Get all subsets of a collection - c#

I am trying to create a method that will return all subsets of a set.
For example if I have the collection 10,20,30 I will like to get the following output
return new List<List<int>>()
{
new List<int>(){10},
new List<int>(){20},
new List<int>(){30},
new List<int>(){10,20},
new List<int>(){10,30},
new List<int>(){20,30},
//new List<int>(){20,10}, that substet already exists
// new List<int>(){30,20}, that subset already exists
new List<int>(){10,20,30}
};
because the collection can also be a collection of strings for instance I want to create a generic method. This is what I have worked out based on this solution.
static void Main(string[] args)
{
Foo<int>(new int[] { 10, 20, 30});
}
static List<List<T>> Foo<T>(T[] set)
{
// Init list
List<List<T>> subsets = new List<List<T>>();
// Loop over individual elements
for (int i = 1; i < set.Length; i++)
{
subsets.Add(new List<T>(){set[i - 1]});
List<List<T>> newSubsets = new List<List<T>>();
// Loop over existing subsets
for (int j = 0; j < subsets.Count; j++)
{
var tempList = new List<T>();
tempList.Add(subsets[j][0]);
tempList.Add(subsets[i][0]);
var newSubset = tempList;
newSubsets.Add(newSubset);
}
subsets.AddRange(newSubsets);
}
// Add in the last element
//subsets.Add(set[set.Length - 1]);
//subsets.Sort();
//Console.WriteLine(string.Join(Environment.NewLine, subsets));
return null;
}
Edit
Sorry that is wrong I still get duplicates...
static List<List<T>> GetSubsets<T>(IEnumerable<T> Set)
{
var set = Set.ToList<T>();
// Init list
List<List<T>> subsets = new List<List<T>>();
subsets.Add(new List<T>()); // add the empty set
// Loop over individual elements
for (int i = 1; i < set.Count; i++)
{
subsets.Add(new List<T>(){set[i - 1]});
List<List<T>> newSubsets = new List<List<T>>();
// Loop over existing subsets
for (int j = 0; j < subsets.Count; j++)
{
var newSubset = new List<T>();
foreach(var temp in subsets[j])
newSubset.Add(temp);
newSubset.Add(set[i]);
newSubsets.Add(newSubset);
}
subsets.AddRange(newSubsets);
}
// Add in the last element
subsets.Add(new List<T>(){set[set.Count - 1]});
//subsets.Sort();
return subsets;
}
Then I could call that method as:


This is a basic algorithm which i used the below technique to make a single player scrabble word solver (the newspaper ones).
Let your set have n elements. Increment an integer starting from 0 to 2^n. For each generater number bitmask each position of the integer. If the i th position of the integer is 1 then select the i th element of the set. For each generated integer from 0 to 2^n doing the above bitmasting and selection will get you all the subsets.
Here is a post: http://phoxis.org/2009/10/13/allcombgen/


Here is an adaptation of the code provided by Marvin Mendes in this answer but refactored into a single method with an iterator block.
public static IEnumerable<IEnumerable<T>> Subsets<T>(IEnumerable<T> source)
{
List<T> list = source.ToList();
int length = list.Count;
int max = (int)Math.Pow(2, list.Count);
for (int count = 0; count < max; count++)
{
List<T> subset = new List<T>();
uint rs = 0;
while (rs < length)
{
if ((count & (1u << (int)rs)) > 0)
{
subset.Add(list[(int)rs]);
}
rs++;
}
yield return subset;
}
}


I know that this question is a little old but i was looking for a answer and dont find any good here, so i want to share this solution that is an adaptation found in this blog: http://praseedp.blogspot.com.br/2010/02/subset-generation-in-c.html
I Only transform the class into a generic class:
public class SubSet<T>
{
private IList<T> _list;
private int _length;
private int _max;
private int _count;
public SubSet(IList<T> list)
{
if (list== null)
throw new ArgumentNullException("lista");
_list = list;
_length = _list.Count;
_count = 0;
_max = (int)Math.Pow(2, _length);
}
public IList<T> Next()
{
if (_count == _max)
{
return null;
}
uint rs = 0;
IList<T> l = new List<T>();
while (rs < _length)
{
if ((_count & (1u << (int)rs)) > 0)
{
l.Add(_list[(int)rs]);
}
rs++;
}
_count++;
return l;
}
}
To use this code you can do like something that:
List<string> lst = new List<string>();
lst.AddRange(new string[] {"A", "B", "C" });
SubSet<string> subs = new SubSet<string>(lst);
IList<string> l = subs.Next();
while (l != null)
{
DoSomething(l);
l = subs.Next();
}
Just remember: this code still be an O(2^n) and if you pass something like 20 elements in the list you will get 2^20= 1048576 subsets!
Edit:
As Servy sugest i add an implementation with interator block to use with Linq an foreach, the new class is like this:
private class SubSet<T> : IEnumerable<IEnumerable<T>>
{
private IList<T> _list;
private int _length;
private int _max;
private int _count;
public SubSet(IEnumerable<T> list)
{
if (list == null)
throw new ArgumentNullException("list");
_list = new List<T>(list);
_length = _list.Count;
_count = 0;
_max = (int)Math.Pow(2, _length);
}
public int Count
{
get { return _max; }
}
private IList<T> Next()
{
if (_count == _max)
{
return null;
}
uint rs = 0;
IList<T> l = new List<T>();
while (rs < _length)
{
if ((_count & (1u << (int)rs)) > 0)
{
l.Add(_list[(int)rs]);
}
rs++;
}
_count++;
return l;
}
public IEnumerator<IEnumerable<T>> GetEnumerator()
{
IList<T> subset;
while ((subset = Next()) != null)
{
yield return subset;
}
}
System.Collections.IEnumerator System.Collections.IEnumerable.GetEnumerator()
{
return GetEnumerator();
}
}
and you now can use it like this:
List<string> lst = new List<string>();
lst.AddRange(new string[] {"A", "B", "C" });
SubSet<string> subs = new SubSet<string>(lst);
foreach(IList<string> l in subs)
{
DoSomething(l);
}
Thanks Servy for the advice.


It Doesn't give duplicate value;
Don't add a value of the int array at the start of the subsets
Correct program is as follows:
class Program
{
static HashSet<List<int>> SubsetMaker(int[] a, int sum)
{
var set = a.ToList<int>();
HashSet<List<int>> subsets = new HashSet<List<int>>();
subsets.Add(new List<int>());
for (int i =0;i<set.Count;i++)
{
//subsets.Add(new List<int>() { set[i]});
HashSet<List<int>> newSubsets = new HashSet<List<int>>();
for (int j = 0; j < subsets.Count; j++)
{
var newSubset = new List<int>();
foreach (var temp in subsets.ElementAt(j))
{
newSubset.Add(temp);
}
newSubset.Add(set[i]);
newSubsets.Add(newSubset);
}
Console.WriteLine("New Subset");
foreach (var t in newSubsets)
{
var temp = string.Join<int>(",", t);
temp = "{" + temp + "}";
Console.WriteLine(temp);
}
Console.ReadLine();
subsets.UnionWith(newSubsets);
}
//subsets.Add(new List<int>() { set[set.Count - 1] });
//subsets=subsets.;
return subsets;
}
static void Main(string[] args)
{
int[] b = new int[] { 1,2,3 };
int suma = 6;
var test = SubsetMaker(b, suma);
Console.WriteLine("Printing final set...");
foreach (var t in test)
{
var temp = string.Join<int>(",", t);
temp = "{" + temp + "}";
Console.WriteLine(temp);
}
Console.ReadLine();
}
}


You don't want to return a set of lists, you want to use java's set type. Set already does part of what you are looking for by holding only one unique element of each type. So you can't add 20 twice for instance. It is an unordered type, so what you might do is write a combinatoric function that creates a bunch of sets and then return a list that includes alist of those.


Get all subsets of a collection of a specific subsetlength:
public static IEnumerable<IEnumerable<T>> GetPermutations<T>(IEnumerable<T> list, int length) where T : IComparable
{
if (length == 1) return list.Select(t => new T[] { t });
return GetPermutations(list, length - 1).SelectMany(t => list.Where(e => t.All(g => g.CompareTo(e) != 0)), (t1, t2) => t1.Concat(new T[] { t2 }));
}
public static IEnumerable<IEnumerable<T>> GetOrderedSubSets<T>(IEnumerable<T> list, int length) where T : IComparable
{
if (length == 1) return list.Select(t => new T[] { t });
return GetOrderedSubSets(list, length - 1).SelectMany(t => list.Where(e => t.All(g => g.CompareTo(e) == -1)), (t1, t2) => t1.Concat(new T[] { t2 }));
}
Testcode:
List<int> set = new List<int> { 1, 2, 3 };
foreach (var x in GetPermutations(set, 3))
{
Console.WriteLine(string.Join(", ", x));
}
Console.WriteLine();
foreach (var x in GetPermutations(set, 2))
{
Console.WriteLine(string.Join(", ", x));
}
Console.WriteLine();
foreach (var x in GetOrderedSubSets(set, 2))
{
Console.WriteLine(string.Join(", ", x));
}
Test results:
1, 2, 3
1, 3, 2
2, 1, 3
2, 3, 1
3, 1, 2
3, 2, 1
1, 2
1, 3
2, 1
2, 3
3, 1
3, 2
1, 2
1, 3
2, 3


A simple algorithm based upon recursion:
private static List<List<int>> GetPowerList(List<int> a)
{
int n = a.Count;
var sublists = new List<List<int>>() { new List<int>() };
for (int i = 0; i < n; i++)
{
for (int j = i; j < n; j++)
{
var first = a[i];
var last = a[j];
if ((j - i) > 1)
{
sublists.AddRange(GetPowerList(a
.GetRange(i + 1, j - i - 1))
.Select(l => l
.Prepend(first)
.Append(last).ToList()));
}
else sublists.Add(a.GetRange(i,j - i + 1));
}
}
return sublists;
}

