# Combinations Algorithm

#### [email protected]

MHF Hall of Honor
I'm looking for an efficient algorithm to take a set $$\displaystyle S=\{s_1,\ldots,s_n\}$$ and return all possible subsets (minus $$\displaystyle \emptyset$$), where order doesn't matter.

Example: Given $$\displaystyle S=\{1,2,3\}$$, the output would be $$\displaystyle \{1,2,3\},\;\{1,2\},\;\{1,3\},\;\{2,3\},\;\{1\},\;\{2\},\;\{3\}$$.

-Thanks!

#### undefined

MHF Hall of Honor
I'm looking for an efficient algorithm to take a set $$\displaystyle S=\{s_1,\ldots,s_n\}$$ and return all possible subsets (minus $$\displaystyle \emptyset$$), where order doesn't matter.

Example: Given $$\displaystyle S=\{1,2,3\}$$, the output would be $$\displaystyle \{1,2,3\},\;\{1,2\},\;\{1,3\},\;\{2,3\},\;\{1\},\;\{2\},\;\{3\}$$.

-Thanks!
The first thing that comes to mind is recursion, in particular depth first search. Are you working in a specific programming language? The basic idea is to start with an empty list and then for each element recusively do: add the element to the list, or don't.

Looks like this in java

Code:
import java.util.ArrayList;

public class Subsets {
static ArrayList<ArrayList<Integer>> subsets = new ArrayList<ArrayList<Integer>>();
static int[] S = {1,2,3};

public static void main(String[] args) {
getSubsets(0, new ArrayList<Integer>());
System.out.println(subsets);
}

static void getSubsets(int x, ArrayList<Integer> list) {
if (x==S.length) {
return;
}
ArrayList<Integer> option1 = new ArrayList<Integer>(list);
getSubsets(x+1, option1);
ArrayList<Integer> option2 = new ArrayList<Integer>(list);
getSubsets(x+1, option2);
}
}

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#### [email protected]

MHF Hall of Honor
I'm coding in Matlab. I can easily convert this code to Matlab, thanks!

#### awkward

MHF Hall of Honor
Another option is to realize that there are 2^n - 1 nonempty subsets, so there is a natural bijection between the subsets and the integers 1 through 2^n - 1. Just count from 1 to 2^n - 1 and convert each number to binary. Say an element is in the subset if its corresponding bit is 1.