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;
}
