private static IEnumerable<List<Edge>> AllTriangles(Graph graph) { List<Vertex> vertices = new List<Vertex>(graph.Vertices); vertices.Sort((x, y) => graph.GetDegree(x).CompareTo(graph.GetDegree(y))); vertices.Reverse(); // From highest to lowest Dictionary<Vertex, int> index = new Dictionary<Vertex, int>(); Dictionary<Vertex, List<Vertex>> A = new Dictionary<Vertex, List<Vertex>>(); for (int i = 0; i < vertices.Count; i++) { index.Add(vertices[i], i); A.Add(graph.Vertices[i], new List<Vertex>()); } foreach (var firstVertex in vertices) { foreach (var firstEdge in graph.GetEdgesForVertex(firstVertex)) { var secondVertex = firstEdge.Other(firstVertex); if (index[firstVertex] < index[secondVertex]) { foreach (var thirdVertex in A[firstVertex].Intersect(A[secondVertex])) { var secondEdge = graph.GetEdgesForVertex(firstVertex).Single(x => x.Other(firstVertex) == thirdVertex); var thirdEdge = graph.GetEdgesForVertex(secondVertex).Single(x => x.Other(secondVertex) == thirdVertex); yield return new List<Edge>(new [] { firstEdge, secondEdge, thirdEdge }); } A[secondVertex].Add(firstVertex); } } } }
/// <summary> /// Runs the Degree Test to reduce the graph. /// </summary> /// <param name="graph">The graph on which to run the test.</param> /// <returns>The reduced graph.</returns> public static ReductionResult RunTest(Graph graph) { // Use some simple rules to reduce the problem. // The rules are: - if a vertex v has degree 1 and is not part of the required nodes, remove the vertex. // - if a vertex v has degree 1 and is part of the required nodes, the one edge it is // connected to has to be in the solution, and as a consequence, the node at the other side // is either a Steiner node or also required. // - (not implemented yet) if a vertex v is required and has degree 2, and thus is connected to 2 edges, // namely e1 and e2, and cost(e1) < cost(e2) and e1 = (u, v) and u is // also a required vertex, then every solution must contain e1 var result = new ReductionResult(); // Remove leaves as long as there are any var leaves = graph.Vertices.Where(v => graph.GetDegree(v) == 1 && !graph.Terminals.Contains(v)).ToList(); while (leaves.Count > 0) { foreach (var leaf in leaves) graph.RemoveVertex(leaf); result.RemovedVertices.AddRange(leaves); leaves = graph.Vertices.Where(v => graph.GetDegree(v) == 1 && !graph.Terminals.Contains(v)).ToList(); } // When a leaf is required, add the node on the other side of its one edge to required nodes var requiredLeaves = graph.Vertices.Where(v => graph.GetDegree(v) == 1 && graph.Terminals.Contains(v)).ToList(); foreach (var requiredLeaf in requiredLeaves) { // Find the edge this leaf is connected to var alsoRequired = graph.GetEdgesForVertex(requiredLeaf).Single().Other(requiredLeaf); if (!graph.Terminals.Contains(alsoRequired)) graph.RequiredSteinerNodes.Add(alsoRequired); } return result; }
private void AddEdgesToMST(Graph mst, List<Edge> edges) { foreach (var edge in edges) { if (edge.WhereBoth(x => mst.GetDegree(x) > 0)) //Both vertices of this edge are in the MST, introducing this edge creates cycle! { var v1 = edge.Either(); var v2 = edge.Other(v1); var components = mst.CreateComponentTable(); if (components[v1] == components[v2]) { // Both are in the same component, and a path exists. // Travel the path to see if adding this edge makes it cheaper var path = Algorithms.DijkstraPath(v1, v2, mst); // However, we can remove a set of edges between two