int outDegree(N node);

  /**
   * Returns true if there is an edge that directly connects {@code nodeU} to {@code nodeV}. This is
   * equivalent to {@code nodes().contains(nodeU) && successors(nodeU).contains(nodeV)}.
   *
   * <p>In an undirected graph, this is equal to {@code hasEdgeConnecting(nodeV, nodeU)}.
   *
   * @since 23.0
   */
  @Override
  boolean hasEdgeConnecting(N nodeU, N nodeV);

  /**

        return AbstractValueGraph.this.degree(node);
      }

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

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

  @Override
  public Optional<V> edgeValue(N nodeU, N nodeV) {

     *
     * <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
     * silently {@link #addNode(Object) add} {@code nodeU} and {@code nodeV} to the graph.
     *
     * @return this {@code Builder} object

    // 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();
  }

    // 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

  @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

   *
   * <p>If {@code nodeU} and {@code nodeV} are not already present in this graph, this method will
   * silently {@link #addNode(Object) add} {@code nodeU} and {@code nodeV} to the graph.
   *
   * <p>If {@code edge} already connects {@code nodeU} to {@code nodeV} (in the specified order if
   * this network {@link #isDirected()}, else in any order), then this method will have no effect.
   *

      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();
      }

      assertThat(graph.nodes()).hasSize(NUM_NODES);

   *
   * <p>If {@code nodeU} and {@code nodeV} are not already present in this graph, this method will
   * silently {@link #addNode(Object) add} {@code nodeU} and {@code nodeV} to the graph.
   *
   * <p>If {@code edge} already connects {@code nodeU} to {@code nodeV} (in the specified order if
   * this network {@link #isDirected()}, else in any order), then this method will have no effect.
   *