1}] from [1,1] and computes {v1} + {v2},
- [{v1}{v2}] from [1,2] and computes {v1} + {v3} and {v2} + {v3}
- [{v1}, {v2}, {v3}] from [1,3] and computes {v4} + {v1}, {v4} + {v2} and {v4} + {v3}, saves it to Block [2,4].

Now we have all the values we need in Block [2,4]. how can i code this concept efficiently in C++ ?

Any help is greatly appreciated. Thanks.

Here is my idea :

I don't know if this is right. Sorry

=========================================================

//Block [0...k][0...n]; 
//Block[i][j] contains i-set groups (for eg :if i = 2 it will have v1v2, v1v3, etc..)

//initially, Block[0][i] = [ ] for 0 <= i <= n and all other Block [i][j] = [ $ ]
// "$" is just a symbol to indicate that no computation is done on that block

algorithm(int k, int n) //k-point skyline groups from n number of points.
{
   if( Block[k][n] != [ $ ] ) return memory[k][n];

   Group = [ ]; //G indicate a collection of vectors
   for( int i = k; i <= n; i++ )
   {
      Group` = algorithm(k-1, i-1);
      for( each v` in Group` )
      {
         Group = Group + (group` + v_i);
      }
   }
memory[k][n] = Group;
return Group;
}

=========================================================

Here is my program for the above mentioned algorithm :

#include <iostream>
#include <iterator>
#include <set>
#include <vector>
#include <map>
#define DIMENSIONS 5 // No. of elements in the vector. eg. v1: 1 1 1 1 1 
using namespace std;

typedef std::set < int > my_set;  // To hold vector id's
typedef std::vector < int > my_vector; // To hold values of the vector's
typedef std::vector < std::pair < my_set, my_vector > > my_vector_pair;
typedef std::map < my_set, my_vector > my_map;
typedef std::vector < vector < std::pair < int,my_map > > > my_pair;
typedef my_map::iterator m_it;

my_vector_pair bases;  // To hold all the initial <id,vector_values> pair
my_map data, G;
my_pair memory;

void print(my_map& data)
{
    for( m_it it(data.begin()) ; it!=data.end(); ++it) 
    {   
        cout << "Id : ";
        copy(it->first.begin(), it->first.end(), ostream_iterator<int>(cout, " "));
        cout << " => value : ";
        copy (it->second.begin(),it->second.end(),ostream_iterator<int>(cout," "));
        cout << endl;
    }
    cout << "---------------------------------------------------------------\n";
}

my_map union_(my_map& G, int p) 
{
    static my_map result;
    my_set id;
    my_vector scores;
    result.clear();
    for (m_it it(G.begin()); it != G.end(); ++it) 
    {
        id = it->first;
        scores = it->second;
        id.insert( bases.at(p-1).first.begin(),bases.at(p-1).first.end() );

            for (int j = 0; j < DIMENSIONS; j++) 
            {
                scores.at(j) += bases.at(p - 1).second.at(j);
            }
            result.insert(make_pair(id, scores));
    }
    return result;
}

my_map algorithm_(int k, int n) {

    unsigned long size = memory.at(n).size();
    for (unsigned long i = 0; i < size; i++) {
        if (memory.at(n).at(i).first == k) {
            return memory.at(n).at(i).second; //if exists in hash table then no need to calculate again
        }
    }
    my_map G_k_1;

    if (k != n)
    {
        G_k_1 = algorithm_(k, n - 1);
        if(G_k_1.size() == 0)
            {
                return G_k_1;
            }
    }
    G_k_1 = algorithm_(k - 1, n - 1);
    if(G_k_1.size() == 0)
    {
        return G_k_1;
    }

    G_k_1 = union_(G_k_1, n);

    if (k != n) {
        for (unsigned long i = 0; i < memory.at(n - 1).size(); i++) {
            if (memory.at(n - 1).at(i).first == k) {
                G_k_1.insert(memory.at(n - 1).at(i).second.begin(), memory.at(n - 1).at(i).second.end());
                memory.at(n - 1).at(i).second.clear();
                break;
            }
        }
    }
    std::pair<int,my_map> temp;
    temp.first = k ;
    temp.second = G_k_1;
    memory.at(n).push_back( temp ); //storing in hash table for further use
    return memory.at(n).back().second;
}


int main()
{
   my_vector v1,v2,v3,v4,v5;
   my_set s1,s2,s3,s4,s5;
   for(int i = 1; i<=5; ++i)
   {
      v1.push_back(1);
      v2.push_back(2);
      v3.push_back(3);
      v4.push_back(4);
      v5.push_back(5);
   }


   s1.insert(1);
   s2.insert(2);
   s3.insert(3);
   s4.insert(4);
   s5.insert(5);

   bases.insert(bases.end(),make_pair(s1,v1));
   bases.insert(bases.end(),make_pair(s2,v2));
   bases.insert(bases.end(),make_pair(s3,v3));
   bases.insert(bases.end(),make_pair(s4,v4));
   bases.insert(bases.end(),make_pair(s5,v5));

