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

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

    new EqualsTester()
        .addEqualityGroup(lexy, Comparators.lexicographical(comparator))
        .addEqualityGroup(Comparators.lexicographical(String.CASE_INSENSITIVE_ORDER))
        .addEqualityGroup(Ordering.natural())
        .testEquals();
  }

  public void testIsInOrder() {

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

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

    new EqualsTester()
        .addEqualityGroup(lexy, Comparators.lexicographical(comparator))
        .addEqualityGroup(Comparators.lexicographical(String.CASE_INSENSITIVE_ORDER))
        .addEqualityGroup(Ordering.natural())
        .testEquals();
  }

  public void testIsInOrder() {

   * 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.
   *

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

    Helpers.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");

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

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

  public void testNullsFirst() {

   * 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]}).
   *

    }
    return builder.toString();
  }

  /**
   * Returns a comparator that compares two {@code byte} arrays <a
   * href="http://en.wikipedia.org/wiki/Lexicographical_order">lexicographically</a>. That is, it
   * compares, using {@link #compare(byte, byte)}), the first pair of values that follow any common
   * prefix, or when one array is a prefix of the other, treats the shorter array as the lesser. For

   *   <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:
   *
   * <pre>{@code
   * for (B b0 : lists.get(0)) {
   *   for (B b1 : lists.get(1)) {
   *     ...