}
    return builder.toString();
  }

  /**
   * Returns a comparator that compares two {@code double} arrays <a
   * href="http://en.wikipedia.org/wiki/Lexicographical_order">lexicographically</a>. That is, it
   * compares, using {@link #compare(double, double)}), 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

    }
    return builder.toString();
  }

  /**
   * Returns a comparator that compares two {@code int} arrays <a
   * href="http://en.wikipedia.org/wiki/Lexicographical_order">lexicographically</a>. That is, it
   * compares, using {@link #compare(int, int)}), 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

func Equal(a, b []byte) bool {
	// Neither cmd/compile nor gccgo allocates for these string conversions.
	return string(a) == string(b)
}

// Compare returns an integer comparing two byte slices lexicographically.
// The result will be 0 if a == b, -1 if a < b, and +1 if a > b.
// A nil argument is equivalent to an empty slice.
func Compare(a, b []byte) int {
	return bytealg.Compare(a, b)
}

               *
               * The next combination is
               *
               * {0, 1, ..., bitToFlip - firstSetBit - 2, bitToFlip, ...}
               *
               * This is lexicographically next if you look at the combinations in descending order
               * e.g. {2, 1, 0}, {3, 1, 0}, {3, 2, 0}, {3, 2, 1}, {4, 1, 0}...
               */

              bits.set(0, bitToFlip - firstSetBit - 1);

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

    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() {

   *
   * <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]}).
   */

    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() {

   * 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]}).
   *