Related

Function to make every word possible for array [duplicate]

I have a list of integers List<int> in my C# program. However, I know the number of items I have in my list only at runtime.
Let us say, for the sake of simplicity, my list is {1, 2, 3}
Now I need to generate all possible combinations as follows.
{1, 2, 3}
{1, 2}
{1, 3}
{2, 3}
{1}
{2}
{3}
Can somebody please help with this?

try this:
static void Main(string[] args)
{
GetCombination(new List<int> { 1, 2, 3 });
}
static void GetCombination(List<int> list)
{
double count = Math.Pow(2, list.Count);
for (int i = 1; i <= count - 1; i++)
{
string str = Convert.ToString(i, 2).PadLeft(list.Count, '0');
for (int j = 0; j < str.Length; j++)
{
if (str[j] == '1')
{
Console.Write(list[j]);
}
}
Console.WriteLine();
}
}

Assuming that all items within the initial collection are distinct, we can try using Linq in order to query; let's generalize the solution:
Code:
public static IEnumerable<T[]> Combinations<T>(IEnumerable<T> source) {
if (null == source)
throw new ArgumentNullException(nameof(source));
T[] data = source.ToArray();
return Enumerable
.Range(0, 1 << (data.Length))
.Select(index => data
.Where((v, i) => (index & (1 << i)) != 0)
.ToArray());
}
Demo:
var data = new char[] { 'A', 'B', 'C' };
var result = Combinations(data);
foreach (var item in result)
Console.WriteLine($"[{string.Join(", ", item)}]");
Outcome:
[]
[A]
[B]
[A, B]
[C]
[A, C]
[B, C]
[A, B, C]
If you want to exclude the initial empty array, put .Range(1, (1 << (data.Length)) - 1) instead of .Range(0, 1 << (data.Length))
Algorithm explanation:
Having a collection of collection.Length distinct items we get 2 ** collection.Length combinations (we can compute it as 1 << collection.Length):
mask - comments
------------------------------------
00..0000 - empty, no items are taken
00..0001 - 1st item taken
00..0010 - 2nd item taken
00..0011 - 1st and 2nd items are taken
00..0100 - 3d item taken
...
11..1111 - all items are taken
To generate all masks we can use direct Enumerable.Range(0, 1 << (data.Length)) Linq query. Now having index mask we should take item from the collection if and only if corresponding bit within index is set to 1:
011001001
^^ ^ ^
take 7, 6, 3, 0-th items from the collection
The code can be
.Select(index => data.Where((v, i) => (index & (1 << i)) != 0)
here for each item (v) in the collection data we check if i-th bit is set in the index (mask).

Here are two generic solutions for strongly typed lists that will return all unique combinations of list members (if you can solve this with simpler code, I salute you):
// Recursive
public static List<List<T>> GetAllCombos<T>(List<T> list)
{
List<List<T>> result = new List<List<T>>();
// head
result.Add(new List<T>());
result.Last().Add(list[0]);
if (list.Count == 1)
return result;
// tail
List<List<T>> tailCombos = GetAllCombos(list.Skip(1).ToList());
tailCombos.ForEach(combo =>
{
result.Add(new List<T>(combo));
combo.Add(list[0]);
result.Add(new List<T>(combo));
});
return result;
}
// Iterative, using 'i' as bitmask to choose each combo members
public static List<List<T>> GetAllCombos<T>(List<T> list)
{
int comboCount = (int) Math.Pow(2, list.Count) - 1;
List<List<T>> result = new List<List<T>>();
for (int i = 1; i < comboCount + 1; i++)
{
// make each combo here
result.Add(new List<T>());
for (int j = 0; j < list.Count; j++)
{
if ((i >> j) % 2 != 0)
result.Last().Add(list[j]);
}
}
return result;
}
// Example usage
List<List<int>> combos = GetAllCombos(new int[] { 1, 2, 3 }.ToList());

This answer uses the same algorithm as ojlovecd and (for his iterative solution) jaolho. The only thing I'm adding is an option to filter the results for a minimum number of items in the combinations. This can be useful, for example, if you are only interested in the combinations that contain at least two items.
Edit: As requested by #user3610374 a filter for the maximum number of items has been added.
Edit 2: As suggested by #stannius the algorithm has been changed to make it more efficient for cases where not all combinations are wanted.
/// <summary>
/// Method to create lists containing possible combinations of an input list of items. This is
/// basically copied from code by user "jaolho" on this thread:
/// http://stackoverflow.com/questions/7802822/all-possible-combinations-of-a-list-of-values
/// </summary>
/// <typeparam name="T">type of the items on the input list</typeparam>
/// <param name="inputList">list of items</param>
/// <param name="minimumItems">minimum number of items wanted in the generated combinations,
/// if zero the empty combination is included,
/// default is one</param>
/// <param name="maximumItems">maximum number of items wanted in the generated combinations,
/// default is no maximum limit</param>
/// <returns>list of lists for possible combinations of the input items</returns>
public static List<List<T>> ItemCombinations<T>(List<T> inputList, int minimumItems = 1,
int maximumItems = int.MaxValue)
{
int nonEmptyCombinations = (int)Math.Pow(2, inputList.Count) - 1;
List<List<T>> listOfLists = new List<List<T>>(nonEmptyCombinations + 1);
// Optimize generation of empty combination, if empty combination is wanted
if (minimumItems == 0)
listOfLists.Add(new List<T>());
if (minimumItems <= 1 && maximumItems >= inputList.Count)
{
// Simple case, generate all possible non-empty combinations
for (int bitPattern = 1; bitPattern <= nonEmptyCombinations; bitPattern++)
listOfLists.Add(GenerateCombination(inputList, bitPattern));
}
else
{
// Not-so-simple case, avoid generating the unwanted combinations
for (int bitPattern = 1; bitPattern <= nonEmptyCombinations; bitPattern++)
{
int bitCount = CountBits(bitPattern);
if (bitCount >= minimumItems && bitCount <= maximumItems)
listOfLists.Add(GenerateCombination(inputList, bitPattern));
}
}
return listOfLists;
}
/// <summary>
/// Sub-method of ItemCombinations() method to generate a combination based on a bit pattern.
/// </summary>
private static List<T> GenerateCombination<T>(List<T> inputList, int bitPattern)
{
List<T> thisCombination = new List<T>(inputList.Count);
for (int j = 0; j < inputList.Count; j++)
{
if ((bitPattern >> j & 1) == 1)
thisCombination.Add(inputList[j]);
}
return thisCombination;
}
/// <summary>
/// Sub-method of ItemCombinations() method to count the bits in a bit pattern. Based on this:
/// https://graphics.stanford.edu/~seander/bithacks.html#CountBitsSetKernighan
/// </summary>
private static int CountBits(int bitPattern)
{
int numberBits = 0;
while (bitPattern != 0)
{
numberBits++;
bitPattern &= bitPattern - 1;
}
return numberBits;
}

Here's a generic solution using recursion
public static ICollection<ICollection<T>> Permutations<T>(ICollection<T> list) {
var result = new List<ICollection<T>>();
if (list.Count == 1) { // If only one possible permutation
result.Add(list); // Add it and return it
return result;
}
foreach (var element in list) { // For each element in that list
var remainingList = new List<T>(list);
remainingList.Remove(element); // Get a list containing everything except of chosen element
foreach (var permutation in Permutations<T>(remainingList)) { // Get all possible sub-permutations
permutation.Add(element); // Add that element
result.Add(permutation);
}
}
return result;
}
I know this is an old post, but someone might find this helpful.

Another solution using Linq and recursion...
static void Main(string[] args)
{
List<List<long>> result = new List<List<long>>();
List<long> set = new List<long>() { 1, 2, 3, 4 };
GetCombination<long>(set, result);
result.Add(set);
IOrderedEnumerable<List<long>> sorted = result.OrderByDescending(s => s.Count);
sorted.ToList().ForEach(l => { l.ForEach(l1 => Console.Write(l1 + " ")); Console.WriteLine(); });
}
private static void GetCombination<T>(List<T> set, List<List<T>> result)
{
for (int i = 0; i < set.Count; i++)
{
List<T> temp = new List<T>(set.Where((s, index) => index != i));
if (temp.Count > 0 && !result.Where(l => l.Count == temp.Count).Any(l => l.SequenceEqual(temp)))
{
result.Add(temp);
GetCombination<T>(temp, result);
}
}
}

This is an improvement of #ojlovecd answer without using strings.
static void Main(string[] args)
{
GetCombination(new List<int> { 1, 2, 3 });
}
private static void GetCombination(List<int> list)
{
double count = Math.Pow(2, list.Count);
for (int i = 1; i <= count - 1; i++)
{
for (int j = 0; j < list.Count; j++)
{
int b = i & (1 << j);
if (b > 0)
{
Console.Write(list[j]);
}
}
Console.WriteLine();
}
}

Firstly, given a set of n elements, you compute all combinations of k elements out of it (nCk). You have to change the value of k from 1 to n to meet your requirement.
See this codeproject article for C# code for generating combinations.
In case, you are interested in developing the combination algorithm by yourself, check this SO question where there are a lot of links to the relevant material.

