使用位掩码生成排列

Question

I am generating all permutations of a string using bitmask.

void recurse(string s, int mask,int taken){

    if(taken == n){
        cout << " ";
        return;
    }
    for(int i = 0; i < n; i++){
        if(((1 << i) & mask) == 0){
            cout << s[i];
            recurse(s, (mask|(1 << i)), taken + 1);
        }
    }
}

In this function n is the length of the string. I am keeping track of how many characters are printed so far using taken variable. In the main function I am calling

recurse(s,0,0);

But this is not working correctly. For input

red

Its output is

red de erd dr dre er

Where I am going wrong?

---

UPDATE // Below code works fine.

void recurse(string s, int mask,int taken, string pref){

    if(taken == n){
        cout << pref <<endl; 
        return;
    }
    for(int i = 0; i < n; i++){
        if(((mask >> i) & 1) == 0){
            recurse(s,(mask | (1 << i)),taken + 1, pref + s[i]);
        }
    }
}

Answer 1

Actually, the questioner provided the answer himself. (Congratulation.)

As I already started to fiddle (couldn't resist) I want to present my solution as well:

#include <iostream>
#include <string>

using namespace std;

void recurse(
  const string &s, unsigned mask = 0, const string &out = string())
{
  size_t n = s.size();
  if (out.size() == n) cout << ' ' << out;
  for (size_t i = 0; i < n; ++i) {
    unsigned bit = 1 << i;
    if (mask & bit) continue;
    recurse(s, mask | bit, out + s[i]);
  }
}

int main()
{
  string test = "red";
  recurse(test);
  cout << endl;
  return 0;
}

Compiled and tested:

 red rde erd edr dre der

recurse() iterates through all characters of s looking for one which is not already marked in the mask as taken. Each found characters is added to output out . Then, the recursive call repeats it for all untaken characters.

Check out the sample code yourself on ideone .

Answer 2

Permutation tree for 'abc'

Your first code was visiting each node of this tree once and printing a character. So it was giving wrong output.

On a different note, you have used some redundant variables.

Instead of n you should use s.size() .

In the second code instead of taken you you should use pref.size() .

Answer 3

Here is another version. It's different from the questioner's code in 2 ways:

1. The parameter taken can be omitted and we can use mask + 1 == (1 << n) instead. It basically checks if bits 1 to n-1 of mask are all 1's. If so, then the recursion depth is n and we print the permutation.

2. Copying the string pref in each iteration can be slow if the size of the string is large. We can instead use a reference.

#include <iostream>
#include <string>

using namespace std;

void recurse(string s, int mask, string &pref);

int n = 3;

int main()
{
    string pref("");
    recurse(string("abc"), 0, pref);
    return 0;
}

void recurse(string s, int mask, string &pref)
{
    if (mask + 1 == (1 << n)) {
        cout << pref << endl;
        return;
    }
    for (int i = 0; i < n; i++) {
        if (((mask >> i) & 1) == 0) {
            pref += s[i];
            recurse(s, (mask | (1 << i)), pref);
            pref.erase(pref.end() - 1);
        }
    }
}

where n is the size of the string. The output is

abc
acb
bac
bca
cab
cba