/*************************/
// Selecting initial vertices
/*************************/
// Returns a vertex on the last level of a Breadth-first search (BFS)
public static int BFS(Graph graph, int init)
{
// BFS working queue
Queue<int> queue = new Queue<int>();
// Vertices that have already been visited
Set<int> visited = new Set<int>();
// Initial vertex is given as input
visited.Add(init);
queue.Enqueue(init);
int current = init;
// While there is still a vertex in the queue...
while (queue.Count > 0)
{
//... select the first vertex and process it
current = queue.Dequeue();
foreach (int w in graph.OpenNeighborhood(current))
{
// Enqueue all neighbors that have not been processed yet
if (!visited.Contains(w))
{
visited.Add(w);
queue.Enqueue(w);
}
}
}
// This is the last vertex that has been processed, thus a vertex that is on the last level of the BFS search
return current;
}
// Constructor for creating a new representative table. We directly fill the table depending on the given sequence.
public RepresentativeTable(Graph graph, Tree tree)
{
// Initialize the table
Table = new Dictionary<BitSet, RepresentativeList>();
Queue<BitSet> queue = new Queue<BitSet>();
queue.Enqueue(tree.Root);
Table[new BitSet(0, graph.Size)] = new RepresentativeList();
int i = 0;
while (queue.Count != 0)
{
BitSet node = queue.Dequeue();
FillTable(graph, node);
FillTable(graph, graph.Vertices - node);
if (tree.LeftChild.ContainsKey(node))
{
queue.Enqueue(tree.LeftChild[node]);
}
if (tree.RightChild.ContainsKey(node))
{
queue.Enqueue(tree.RightChild[node]);
}
}
}
// Uses depth-first search to check if the graph induced by the subgraph given as a parameter is connected
// In other words, we retreive all edges from the original graph and check the subgraph for connectedness
public static bool Connected(Graph graph, BitSet subgraph)
{
// Vertices that are visited
Set<int> visited = new Set<int>();
// Stack of vertices yet to visit
Stack<int> stack = new Stack<int>();
// Initial vertex
int s = subgraph.First();
stack.Push(s);
// Continue while there are vertices on the stack
while (stack.Count > 0)
{
int v = stack.Pop();
// If we have not encountered this vertex before, then we check for all neighbors if they are part of the subgraph
// If a neighbor is part of the subgraph it means that we have to push it on the stack to explore it at a later stage
if (!visited.Contains(v))
{
visited.Add(v);
foreach (int w in graph.OpenNeighborhood(v))
if (subgraph.Contains(w))
stack.Push(w);
}
}
// If we visited an equal number of vertices as there are vertices in the subgraph then the subgraph is connected
return visited.Count == subgraph.Count;
}
// Counts the number of independent sets in a graph, such that:
// - The vertices in P are legal choices for our IS, initially set to all vertices in the graph
// - None of the vertices in X are used, initially empty
// The performDFS boolean is used to check if we should perform a check for connectedness on this level of the recursion
private static long Compute(Graph graph, BitSet P, BitSet X, bool performDFS)
{
// Base case, when P and X are both empty we cannot expand the IS
if (P.IsEmpty && X.IsEmpty)
return 1;
// Base case, if a vertex w in X has no neighbor in P, then it means that this IS will never get maximal
// since we could always include w. Thus, the IS will not be valid and we return 0.
foreach (int w in X)
if ((graph.OpenNeighborhood(w) * P).IsEmpty)
return 0;
long count = 0;
// If a DFS is needed we check if the graph induced by (P + X) is still connected.
// If the graph is disconnected, in components c1,...,cn then we can simply count the IS of all these components
// after which we simply multiply these numbers.
if (performDFS)
{
if (!DepthFirstSearch.Connected(graph, P + X))
{
count = 1;
foreach (BitSet component in DepthFirstSearch.ConnectedComponents(graph, P + X))
count *= Compute(graph, component * P, component * X, false);
return count;
}
}
// Select a pivot in P to branch on
// In this case we pick the vertex with the largest degree
int maxDegree = -1; ;
int pivot = -1;
foreach (int u in P)
{
int deg = graph.Degree(u);
if (deg > maxDegree)
{
maxDegree = deg;
pivot = u;
}
}
// There should always be a pivot after the selection procedure
if (pivot == -1)
throw new Exception("Pivot has not been selected");
// We branch on the pivot, one branch we include the pivot in the IS.