protected List<List<T>> AllCombos<T>(Func<List<T>, List<T>, bool> comparer, params T[] items)
{
List<List<T>> results = new List<List<T>>();
List<T> workingWith = items.ToList();
results.Add(workingWith);
items.ToList().ForEach((x) =>
{
results.Add(new List<T>() { x });
});
for (int i = 0; i < workingWith.Count(); i++)
{
T removed = workingWith[i];
workingWith.RemoveAt(i);
List<List<T>> nextResults = AllCombos2(comparer, workingWith.ToArray());
results.AddRange(nextResults);
workingWith.Insert(i, removed);
}
results = results.Where(x => x.Count > 0).ToList();
for (int i = 0; i < results.Count; i++)
{
List<T> list = results[i];
if (results.Where(x => comparer(x, list)).Count() > 1)
{
results.RemoveAt(i);
}
}
return results;
}
protected List<List<T>> AllCombos2<T>(Func<List<T>, List<T>, bool> comparer, params T[] items)
{
List<List<T>> results = new List<List<T>>();
List<T> workingWith = items.ToList();
if (workingWith.Count > 1)
{
results.Add(workingWith);
}
for (int i = 0; i < workingWith.Count(); i++)
{
T removed = workingWith[i];
workingWith.RemoveAt(i);
List<List<T>> nextResults = AllCombos2(comparer, workingWith.ToArray());
results.AddRange(nextResults);
workingWith.Insert(i, removed);
}
results = results.Where(x => x.Count > 0).ToList();
for (int i = 0; i < results.Count; i++)
{
List<T> list = results[i];
if (results.Where(x => comparer(x, list)).Count() > 1)
{
results.RemoveAt(i);
}
}
return results;
}
This worked for me, it's slightly more complex and actually takes a comparer callback function, and it's actually 2 functions, the difference being that the AllCombos adds the single item lists explicitly. It is very raw and can definitely be trimmed down but it gets the job done. Any refactoring suggestions are welcome. Thanks,

public class CombinationGenerator{
private readonly Dictionary<int, int> currentIndexesWithLevels = new Dictionary<int, int>();
private readonly LinkedList<List<int>> _combinationsList = new LinkedList<List<int>>();
private readonly int _combinationLength;
public CombinationGenerator(int combinationLength)
{
_combinationLength = combinationLength;
}
private void InitializeLevelIndexes(List<int> list)
{
for (int i = 0; i < _combinationLength; i++)
{
currentIndexesWithLevels.Add(i+1, i);
}
}
private void UpdateCurrentIndexesForLevels(int level)
{
int index;
if (level == 1)
{
index = currentIndexesWithLevels[level];
for (int i = level; i < _combinationLength + 1; i++)
{
index = index + 1;
currentIndexesWithLevels[i] = index;
}
}
else
{
int previousLevelIndex;
for (int i = level; i < _combinationLength + 1; i++)
{
if (i > level)
{
previousLevelIndex = currentIndexesWithLevels[i - 1];
currentIndexesWithLevels[i] = previousLevelIndex + 1;
}
else
{
index = currentIndexesWithLevels[level];
currentIndexesWithLevels[i] = index + 1;
}
}
}
}
public void FindCombinations(List<int> list, int level, Stack<int> stack)
{
int currentIndex;
InitializeLevelIndexes(list);
while (true)
{
currentIndex = currentIndexesWithLevels[level];
bool levelUp = false;
for (int i = currentIndex; i < list.Count; i++)
{
if (level < _combinationLength)
{
currentIndex = currentIndexesWithLevels[level];
MoveToUpperLevel(ref level, stack, list, currentIndex);
levelUp = true;
break;
}
levelUp = false;
stack.Push(list[i]);
if (stack.Count == _combinationLength)
{
AddCombination(stack);
stack.Pop();
}
}
if (!levelUp)
{
MoveToLowerLevel(ref level, stack, list, ref currentIndex);
while (currentIndex >= list.Count - 1)
{
if (level == 1)
{
AdjustStackCountToCurrentLevel(stack, level);
currentIndex = currentIndexesWithLevels[level];
if (currentIndex >= list.Count - 1)
{
return;
}
UpdateCurrentIndexesForLevels(level);
}
else
{
MoveToLowerLevel(ref level, stack, list, ref currentIndex);
}
}
}
}
}
private void AddCombination(Stack<int> stack)
{
List<int> listNew = new List<int>();
listNew.AddRange(stack);
_combinationsList.AddLast(listNew);
}
private void MoveToUpperLevel(ref int level, Stack<int> stack, List<int> list, int index)
{
stack.Push(list[index]);
level++;
}
private void MoveToLowerLevel(ref int level, Stack<int> stack, List<int> list, ref int currentIndex)
{
if (level != 1)
{
level--;
}
AdjustStackCountToCurrentLevel(stack, level);
UpdateCurrentIndexesForLevels(level);
currentIndex = currentIndexesWithLevels[level];
}
private void AdjustStackCountToCurrentLevel(Stack<int> stack, int currentLevel)
{
while (stack.Count >= currentLevel)
{
if (stack.Count != 0)
stack.Pop();
}
}
public void PrintPermutations()
{
int count = _combinationsList.Where(perm => perm.Count() == _combinationLength).Count();
Console.WriteLine("The number of combinations is " + count);
}
}

We can use recursion for combination/permutation problems involving string or integers.
public static void Main(string[] args)
{
IntegerList = new List<int> { 1, 2, 3, 4 };
PrintAllCombination(default(int), default(int));
}
public static List<int> IntegerList { get; set; }
public static int Length { get { return IntegerList.Count; } }
public static void PrintAllCombination(int position, int prefix)
{
for (int i = position; i < Length; i++)
{
Console.WriteLine(prefix * 10 + IntegerList[i]);
PrintAllCombination(i + 1, prefix * 10 + IntegerList[i]);
}
}

What about
static void Main(string[] args)
{
Combos(new [] { 1, 2, 3 });
}
static void Combos(int[] arr)
{
for (var i = 0; i <= Math.Pow(2, arr.Length); i++)
{
Console.WriteLine();
var j = i;
var idx = 0;
do
{
if ((j & 1) == 1) Console.Write($"{arr[idx]} ");
} while ((j >>= 1) > 0 && ++idx < arr.Length);
}
}

A slightly more generalised version for Linq using C# 7. Here filtering by items that have two elements.
static void Main(string[] args)
{
foreach (var vals in Combos(new[] { "0", "1", "2", "3" }).Where(v => v.Skip(1).Any() && !v.Skip(2).Any()))
Console.WriteLine(string.Join(", ", vals));
}
static IEnumerable<IEnumerable<T>> Combos<T>(T[] arr)
{
IEnumerable<T> DoQuery(long j, long idx)
{
do
{
if ((j & 1) == 1) yield return arr[idx];
} while ((j >>= 1) > 0 && ++idx < arr.Length);
}
for (var i = 0; i < Math.Pow(2, arr.Length); i++)
yield return DoQuery(i, 0);
}

Here is how I did it.
public static List<List<int>> GetCombination(List<int> lst, int index, int count)
{
List<List<int>> combinations = new List<List<int>>();
List<int> comb;
if (count == 0 || index == lst.Count)
{
return null;
}
for (int i = index; i < lst.Count; i++)
{
comb = new List<int>();
comb.Add(lst.ElementAt(i));
combinations.Add(comb);
var rest = GetCombination(lst,i + 1, count - 1);
if (rest != null)
{
foreach (var item in rest)
{
combinations.Add(comb.Union(item).ToList());
}
}
}
return combinations;
}
You call it as :
List<int> lst= new List<int>(new int[]{ 1, 2, 3, 4 });
var combinations = GetCombination(lst, 0, lst.Length)

I just run into a situation where I needed to do this, this is what I came up with:
private static List<string> GetCombinations(List<string> elements)
{
List<string> combinations = new List<string>();
combinations.AddRange(elements);
for (int i = 0; i < elements.Count - 1; i++)
{
combinations = (from combination in combinations
join element in elements on 1 equals 1
let value = string.Join(string.Empty, $"{combination}{element}".OrderBy(c => c).Distinct())
select value).Distinct().ToList();
}
return combinations;
}
It may be not too efficient, and it sure has room for improvement, but gets the job done!
List<string> elements = new List<string> { "1", "2", "3" };
List<string> combinations = GetCombinations(elements);
foreach (string combination in combinations)
{
System.Console.Write(combination);
}

This is an improved version based on the answer from ojlovecd using extension methods:
public static class ListExtensions
{
public static IEnumerable<List<T>> GetCombinations<T>(
this List<T> valuesToCombine)
{
var count = Math.Pow(2, valuesToCombine.Count);
for(var i = 1; i <= count; i++)
{
var itemFlagList = i.ToBinaryString(valuesToCombine.Count())
.Select(x => x == '1').ToList();
yield return GetCombinationByFlagList(valuesToCombine, itemFlagList)
.ToList();
}
}
private static IEnumerable<T> GetCombinationByFlagList<T>(
List<T> valuesToCombine, List<bool> flagList)
{
for (var i = 0; i < valuesToCombine.Count; i++)
{
if (!flagList[i]) continue;
yield return valuesToCombine.ElementAt(i);
}
}
}
public static class IntegerExtensions
{
public static string ToBinaryString(this int value, int length)
{
return Convert.ToString(value, 2).ToString().PadLeft(length, '0');
}
}
Usage:
var numbersList = new List<int>() { 1, 2, 3 };
var combinations = numbersList.GetCombinations();
foreach (var combination in combinations)
{
System.Console.WriteLine(string.Join(",", combination));
}
Output:
3
2
2,3
1
1,3
1,2
1,2,3
The idea is to basically use some flags to keep track of which items were already added to the combination. So in case of 1, 2 & 3, the following binary strings are generated in order to indicate whether an item should be included or excluded:
001, 010, 011, 100, 101, 110 & 111

