g2.putEdge(N1, N2);

    assertThat(graph).isEqualTo(g2);
  }

  // Node/edge sets and node/edge connections are the same, but edge order differs.
  // In this case the graphs are considered equivalent; the edge add orderings are irrelevant.
  @Test
  public void equivalent_edgeAddOrdersDiffer() {
    GraphBuilder<Integer> builder = GraphBuilder.from(graph);

  }

  /**
   * Parses a trie node and returns the number of characters consumed.
   *
   * @param stack The prefixes that precede the characters represented by this node. Each entry of
   *     the stack is in reverse order.
   * @param encoded The serialized trie.
   * @param start An index in the encoded serialized trie to begin reading characters from.
   * @param builder A map builder to which all entries will be added.

   * depend on the element values, such as a {@link SortedSet}, or on the insertion ordering, such
   * as a {@link LinkedHashSet}. All list tests and sorted-collection tests automatically specify
   * this feature.
   */
  KNOWN_ORDER,

  /**
   * Indicates that a collection has a different {@link Object#toString} representation than most
   * collections. If not specified, the collection tests will examine the value returned by {@link

         * NOTE: This is primarily to keep backward-compat with Maven 2.x which did not validate that dependencies are
         * unique within a single POM. Upon multiple declarations, 2.x just kept the last one but retained the order of
         * the first occurrence. So when we're in lenient/compat mode, we have to deal with such broken POMs and mimic

  }

  /**
   * Compares the specified object to this multimap for equality.
   *
   * <p>Two {@code ListMultimap} instances are equal if, for each key, they contain the same values
   * in the same order. If the value orderings disagree, the multimaps will not be considered equal.
   */
  @Override
  public boolean equals(@CheckForNull Object object) {
    return super.equals(object);
  }

    @Override
    public Character[] createArray(int length) {
      return new Character[length];
    }

    /** Returns the original element list, unchanged. */
    @Override
    public List<Character> order(List<Character> insertionOrder) {
      return insertionOrder;
    }
  }

  public static class SampleChars extends SampleElements<Character> {
    public SampleChars() {

    @Override
    public Double[] createArray(int length) {
      return new Double[length];
    }

    /** Returns the original element list, unchanged. */
    @Override
    public List<Double> order(List<Double> insertionOrder) {
      return insertionOrder;
    }
  }

  public static class SampleDoubles extends SampleElements<Double> {
    public SampleDoubles() {

    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() {
    assertFalse(Comparators.isInOrder(asList(5, 3, 0, 9), Ordering.natural()));

    @Override
    public Character[] createArray(int length) {
      return new Character[length];
    }

    /** Returns the original element list, unchanged. */
    @Override
    public List<Character> order(List<Character> insertionOrder) {
      return insertionOrder;
    }
  }

  public static class SampleChars extends SampleElements<Character> {
    public SampleChars() {

  }

  /**
   * Compares the specified object to this multimap for equality.
   *
   * <p>Two {@code ListMultimap} instances are equal if, for each key, they contain the same values
   * in the same order. If the value orderings disagree, the multimaps will not be considered equal.
   */
  @Override
  public boolean equals(@CheckForNull Object object) {
    return super.equals(object);
  }