// This might possibly disconnect the subgraph G(P + X), thus we set the performDFS boolean to true.
count = Compute(graph, P - graph.ClosedNeighborhood(pivot), X - graph.OpenNeighborhood(pivot), true);
// One branch we exclude the pivot of the IS. This will not cause the graph to get possibly disconnected
count += Compute(graph, P - pivot, X + pivot, false);
return count;
}
/*************************/
// Computing Exact Decompositions
/*************************/
public static Decomposition Compute(Graph graph)
{
Initialize(graph.Size);
// Construct the table with the correct width values
ConstructTree(graph, graph.Vertices);
// Retrieve a sequence that will result in our bound not being exceeded
Tree tree = RetrieveTree(graph);
return new Decomposition(graph, tree);
}
/*************************/
// Computing Exact Linear Decompositions
/*************************/
public static LinearDecomposition ExactlinearDecomposition(Graph graph)
{
Initialize(graph.Size);
// Construct the table with the correct width values
ConstructSequence(graph, graph.Vertices);
// Retrieve a sequence that will result in our bound not being exceeded
List<int> sequence = RetrieveSequence(graph, graph.Vertices);
return new LinearDecomposition(graph, sequence);
}
private static List<int> ComputeSequence(Graph graph, BitSet connectedComponent, CandidateStrategy candidateStrategy, int init, out int value)
{
int n = graph.Size;
List<int> sequence = new List<int>() { init };
BitSet left = new BitSet(0, n) { init };
BitSet right = connectedComponent - init;
// Initially we store the empty set and the set with init as the representative, ie N(init) * right
Set<BitSet> UN_left = new Set<BitSet>() { new BitSet(0, n), graph.OpenNeighborhood(init) * right };
value = int.MinValue;
while (!right.IsEmpty)
{
Set<BitSet> UN_chosen = new Set<BitSet>();
int chosen = Heuristics.TrivialCases(graph, left, right);
if (chosen != -1)
{
UN_chosen = IncrementUN(graph, left, UN_left, chosen);
}
// If chosen has not been set it means that no trivial case was found
// Depending on the criteria for the next vertex we call a different algorithm
else
{
BitSet candidates = Heuristics.Candidates(graph, left, right, candidateStrategy);
int min = int.MaxValue;
foreach (int v in candidates)
{
Set<BitSet> UN_v = IncrementUN(graph, left, UN_left, v);
if (UN_v.Count < min)
{
chosen = v;
UN_chosen = UN_v;
min = UN_v.Count;
}
}
}
// This should never happen
if (chosen == -1)
throw new Exception("No vertex is chosen for next step in the heuristic");
// Add/remove the next vertex in the appropiate sets
sequence.Add(chosen);
left.Add(chosen);
right.Remove(chosen);
UN_left = UN_chosen;
value = Math.Max(UN_chosen.Count, value);
}
return sequence;
}
// This constructor returns a new bipartite graph by putting all vertices in 'left' on one side, and 'right' on the other side
// There will be an edge between two vertices if there was an edge in the original graph
public BipartiteGraph(Graph graph, BitSet _left, BitSet _right)
: this(_left, _right, _left.Count + _right.Count)
{
Dictionary<int, int> mapping = new Dictionary<int, int>();
int i = 0;
foreach (int v in left + right)
mapping[v] = i++;
foreach (int v in left)
foreach (int w in graph.OpenNeighborhood(v) * right)
Connect(mapping[v], mapping[w]);
}
private void FillTable(Graph graph, List<int> sequence)
{
int n = graph.Size;
// Processed vertices
BitSet left = new BitSet(0, n);
// Unprocessed vertices
BitSet right = graph.Vertices;
// Lists of representatives that we keep track of on each level of the algorithm
LinearRepresentativeList reps = new LinearRepresentativeList();
// Initial insertions, the empty set always has the empty neighborhood set as a representative initially
reps.Update(new BitSet(0, n), new BitSet(0, n));
for (int i = 0; i < sequence.Count; i++)
{
/// We give v the possibility to be a representative instead of being contained in neighborhoods
int v = sequence[i];
// Actually move v from one set to the other set
left.Add(v);
right.Remove(v);
// We don't want to disturb any pointers so we create new empty datastructures to save everything for the next iteration
LinearRepresentativeList nextReps = new LinearRepresentativeList();
// We are iterating over all representatives saved inside the list of the previous step. For each entry there are only two possibilities to create a new neighborhood
foreach (BitSet representative in reps)
{
// Case 1: The neighborhood possibly contained v (thus v has to be removed), but is still valid
BitSet neighborhood = reps.GetNeighborhood(representative) - v;
nextReps.Update(representative, neighborhood);
// Case 2: The union of the old neighborhood, together with v's neighborhood, creates a new entry. The representative is uniond together with v and saved.
BitSet representative_ = representative + v;
BitSet neighborhood_ = neighborhood + (graph.OpenNeighborhood(v) * right);
nextReps.Update(representative_, neighborhood_);
}
// Update the values for the next iteration
reps = nextReps;
// Save the maximum size that we encounter during all iterations; this will be the boolean dimension of the graph.