I'd like to suggest an approach that I find to be quite intuitive and easy to read. (Note: It is slower than the currently accepted solution.)
It is built on the idea that for each integer in the list, we need to extend the so-far-aggregated resulting combination list with
all currently existing combinations, each extended with the given integer
a single "combination" of that integer alone
Here, I am using .Aggregate() with a seed that is an IEnumerable<IEnumerable<int>> containing a single, empty collection entry. That empty entry lets us easily do the two steps above simultaneously. The empty collection entry can be skipped after the resulting combination collection has been aggregated.
It goes like this:
var emptyCollection = Enumerable.Empty<IEnumerable<int>>();
var emptyCombination = Enumerable.Empty<int>();
IEnumerable<int[]> combinations = list
.Aggregate(emptyCollection.Append(emptyCombination),
( acc, next ) => acc.Concat(acc.Select(entry => entry.Append(next))))
.Skip(1) // skip the initial, empty combination
.Select(comb => comb.ToArray());
For each entry in the input integer list { 1, 2, 3 }, the accumulation progresses as follows:
next = 1
{ { } }.Concat({ { }.Append(1) })
{ { } }.Concat({ { 1 } })
{ { }, { 1 } } // acc
next = 2
{ { }, { 1 } }.Concat({ { }.Append(2), { 1 }.Append(2) })
{ { }, { 1 } }.Concat({ { 2 }, { 1, 2 } })
{ { }, { 1 }, { 2 }, { 1, 2 } } // acc
next = 3
{ { }, { 1 }, { 2 }, { 1, 2 } }.Concat({ { }.Append(3), { 1 }.Append(3), { 2 }.Append(3), { 1, 2 }.Append(3) })
{ { }, { 1 }, { 2 }, { 1, 2 } }.Concat({ { 3 }, { 1, 3 }, { 2, 3 }, { 1, 2, 3 } })
{ { }, { 1 }, { 2 }, { 1, 2 }, { 3 }, { 1, 3 }, { 2, 3 }, { 1, 2, 3 } } // acc
Skipping the first (empty) entry, we are left with the following collection:
1
2
1 2
3
1 3
2 3
1 2 3
, which can easily be ordered by collection length and collection entry sum for a clearer overview.
Example fiddle here.

Some of the solutions here are truly ingenious; especially the ones that use bitmaps.
But I found that in practice these algos
aren't easy to modify if a specific range of lengths needed (e.g. all variations of 3 to 5 choices from an input set of 8 elements)
can't handle LARGE input lists (and return empty or singleton results instead of throwing exception); and
can be tricky to debug.
So I decided to write something not as clever as the other people here.
My more basic approach recognises that the set of Variations(1 to maxLength) is simply a UNION of all fixed-length Variations of each length 1 to maxLength:
i.e
Variations(1 to maxLength) = Variations(1) + Variations(2) + ... + Variations(maxLength)
So you can do a "choose K from N" for each required length (for each K in (1, 2, 3, ..., maxLength)) and then just do a Union of these separate results to yield a List of Lists.
This resulting code intends to be easy to understand and to maintain:
/// <summary>
/// Generates ALL variations of length between minLength and maxLength (inclusive)
/// Relies on Combinatorics library to generate each set of Variations
/// Nuget https://www.nuget.org/packages/Combinatorics/
/// Excellent more general references (without this solution):
/// https://www.codeproject.com/Articles/26050/Permutations-Combinations-and-Variations-using-C-G
/// Self-authored solution.
/// </summary>
/// <typeparam name="T">Any type without any constraints.</typeparam>
/// <param name="sourceList">The source list of elements to be combined.</param>
/// <param name="minLength">The minimum length of variation required.</param>
/// <param name="maxLength">The maximum length of variation required.</param>
/// <returns></returns>
public static List<List<T>> GenerateVariations<T>(this IEnumerable<T> sourceList, int minLength, int maxLength)
{
List<List<T>> finalUnion = new();
foreach (int length in Enumerable.Range(minLength, maxLength))
{
Variations<T> variations = new Variations<T>(sourceList, length, GenerateOption.WithoutRepetition);
foreach (var variation in variations)
{
var list = variation.ToList<T>();
finalUnion.Add(list);
}
}
Debug.WriteLine(sourceList.Count() + " source " + typeof(T).Name + " yielded " + finalUnion.Count());
return finalUnion;
}
Happy to receive comments (good and bad). Maybe there's a more succint way to achieve this in LINQ? Maybe the really smart people here can marry their approach with my more basic one?

Please find very very simple solution without recursion and which dont eat RAM.
Unique Combinations

Permutations with constant prefix numbers

I have an array of integers where each value will have distinct meanings.The first value means the length of permutation, the second value represent the length of initial prefix and rest of integers are single integer that make up prefix of all permutations.
For e.g. if the array has elements {5,2,1,4}
where 5 is the number of elements in the permutation array.
and 2 is the length of the integer that will makeup the first 2 elements prefix in the array permutation. 1,4 are the prefix integers i.e. length 2 in 5 element permutation combination so missing elements are 2,3,5 where 1&4 being common prefix across all permutations as below
[14235][14253][14325][14352][14523][14532] where input array is {5,2,1,4}
How to achieve this?
I have below code to get the permutation of one missing elements 2,3 & 5 but I am not getting how to program the entire the solution
static void Main(string[] args)
{
int output;
int ip1;
ip1 = Convert.ToInt32(Console.ReadLine());
int ip2_size = 0;
ip2_size = Convert.ToInt32(Console.ReadLine());
int[] ip2 = new int[ip2_size];
int ip2_item;
for (int ip2_i = 0; ip2_i < ip2_size; ip2_i++)
{
ip2_item = Convert.ToInt32(Console.ReadLine());
ip2[ip2_i] = ip2_item;
}
output = correctResult(ip1, ip2);
Console.WriteLine(output);
}
static int correctResult(int n, int[] arr)
{
int permLength = 0;
int prefLength = 0;
int result = 0;
permLength = n;
prefLength = arr.Length;
int[] permArray = new int[permLength];
int len = 0;
var missingNum = Enumerable.Range(1,
permLength).Except(arr).ToArray<int>();
if (permLength < (missingNum.Length + len))
{
result = -1;
}
else
{
for (int i = 0; i < missingNum.Length; i++)
{
permArray[prefLength + i] = missingNum[i];
}
result = permute(missingNum, 0, missingNum.Length - 1);
}
return result;
}
static int permute(int[] arry, int i, int n)
{
int j;
if (i == n)
{
int s1, s2;
s1 = s2 = 0;
for (int a = 0; a < n - 1; a++)
{
for (int b = a + 1; b < n; b++)
{
if (arry[a] > arry[b])
{
s1++;
}
}
s2 = s2 + Math.Max(0, a + 1 - arry[a]);
}
int count = 0;
if (s1 == s2)
count++;
return count;
}
else
{
int count = 0;
for (j = i; j <= n; j++)
{
swap(ref arry[i], ref arry[j]);
count += permute(arry, i + 1, n);
swap(ref arry[i], ref arry[j]);
}
return count;
}
}
static void swap(ref int a, ref int b)
{
int tmp;
tmp = a;
a = b;
b = tmp;
}

Try solving this with immutable types, its easier to reason about them. If, after solving the problem, you have a performance goal you haven't met then you can start trying to optimize the code.
Consider the following approach with an immutable stack that keeps track of the current permutation:
static IEnumerable<IEnumerable<int>> GetPermutations(IList<int> input)
{
if (input == null)
throw new ArgumentNullException(nameof(input));
if (input.Count < 2)
throw new ArgumentException("Input does not have a valid format.");
var setSize = input[0];
var prefixSize = input[1];
if (prefixSize != input.Count - 2)
throw new ArgumentException("Input does not have a valid format.");
if (input.Skip(2).Any(i => i > setSize)) //we are assuming, per example, that valid range starts at 1.
throw new ArgumentException("Input does not have a valid format.");
//Ok, we've got a valid input, interesting stuff starts here.
var prefix = input.Skip(2).ToArray();
var initialSet = Enumerable.Range(1, setSize)
.Except(prefix)
.ToArray();
foreach (var p in getPermutations(ImmutableStack<int>.Empty, initialSet))
{
yield return prefix.Concat(p);
}
IEnumerable<IEnumerable<int>> getPermutations(ImmutableStack<int> permutation, IEnumerable<int> set)
{
if (permutation.Count == setSize - prefixSize)
{
yield return permutation;
}
else
{
foreach (var i in set)
{
foreach (var p in getPermutations(permutation.Push(i), set.Except(new[] { i })))
{
yield return p;
}
}
}
}
}
And that is it, solving your problem was about 10-12 lines of real code (not considering input validation). Note that I am using some c#7 features here, but its easily translatable to previous versions of the language. Also I'd like to underline the argument validation we are doing upfront; make sure you have a valid input before trying out anything.
For ImmutableStack<T> you can use the one in System.Collections.Immutable (you have to download the NuGet package) or implement your own, its simple:
private class ImmutableStack<T>: IEnumerable<T>
{
public static readonly ImmutableStack<T> Empty = new ImmutableStack<T>();
private readonly T head;
private readonly ImmutableStack<T> tail;
private ImmutableStack() { }
private ImmutableStack(T head, ImmutableStack<T> tail)
{
Debug.Assert(tail != null);
this.head = head;
this.tail = tail;
Count = tail.Count + 1;
}
public int Count { get; }
public T Peek() =>
this != Empty ? head : throw new InvalidOperationException("Empty stack.");
public ImmutableStack<T> Pop() =>
this != Empty ? tail : throw new InvalidOperationException("Empty stack.");
public ImmutableStack<T> Push(T item) => new ImmutableStack<T>(item, this);
public IEnumerator<T> GetEnumerator()
{
var current = this;
while (current != Empty)
{
yield return current.head;
current = current.tail;
}
}
IEnumerator IEnumerable.GetEnumerator() => GetEnumerator();
}
If you use the collections in System.Collections.Immutable, then you'll probably want to use some kind of immutable set for initalSet and set.

