Generate all combinations for a list of strings

Question

I want to generate a list of all possible combinations of a list of strings (it's actually a list of objects, but for simplicity we'll use strings). I need this list so that I can test every possible combination in a unit test.

So for example if I have a list of:

  var allValues = new List<string>() { "A1", "A2", "A3", "B1", "B2", "C1" }

I need a List<List<string>> with all combinations like:

  A1
  A2
  A3
  B1
  B2
  C1
  A1 A2
  A1 A2 A3
  A1 A2 A3 B1
  A1 A2 A3 B1 B2
  A1 A2 A3 B1 B2 C1
  A1 A3
  A1 A3 B1
  etc...

A recursive function is probably the way to do it to get all combinations, but it seems harder than I imagined.

Any pointers?

Thank you.

EDIT: two solutions, with or without recursion:

public class CombinationGenerator<T>
{
    public IEnumerable<List<T>> ProduceWithRecursion(List<T> allValues) 
    {
        for (var i = 0; i < (1 << allValues.Count); i++)
        {
            yield return ConstructSetFromBits(i).Select(n => allValues[n]).ToList();
        }
    }

    private IEnumerable<int> ConstructSetFromBits(int i)
    {
        var n = 0;
        for (; i != 0; i /= 2)
        {
            if ((i & 1) != 0) yield return n;
            n++;
        }
    }

    public List<List<T>> ProduceWithoutRecursion(List<T> allValues)
    {
        var collection = new List<List<T>>();
        for (int counter = 0; counter < (1 << allValues.Count); ++counter)
        {
            List<T> combination = new List<T>();
            for (int i = 0; i < allValues.Count; ++i)
            {
                if ((counter & (1 << i)) == 0)
                    combination.Add(allValues[i]);
            }

            // do something with combination
            collection.Add(combination);
        }
        return collection;
    }
}

I know this isn't quite what you were looking for but Microsoft has this system in beta that will auto generate inputs combinations for you. It is called Pex: http://research.microsoft.com/en-us/projects/pex/ — Christopher Rathermel, May 09 '12 at 11:51
possible duplicate of [Listing all permutations of a string/integer](http://stackoverflow.com/questions/756055/listing-all-permutations-of-a-string-integer) — Oliver, May 09 '12 at 12:13
You don't need recursion: http://stackoverflow.com/questions/10331229/the-different-combinations-of-a-vectors-values/10331312#10331312 — Henrik, May 09 '12 at 12:29
You are looking for all subsets of a set not all combinations or permutations of the set. — stuckintheshuck, Sep 05 '13 at 00:32
Disclaimer: This may destroy the known universe. For more information about this, check: https://en.wikipedia.org/wiki/The_Nine_Billion_Names_of_God — Leandro Bardelli, Mar 06 '23 at 20:47

Vlad · Accepted Answer · 2012-05-09T12:25:29.443

You can make in manually, using the fact that n-bit binary number naturally corresponds to a subset of n-element set.

private IEnumerable<int> constructSetFromBits(int i)
{
    for (int n = 0; i != 0; i /= 2, n++)
    {
        if ((i & 1) != 0)
            yield return n;
    }
}

List<string> allValues = new List<string>()
        { "A1", "A2", "A3", "B1", "B2", "C1" };

private IEnumerable<List<string>> produceEnumeration()
{
    for (int i = 0; i < (1 << allValues.Count); i++)
    {
        yield return
            constructSetFromBits(i).Select(n => allValues[n]).ToList();
    }
}

public List<List<string>> produceList()
{
    return produceEnumeration().ToList();
}

Tim Schmelter · Answer 2 · 2012-05-09T13:25:01.193

If you want all variations, have a look at this project to see how it's implemented.

http://www.codeproject.com/Articles/26050/Permutations-Combinations-and-Variations-using-C-G

But you can use it since it's open source under CPOL.

For example:

var allValues = new List<string>() { "A1", "A2", "A3", "B1", "B2", "C1" };
List<String> result = new List<String>();
var indices = Enumerable.Range(1, allValues.Count);
foreach (int lowerIndex in indices)
{
    var partVariations = new Facet.Combinatorics.Variations<String>(allValues, lowerIndex);
    result.AddRange(partVariations.Select(p => String.Join(" ", p)));
}

var length = result.Count;  // 1956

score 0 · Answer 3 · answered Oct 15 '16 at 07:02

One more recursive solution. From AllCombinations in below code, you will get all possible combinations. Logic:

Starting with one element.
Generate all possible combinations with it.
Move to next element and begin with step 2 again.

Code:

public class Combination<T>
{
    private IEnumerable<T> list { get; set; }
    private int length;
    private List<IEnumerable<T>> _allCombination;

    public Combination(IEnumerable<T> _list)
    {
        list = _list;
        length = _list.Count();

        _allCombination = new List<IEnumerable<T>>();
    }

    public IEnumerable<IEnumerable<T>> AllCombinations
    {
        get
        {
            GenerateCombination(default(int), Enumerable.Empty<T>());

            return _allCombination;
        }
    }

    private void GenerateCombination(int position, IEnumerable<T> previousCombination)
    {
        for (int i = position; i < length; i++)
        {
            var currentCombination = new List<T>();
            currentCombination.AddRange(previousCombination);
            currentCombination.Add(list.ElementAt(i));

            _allCombination.Add(currentCombination);

            GenerateCombination(i + 1, currentCombination);

        }
    }
}

Generate all combinations for a list of strings

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