How to generate the power-set of a given List?
I'm trying to generate a collection of all 2^N - 1 possible combinations of a given List of length N. The collection will map the number of elements in a combination to an ordered list of combinations containing combinations of the specific length. For instance, for the List:
[A, B, C, D]
I want to generate the map:
{
1 -> [{A}, {B}, {C}, {D}]
2 -> [{A, B}, {A, C}, {A, D}, {B, C}, {B, D}, {C, D}]
3 -> [{A, B, C}, {A, B, D}, {A, C, D}, {B, C, D}]
4 -> [{A, B, C, D}]
}
The generated database should maintain the original order (where [] represents an ordered series (List), and {} represents an un-ordered group (Set)), and run as fast as possible.
I was struggling with some recursive code all day (I know the implementation should be recursive) but couldn't get to the bottom of it.
Is there a reference I can use/a ready implementation of such algorithm?
Solution 1:
What you're looking for is essentially the power set (minus perhaps the empty set). Guava actually has a method for this: Sets.powerSet(). You can view the source of the Sets class to see how the method is implemented if you want to write it yourself; you might need to modify it to return a List instead of a Set since you want to preserve order, although this change should not be too drastic. Once you have the power set, it should be trivial to iterate over it and construct the map you want.
Solution 2:
What you're asking is generating all possible subsets of a set. You can think of it as iterating over all possible binary arrays of size N (the size of your list):
000000...000
000000...001
000000...010
000000...011
etc.
Why is that? The answer is simple: 1 indicates that an element exists in a subset, while 0 indicates that it is absent.
So, the basic algorithm is obvious:
s = [A, B, C, D]
for i=0 to 2^N-1:
b = convert_number_to_bin_array(i)
ss = calculate_subset_based_on_bin_array(s, b)
print ss
Where calculate_subset_based_on_bin_array iterates on b and s and selects elements from s[current] where b[current] = 1.
The above will print out all existing subsets. You can adapt this algorithm in order to get the map that you've asked for in your question.
Solution 3:
static Map<Integer, List<LinkedList<Integer>>> powerset = new HashMap<>();
public static void main(String[] args) throws IOException {
powerset(Arrays.asList(1, 2, 3));
for (Integer key : powerset.keySet()) {
System.out.print(key + " -> ");
System.out.println(Arrays.toString(powerset.get(key).toArray()));
}
}
static void powerset(List<Integer> src) {
powerset(new LinkedList<>(), src);
}
private static void powerset(LinkedList<Integer> prefix, List<Integer> src) {
if (src.size() > 0) {
prefix = new LinkedList<>(prefix); //create a copy to not modify the orig
src = new LinkedList<>(src); //copy
Integer curr = src.remove(0);
collectResult(prefix, curr);
powerset(prefix, src);
prefix.add(curr);
powerset(prefix, src);
}
}
private static void collectResult(LinkedList<Integer> prefix, Integer curr) {
prefix = new LinkedList<>(prefix); //copy
prefix.add(curr);
List<LinkedList<Integer>> addTo;
if (powerset.get(prefix.size()) == null) {
List<LinkedList<Integer>> newList = new LinkedList<>();
addTo = newList;
} else {
addTo = powerset.get(prefix.size());
}
addTo.add(prefix);
powerset.put(prefix.size(), addTo);
}
OUTPUT
1 -> [[1], [2], [3]]
2 -> [[2, 3], [1, 2], [1, 3]]
3 -> [[1, 2, 3]]
Solution 4:
Here is a code I have tested to generate all possible combinations out of given array:
enter code here
import java.util.Arrays;
public class PasswordGen {
static String[] options = { "a", "b", "c", "d" };
static String[] places = new String[options.length];
static int count;
public static void main(String[] args) {
// Starting with initial position of a i.e. 0
sequence(0, places.clone());
}
private static void sequence(int level, String[] holder) {
if (level >= options.length) {
// combination complete
System.out.println("" + (++count) + " Combination "
+ Arrays.toString(holder));
return;
}
String val = options[level];
String[] inrHolder = null;
for (int c = 0; c < holder.length; c++) {
inrHolder = holder.clone();
if (inrHolder[c] == null) {
inrHolder[c] = val;
sequence(level + 1, inrHolder.clone());
}
}
return;
}
}
Solution 5:
There are multiple ways to solve this problem, but the easiest way IMHO is to use recursion. Below I have provided the iterative and recursive approaches to solving the problem of generating all the combinations of a set. The general idea for both approaches is to generate the set of indexes which belong to the elements you want to select.
