Permutation algorithm C++

Question

I try to translate an algorithm that generates all permutations of k out of n in C++ :

public void calculerEquipeTOT(ArrayList<Nageur> L, ArrayList<Nageur> F, int k) {

    if (k == 0) {
        if (calculerPointsTOT(L) > this.pointsMeilleureEquipe){
            this.meilleureEquipe = L;
            this.pointsMeilleureEquipe = calculerPointsTOT(meilleureEquipe);
        }
    } else {
        for (Nageur x : F) {
            ArrayList<Nageur> G = new ArrayList<Nageur>(F);
            G.remove(G.indexOf(x));
            ArrayList<Nageur> L2 = new ArrayList<Nageur>(L);
            L2.add(x);
            calculerEquipeTOT(L2, G, k - 1);
        }
    }
}

My problem is that the Lists could be Objects list and I don't know how to remove the x of the L2 list... I am not a C++ specialist, I managed it in Java but I have to do it in C++.

Because I understand that std::next_permutation works on sorted array, and I want to use an Object List like vector<People> for example, no ? — Terry Blain
– Terry Blain, Commented Jun 7, 2015 at 21:58
@TerryBlain No. std::next_permutation will help you in calculating all permutations of a range: std::vector<People> ps = ...; do { doSomething(ps); } while ( std::next_permutation(ps.begin(), ps.end()) );. See en.cppreference.com/w/cpp/algorithm/next_permutation for an example and detailed reference. — stefan
– stefan, Commented Jun 7, 2015 at 22:01
Ok thank you @stefan but in fact I had to find all k permutation of n... — Terry Blain
– Terry Blain, Commented Jun 7, 2015 at 22:04
@TerryBlain I don't think you actually mean "permutation", but "subset". If I understand correctly, you're searching for all subsets of size k of a set S that has size n, right? In that case, permutation won't get you anywhere near the right solution. — stefan
– stefan, Commented Jun 7, 2015 at 22:07

Vlad from Moscow · Accepted Answer · 2015-06-07 22:26:27Z

3

I have transliterated your function and I have gotten the following

#include <iostream>
#include <list>
#include <iterator>

void arrangements( std::list<char> l, std::list<char> f, size_t k ) 
{
    if ( k == 0 ) 
    {
        for ( char c : l ) std::cout << c << ' ';
        std::cout << std::endl;
    } 
    else 
    {
        for ( auto it = f.begin(); it != f.end(); ++it )
        {
            std::list<char> g( f.begin(), it );
            g.insert( g.end(), std::next( it ), f.end() );

            std::list<char> l2( l );
            l2.push_back( *it );

            arrangements( l2, g , k-1 );
        }
    }
}

int main()
{
    std::list<char> f = { 'A', 'B', 'C', 'D' };

    arrangements( std::list<char>(), f, 2 );
}

The program output is

A B 
A C 
A D 
B A 
B C 
B D 
C A 
C B 
C D 
D A 
D B 
D C

I do not know whether it is what you want to get.

If to call the function with k equal to 3 then the program output will be

A B C 
A B D 
A C B 
A C D 
A D B 
A D C 
B A C 
B A D 
B C A 
B C D 
B D A 
B D C 
C A B 
C A D 
C B A 
C B D 
C D A 
C D B 
D A B 
D A C 
D B A 
D B C 
D C A 
D C B

edited Jun 7, 2015 at 22:26

answered Jun 7, 2015 at 22:19

Vlad from Moscow

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12 Comments

Terry Blain Over a year ago

This is exactly what I want (I have updated the topic with my Java code, but I don't know if it is the most efficient way to do it)

Vlad from Moscow Over a year ago

@Terry Blain I also do not know.:)

Vlad from Moscow Over a year ago

@Terry Blain The C++ algorithm is not efficient. it would be better do not create a new list G and use the same list F in the recursion calls.:)

Vlad from Moscow Over a year ago

@Terry Blain Also it seems it would be better to use method slice of the list and pass the list by reference. I simply quickly translated what you initially showed in pseudo code . So you can improve the code yourself.:)

Terry Blain Over a year ago

Yes I am interested by better performance.. What do you mean exactly ? Thank you btw !

|

Terry Blain · Accepted Answer · 2015-06-10 07:22:47Z

I found a way to do what I want using next_permutation() from standard library and also another next_combination() from this article : http://www.codeguru.com/cpp/cpp/algorithms/combinations/article.php/c5117/Combinations-in-C.htm

My solution :

int main(int argc, const char * argv[]) {

    cout << "Hello, World!\n";
    int nb = 0;

    int tab1[] = {0,1,2,3};
    vector<int> n (tab1, tab1+sizeof tab1 / sizeof tab1[0]);
    int tab2[] = {0,1};
    vector<int> r (tab2, tab2+sizeof tab2 / sizeof tab2[0]);

    sort (n.begin(), n.end());
    do
    {
        sort(r.begin(),r.end());
        //do your processing on the new combination here
        vector<int> r2 = r;
        do
        {
            //do your processing on the new permutation here
            nb++;
            display_vector(r2);
            cout << endl;
        }
        while(next_permutation(r2.begin(),r2.end()));
    }
    while(next_combination(n.begin(),n.end(),r.begin(),r.end() ));

    cout << "Number of possibilities = " << nb << endl;

    return 0;
}

That displays :

Hello, World!
0 1 
1 0 
0 2 
2 0 
0 3 
3 0 
1 2 
2 1 
1 3 
3 1 
2 3 
3 2 
Number of possibilities = 12

It takes less than 1s to find all 10 permutations out of 12 on my computer... I don't know if this is good algorithm but it's faster than my previous algorithm in Java.

If someone see how to improve and optimize it I am interested ! :)

Monir · Accepted Answer · 2022-12-04 13:14:53Z

Call permute(nums) and pass nums as a vector that will return another vector containing all permuted numbers.

void getPermutation(vector<int>& nums,vector<int> current_indices, vector<vector<int>>& permutation)
    {
        if (nums.size()==current_indices.size()) 
        {
            vector <int> temp;
            for(auto index:current_indices)
                temp.push_back(nums[index]);
            permutation.push_back(temp);

            return;
        }

        vector <int> remaining_indices;
        for (int i=0;i<nums.size();i++)
        {
            if(std::find(current_indices.begin(), current_indices.end(), i) != current_indices.end()) 
            {
                //do nothing
            } 
            else 
            {
                remaining_indices.push_back(i);
            }
        }

        while (remaining_indices.size()>0)
        {
            vector<int> temp = current_indices;
            temp.push_back(remaining_indices[0]);
            getPermutation(nums,temp,permutation);
            remaining_indices.erase(remaining_indices.begin());
        }

        return;

    }
    vector<vector<int>> permute(vector<int>& nums) {

      vector<vector<int>> permutation;
      vector<int> current_indices; 
      getPermutation( nums,current_indices, permutation);
      
      return permutation;

        
    }

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