Given an array of variable dimensions.... Eg array={1,2,4,5}
I need a way to generale all possible combinations and subset of the array.
Given an array of n elements I need to have all subsets (all subsets of 1 element, all subset of 2 elements, all subset of n elements) an of each subset all possible permutations.
For example result should be:
{1}
{2}
{4}
{5}
{1,2}
{1,4}
{1,5}
{2,1}
{2,4}
{2,5}
....
....
{1,2,4,5}
{1,2,5,4}
{1,4,2,5}
{1,5,2,4}
{1,5,4,2}
{2,1,4,5}
{2,1,5,4}
....
....
{5,1,4,2}
{5,1,2,4}
{5,2,4,1}
....
....
etc...
ALL combination!
Is there a quick way? I don't have idea....
You should apply 2 steps:
This implementation uses some utility classes from a combinatorics project. The output also contains the empty set {}
and is not ordered by the size, but this may easily be done as a postprocessing step.
import java.util.ArrayList;
import java.util.Arrays;
import java.util.Collections;
import java.util.Iterator;
import java.util.List;
import java.util.NoSuchElementException;
public class AllCombinations {
public static void main(String[] args) {
List<Integer> list = Arrays.asList(1,2,4,5);
PowerSetIterable<Integer> powerSet =
new PowerSetIterable<Integer>(list);
for (List<Integer> subset : powerSet)
{
PermutationIterable<Integer> permutations =
new PermutationIterable<Integer>(subset);
for (List<Integer> permutation : permutations) {
System.out.println(permutation);
}
}
}
}
//From https://github.com/javagl/Combinatorics
class PowerSetIterable<T> implements Iterable<List<T>> {
private final List<T> input;
private final int numElements;
public PowerSetIterable(List<T> input) {
this.input = input;
numElements = 1 << input.size();
}
@Override
public Iterator<List<T>> iterator() {
return new Iterator<List<T>>() {
private int current = 0;
@Override
public boolean hasNext() {
return current < numElements;
}
@Override
public List<T> next() {
if (!hasNext()) {
throw new NoSuchElementException("No more elements");
}
List<T> element = new ArrayList<T>();
for (int i = 0; i < input.size(); i++) {
long b = 1 << i;
if ((current & b) != 0) {
element.add(input.get(i));
}
}
current++;
return element;
}
@Override
public void remove() {
throw new UnsupportedOperationException(
"May not remove elements from a power set");
}
};
}
}
//From https://github.com/javagl/Combinatorics
class PermutationIterable<T> implements Iterable<List<T>> {
public static int factorial(int n) {
int f = 1;
for (int i = 2; i <= n; i++) {
f = f * i;
}
return f;
}
private final List<T> input;
private final int numPermutations;
public PermutationIterable(List<T> input) {
this.input = input;
numPermutations = factorial(input.size());
}
@Override
public Iterator<List<T>> iterator() {
if (input.size() == 0) {
return Collections.<List<T>> singletonList(
Collections.<T> emptyList()).iterator();
}
return new Iterator<List<T>>() {
private int current = 0;
@Override
public boolean hasNext() {
return current < numPermutations;
}
@Override
public List<T> next() {
if (!hasNext()) {
throw new NoSuchElementException("No more elements");
}
// Adapted from http://en.wikipedia.org/wiki/Permutation
List<T> result = new ArrayList<T>(input);
int factorial = numPermutations / input.size();
for (int i = 0; i < result.size() - 1; i++) {
int tempIndex = (current / factorial) % (result.size() - i);
T temp = result.get(i + tempIndex);
for (int j = i + tempIndex; j > i; j--) {
result.set(j, result.get(j - 1));
}
result.set(i, temp);
factorial /= (result.size() - (i + 1));
}
current++;
return result;
}
@Override
public void remove() {
throw new UnsupportedOperationException(
"May not remove elements from a permutation");
}
};
}
}
First, you need to find all the subsets of the array which are 2^n sets (including the empty set). Then once you find the subsets, loop through each of them and compute it permutation using a simple recursion which you can easily find online.
The easiest way I know is to loop over i
from 1
to 2^n - 1
, where n is the size of the array.
The 1s in the bit pattern of i
tell you which elements to select.
eg: With the array [4, 28, 37, 135]
on the 10th loop:
10 == 1010b
1, 0, 1, 0 tells you to select the first and third elements of the array: [4, 37]
.
Now that you have all combinations of elements in the array, you need to get all permutations, which can be done with some simple recursion.
pseudocode:
function getPermutations(arr)
{
if length of arr == 1 {
return [arr]
} else {
for i = 0 to highest index of arr {
sub_arr = copy of arr
remove element i from sub_arr
perms = getPermutations(sub_arr)
for each perm in perms {
insert arr[i] at beginning of perm
}
return perms
}
}
}
I'm providing the first solution it came to my mind for finding all subsets given a List (not the permutations, only the subsets). The method subSets gives all subsets of a specific size, while allSubSets iterates over the sizes. Once you have a list of all subsets, you can implement a permutation function that iterates over this list.
public class Subsets<T> {
public List<List<T>> allSubSets(List<T> list) {
List<List<T>> out = new ArrayList<List<T>>();
for(int i=1; i<=list.size(); i++) {
List<List<T>> outAux = this.subSets(list, i);
out.addAll(outAux);
}
return out;
}
private List<List<T>> subSets(List<T> list, int size) {
List<List<T>> out = new ArrayList<List<T>>();
for(int i=0; i<list.size()-size+1;i++) {
List<T> subset = new ArrayList<T>();
for (int j=i;j<i+size-1;j++) {
subset.add(list.get(j));
}
if (!(size==1 && i>0)) {
for (int j=i+size-1;j<list.size();j++) {
List<T> newsubset = new ArrayList<T>(subset);
newsubset.add(list.get(j));
out.add(newsubset);
}
}
}
return out;
}
}
To use it:
Subsets<Integer> aux = new Subsets<Integer>();
List<Integer> list = new ArrayList<Integer>();
list.add(1);
list.add(2);
list.add(3);
list.add(4);
list.add(5);
List<List<Integer>> out = aux.allSubSets(list);
System.out.println(out);
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