You can rewrite your permute method (based on this answer):
private static IEnumerable<IEnumerable<T>> Permute<T>(List<T> prefix, List<T> suffix)
{
for (var i = 0; i < suffix.Count; ++i)
{
var newPrefix = new List<T>(prefix) {suffix[i]};
var newSuffix = new List<T>(suffix.Take(i).Concat(suffix.Skip(i + 1)));
if (newSuffix.Count == 0)
{
yield return newPrefix;
continue;
}
foreach (var permutation in Permute(newPrefix, newSuffix))
yield return permutation;
}
}
And use it like this:
public static void PrintAllPermutations()
{
var input = new[] {5, 2, 1, 4};
var prefix = input.Skip(2).Take(input[1]).ToList();
var suffx = Enumerable.Range(1, input[0]).Except(prefix).ToList();
foreach (var permutation in Permute(prefix, suffx))
Console.WriteLine(string.Join(", ", permutation));
}
Reult would be:
1, 4, 2, 3, 5
1, 4, 2, 5, 3
1, 4, 3, 2, 5
1, 4, 3, 5, 2
1, 4, 5, 2, 3
1, 4, 5, 3, 2

c# list permutations with limited length

I have a list of Offers, from which I want to create "chains" (e.g. permutations) with limited chain lengths.
I've gotten as far as creating the permutations using the Kw.Combinatorics project.
However, the default behavior creates permutations in the length of the list count. I'm not sure how to limit the chain lengths to 'n'.
Here's my current code:
private static List<List<Offers>> GetPerms(List<Offers> list, int chainLength)
{
List<List<Offers>> response = new List<List<Offers>>();
foreach (var row in new Permutation(list.Count).GetRows())
{
List<Offers> innerList = new List<Offers>();
foreach (var mix in Permutation.Permute(row, list))
{
innerList.Add(mix);
}
response.Add(innerList);
innerList = new List<Offers>();
}
return response;
}
Implemented by:
List<List<AdServer.Offers>> lst = GetPerms(offers, 2);
I'm not locked in KWCombinatorics if someone has a better solution to offer.

Here's another implementation which I think should be faster than the accepted answer (and it's definitely less code).
public static IEnumerable<IEnumerable<T>> GetVariationsWithoutDuplicates<T>(IList<T> items, int length)
{
if (length == 0 || !items.Any()) return new List<List<T>> { new List<T>() };
return from item in items.Distinct()
from permutation in GetVariationsWithoutDuplicates(items.Where(i => !EqualityComparer<T>.Default.Equals(i, item)).ToList(), length - 1)
select Prepend(item, permutation);
}
public static IEnumerable<IEnumerable<T>> GetVariations<T>(IList<T> items, int length)
{
if (length == 0 || !items.Any()) return new List<List<T>> { new List<T>() };
return from item in items
from permutation in GetVariations(Remove(item, items).ToList(), length - 1)
select Prepend(item, permutation);
}
public static IEnumerable<T> Prepend<T>(T first, IEnumerable<T> rest)
{
yield return first;
foreach (var item in rest) yield return item;
}
public static IEnumerable<T> Remove<T>(T item, IEnumerable<T> from)
{
var isRemoved = false;
foreach (var i in from)
{
if (!EqualityComparer<T>.Default.Equals(item, i) || isRemoved) yield return i;
else isRemoved = true;
}
}
On my 3.1 GHz Core 2 Duo, I tested with this:
public static void Test(Func<IList<int>, int, IEnumerable<IEnumerable<int>>> getVariations)
{
var max = 11;
var timer = System.Diagnostics.Stopwatch.StartNew();
for (int i = 1; i < max; ++i)
for (int j = 1; j < i; ++j)
getVariations(MakeList(i), j).Count();
timer.Stop();
Console.WriteLine("{0,40}{1} ms", getVariations.Method.Name, timer.ElapsedMilliseconds);
}
// Make a list that repeats to guarantee we have duplicates
public static IList<int> MakeList(int size)
{
return Enumerable.Range(0, size/2).Concat(Enumerable.Range(0, size - size/2)).ToList();
}
Unoptimized
GetVariations 11894 ms
GetVariationsWithoutDuplicates 9 ms
OtherAnswerGetVariations 22485 ms
OtherAnswerGetVariationsWithDuplicates 243415 ms
With compiler optimizations
GetVariations 9667 ms
GetVariationsWithoutDuplicates 8 ms
OtherAnswerGetVariations 19739 ms
OtherAnswerGetVariationsWithDuplicates 228802 ms

You're not looking for a permutation, but for a variation. Here is a possible algorithm. I prefer iterator methods for functions that can potentially return very many elements. This way, the caller can decide if he really needs all elements:
IEnumerable<IList<T>> GetVariations<T>(IList<T> offers, int length)
{
var startIndices = new int[length];
var variationElements = new HashSet<T>(); //for duplicate detection
while (startIndices[0] < offers.Count)
{
var variation = new List<T>(length);
var valid = true;
for (int i = 0; i < length; ++i)
{
var element = offers[startIndices[i]];
if (variationElements.Contains(element))
{
valid = false;
break;
}
variation.Add(element);
variationElements.Add(element);
}
if (valid)
yield return variation;
//Count up the indices
startIndices[length - 1]++;
for (int i = length - 1; i > 0; --i)
{
if (startIndices[i] >= offers.Count)
{
startIndices[i] = 0;
startIndices[i - 1]++;
}
else
break;
}
variationElements.Clear();
}
}
The idea for this algorithm is to use a number in offers.Count base. For three offers, all digits are in the range 0-2. We then basically increment this number step by step and return the offers that reside at the specified indices. If you want to allow duplicates, you can remove the check and the HashSet<T>.
Update
Here is an optimized variant that does the duplicate check on the index level. In my tests it is a lot faster than the previous variant:
IEnumerable<IList<T>> GetVariations<T>(IList<T> offers, int length)
{
var startIndices = new int[length];
for (int i = 0; i < length; ++i)
startIndices[i] = i;
var indices = new HashSet<int>(); // for duplicate check
while (startIndices[0] < offers.Count)
{
var variation = new List<T>(length);
for (int i = 0; i < length; ++i)
{
variation.Add(offers[startIndices[i]]);
}
yield return variation;
//Count up the indices
AddOne(startIndices, length - 1, offers.Count - 1);
//duplicate check
var check = true;
while (check)
{
indices.Clear();
for (int i = 0; i <= length; ++i)
{
if (i == length)
{
check = false;
break;
}
if (indices.Contains(startIndices[i]))
{
var unchangedUpTo = AddOne(startIndices, i, offers.Count - 1);
indices.Clear();
for (int j = 0; j <= unchangedUpTo; ++j )
{
indices.Add(startIndices[j]);
}
int nextIndex = 0;
for(int j = unchangedUpTo + 1; j < length; ++j)
{
while (indices.Contains(nextIndex))
nextIndex++;
startIndices[j] = nextIndex++;
}
break;
}
indices.Add(startIndices[i]);
}
}
}
}
int AddOne(int[] indices, int position, int maxElement)
{
//returns the index of the last element that has not been changed
indices[position]++;
for (int i = position; i > 0; --i)
{
if (indices[i] > maxElement)
{
indices[i] = 0;
indices[i - 1]++;
}
else
return i;
}
return 0;
}

If I got you correct here is what you need
this will create permutations based on the specified chain limit
public static List<List<T>> GetPerms<T>(List<T> list, int chainLimit)
{
if (list.Count() == 1)
return new List<List<T>> { list };
return list
.Select((outer, outerIndex) =>
GetPerms(list.Where((inner, innerIndex) => innerIndex != outerIndex).ToList(), chainLimit)
.Select(perms => (new List<T> { outer }).Union(perms).Take(chainLimit)))
.SelectMany<IEnumerable<IEnumerable<T>>, List<T>>(sub => sub.Select<IEnumerable<T>, List<T>>(s => s.ToList()))
.Distinct(new PermComparer<T>()).ToList();
}
class PermComparer<T> : IEqualityComparer<List<T>>
{
public bool Equals(List<T> x, List<T> y)
{
return x.SequenceEqual(y);
}
public int GetHashCode(List<T> obj)
{
return (int)obj.Average(o => o.GetHashCode());
}
}
and you'll call it like this
List<List<AdServer.Offers>> lst = GetPerms<AdServer.Offers>(offers, 2);
I made this function is pretty generic so you may use it for other purpose too
eg
List<string> list = new List<string>(new[] { "apple", "banana", "orange", "cherry" });
List<List<string>> perms = GetPerms<string>(list, 2);
result

All Possible Combinations of a list of Values

I have a list of integers List<int> in my C# program. However, I know the number of items I have in my list only at runtime.
Let us say, for the sake of simplicity, my list is {1, 2, 3}
Now I need to generate all possible combinations as follows.
{1, 2, 3}
{1, 2}
{1, 3}
{2, 3}
{1}
{2}
{3}
Can somebody please help with this?

try this:
static void Main(string[] args)
{
GetCombination(new List<int> { 1, 2, 3 });
}
static void GetCombination(List<int> list)
{
double count = Math.Pow(2, list.Count);
for (int i = 1; i <= count - 1; i++)
{
string str = Convert.ToString(i, 2).PadLeft(list.Count, '0');
for (int j = 0; j < str.Length; j++)
{
if (str[j] == '1')
{
Console.Write(list[j]);
}
}
Console.WriteLine();
}
}

Assuming that all items within the initial collection are distinct, we can try using Linq in order to query; let's generalize the solution:
Code:
public static IEnumerable<T[]> Combinations<T>(IEnumerable<T> source) {
if (null == source)
throw new ArgumentNullException(nameof(source));
T[] data = source.ToArray();
return Enumerable
.Range(0, 1 << (data.Length))
.Select(index => data
.Where((v, i) => (index & (1 << i)) != 0)
.ToArray());
}
Demo:
var data = new char[] { 'A', 'B', 'C' };
var result = Combinations(data);
foreach (var item in result)
Console.WriteLine($"[{string.Join(", ", item)}]");
Outcome:
[]
[A]
[B]
[A, B]
[C]
[A, C]
[B, C]
[A, B, C]
If you want to exclude the initial empty array, put .Range(1, (1 << (data.Length)) - 1) instead of .Range(0, 1 << (data.Length))
Algorithm explanation:
Having a collection of collection.Length distinct items we get 2 ** collection.Length combinations (we can compute it as 1 << collection.Length):
mask - comments
------------------------------------
00..0000 - empty, no items are taken
00..0001 - 1st item taken
00..0010 - 2nd item taken
00..0011 - 1st and 2nd items are taken
00..0100 - 3d item taken
...
11..1111 - all items are taken
To generate all masks we can use direct Enumerable.Range(0, 1 << (data.Length)) Linq query. Now having index mask we should take item from the collection if and only if corresponding bit within index is set to 1:
011001001
^^ ^ ^
take 7, 6, 3, 0-th items from the collection
The code can be
.Select(index => data.Where((v, i) => (index & (1 << i)) != 0)
here for each item (v) in the collection data we check if i-th bit is set in the index (mask).