nodes // When going from T1 to T3 // E.g.: T1 - v2 - v3 - v4 - T2 - v5 - v6 - T3 // The edges between T1, v2, v3, v4, T2 cost more than the path between T1 and T3. // Removing those edges, and adding the new edge from T1 to T3, also connects // T1 to T2, so still a tree! var from = path.Start; var last = path.Start; int betweenTerminals = 0; var subtractedPath = new Path(path.Start); List<Edge> edgesInSubtraction = new List<Edge>(); Dictionary<Edge, List<Edge>> subtractions = new Dictionary<Edge, List<Edge>>(); for (int i = 0; i < path.Edges.Count; i++) { var pe = path.Edges[i]; betweenTerminals += pe.Cost; last = pe.Other(last); edgesInSubtraction.Add(pe); if (mst.GetDegree(last) > 2 || mst.Terminals.Contains(last) || i == path.Edges.Count - 1) { var subtractedEdge = new Edge(from, last, betweenTerminals); subtractions.Add(subtractedEdge, edgesInSubtraction); edgesInSubtraction = new List<Edge>(); subtractedPath.Edges.Add(subtractedEdge); from = last; betweenTerminals = 0; } } var mostCostly = subtractedPath.Edges[0]; for (int i = 1; i < subtractedPath.Edges.Count; i++) { if (subtractedPath.Edges[i].Cost > mostCostly.Cost) mostCostly = subtractedPath.Edges[i]; } if (mostCostly.Cost >= edge.Cost) { foreach (var e in subtractions[mostCostly]) mst.RemoveEdge(e, false); mst.AddEdge(edge); } } else // Connect the two disconnected components! mst.AddEdge(edge); } else mst.AddEdge(edge); } }
public static Graph RunSolver(Graph graph) { var solution = new Graph(graph.Vertices); DijkstraState state = new DijkstraState(); // Create the states needed for every execution of the Dijkstra algorithm foreach (var terminal in graph.Terminals) state.AddVertexToInterleavingDijkstra(terminal, graph); // Initialize Vertex currentVertex = state.GetNextVertex(); FibonacciHeap<int, Vertex> labels = state.GetLabelsFibonacciHeap(); HashSet<Vertex> visited = state.GetVisitedHashSet(); Dictionary<Vertex, Path> paths = state.GetPathsFound(); Dictionary<Vertex, FibonacciHeap<int, Vertex>.Node> nodes = state.GetNodeMapping(); Dictionary<Vertex, Edge> comingFrom = state.GetComingFromDictionary(); Dictionary<Vertex, int> components = solution.CreateComponentTable(); Dictionary<Vertex, double> terminalFValues = CreateInitialFValuesTable(graph); int maxLoopsNeeded = graph.Terminals.Count * graph.NumberOfVertices; int loopsDone = 0; int updateInterval = 100; int longestPath = graph.Terminals.Max(x => Algorithms.DijkstraToAll(x, graph).Max(y => y.Value)); while (state.GetLowestLabelVertex() != null) { if (loopsDone % updateInterval == 0) Console.Write("\rRunning IDA... {0:0.0}% \r", 100.0 * loopsDone / maxLoopsNeeded); loopsDone++; if (state.GetLowestLabelVertex() != currentVertex) { // Interleave. Switch to the Dijkstra procedure of the vertex which currently has the lowest distance. state.SetLabelsFibonacciHeap(labels); state.SetVisitedHashSet(visited); state.SetPathsFound(paths); state.SetComingFromDictionary(comingFrom); currentVertex = state.GetNextVertex(); labels = state.GetLabelsFibonacciHeap(); visited = state.GetVisitedHashSet(); paths = state.GetPathsFound(); nodes = state.GetNodeMapping(); comingFrom = state.GetComingFromDictionary(); } // Do one loop in Dijkstra algorithm var currentNode = labels.ExtractMin(); var current = currentNode.Value; if (currentNode.Key > longestPath / 2) break; //Travelled across the half of longest distance. No use in going further. // Consider all edges ending in unvisited