* For example, a lexicographical natural ordering over integers considers {@code [] < [1] < [1,
   * 1] < [1, 2] < [2]}.
   *
   * <p>Note that {@code Collections.reverseOrder(lexicographical(comparator))} is not equivalent to
   * {@code lexicographical(Collections.reverseOrder(comparator))} (consider how each would order
   * {@code [1]} and {@code [1, 1]}).
   */

   *
   * <p><i>Notes:</i> This is an implementation of the algorithm for Lexicographical Permutations
   * Generation, described in Knuth's "The Art of Computer Programming", Volume 4, Chapter 7,
   * Section 7.2.1.2. The iteration order follows the lexicographical order. This means that the
   * first permutation will be in ascending order, and the last will be in descending order.
   *

   *
   * <p><i>Notes:</i> This is an implementation of the algorithm for Lexicographical Permutations
   * Generation, described in Knuth's "The Art of Computer Programming", Volume 4, Chapter 7,
   * Section 7.2.1.2. The iteration order follows the lexicographical order. This means that the
   * first permutation will be in ascending order, and the last will be in descending order.
   *

   * For example, a lexicographical natural ordering over integers considers {@code [] < [1] < [1,
   * 1] < [1, 2] < [2]}.
   *
   * <p>Note that {@code Collections.reverseOrder(lexicographical(comparator))} is not equivalent to
   * {@code lexicographical(Collections.reverseOrder(comparator))} (consider how each would order
   * {@code [1]} and {@code [1, 1]}).
   */

   * but not the other, the shorter iterable is considered to be less than the longer one. For
   * example, a lexicographical natural ordering over integers considers {@code [] < [1] < [1, 1] <
   * [1, 2] < [2]}.
   *
   * <p>Note that {@code ordering.lexicographical().reverse()} is not equivalent to {@code
   * ordering.reverse().lexicographical()} (consider how each would order {@code [1]} and {@code [1,
   * 1]}).
   *

   * but not the other, the shorter iterable is considered to be less than the longer one. For
   * example, a lexicographical natural ordering over integers considers {@code [] < [1] < [1, 1] <
   * [1, 2] < [2]}.
   *
   * <p>Note that {@code ordering.lexicographical().reverse()} is not equivalent to {@code
   * ordering.reverse().lexicographical()} (consider how each would order {@code [1]} and {@code [1,
   * 1]}).
   *

    ImmutableList<String> b = ImmutableList.of("b");

    testComparator(lexy, empty, a, aa, ab, b);

    new EqualsTester()
        .addEqualityGroup(lexy, ordering.lexicographical())
        .addEqualityGroup(numberOrdering.lexicographical())
        .addEqualityGroup(Ordering.natural())
        .testEquals();
  }

  public void testNullsFirst() {

    ImmutableList<String> b = ImmutableList.of("b");

    testComparator(lexy, empty, a, aa, ab, b);

    new EqualsTester()
        .addEqualityGroup(lexy, ordering.lexicographical())
        .addEqualityGroup(numberOrdering.lexicographical())
        .addEqualityGroup(Ordering.natural())
        .testEquals();
  }

  public void testNullsFirst() {

    assertEquals(HashMultiset.create(asList("a")), multiset);
  }

  public void testCreation_arrayOfArray() {
    Comparator<String[]> comparator =
        Ordering.natural().lexicographical().onResultOf(Arrays::asList);
    String[] array = new String[] {"a"};
    Multiset<String[]> multiset = ImmutableSortedMultiset.orderedBy(comparator).add(array).build();

   *   <li>{@code ImmutableList.of(2, "A")}
   *   <li>{@code ImmutableList.of(2, "B")}
   *   <li>{@code ImmutableList.of(2, "C")}
   * </ul>
   *
   * <p>The result is guaranteed to be in the "traditional", lexicographical order for Cartesian
   * products that you would get from nesting for loops:
   *
   * {@snippet :
   * for (B b0 : lists.get(0)) {
   *   for (B b1 : lists.get(1)) {
   *     ...