Generating N choose K Permutations in C++ [duplicate] - c++

Question

This question already has answers here:
Implementation of Permutation, Combinations and PowerSet in C++ [duplicate]
(2 answers)
Closed 8 years ago.
I have a function that receives n and k to create all the possible permutations of n choose k, and while it works for most combinations like 5 choose 3 or 3 choose 2, it doesn't for for others like 4 choose 2. I need some help finding and understanding the bug. Thanks for looking.
The function:
void PermGenerator(int n, int k)
{
int d[] = {1,2,3,4,5,6,7,8,9};
sort (d, d+n);
cout << "These are the Possible Permutations: " << endl;
do
{
for (int i = 0; i < k; i++)
{
cout << d[i] << " ";
if (i == k-1) cout << endl;
}
} while (next_permutation(d, d+n));
}
I'm using the next_permutation function. cplusplus
When I try 4 choose 2, I should be getting 12 permutations, instead I get this:
1 2
1 2
1 3
1 3
1 4
1 4
2 1
2 1
2 3
2 3
2 4
2 4
3 1
3 1
3 2
3 2
3 4
3 4
4 1
4 1
4 2
4 2
4 3
4 3
Whereas, 3 choose 2 works perfectly with 6 possible permutations:
1 2
1 3
2 1
2 3
3 1
3 2

Answer 1

The first k values are repeated n-k factorial times. Here is an easy, although inefficient, way to avoid the repetition:
int Factorial(int n)
{
int result = 1;
while (n>1) {
result *= n--;
}
return result;
}
void PermGenerator(int n, int k)
{
std::vector<int> d(n);
std::iota(d.begin(),d.end(),1);
cout << "These are the Possible Permutations: " << endl;
int repeat = Factorial(n-k);
do
{
for (int i = 0; i < k; i++)
{
cout << d[i] << " ";
}
cout << endl;
for (int i=1; i!=repeat; ++i)
{
next_permutation(d.begin(),d.end());
}
} while (next_permutation(d.begin(),d.end()));
}
However, there is an even easier and more efficient way to do it using std::reverse (from https://stackoverflow.com/a/2616837/951890)
void PermGenerator(int n, int k)
{
std::vector<int> d(n);
std::iota(d.begin(),d.end(),1);
cout << "These are the Possible Permutations: " << endl;
do
{
for (int i = 0; i < k; i++)
{
cout << d[i] << " ";
}
cout << endl;
std::reverse(d.begin()+k,d.end());
} while (next_permutation(d.begin(),d.end()));
}
The trick here is to realize that the last permutation is just the reverse of the first permutation, so by reversing the last n-k elements, you automatically skip to the last permutation of those elements.

Answer 2

You may use the following:
template <typename T>
void Combination(const std::vector<T>& v, std::size_t count)
{
assert(count <= v.size());
std::vector<bool> bitset(v.size() - count, 0);
bitset.resize(v.size(), 1);
do {
for (std::size_t i = 0; i != v.size(); ++i) {
if (bitset[i]) {
std::cout << v[i] << " ";
}
}
std::cout << std::endl;
} while (std::next_permutation(bitset.begin(), bitset.end()));
}
Live example

Answer 3

You output first k members of every n! permutations.
4! = 24 permutations. First two permutations are
1,2,3,4
1,2,4,3
and you have got 1,2 and 1,2
To get combinations (4,2), you might, for example, use vector
{0,0,1,1}
permute it, and output indexes of 1's