Generate all permutations of several lists in Java

Question

I have n lists, for example:

L_1 = [a_11, a_12, ...]
L_2 = [a_21, a_22, ...]
...
L_n = [a_n1, a_n2, ...]

where ith list has k_i elements. And now, I want to generate all n-elements list, where ith element is from L_i, I mean:

[a_11, a_21, ..., a_n1]
[a_11, a_21, ..., a_n2]
...
[a_11, a_22, ..., a_n1]
[a_11, a_22, ..., a_n2]
...
[a_12, a_21, ..., a_n1]
[a_12, a_21, ..., a_n2]
...
[a_12, a_22, ..., a_n1]
[a_12, a_22, ..., a_n2]
...

The total number of lists shoulbe be equal to k_1*k_2*...k_n. Could you describe pseudo-code of this algorithm or use Java code? I can do this using nested for-loops when number of lists is hardcoded, but I'm completely blocked when n is customizable at runtime.

score 5 · Answer 1 · answered Mar 20 '15 at 17:49

Ok, I implemented this algorithm.

import com.google.common.collect.Lists;

import java.util.ArrayList;
import java.util.Collections;
import java.util.List;

public class PermutationGenerator {
    private List<List<Integer>> result;
    private List<List<Integer>> data;

    public List<List<Integer>> permutate(List<List<Integer>> data) {
        this.data = data;
        this.result = Lists.newArrayList();

        List<Integer> integers = new ArrayList<Integer>(Collections.nCopies(data.size(), 0));
        foo(0, data.size() - 1, integers);
        return result;
    }

    private void foo(Integer index, Integer maxIndex, List<Integer> output) {
        List<Integer> list = data.get(index);
        for (int i = 0; i < list.size(); i++) {
            output.set(index, list.get(i));
            if (index == maxIndex) {
                result.add(Lists.newArrayList(output));
            } else {
                foo(index + 1, maxIndex, output);
            }
        }
    }
}

Test class:

import com.google.common.collect.Lists;
import org.junit.Test;

import java.util.Arrays;
import java.util.List;

public class PermutationGeneratorTest {
    @Test
    public void test() throws Exception {
        // given
        PermutationGenerator pg = new PermutationGenerator();

        List<Integer> list1 = Lists.newArrayList(1, 2, 3);
        List<Integer> list2 = Lists.newArrayList(4, 5);
        List<Integer> list3 = Lists.newArrayList(6, 7, 8, 9);

        List<List<Integer>> input = Lists.newArrayList(list1, list2, list3);

        // when
        List<List<Integer>> output = pg.permutate(input);

        // then
        print(output);
    }

    private void print(List<List<Integer>> output) {
        for (List<Integer> list : output) {
            System.out.println(Arrays.toString(list.toArray()));
        }
        System.out.println("TOTAL: " + output.size());
    }
}

Output:

[1, 4, 6]
[1, 4, 7]
[1, 4, 8]
[1, 4, 9]
[1, 5, 6]
[1, 5, 7]
[1, 5, 8]
[1, 5, 9]
[2, 4, 6]
[2, 4, 7]
[2, 4, 8]
[2, 4, 9]
[2, 5, 6]
[2, 5, 7]
[2, 5, 8]
[2, 5, 9]
[3, 4, 6]
[3, 4, 7]
[3, 4, 8]
[3, 4, 9]
[3, 5, 6]
[3, 5, 7]
[3, 5, 8]
[3, 5, 9]
TOTAL: 24

dhke · Accepted Answer · 2015-03-20T19:19:06.500

As you have already found out yourself, the usual trick is to think of the lists a non-uniform version of the g-adic numbers and do carry increment on the list index positions:

When you have n lists, you have n index positions in those lists:

index_pos = [i0, ..., in-1]

The trick is now as follows:

start with index_pos = [0, 0, ...]
increment index_pos[0].
- If the result is larger or equal to lists[0].size(), set index_pos[0] = 0 and increment index_pos[1].
- if index_pos[1] is larger than or equal to lists[1].size() ... and so on
You are done when index_pos[n - 1] overflows

An non-recursive solution in Java would be like

public static <T> void permute(
    final List<List<T>> lists,
    final Consumer<List<T>> consumer
)
{
     final int[] index_pos = new int[lists.size()];

     final int last_index = lists.size() - 1;
     final List<T> permuted = new ArrayList<T>(lists.size());

     for (int i = 0; i < lists.size(); ++i) {
         permuted.add(null);
     } 

     while (index_pos[last_index] < lists.get(last_index).size()) {
         for (int i = 0; i < lists.size(); ++i) {
            permuted.set(i, lists.get(i).get(index_pos[i]));
         }  
         consumer.accept(permuted);

         for (int i = 0; i < lists.size(); ++i) {
             ++index_pos[i];
             if (index_pos[i] < lists.get(i).size()) {
                /* stop at first element without overflow */
                break;
             } else if (i < last_index) {
                index_pos[i] = 0;
             }  
         }
     }
}

Usage example:

public static void main(String[] args)
{
    final List<List<Integer>> lists = new ArrayList<List<Integer>>();

    final List<Integer> list0 = new ArrayList<Integer>();
    list0.add(0);
    list0.add(1);
    list0.add(2); 
    list0.add(4);
    lists.add(list0);
    lists.add(list0);
    lists.add(list0);

    permute(lists, (permutation -> System.out.println(permutation)));
}

Thanks a lot :) I need to learn a lot about new features in java 8 — It'sMe, Mar 20 '15 at 19:50
Instead of the `Consumer` you could also use plain old `Callable`. I could not resist putting in the Lambda, though ;-) — dhke, Mar 20 '15 at 19:52
`permute(lists, (permutation -> System.out.println(permutation)));` could be written as `permute(lists, System.out::println);` since `::println` is the method reference of `println(...)` which is automatically a consumer. — thatsIch, Mar 01 '16 at 12:43

Generate all permutations of several lists in Java

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