all combinations of k elements out of n - c++

Question

Can somebody provide me a link or pseudocode of a function for finding all combinations of k elements out of n? possibly in STL. I don't need to compute n choose k, I need to list all vectors of numbers of size k.
Thanks

Answer 1

In C++ given the following routine:
template <typename Iterator>
inline bool next_combination(const Iterator first, Iterator k, const Iterator last)
{
/* Credits: Thomas Draper */
if ((first == last) || (first == k) || (last == k))
return false;
Iterator itr1 = first;
Iterator itr2 = last;
++itr1;
if (last == itr1)
return false;
itr1 = last;
--itr1;
itr1 = k;
--itr2;
while (first != itr1)
{
if (*--itr1 < *itr2)
{
Iterator j = k;
while (!(*itr1 < *j)) ++j;
std::iter_swap(itr1,j);
++itr1;
++j;
itr2 = k;
std::rotate(itr1,j,last);
while (last != j)
{
++j;
++itr2;
}
std::rotate(k,itr2,last);
return true;
}
}
std::rotate(first,k,last);
return false;
}
You can then proceed to do the following:
// 9-choose-3
std::string s = "123456789";
std::size_t k = 3;
do
{
std::cout << std::string(s.begin(),s.begin() + k) << std::endl;
}
while(next_combination(s.begin(),s.begin() + k,s.end()));
Or for a std::vector of int's:
// 5-choose-3
std::size_t n = 5;
std::size_t k = 3;
std::vector<int> ints;
for (int i = 0; i < n; ints.push_back(i++));
do
{
for (int i = 0; i < k; ++i)
{
std::cout << ints[i];
}
std::cout << "\n";
}
while(next_combination(ints.begin(),ints.begin() + k,ints.end()));

Answer 2

http://howardhinnant.github.io/combinations.html
Search for "for_each_combination". If you find something faster, please let me know. Unlike other algorithms I often see, this one doesn't require the element type to be LessThanComparable.

Answer 3

Create an auxiliary vector with n - k zeros followed by k ones. A zero means the element in the original container is not included, whereas one means the element is included.
Now use std::next_permutation on the auxiliary vector to get the next combination.

Answer 4

Here is a lazy example of pseudocode that can get the job done...
void nChooseK(array[n],k){
recurse("",array[n],k);
}
void recurse(initialString,array[n],k){
if(k == 0){
print initialString;
return;
}
for(i=0;i<n;i++){
tmpArray = array[0...i-1]+array[i+1...];//the array without the object to remove
recurse(initialString + array[i], tmpArray,k-1)
}
}

Answer 5

You could use std::next_permutation but it is n! and not n choose k. You could filter them after you created them. But this solution is O(n!), not really perfect. Here is the trial and error solution:
int factorial(int value)
{
int result = 1;
for(int i = 1; i <= value; i++)
{
result *= i;
}
return result;
}
std::set<std::set<int>> binomial_coefficient(std::vector<int> input, int k)
{
std::set<std::set<int>> solutions;
for(unsigned int i = 0; i < factorial(input.size()); i++)
{
std::next_permutation(input.begin(), input.end());
solutions.insert(std::set<int>(input.begin(), input.begin() + k));
}
return solutions;
}