Although you did not precisely describe what you are looking for, and attempt to answer: It seems like you are just looking for all 3-element subsets of the input (1,2,3,5,6,7). Each subset is the first vector of one solution, and the respective remaining elements the other vector.
Here is an example how this may be computed, based on a ChoiceIterable utility class that I wrote a while ago:
import java.util.ArrayList;
import java.util.Arrays;
import java.util.Iterator;
import java.util.LinkedHashSet;
import java.util.List;
import java.util.NoSuchElementException;
import java.util.Set;
public class CombinationsOfVectors
{
public static void main(String[] args)
{
List<Integer> input = Arrays.asList(1,2,3,5,6,7);
ChoiceIterable<Integer> c = new ChoiceIterable<Integer>(3, input);
for (List<Integer> v0 : c)
{
Set<Integer> s = new LinkedHashSet<Integer>(input);
s.removeAll(v0);
List<Integer> v1 = new ArrayList<Integer>(s);
System.out.println(v0+" and "+v1);
}
}
}
// From https://github.com/javagl/Combinatorics/blob/master/src/
// main/java/de/javagl/utils/math/combinatorics/ChoiceIterable.java
// See the GitHub repo for a commented version
class ChoiceIterable<T> implements Iterable<List<T>>
{
private final List<T> input;
private final int sampleSize;
private final long numElements;
public ChoiceIterable(int sampleSize, List<T> input)
{
this.sampleSize = sampleSize;
this.input = input;
long nf = factorial(input.size());
long kf = factorial(sampleSize);
long nmkf = factorial(input.size() - sampleSize);
long divisor = kf * nmkf;
long result = nf / divisor;
numElements = result;
}
private static long factorial(int n)
{
long f = 1;
for (long i = 2; i <= n; i++)
{
f = f * i;
}
return f;
}
@Override
public Iterator<List<T>> iterator()
{
return new Iterator<List<T>>()
{
private int current = 0;
private final int chosen[] = new int[sampleSize];
{
for (int i = 0; i < sampleSize; i++)
{
chosen[i] = i;
}
}
@Override
public boolean hasNext()
{
return current < numElements;
}
@Override
public List<T> next()
{
if (!hasNext())
{
throw new NoSuchElementException("No more elements");
}
List<T> result = new ArrayList<T>(sampleSize);
for (int i = 0; i < sampleSize; i++)
{
result.add(input.get(chosen[i]));
}
current++;
if (current < numElements)
{
increase(sampleSize - 1, input.size() - 1);
}
return result;
}
private void increase(int n, int max)
{
if (chosen[n] < max)
{
chosen[n]++;
for (int i = n + 1; i < sampleSize; i++)
{
chosen[i] = chosen[i - 1] + 1;
}
}
else
{
increase(n - 1, max - 1);
}
}
@Override
public void remove()
{
throw new UnsupportedOperationException(
"May not remove elements from a choice");
}
};
}
}
The output in this example will be
[1, 2, 3] and [5, 6, 7]
[1, 2, 5] and [3, 6, 7]
[1, 2, 6] and [3, 5, 7]
[1, 2, 7] and [3, 5, 6]
[1, 3, 5] and [2, 6, 7]
[1, 3, 6] and [2, 5, 7]
[1, 3, 7] and [2, 5, 6]
[1, 5, 6] and [2, 3, 7]
[1, 5, 7] and [2, 3, 6]
[1, 6, 7] and [2, 3, 5]
[2, 3, 5] and [1, 6, 7]
[2, 3, 6] and [1, 5, 7]
[2, 3, 7] and [1, 5, 6]
[2, 5, 6] and [1, 3, 7]
[2, 5, 7] and [1, 3, 6]
[2, 6, 7] and [1, 3, 5]
[3, 5, 6] and [1, 2, 7]
[3, 5, 7] and [1, 2, 6]
[3, 6, 7] and [1, 2, 5]
[5, 6, 7] and [1, 2, 3]
If this is not what you have been looking for, you should describe more clearly and precisely what the intended result is.
(E.g. whether or not
[1, 2, 3] and [5, 6, 7]
and
[5, 6, 7] and [1, 2, 3]
count as different results is up to you, but you may filter the results accordingly)