For the recursive approach, you want to keep track of three pieces of information: A boolean array indicating whether an item was selected or not, the position you're at in your list of items and a variable r tracking the number of elements remaining to select. Then as long as r != 0 (meaning you still need to select r elements to finish this combination) you backtrack selecting an element you have not yet selected and move forward in the array.
The code shown below was taken from my github repo.
/**
* Here we present two methods (recursive and iterative) of generating all
* the combinations of a set by choosing only r of n elements.
*
* Time Complexity: O( n choose r )
*
* @author William Fiset, Micah Stairs
**/
public class Combinations {
// This method finds all the combinations of size r in a set
public static void combinationsChooseR(int[] set, int r) {
if (r < 0) return;
if (set == null) return;
boolean [] used = new boolean[set.length];
combinations(set, r, 0, used);
}
// To find all the combinations of size r we need to recurse until we have
// selected r elements (aka r = 0), otherwise if r != 0 then we need to select
// an element which is found after the position of our last selected element
private static void combinations(int[] set, int r, int at, boolean[] used) {
final int N = set.length;
// If there are more elements left to select than what are available
// This is a short circuiting optimization we can take advantage of
int elementsLeftToPick = N - at;
if (elementsLeftToPick < r) return;
// We selected 'r' elements so we found a valid subset!
if (r == 0) {
System.out.print("{ ");
for (int i = 0; i < N; i++)
if (used[i])
System.out.print(set[i] + " ");
System.out.println("}");
} else {
for (int i = at; i < N; i++) {
// Try including this element
used[i] = true;
combinations(set, r - 1, i + 1, used);
// Backtrack and try the instance where we did not include this element
used[i] = false;
}
}
}
// Use this method in combination with a do while loop to generate all the
// combinations of a set choose r in a iterative fashion. This method returns
// false once the last combination has been generated.
// NOTE: Originally the selection needs to be initialized to {0,1,2,3 ... r-1}
public static boolean nextCombination(int[] selection, int N, int r) {
if (r > N) throw new IllegalArgumentException("r must be <= N");
int i = r - 1;
while (selection[i] == N - r + i)
if (--i < 0) return false;
selection[i]++;
for (int j = i + 1; j < r; j++) selection[j] = selection[i] + j - i;
return true;
}
public static void main(String[] args) {
// Recursive approach
int R = 3;
int[] set = {1,2,3,4,5};
combinationsChooseR(set, R);
// prints:
// { 1 2 3 }
// { 1 2 4 }
// { 1 2 5 }
// { 1 3 4 }
// { 1 3 5 }
// { 1 4 5 }
// { 2 3 4 }
// { 2 3 5 }
// { 2 4 5 }
// { 3 4 5 }
// Suppose we want to select all combinations of colors where R = 3
String[] colors = {"red", "purple", "green", "yellow", "blue", "pink"};
R = 3;
// Initialize the selection to be {0, 1, ... , R-1}
int[] selection = { 0, 1, 2 };
do {
// Print each combination
for (int index : selection)
System.out.print(colors[index] + " ");
System.out.println();
} while( nextCombination(selection, colors.length, R) );
// prints:
// red purple green
// red purple yellow
// red purple blue
// red purple pink
// red green yellow
// red green blue
// red green pink
// red yellow blue
// red yellow pink
// red blue pink
// purple green yellow
// purple green blue
// purple green pink
// purple yellow blue
// purple yellow pink
// purple blue pink
// green yellow blue
// green yellow pink
// green blue pink
// yellow blue pink
}
}