public static void BidirectionalContainsEdgeAssertions(
            [NotNull] IBidirectionalGraph <int, IEdge <int> > graph,
            [NotNull] IEdge <int> e12,
            [NotNull] IEdge <int> f12,
            [CanBeNull] IEdge <int> e21,
            [CanBeNull] IEdge <int> f21)
        {
            Assert.AreEqual(0, graph.InDegree(1));
            Assert.AreEqual(1, graph.OutDegree(1));
            Assert.AreEqual(1, graph.InDegree(2));
            Assert.AreEqual(0, graph.OutDegree(2));
            Assert.AreEqual(1, graph.OutEdges(1).Count());
            Assert.AreEqual(1, graph.InEdges(2).Count());

            // e12 must be present in u, because we added it.
            Assert.IsTrue(graph.ContainsEdge(e12));

            // f12 is also in u, because e12 == f12.
            Assert.IsTrue(graph.ContainsEdge(f12));

            // e21 and f21 are not in u, because it's a directed graph.
            if (e21 != null)
            {
                Assert.IsFalse(graph.ContainsEdge(e21));
            }
            if (f21 != null)
            {
                Assert.IsFalse(graph.ContainsEdge(f21));
            }

            // There must be an edge between vertices 1, 2.
            Assert.IsTrue(graph.ContainsEdge(1, 2));

            // No edge between vertices 2, 1, because the graph is directed.
            Assert.IsFalse(graph.ContainsEdge(2, 1));

            // ContainsEdge(1, 3) raises contracts violation in IncidenceGraphContract, because 3 is not in the graph.
            // obviously no edge between vertices 1, 3, as vertex 3 is not even present in the graph.
            // Assert.IsFalse(g.ContainsEdge(1, 3));
        }
        public static void ContainsEdgeAssertions(IBidirectionalGraph<int, IEdge<int>> g,
            IEdge<int> e12,
            IEdge<int> f12,
            IEdge<int> e21,
            IEdge<int> f21)
        {
            Assert.AreEqual(0, g.InDegree(1));
            Assert.AreEqual(1, g.OutDegree(1));
            Assert.AreEqual(1, g.InDegree(2));
            Assert.AreEqual(0, g.OutDegree(2));
            Assert.AreEqual(1, g.OutEdges(1).Count());
            Assert.AreEqual(1, g.InEdges(2).Count());

            // e12 must be present in u, because we added it.
            Assert.IsTrue(g.ContainsEdge(e12));

            // f12 is also in u, because e12 == f12.
            Assert.IsTrue(g.ContainsEdge(f12));

            // e21 and f21 are not in u, because it's a directed graph.
            if (e21 != null) Assert.IsFalse(g.ContainsEdge(e21));
            if (f21 != null) Assert.IsFalse(g.ContainsEdge(f21));

            // there must be an edge between vertices 1, 2.
            Assert.IsTrue(g.ContainsEdge(1, 2));

            // no edge between vertices 2, 1, because the graph is directed.
            Assert.IsFalse(g.ContainsEdge(2, 1));

            // ContainsEdge(1, 3) raises contracts violation in IIncidenceGraphContract, because 3 is not in the graph.
            // obviously no edge between vertices 1, 3, as vertex 3 is not even present in the graph.
            // Assert.IsFalse(g.ContainsEdge(1, 3));
        }