Calculate Combination based on position

I 开发者_高级运维have combinations like this:

1,2,3,4 //index 0

1,2,3,5 //index 1

1,2,3,6 //index 2

and so on until 7,8,9,10

So this will be n=10 k=4 from combinatorics

How calculate combination by index

For example when my index==1

myCmb = func(index)

returns 1,2,3,5

this is example i need this for bigest numbers, and for more params and without (if this possible) many loops

i find something like this to obtain position: http://answers.yahoo.com/question/index?qid=20110320070039AA045ib

I want now reverse this...

I programming in C++ Thanks for any sugesstions or help

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It seems like you want to find the k-combination for a given number.

Following the example, here's something that should work:

#include <cstddef>
#include <iostream>
#include <string>
#include <boost/lexical_cast.hpp>
#include <boost/math/special_functions/binomial.hpp>


std::size_t Choose(double n, double k) {
  using boost::math::binomial_coefficient;
  if (n < k) return 0;
  return static_cast<std::size_t>(binomial_coefficient<double>(n, k));
}

// Returns the largest n such that Choose(n, k) <= pos.
int CombinationElement(int k, int pos) {
  int n = k;
  int coeff = 1, prev_coeff = 0;

  while (coeff <= pos) {
    coeff = Choose(++n, k);
    prev_coeff = coeff;
  }

  return n - 1;
}

// Returns an k-combination at position pos.
std::vector<int> Combination(int k, int pos) {
  std::vector<int> combination;
  for (; k > 0; --k) {
    int n = CombinationElement(k, pos);
    combination.push_back(n);
    pos -= Choose(n, k);
  }
  return combination;
}

int main(int argc, char** argv) {
  using std::cout;
  using std::endl;

  if (argc != 3) {
    cout << "Usage:  $ " << argv[0] << " K POS" << endl;
    cout << "Prints the K-combination at position POS." << endl;
    return 1;
  }

  int k = boost::lexical_cast<int>(argv[1]);
  int pos = boost::lexical_cast<int>(argv[2]);

  std::vector<int> combination = Combination(k, pos);

  for (int i = 0; i < k; i++)
    cout << combination[i] << " ";
  cout << std::endl;
}

Note, for convenience, the code depends on Boost to calculate binomial coefficients (boost::math::binomial_coefficient<T>), and to parse strings into integers (boost::lexical_cast).

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Here is an implementation in Mathematica, from the package Combinatorica. The semantics are fairly generic, so I think it is helpful. Please leave a comment if you need anything explained.

UnrankKSubset::usage = "UnrankKSubset[m, k, l] gives the mth k-subset of set l, listed in lexicographic order."

UnrankKSubset[m_Integer, 1, s_List] := {s[[m + 1]]}
UnrankKSubset[0, k_Integer, s_List] := Take[s, k]
UnrankKSubset[m_Integer, k_Integer, s_List] := 
       Block[{i = 1, n = Length[s], x1, u, $RecursionLimit = Infinity}, 
             u = Binomial[n, k]; 
             While[Binomial[i, k] < u - m, i++]; 
             x1 = n - (i - 1); 
             Prepend[UnrankKSubset[m - u + Binomial[i, k], k-1, Drop[s, x1]], s[[x1]]]
       ]

Usage is like:

UnrankKSubset[1, 4, {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}]

   {1, 2, 3, 5}

As you can see this function operates on sets.

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Below is my attempt to explain the code above.

UnrankKSubset is a recursive function with three arguments, called (m, k, s):

1. m an Integer, the "rank" of the combination in lexigraphical order, starting from zero.
2. k an Integer, the number of elements in each combination
3. s a List, the elements from which to assemble combinations

There are two boundary conditions on the recursion:

1. for any rank m, and any list s, if the number of elements in each combination k is 1, then:

   return the m + 1 element of the list s, itself in a list.

   (+ 1 is needed because Mathematica indexes from one, rather than zero. I believe in C++ this would be s[m] )

2. if rank m is 0 then for any k and any s:

   return the first k elements of s

The main recursive function, for any other arguments than ones specified above:

local variables: (i, n, x1, u)

Binomial is binomial coefficient: Binomial[7, 5] = 21

Do:

i = 1
n = Length[s]
u = Binomial[n, k]
While[Binomial[i, k] < u - m, i++];
x1 = n - (i - 1);

Then return:

Prepend[
  UnrankKSubset[m - u + Binomial[i, k], k - 1, Drop[s, x1]], 
  s[[x1]]
]

That is, "prepend" the x1 element of list s (remember Mathematica indexes from one, so I believe this would be the x1 - 1 index in C++) to the list returned by the recursively called UnrankKSubset function with the arguments:

• m - u + Binomial[i, k]
• k - 1
• Drop[s, x1]

Drop[s, x1] is the rest of list s with the first x1 elements removed.

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If anything above is not understandable, or if what you wanted was an explanation of the algorithm, rather than an explanation of the code, please leave a comment and I will try again.