neighbours var edges = graph.GetEdgesForVertex(current).Where(x => !visited.Contains(x.Other(current))); // Update labels on the other end foreach (var edge in edges) { if (currentNode.Key + edge.Cost < nodes[edge.Other(current)].Key) { labels.DecreaseKey(nodes[edge.Other(current)], currentNode.Key + edge.Cost); comingFrom[edge.Other(current)] = edge; } } visited.Add(current); if (current != currentVertex) { // Travel back the new path List<Edge> pathEdges = new List<Edge>(); Vertex pathVertex = current; while (pathVertex != currentVertex) { pathEdges.Add(comingFrom[pathVertex]); pathVertex = comingFrom[pathVertex].Other(pathVertex); } pathEdges.Reverse(); Path path = new Path(currentVertex); path.Edges.AddRange(pathEdges); paths[current] = path; } // Find matching endpoints from two different terminals var mutualEnd = state.FindPathsEndingInThisVertex(current); if (mutualEnd.Count() > 1) { var terminals = mutualEnd.Select(x => x.Start).ToList(); // Step 1. Calculate new heuristic function value for this shared point. // f(x) = (Cost^2)/(NumberOfTerminals^3) var f1 = Math.Pow(mutualEnd.Sum(p => p.TotalCost), 2) / Math.Pow(terminals.Count, 3); var f2 = Math.Pow(mutualEnd.Sum(p => p.TotalCost), 1) / Math.Pow(terminals.Count, 2); var f3 = Math.Pow(mutualEnd.Sum(p => p.TotalCost), 3) / Math.Pow(terminals.Count, 2); var terminalsAvgF = terminals.Select(x => terminalFValues[x]).Average(); var terminalsMinF = terminals.Select(x => terminalFValues[x]).Min(); var f = (new[] { f1, f2, f3 }).Max(); Debug.WriteLine("F value: {0}, Fmin: {3} - Connecting terminals: {1} via {2}", f, string.Join(", ", terminals.Select(x => x.VertexName)), current.VertexName, terminalsMinF); // Do not proceed if f > avgF AND working in same component if (terminals.Select(x => components[x]).Distinct().Count() == 1 && f > terminalsMinF) continue; Debug.WriteLine("Proceeding with connection..."); // Step 2. Disconnect terminals in mutual component. foreach (var group in terminals.GroupBy(x => components[x])) { if (group.Count() <= 1) continue; HashSet<Edge> remove = new HashSet<Edge>(); var sameComponentTerminals = group.ToList(); for (int i = 0; i < sameComponentTerminals.Count-1; i++) { for (int j = i+1; j< sameComponentTerminals.Count; j++) { var removePath = Algorithms.DijkstraPath(sameComponentTerminals[i], sameComponentTerminals[j], solution); foreach (var e in removePath.Edges) remove.Add(e); } } foreach (var e in remove) solution.RemoveEdge(e, false); } components = solution.CreateComponentTable(); // Step 3. Reconnect all now disconnected terminals via shared endpoint foreach (var t in terminals) { var path = Algorithms.DijkstraPath(t, current, graph); foreach (var edge in path.Edges) solution.AddEdge(edge); // Update f value terminalFValues[t] = f; } components = solution.CreateComponentTable(); } } // If this solution is connected, take MST if (graph.Terminals.Select(x => components[x]).Distinct().Count() == 1) { // Clean up! foreach (var vertex in solution.Vertices.Where(x => solution.GetDegree(x) == 0).ToList()) solution.RemoveVertex(vertex); int componentNumber = graph.Terminals.Select(x => components[x]).Distinct().Single(); foreach (var vertex in components.Where(x => x.Value != componentNumber).Select(x => x.Key).ToList()) solution.RemoveVertex(vertex); solution = Algorithms.Kruskal(solution); return solution; } // If the solution is not connected, it is not a good solution. return null; }