Here are two generic solutions for strongly typed lists that will return all unique combinations of list members (if you can solve this with simpler code, I salute you):
// Recursive
public static List<List<T>> GetAllCombos<T>(List<T> list)
{
List<List<T>> result = new List<List<T>>();
// head
result.Add(new List<T>());
result.Last().Add(list[0]);
if (list.Count == 1)
return result;
// tail
List<List<T>> tailCombos = GetAllCombos(list.Skip(1).ToList());
tailCombos.ForEach(combo =>
{
result.Add(new List<T>(combo));
combo.Add(list[0]);
result.Add(new List<T>(combo));
});
return result;
}
// Iterative, using 'i' as bitmask to choose each combo members
public static List<List<T>> GetAllCombos<T>(List<T> list)
{
int comboCount = (int) Math.Pow(2, list.Count) - 1;
List<List<T>> result = new List<List<T>>();
for (int i = 1; i < comboCount + 1; i++)
{
// make each combo here
result.Add(new List<T>());
for (int j = 0; j < list.Count; j++)
{
if ((i >> j) % 2 != 0)
result.Last().Add(list[j]);
}
}
return result;
}
// Example usage
List<List<int>> combos = GetAllCombos(new int[] { 1, 2, 3 }.ToList());

This answer uses the same algorithm as ojlovecd and (for his iterative solution) jaolho. The only thing I'm adding is an option to filter the results for a minimum number of items in the combinations. This can be useful, for example, if you are only interested in the combinations that contain at least two items.
Edit: As requested by #user3610374 a filter for the maximum number of items has been added.
Edit 2: As suggested by #stannius the algorithm has been changed to make it more efficient for cases where not all combinations are wanted.
/// <summary>
/// Method to create lists containing possible combinations of an input list of items. This is
/// basically copied from code by user "jaolho" on this thread:
/// http://stackoverflow.com/questions/7802822/all-possible-combinations-of-a-list-of-values
/// </summary>
/// <typeparam name="T">type of the items on the input list</typeparam>
/// <param name="inputList">list of items</param>
/// <param name="minimumItems">minimum number of items wanted in the generated combinations,
/// if zero the empty combination is included,
/// default is one</param>
/// <param name="maximumItems">maximum number of items wanted in the generated combinations,
/// default is no maximum limit</param>
/// <returns>list of lists for possible combinations of the input items</returns>
public static List<List<T>> ItemCombinations<T>(List<T> inputList, int minimumItems = 1,
int maximumItems = int.MaxValue)
{
int nonEmptyCombinations = (int)Math.Pow(2, inputList.Count) - 1;
List<List<T>> listOfLists = new List<List<T>>(nonEmptyCombinations + 1);
// Optimize generation of empty combination, if empty combination is wanted
if (minimumItems == 0)
listOfLists.Add(new List<T>());
if (minimumItems <= 1 && maximumItems >= inputList.Count)
{
// Simple case, generate all possible non-empty combinations
for (int bitPattern = 1; bitPattern <= nonEmptyCombinations; bitPattern++)
listOfLists.Add(GenerateCombination(inputList, bitPattern));
}
else
{
// Not-so-simple case, avoid generating the unwanted combinations
for (int bitPattern = 1; bitPattern <= nonEmptyCombinations; bitPattern++)
{
int bitCount = CountBits(bitPattern);
if (bitCount >= minimumItems && bitCount <= maximumItems)
listOfLists.Add(GenerateCombination(inputList, bitPattern));
}
}
return listOfLists;
}
/// <summary>
/// Sub-method of ItemCombinations() method to generate a combination based on a bit pattern.
/// </summary>
private static List<T> GenerateCombination<T>(List<T> inputList, int bitPattern)
{
List<T> thisCombination = new List<T>(inputList.Count);
for (int j = 0; j < inputList.Count; j++)
{
if ((bitPattern >> j & 1) == 1)
thisCombination.Add(inputList[j]);
}
return thisCombination;
}
/// <summary>
/// Sub-method of ItemCombinations() method to count the bits in a bit pattern. Based on this:
/// https://graphics.stanford.edu/~seander/bithacks.html#CountBitsSetKernighan
/// </summary>
private static int CountBits(int bitPattern)
{
int numberBits = 0;
while (bitPattern != 0)
{
numberBits++;
bitPattern &= bitPattern - 1;
}
return numberBits;
}

Here's a generic solution using recursion
public static ICollection<ICollection<T>> Permutations<T>(ICollection<T> list) {
var result = new List<ICollection<T>>();
if (list.Count == 1) { // If only one possible permutation
result.Add(list); // Add it and return it
return result;
}
foreach (var element in list) { // For each element in that list
var remainingList = new List<T>(list);
remainingList.Remove(element); // Get a list containing everything except of chosen element
foreach (var permutation in Permutations<T>(remainingList)) { // Get all possible sub-permutations
permutation.Add(element); // Add that element
result.Add(permutation);
}
}
return result;
}
I know this is an old post, but someone might find this helpful.

Another solution using Linq and recursion...
static void Main(string[] args)
{
List<List<long>> result = new List<List<long>>();
List<long> set = new List<long>() { 1, 2, 3, 4 };
GetCombination<long>(set, result);
result.Add(set);
IOrderedEnumerable<List<long>> sorted = result.OrderByDescending(s => s.Count);
sorted.ToList().ForEach(l => { l.ForEach(l1 => Console.Write(l1 + " ")); Console.WriteLine(); });
}
private static void GetCombination<T>(List<T> set, List<List<T>> result)
{
for (int i = 0; i < set.Count; i++)
{
List<T> temp = new List<T>(set.Where((s, index) => index != i));
if (temp.Count > 0 && !result.Where(l => l.Count == temp.Count).Any(l => l.SequenceEqual(temp)))
{
result.Add(temp);
GetCombination<T>(temp, result);
}
}
}

This is an improvement of #ojlovecd answer without using strings.
static void Main(string[] args)
{
GetCombination(new List<int> { 1, 2, 3 });
}
private static void GetCombination(List<int> list)
{
double count = Math.Pow(2, list.Count);
for (int i = 1; i <= count - 1; i++)
{
for (int j = 0; j < list.Count; j++)
{
int b = i & (1 << j);
if (b > 0)
{
Console.Write(list[j]);
}
}
Console.WriteLine();
}
}

Firstly, given a set of n elements, you compute all combinations of k elements out of it (nCk). You have to change the value of k from 1 to n to meet your requirement.
See this codeproject article for C# code for generating combinations.
In case, you are interested in developing the combination algorithm by yourself, check this SO question where there are a lot of links to the relevant material.

protected List<List<T>> AllCombos<T>(Func<List<T>, List<T>, bool> comparer, params T[] items)
{
List<List<T>> results = new List<List<T>>();
List<T> workingWith = items.ToList();
results.Add(workingWith);
items.ToList().ForEach((x) =>
{
results.Add(new List<T>() { x });
});
for (int i = 0; i < workingWith.Count(); i++)
{
T removed = workingWith[i];
workingWith.RemoveAt(i);
List<List<T>> nextResults = AllCombos2(comparer, workingWith.ToArray());
results.AddRange(nextResults);
workingWith.Insert(i, removed);
}
results = results.Where(x => x.Count > 0).ToList();
for (int i = 0; i < results.Count; i++)
{
List<T> list = results[i];
if (results.Where(x => comparer(x, list)).Count() > 1)
{
results.RemoveAt(i);
}
}
return results;
}
protected List<List<T>> AllCombos2<T>(Func<List<T>, List<T>, bool> comparer, params T[] items)
{
List<List<T>> results = new List<List<T>>();
List<T> workingWith = items.ToList();
if (workingWith.Count > 1)
{
results.Add(workingWith);
}
for (int i = 0; i < workingWith.Count(); i++)
{
T removed = workingWith[i];
workingWith.RemoveAt(i);
List<List<T>> nextResults = AllCombos2(comparer, workingWith.ToArray());
results.AddRange(nextResults);
workingWith.Insert(i, removed);
}
results = results.Where(x => x.Count > 0).ToList();
for (int i = 0; i < results.Count; i++)
{
List<T> list = results[i];
if (results.Where(x => comparer(x, list)).Count() > 1)
{
results.RemoveAt(i);
}
}
return results;
}
This worked for me, it's slightly more complex and actually takes a comparer callback function, and it's actually 2 functions, the difference being that the AllCombos adds the single item lists explicitly. It is very raw and can definitely be trimmed down but it gets the job done. Any refactoring suggestions are welcome. Thanks,

