CODEWITHSUNDEEP
algorithmcombinations
UnTraDe
UnTraDe

Reputation: 3867

Getting word combinations from group of letters

I searched over the internet a lot how to do this but I didn't came up with something I completely understood.

Im trying to generate all the possible combinations from an array of letters by specifying the amount of letters in each group, for example:

letters: A, B, C

length: 2

result: AB, AC, BC

(I know there are: BA, CA and CB too but I only need to get the groups the order isn't matter.)

example 2:

letters: A, B, C, D

length: 3

result: ABC, ACD, BCD, CDA, DAB

and etc…

I intend to implement that algorithm in C++ but examples in C#, Java or Javascript are welcome as well.

Upvotes: 1

Views: 2123

Answers (4)

Konchog
Konchog

Reputation: 2188

This is known as permutation, there are lots of solutions for it. Here is a non-recursive one that I wrote which is super fast. (If you are on Windows, you may need to look-up _BitScanReverse instead of using __builtin_ctz).

#include <iostream>
#include <cstdlib>
using namespace std;

void perm(unsigned &v) { //v must be initialised with t = ( 2 << N ) - 1;
    unsigned t = v | (v - 1);
    v = (t + 1) | (((~t & -~t) - 1) >> (__builtin_ctz(v) + 1));
}

int main (int argc, char *argv[]) {
    unsigned n = argc > 1 ? atoi(argv[1]) : 3; //bins:   0..31
    unsigned k = argc > 2 ? atoi(argv[2]) : 2; //things: 0 .. bins.
    unsigned m = (1 << n) - 1; //bitmask is size of n (bins).
    unsigned v = (1 << k) - 1; //start value - initial allocation.
    do {
        cout << bitset<31>(v & m) << endl;
        perm(v);
    } while (v < m);
    return 0;
}

In your question you suggest - letters: A, B, C length: 2 as an example. So, this code would generate (passing 3 2 as arguments) (which I have commented)

ABC
011 //BC
101 //AC
110 //AB

Upvotes: 1

android927
android927

Reputation: 293

This should work if you tweak it a little:

void r_nCr(const unsigned int &startNum, const unsigned int &bitVal, const unsigned int &testNum) // Should be called with arguments (2^r)-1, 2^(r-1), 2^(n-1)
{
    unsigned int n = (startNum - bitVal) << 1;
    n += bitVal ? 1 : 0;

    for (unsigned int i = log2(testNum) + 1; i > 0; i--) // Prints combination as a series of 1s and 0s
        cout << (n >> (i - 1) & 1);
    cout << endl;

    if (!(n & testNum) && n != startNum)
        r_nCr(n, bitVal, testNum);

    if (bitVal && bitVal < testNum)
        r_nCr(startNum, bitVal >> 1, testNum);
}

Explanation here.

Upvotes: 0

user unknown
user unknown

Reputation: 36250

If you order them in a reproducable way, you will find an algorithm to produce them more easily:

Let's make it not too easy, take 3 of 5: 

e d c b a 
---------
    x x x abc
  x   x x abd
x     x x abe
  x x   x acd 
x   x   x ace
x x     x ade
  x x x   bcd
x   x x   bce
x x   x   bde 
x x x     cde

Upvotes: 1

Kodra
Kodra

Reputation: 1626

Seems like a good fit for recursion. Take each element, and prepend it to the remaining combinations until a given depth is met.

static List<String> func(List<String> a, Int32 depth)
{
    List<String> retVal = new List<String>();
    if (depth <= 0)
    {
        return retVal;
    }
    for (int i = 0; i < a.Count; i++)
    {
        String element = a[i];

        List<String> temp = new List<String>(a);
        temp.RemoveAt(i);
        List<String> resultset = func(temp, depth - 1);
        if (resultset.Count == 0)
        {
            retVal.Add(element);
        }
        else
        {

            foreach (String result in resultset)
            {
                retVal.Add(element + result);
            }
        }
    }
    return retVal;
}

Upvotes: 1

Related Questions