Re: Create a permutation list

From:
"Francesco S. Carta" <entuland@gmail.com>
Newsgroups:
comp.lang.c++
Date:
Tue, 6 Oct 2009 03:00:22 -0700 (PDT)
Message-ID:
<7b31b7ef-ca68-46d0-93a2-f3d25619a0af@a6g2000vbp.googlegroups.com>
On 30 Set, 22:08, Kai-Uwe Bux <jkherci...@gmx.net> wrote:

Jerry Coffin wrote:

In article <6273cccb-0511-4cb5-acff-
f188e37b6...@l9g2000yqi.googlegroups.com>, markred...@gmail.com
says...

Hello world,
I'm looking for an algorithm in order to perform this operation: I
have a set of N possible values and I want to generate all the
possibile permutations composed by M slots without repetitions and not
considering the order.

Note: I don't know what are the values of N and M befor the software
starts, so i have to implement it at run-time.

For example, if N=4 with values {1 , 2 , 3 , 4} and M=3, I will ob=

tain

only 4 combinations, for example:
(1,2,3) [or (3,2,1), or (2,1,3), the order doesn't matter]
(1,3,4)
(1,2,4)
(2,3,4)

Can you describe an algorithm? (I don't need code, I hope I will able
to develop it!)


From a C++ viewpoint, the proper answer would probably be
std::next_permutation (and possibly std::prev_permutation).


I guess, there is a confusion caused by the OP's use of "permutation" in =

the

subject line. The example of the post, however, makes is pretty clear tha=

t

the OP really wants to iterate through the list of M-element subsets of a=

n

N-element set. He even points out that order does not matter.

Hint to the OP: think of the numbers 0, 1, 2, ..., 2^N - 1. These are
exactly the numbers that can be encoded by N bits, i.e., they correspond =

to

subsets of an N-element set (the 1-bits indicate the chosen elements). Yo=

u

could now create a next_M_subset() function by answering the following
question:

  Given a number k in [0,2^N) that has M set bits, what is the
  next number that has M bits set?

E.g.: N = 5, M = 3 would yield the following order:

  00111
  01011
  01110
  10011
  10101
  10110
  11001
  11010
  11100


I suppose the OP wasn't after any homework, and even in such case,
that should be past due after a week (this sentence is about the fact
that I'm posting a working algorithm, nothing more).

Whatever, here is my implementation of the above suggestion:
-------
#include <iostream>
#include <sstream>
#include <vector>
#include <iomanip>

using namespace std;

typedef vector<unsigned> Combo;
typedef vector<Combo> Combos;

Combos combinations(unsigned slots, unsigned elements) {
    Combos combos;
    if (slots && slots <= elements) {
        vector<bool> flags;
        Combo combo;
        for(unsigned i = 0; i < elements; ++i) {
            flags.push_back(i < slots);
        }
        do {
            for(unsigned i = 0, e = flags.size(); i < e; ++i) {
                if(flags[i]) combo.push_back(i);
            }
            combos.push_back(combo);
            combo.clear();
        } while (prev_permutation(flags.begin(), flags.end()));
    }
    return combos;
}

string qty(unsigned number, const string& noun) {
    ostringstream oss;
    unsigned pos = noun.find("|");
    oss << number;
    oss << noun.substr(0, pos);
    if(number != 1 && pos < noun.size()-1) {
        oss << noun.substr(pos+1);
    }
    return oss.str();
}

ostream& operator<<(ostream& os, const Combo& combo) {
    os << "{ ";
    if(unsigned combo_size = combo.size()) {
        for(unsigned i = 0; i < combo_size -1; ++i) {
            os << combo[i] << ", ";
        }
        os << combo.back() << " ";
    }
    os << "}";
    return os;
}

ostream& operator<<(ostream& os, const Combos& combos) {
    os << qty(combos.size(), " combination|s") << endl;
    for(unsigned i = 0, e = combos.size(); i < e; ++i) {
        os << " " << setw(4) << (i+1) << ": " << combos[i] << endl;
    }
    return os;
}

void print_combinations(unsigned slots, unsigned elements) {
    cout << " " << qty(elements, " element|s");
    cout << ", arranged in\n " << qty(slots, " slot|s");
    cout << ", generate" << ( elements == 1 ? "s\n " : "\n " );
    cout << combinations(slots, elements) << endl;
}

int main() {
    print_combinations(0, 0);
    print_combinations(0, 1);
    print_combinations(1, 0);
    print_combinations(1, 1);
    print_combinations(4, 5);
    return 0;
}
-------

Like it?

--
 Francesco S. Carta, http://fscode.altervista.org

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