public class CombinationGenerator{
private readonly Dictionary<int, int> currentIndexesWithLevels = new Dictionary<int, int>();
private readonly LinkedList<List<int>> _combinationsList = new LinkedList<List<int>>();
private readonly int _combinationLength;
public CombinationGenerator(int combinationLength)
{
_combinationLength = combinationLength;
}
private void InitializeLevelIndexes(List<int> list)
{
for (int i = 0; i < _combinationLength; i++)
{
currentIndexesWithLevels.Add(i+1, i);
}
}
private void UpdateCurrentIndexesForLevels(int level)
{
int index;
if (level == 1)
{
index = currentIndexesWithLevels[level];
for (int i = level; i < _combinationLength + 1; i++)
{
index = index + 1;
currentIndexesWithLevels[i] = index;
}
}
else
{
int previousLevelIndex;
for (int i = level; i < _combinationLength + 1; i++)
{
if (i > level)
{
previousLevelIndex = currentIndexesWithLevels[i - 1];
currentIndexesWithLevels[i] = previousLevelIndex + 1;
}
else
{
index = currentIndexesWithLevels[level];
currentIndexesWithLevels[i] = index + 1;
}
}
}
}
public void FindCombinations(List<int> list, int level, Stack<int> stack)
{
int currentIndex;
InitializeLevelIndexes(list);
while (true)
{
currentIndex = currentIndexesWithLevels[level];
bool levelUp = false;
for (int i = currentIndex; i < list.Count; i++)
{
if (level < _combinationLength)
{
currentIndex = currentIndexesWithLevels[level];
MoveToUpperLevel(ref level, stack, list, currentIndex);
levelUp = true;
break;
}
levelUp = false;
stack.Push(list[i]);
if (stack.Count == _combinationLength)
{
AddCombination(stack);
stack.Pop();
}
}
if (!levelUp)
{
MoveToLowerLevel(ref level, stack, list, ref currentIndex);
while (currentIndex >= list.Count - 1)
{
if (level == 1)
{
AdjustStackCountToCurrentLevel(stack, level);
currentIndex = currentIndexesWithLevels[level];
if (currentIndex >= list.Count - 1)
{
return;
}
UpdateCurrentIndexesForLevels(level);
}
else
{
MoveToLowerLevel(ref level, stack, list, ref currentIndex);
}
}
}
}
}
private void AddCombination(Stack<int> stack)
{
List<int> listNew = new List<int>();
listNew.AddRange(stack);
_combinationsList.AddLast(listNew);
}
private void MoveToUpperLevel(ref int level, Stack<int> stack, List<int> list, int index)
{
stack.Push(list[index]);
level++;
}
private void MoveToLowerLevel(ref int level, Stack<int> stack, List<int> list, ref int currentIndex)
{
if (level != 1)
{
level--;
}
AdjustStackCountToCurrentLevel(stack, level);
UpdateCurrentIndexesForLevels(level);
currentIndex = currentIndexesWithLevels[level];
}
private void AdjustStackCountToCurrentLevel(Stack<int> stack, int currentLevel)
{
while (stack.Count >= currentLevel)
{
if (stack.Count != 0)
stack.Pop();
}
}
public void PrintPermutations()
{
int count = _combinationsList.Where(perm => perm.Count() == _combinationLength).Count();
Console.WriteLine("The number of combinations is " + count);
}
}

We can use recursion for combination/permutation problems involving string or integers.
public static void Main(string[] args)
{
IntegerList = new List<int> { 1, 2, 3, 4 };
PrintAllCombination(default(int), default(int));
}
public static List<int> IntegerList { get; set; }
public static int Length { get { return IntegerList.Count; } }
public static void PrintAllCombination(int position, int prefix)
{
for (int i = position; i < Length; i++)
{
Console.WriteLine(prefix * 10 + IntegerList[i]);
PrintAllCombination(i + 1, prefix * 10 + IntegerList[i]);
}
}

What about
static void Main(string[] args)
{
Combos(new [] { 1, 2, 3 });
}
static void Combos(int[] arr)
{
for (var i = 0; i <= Math.Pow(2, arr.Length); i++)
{
Console.WriteLine();
var j = i;
var idx = 0;
do
{
if ((j & 1) == 1) Console.Write($"{arr[idx]} ");
} while ((j >>= 1) > 0 && ++idx < arr.Length);
}
}

A slightly more generalised version for Linq using C# 7. Here filtering by items that have two elements.
static void Main(string[] args)
{
foreach (var vals in Combos(new[] { "0", "1", "2", "3" }).Where(v => v.Skip(1).Any() && !v.Skip(2).Any()))
Console.WriteLine(string.Join(", ", vals));
}
static IEnumerable<IEnumerable<T>> Combos<T>(T[] arr)
{
IEnumerable<T> DoQuery(long j, long idx)
{
do
{
if ((j & 1) == 1) yield return arr[idx];
} while ((j >>= 1) > 0 && ++idx < arr.Length);
}
for (var i = 0; i < Math.Pow(2, arr.Length); i++)
yield return DoQuery(i, 0);
}

Here is how I did it.
public static List<List<int>> GetCombination(List<int> lst, int index, int count)
{
List<List<int>> combinations = new List<List<int>>();
List<int> comb;
if (count == 0 || index == lst.Count)
{
return null;
}
for (int i = index; i < lst.Count; i++)
{
comb = new List<int>();
comb.Add(lst.ElementAt(i));
combinations.Add(comb);
var rest = GetCombination(lst,i + 1, count - 1);
if (rest != null)
{
foreach (var item in rest)
{
combinations.Add(comb.Union(item).ToList());
}
}
}
return combinations;
}
You call it as :
List<int> lst= new List<int>(new int[]{ 1, 2, 3, 4 });
var combinations = GetCombination(lst, 0, lst.Length)

I just run into a situation where I needed to do this, this is what I came up with:
private static List<string> GetCombinations(List<string> elements)
{
List<string> combinations = new List<string>();
combinations.AddRange(elements);
for (int i = 0; i < elements.Count - 1; i++)
{
combinations = (from combination in combinations
join element in elements on 1 equals 1
let value = string.Join(string.Empty, $"{combination}{element}".OrderBy(c => c).Distinct())
select value).Distinct().ToList();
}
return combinations;
}
It may be not too efficient, and it sure has room for improvement, but gets the job done!
List<string> elements = new List<string> { "1", "2", "3" };
List<string> combinations = GetCombinations(elements);
foreach (string combination in combinations)
{
System.Console.Write(combination);
}

This is an improved version based on the answer from ojlovecd using extension methods:
public static class ListExtensions
{
public static IEnumerable<List<T>> GetCombinations<T>(
this List<T> valuesToCombine)
{
var count = Math.Pow(2, valuesToCombine.Count);
for(var i = 1; i <= count; i++)
{
var itemFlagList = i.ToBinaryString(valuesToCombine.Count())
.Select(x => x == '1').ToList();
yield return GetCombinationByFlagList(valuesToCombine, itemFlagList)
.ToList();
}
}
private static IEnumerable<T> GetCombinationByFlagList<T>(
List<T> valuesToCombine, List<bool> flagList)
{
for (var i = 0; i < valuesToCombine.Count; i++)
{
if (!flagList[i]) continue;
yield return valuesToCombine.ElementAt(i);
}
}
}
public static class IntegerExtensions
{
public static string ToBinaryString(this int value, int length)
{
return Convert.ToString(value, 2).ToString().PadLeft(length, '0');
}
}
Usage:
var numbersList = new List<int>() { 1, 2, 3 };
var combinations = numbersList.GetCombinations();
foreach (var combination in combinations)
{
System.Console.WriteLine(string.Join(",", combination));
}
Output:
3
2
2,3
1
1,3
1,2
1,2,3
The idea is to basically use some flags to keep track of which items were already added to the combination. So in case of 1, 2 & 3, the following binary strings are generated in order to indicate whether an item should be included or excluded:
001, 010, 011, 100, 101, 110 & 111

I'd like to suggest an approach that I find to be quite intuitive and easy to read. (Note: It is slower than the currently accepted solution.)
It is built on the idea that for each integer in the list, we need to extend the so-far-aggregated resulting combination list with
all currently existing combinations, each extended with the given integer
a single "combination" of that integer alone
Here, I am using .Aggregate() with a seed that is an IEnumerable<IEnumerable<int>> containing a single, empty collection entry. That empty entry lets us easily do the two steps above simultaneously. The empty collection entry can be skipped after the resulting combination collection has been aggregated.
It goes like this:
var emptyCollection = Enumerable.Empty<IEnumerable<int>>();
var emptyCombination = Enumerable.Empty<int>();
IEnumerable<int[]> combinations = list
.Aggregate(emptyCollection.Append(emptyCombination),
( acc, next ) => acc.Concat(acc.Select(entry => entry.Append(next))))
.Skip(1) // skip the initial, empty combination
.Select(comb => comb.ToArray());
For each entry in the input integer list { 1, 2, 3 }, the accumulation progresses as follows:
next = 1
{ { } }.Concat({ { }.Append(1) })
{ { } }.Concat({ { 1 } })
{ { }, { 1 } } // acc
next = 2
{ { }, { 1 } }.Concat({ { }.Append(2), { 1 }.Append(2) })
{ { }, { 1 } }.Concat({ { 2 }, { 1, 2 } })
{ { }, { 1 }, { 2 }, { 1, 2 } } // acc
next = 3
{ { }, { 1 }, { 2 }, { 1, 2 } }.Concat({ { }.Append(3), { 1 }.Append(3), { 2 }.Append(3), { 1, 2 }.Append(3) })
{ { }, { 1 }, { 2 }, { 1, 2 } }.Concat({ { 3 }, { 1, 3 }, { 2, 3 }, { 1, 2, 3 } })
{ { }, { 1 }, { 2 }, { 1, 2 }, { 3 }, { 1, 3 }, { 2, 3 }, { 1, 2, 3 } } // acc
Skipping the first (empty) entry, we are left with the following collection:
1
2
1 2
3
1 3
2 3
1 2 3
, which can easily be ordered by collection length and collection entry sum for a clearer overview.
Example fiddle here.