   my_set empty_set;
   my_vector empty_group(DIMENSIONS);
   G.insert(make_pair(empty_set,empty_group));

   vector<std::pair<int,my_map> > empty_element;
   empty_element.push_back(make_pair(0,G));
   for (int i = 0; i <= 5; i++) {  // 5 is the total number od vectors : v1,v2,v3,v4,v5
       memory.push_back(empty_element);
   }



   data.insert(bases.begin(),bases.end());
   cout << endl << "The intial set of vectors are : " << endl;
   print ( data );

   int k;
   cout << "N = 5 " << endl << "Enter the value of k : ";
   cin >> k;

   cout << "The values for N choose k are : " << endl;
   data = algorithm_(k,5); 

   print ( data ); 

}

If you run the program you know what I wanted to achieve and in what way. This ALGORITHM (not program) might not be efficient for a smaller number of vectors but it will be when N > 50k and k ~ 10. I'm aware the implementation of the algorithm (my program) is very inefficient. Is there any way to improve it ? I think the same algorithm can be implemented in a much more elegant fashion. Any help is much appreciated. Thanks.

My apologies for misunderstanding your answer previously, I didn't really understand what you were trying to do in your post, I thought you were just looking for a non-recursive way of computing nCk :P

I've created a class, CombinationGenerator to produce the combinations of vectors, which I believe is what you want. It works by producing a vector of ints representing the indices of the elements to aggregate (I've included a main function below which should help to explain it programmatically).

Here is the header file: http://pastebin.com/F5x4WKD9

And the source file: http://pastebin.com/CTV1PLRb

And here is a sample main function:

typedef std::vector<int> vecInt;

int main() {

    // We have a deque containing 3 elements (try using experimenting with data
    // types to test space complexity, std::set or std::unordered_set might be an option)
    vecInt vec1;
    for( int i = 0; i < 3; i++ )
    {
        vec1.push_back(1);
    }
    vecInt vec2;
    for( int i = 0; i < 3; i++ )
    {
        vec2.push_back(2);
    }
    vecInt vec3;
    for( int i = 0; i < 3; i++ )
    {
        vec3.push_back(3);
    }
    vecInt vec4;
    for( int i = 0; i < 3; i++ )
    {
        vec4.push_back(4);
    }
    vecInt vec5;
    for( int i = 0; i < 3; i++ )
    {
        vec5.push_back(5);
    }

    std::deque<std::vector<int>> dequeVecs;
    dequeVecs.push_back( vec1 );
    dequeVecs.push_back( vec2 );
    dequeVecs.push_back( vec3 );
    dequeVecs.push_back( vec4 );
    dequeVecs.push_back( vec5 );

    // Create our CombinationGenerator:
    CombinationGenerator* gen = new CombinationGenerator();

    g_pCombinationGen = gen;

    gen = NULL;

    unsigned long long size = g_pCombinationGen->ComputeBinomialCoefficient( dequeVecs.size(), 2 );

    std::vector<int> currCombination;

    g_pCombinationGen->Initialize( dequeVecs.size(), 2, size );

    while( !g_pCombinationGen->IsFinished() )
    {
        currCombination = g_pCombinationGen->NextCombination();

        std::vector<int> result;
        for( int i = 0; i < dequeVecs[0].size(); i++ )
        {
            result.push_back( dequeVecs[currCombination[0]][i] + dequeVecs[currCombination[1]][i] );
        }

        std::cout << "(";
        for( int i = 0; i < result.size(); i++ )
        {
            std::cout << result[i];
        }
        std::cout << ")" << std::endl;

    }

    return 0;

}

Whilst this may look rather large, if you analyze the space usage of it (let's assume that you're using n = 50,000 and k = 1000:

There are 50,000 vectors each containing 3 ints (let's assume a reasonably harsh overhead per vector of 32bytes, it's normally around 20 on most implementations): So, (50,000 * 3 * 4) + (50,000 * 32) = 2,200,000 Bytes

You're then containing this in a deque, which we will also assume has an overhead of 32bytes: 2,200,000 + 32 = 2,200,032 Bytes

We're also having an instance running of the Combination Generator, this has 5 member variables, two ints, two long longs, and a vector containing k ints (in this case 1000), so: 2,200,032 + (2*4) + (2*8) + (1000*4) + 32 = 2,204,056 Bytes

We also have the vector containing the result for each iteration again with k ints: 2,204,056 + (1000*4) + 32 = 2,208,088 Bytes

As you can see, this is far below your 4GB of memory. NOTE: It would not be possible to store each of these vectors in memory regardless of what implementation you used, as there would be over 9.94 x 10^2126 vectors containing results. Even if you chose a small value of k such as 10, you would still have over 2.69 x 10^40.

I hope this time I've understood what it is you're asking for! If not, I'll try and understand again what it is you're trying to achieve. :)