ComparableSubClass node8 = new ComparableSubClass(8);

    graph.addNode(node4);
    graph.addNode(node2);
    graph.addNode(node6);
    graph.addNode(node8);

    assertThat(graph.nodeOrder().comparator()).isEqualTo(Ordering.natural());
    assertThat(graph.nodes()).containsExactly(node2, node4, node6, node8).inOrder();
  }

  private static void addNodes(MutableGraph<Integer> graph) {

      AbstractGraphTest.validateGraph(graph);

      Collections.shuffle(nodeList, gen);
      for (Integer node : nodeList) {
        assertThat(graph.addNode(node)).isTrue();
      }
      Collections.shuffle(edgeList, gen);
      for (EndpointPair<Integer> edge : edgeList) {
        assertThat(graph.putEdge(edge.nodeU(), edge.nodeV())).isTrue();
      }

    checkNotNull(nodeU);
    checkNotNull(nodeV);
    return nodes().contains(nodeU) && successors(nodeU).contains(nodeV);
  }

  @Override
  public boolean hasEdgeConnecting(EndpointPair<N> endpoints) {
    checkNotNull(endpoints);
    if (!isOrderingCompatible(endpoints)) {
      return false;
    }
    N nodeU = endpoints.nodeU();
    N nodeV = endpoints.nodeV();

        // We already know that nodeU equals other.nodeU. Combined with the earlier statement,
        // and the transitive property of equality, this implies that nodeU equals nodeV.
        // If nodeU equals nodeV, condition1 == condition2, so checking condition1 is sufficient.
        return nodeV().equals(other.nodeV());
      }

  public Optional<V> edgeValue(N nodeU, N nodeV) {
    return delegate().edgeValue(nodeU, nodeV);
  }

  @Override
  public Optional<V> edgeValue(EndpointPair<N> endpoints) {
    return delegate().edgeValue(endpoints);
  }

  @Override
  public @Nullable V edgeValueOrDefault(N nodeU, N nodeV, @Nullable V defaultValue) {
    return delegate().edgeValueOrDefault(nodeU, nodeV, defaultValue);
  }

  @Override

    return delegate().degree(node);
  }

  @Override
  public int inDegree(N node) {
    return delegate().inDegree(node);
  }

  @Override
  public int outDegree(N node) {
    return delegate().outDegree(node);
  }

  @Override
  public Set<E> edgesConnecting(N nodeU, N nodeV) {
    return delegate().edgesConnecting(nodeU, nodeV);
  }

  @Override

    // whatever ordering the graph's nodes do, so ImmutableSortedMap is unnecessary even if the
    // input nodes are sorted.
    ImmutableMap.Builder<N, GraphConnections<N, V>> nodeConnections = ImmutableMap.builder();
    for (N node : graph.nodes()) {
      nodeConnections.put(node, connectionsOf(graph, node));
    }
    return nodeConnections.buildOrThrow();
  }

  @Override
  public int degree(N node) {
    return delegate().degree(node);
  }

  @Override
  public int inDegree(N node) {
    return delegate().inDegree(node);
  }

  @Override
  public int outDegree(N node) {
    return delegate().outDegree(node);
  }

  @Override
  public boolean hasEdgeConnecting(N nodeU, N nodeV) {
    return delegate().hasEdgeConnecting(nodeU, nodeV);
  }

  @Override

    assertThat(unordered.isOrdered()).isFalse();
    assertThat(unordered).containsExactly("node", "node");
    assertThat(unordered.nodeU()).isEqualTo("node");
    assertThat(unordered.nodeV()).isEqualTo("node");
    assertThat(unordered.adjacentNode("node")).isEqualTo("node");
    assertThat(unordered.toString()).isEqualTo("[node, node]");
  }

  @Test

   */
  @CanIgnoreReturnValue
  boolean addNode(N node);

  /**
   * Adds an edge connecting {@code nodeU} to {@code nodeV} if one is not already present.
   *
   * <p>If the graph is directed, the resultant edge will be directed; otherwise, it will be
   * undirected.
   *
   * <p>If {@code nodeU} and {@code nodeV} are not already present in this graph, this method will