Some of the solutions here are truly ingenious; especially the ones that use bitmaps.
But I found that in practice these algos
aren't easy to modify if a specific range of lengths needed (e.g. all variations of 3 to 5 choices from an input set of 8 elements)
can't handle LARGE input lists (and return empty or singleton results instead of throwing exception); and
can be tricky to debug.
So I decided to write something not as clever as the other people here.
My more basic approach recognises that the set of Variations(1 to maxLength) is simply a UNION of all fixed-length Variations of each length 1 to maxLength:
i.e
Variations(1 to maxLength) = Variations(1) + Variations(2) + ... + Variations(maxLength)
So you can do a "choose K from N" for each required length (for each K in (1, 2, 3, ..., maxLength)) and then just do a Union of these separate results to yield a List of Lists.
This resulting code intends to be easy to understand and to maintain:
/// <summary>
/// Generates ALL variations of length between minLength and maxLength (inclusive)
/// Relies on Combinatorics library to generate each set of Variations
/// Nuget https://www.nuget.org/packages/Combinatorics/
/// Excellent more general references (without this solution):
/// https://www.codeproject.com/Articles/26050/Permutations-Combinations-and-Variations-using-C-G
/// Self-authored solution.
/// </summary>
/// <typeparam name="T">Any type without any constraints.</typeparam>
/// <param name="sourceList">The source list of elements to be combined.</param>
/// <param name="minLength">The minimum length of variation required.</param>
/// <param name="maxLength">The maximum length of variation required.</param>
/// <returns></returns>
public static List<List<T>> GenerateVariations<T>(this IEnumerable<T> sourceList, int minLength, int maxLength)
{
List<List<T>> finalUnion = new();
foreach (int length in Enumerable.Range(minLength, maxLength))
{
Variations<T> variations = new Variations<T>(sourceList, length, GenerateOption.WithoutRepetition);
foreach (var variation in variations)
{
var list = variation.ToList<T>();
finalUnion.Add(list);
}
}
Debug.WriteLine(sourceList.Count() + " source " + typeof(T).Name + " yielded " + finalUnion.Count());
return finalUnion;
}
Happy to receive comments (good and bad). Maybe there's a more succint way to achieve this in LINQ? Maybe the really smart people here can marry their approach with my more basic one?

Please find very very simple solution without recursion and which dont eat RAM.
Unique Combinations

How to get all subsets of an array?

Given an array: [dog, cat, mouse]
what is the most elegant way to create:
[,,]
[,,mouse]
[,cat,]
[,cat,mouse]
[dog,,]
[dog,,mouse]
[dog,cat,]
[dog,cat,mouse]
I need this to work for any sized array.
This is essentially a binary counter, where array indices represent bits. This presumably lets me use some bitwise operation to count, but I can't see a nice way of translating this to array indices though.

Elegant? Why not Linq it.
public static IEnumerable<IEnumerable<T>> SubSetsOf<T>(IEnumerable<T> source)
{
if (!source.Any())
return Enumerable.Repeat(Enumerable.Empty<T>(), 1);
var element = source.Take(1);
var haveNots = SubSetsOf(source.Skip(1));
var haves = haveNots.Select(set => element.Concat(set));
return haves.Concat(haveNots);
}

string[] source = new string[] { "dog", "cat", "mouse" };
for (int i = 0; i < Math.Pow(2, source.Length); i++)
{
string[] combination = new string[source.Length];
for (int j = 0; j < source.Length; j++)
{
if ((i & (1 << (source.Length - j - 1))) != 0)
{
combination[j] = source[j];
}
}
Console.WriteLine("[{0}, {1}, {2}]", combination[0], combination[1], combination[2]);
}

You can use the BitArray class to easily access the bits in a number:
string[] animals = { "Dog", "Cat", "Mouse" };
List<string[]> result = new List<string[]>();
int cnt = 1 << animals.Length;
for (int i = 0; i < cnt; i++) {
string[] item = new string[animals.Length];
BitArray b = new BitArray(i);
for (int j = 0; j < item.Length; j++) {
item[j] = b[j] ? animals[j] : null;
}
result.Add(item);
}

static IEnumerable<IEnumerable<T>> GetSubsets<T>(IList<T> set)
{
var state = new BitArray(set.Count);
do
yield return Enumerable.Range(0, state.Count)
.Select(i => state[i] ? set[i] : default(T));
while (Increment(state));
}
static bool Increment(BitArray flags)
{
int x = flags.Count - 1;
while (x >= 0 && flags[x]) flags[x--] = false ;
if (x >= 0) flags[x] = true;
return x >= 0;
}
Usage:
foreach(var strings in GetSubsets(new[] { "dog", "cat", "mouse" }))
Console.WriteLine(string.Join(", ", strings.ToArray()));

Guffa's answer had the basic functionality that I was searching, however the line with
BitArray b = new BitArray(i);
did not work for me, it gave an ArgumentOutOfRangeException. Here's my slightly adjusted and working code:
string[] array = { "A", "B", "C","D" };
int count = 1 << array.Length; // 2^n
for (int i = 0; i < count; i++)
{
string[] items = new string[array.Length];
BitArray b = new BitArray(BitConverter.GetBytes(i));
for (int bit = 0; bit < array.Length; bit++) {
items[bit] = b[bit] ? array[bit] : "";
}
Console.WriteLine(String.Join("",items));
}

Here's a solution similar to David B's method, but perhaps more suitable if it's really a requirement that you get back sets with the original number of elements (even if empty):.
static public List<List<T>> GetSubsets<T>(IEnumerable<T> originalList)
{
if (originalList.Count() == 0)
return new List<List<T>>() { new List<T>() };
var setsFound = new List<List<T>>();
foreach (var list in GetSubsets(originalList.Skip(1)))
{
setsFound.Add(originalList.Take(1).Concat(list).ToList());
setsFound.Add(new List<T>() { default(T) }.Concat(list).ToList());
}
return setsFound;
}
If you pass in a list of three strings, you'll get back eight lists with three elements each (but some elements will be null).

Here's an easy-to-follow solution along the lines of your conception:
private static void Test()
{
string[] test = new string[3] { "dog", "cat", "mouse" };
foreach (var x in Subsets(test))
Console.WriteLine("[{0}]", string.Join(",", x));
}
public static IEnumerable<T[]> Subsets<T>(T[] source)
{
int max = 1 << source.Length;
for (int i = 0; i < max; i++)
{
T[] combination = new T[source.Length];
for (int j = 0; j < source.Length; j++)
{
int tailIndex = source.Length - j - 1;
combination[tailIndex] =
((i & (1 << j)) != 0) ? source[tailIndex] : default(T);
}
yield return combination;
}
}

This is a small change to Mehrdad's solution above:
static IEnumerable<T[]> GetSubsets<T>(T[] set) {
bool[] state = new bool[set.Length+1];
for (int x; !state[set.Length]; state[x] = true ) {
yield return Enumerable.Range(0, state.Length)
.Where(i => state[i])
.Select(i => set[i])
.ToArray();
for (x = 0; state[x]; state[x++] = false);
}
}
or with pointers
static IEnumerable<T[]> GetSubsets<T>(T[] set) {
bool[] state = new bool[set.Length+1];
for (bool *x; !state[set.Length]; *x = true ) {
yield return Enumerable.Range(0, state.Length)
.Where(i => state[i])
.Select(i => set[i])
.ToArray();
for (x = state; *x; *x++ = false);
}
}

I'm not very familiar with C# but I'm sure there's something like:
// input: Array A
foreach S in AllSubsetsOf1ToN(A.Length):
print (S.toArray().map(lambda x |> A[x]));
Ok, I've been told the answer above won't work. If you value elegance over efficiency, I would try recursion, in my crappy pseudocode:
Array_Of_Sets subsets(Array a)
{
if (a.length == 0)
return [new Set();] // emptyset
return subsets(a[1:]) + subsets(a[1:]) . map(lambda x |> x.add a[0])
}

Here is a variant of mqp's answer, that uses as state a BigInteger instead of an int, to avoid overflow for collections containing more than 30 elements:
using System.Numerics;
public static IEnumerable<IEnumerable<T>> GetSubsets<T>(IList<T> source)
{
BigInteger combinations = BigInteger.One << source.Count;
for (BigInteger i = 0; i < combinations; i++)
{
yield return Enumerable.Range(0, source.Count)
.Select(j => (i & (BigInteger.One << j)) != 0 ? source[j] : default);
}
}

Easy to understand version (with descriptions)
I assumed that source = {1,2,3,4}
public static IEnumerable<IEnumerable<T>> GetSubSets<T>(IEnumerable<T> source)
{
var result = new List<IEnumerable<T>>() { new List<T>() }; // empty cluster added
for (int i = 0; i < source.Count(); i++)
{
var elem = source.Skip(i).Take(1);
// for elem = 2
// and currently result = [ [],[1] ]
var matchUps = result.Select(x => x.Concat(elem));
//then matchUps => [ [2],[1,2] ]
result = result.Concat(matchUps).ToList();
// matchUps and result concat operation
// finally result = [ [],[1],[2],[1,2] ]
}
return result;
}

The way this is written, it is more of a Product (Cartesian product) rather than a list of all subsets.
You have three sets: (Empty,"dog"), (Empty,"cat"),(Empty,"mouse").
There are several posts on general solutions for products. As noted though, since you really just have 2 choices for each axis a single bit can represent the presence or not of the item.
So the total set of sets is all numbers from 0 to 2^N-1. If N < 31 an int will work.

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