MaxDimension = Math.Max(MaxDimension, reps.Count);
// Save the representatives at the current cut in the table
Table[left.Copy()] = reps;
}
}
// Builds a neighborhood for a set X from the ground on up, without relying on what has been saved previously in O(n^2) time
public dNeighborhood(BitSet X, BitSet vector, Graph graph)
{
// O(n) time copy
Vector = vector.Copy();
Occurrences = new Dictionary<int, int>();
// Loops in O(|Vector|) time over all vertices in the vector
foreach (int v in Vector)
{
// Bitset operation of O(n) time
BitSet nv = graph.OpenNeighborhood(v) * X;
Occurrences[v] = Math.Min(nv.Count, dValue);
}
}
// Given a vertex w, we can update the dNeighborhood of a set X to reflect the dNeighborhood of the set X + w in O(n) time
public dNeighborhood CopyAndUpdate(Graph graph, int w)
{
// initialize an empty dNeighborhood in O(|Vector|) time
dNeighborhood nx = new dNeighborhood(Vector);
// Copy all values in O(|Vector|) time
foreach (int v in Vector)
nx.Occurrences[v] = Occurrences[v];
// Foreach vertex in N(w) * Vector they will appear one extra time in the multiset
// This operation take O(n) time because of the bitset operation
foreach (int v in graph.OpenNeighborhood(w) * Vector)
nx.Occurrences[v] = Math.Min(dValue, nx.Occurrences[v] + 1);
return nx;
}
// Implementation of Algorithm 1 of 'Fast dynamic programming for locally checkable vertex subset and vertex partitioning problems' (Bui-Xuan et al.)
// Used to compute the representatives and their corresponding dNeighborhoods for a given node of the decomposition tree
private void FillTable(Graph graph, BitSet cut)
{
int n = graph.Size;
BitSet _cut = graph.Vertices - cut;
// Lists of representatives that we keep track of on each level of the algorithm
RepresentativeList representatives = new RepresentativeList();
RepresentativeList lastLevel = new RepresentativeList();
// Initial insertions, the empty set always has the empty neighborhood set as a representative initially
dNeighborhood dInitial = new dNeighborhood(_cut);
representatives.Update(new BitSet(0, n), dInitial);
lastLevel.Update(new BitSet(0, n), dInitial);
while (lastLevel.Count > 0)
{
RepresentativeList nextLevel = new RepresentativeList();
foreach (BitSet r in lastLevel)
{
foreach (int v in cut)
{
// avoid that r_ = r, since we already saved all sets of that size
if (r.Contains(v))
continue;
BitSet r_ = r + v;
dNeighborhood dn = representatives.GetNeighborhood(r).CopyAndUpdate(graph, v);
if (!representatives.ContainsNeighborhood(dn) && !dn.Equals(representatives.GetNeighborhood(r)))
{
nextLevel.Update(r_, dn);
representatives.Update(r_, dn);
}
}
}
// Update the values for the next iteration
lastLevel = nextLevel;
}
// Save the representatives at the current cut in the table
Table[cut.Copy()] = representatives;
// Save the maximum size that we encounter during all iterations; this will be the boolean dimension of the graph is d = 1.
MaxDimension = Math.Max(MaxDimension, representatives.Count);
}
// Returns all connected components in a certain subgraph, where we define a subgraph by the vertices that are contained in it
// We apply multiple dfs searches to find all connected parts of the graph
public static List<BitSet> ConnectedComponents(Graph graph, BitSet subgraph)
{
// Each connected component is defined as a bitset, thus the result is a list of these bitsets
List<BitSet> result = new List<BitSet>();
// Working stack for the dfs algorithm
Stack<int> stack = new Stack<int>();
// Each vertex of the subgraph will either be explored, or it will be the starting point of a new dfs search
BitSet todo = subgraph.Copy();
while (!todo.IsEmpty)
{
int s = todo.First();
stack.Push(s);
// Start tracking the new component
BitSet component = new BitSet(0, graph.Size);
// Default dfs exploring part of the graph
while (stack.Count > 0)
{
int v = stack.Pop();
// If we have not encountered this vertex before (meaning it isn't in this component), then we check for all neighbors if they are part of the subgraph
// If a neighbor is part of the subgraph it means that we have to push it on the stack to explore it at a later stage for this component
if (!component.Contains(v))
{
component.Add(v);
// Remove this vertex from the 'todo' list, since it can never be the starting point of a new component
todo.Remove(v);
foreach (int w in graph.OpenNeighborhood(v))
if (subgraph.Contains(w))
stack.Push(w);
}
}
// The whole connected component has been found, so we can add it to the list of results
result.Add(component);
}
return result;
}
// Counts the number of independent sets in a graph, such that:
// - All vertices in R are included in the independent set, initially empty
// - Some of the vertices in P are included in the independent set, initially all vertices of the graph
// - None of the vertices in X are included in the independent set, initially empty
private static int Compute(Graph graph, BitSet R, BitSet P, BitSet X)
{
// Base case, when P and X are both empty we cannot expand the IS
if (P.IsEmpty && X.IsEmpty)
return 1;
int count = 0;
BitSet copy = P.Copy();
// Foreach vertex v in P we include it in the IS and compute how many maximal IS will include v by going into recursion.
foreach (int v in copy)
{
count += Compute(graph, R + v, P - graph.ClosedNeighborhood(v), X - graph.OpenNeighborhood(v));
P.Remove(v);
X.Add(v);
}
return count;
}
public static LinearDecomposition Compute(Graph graph, CandidateStrategy candidateStrategy, InitialVertexStrategy initialVertexStrategy)
{
List<BitSet> connectedComponents = DepthFirstSearch.ConnectedComponents(graph);
List<int> sequence = new List<int>();
int tempValue;
foreach (BitSet connectedComponent in connectedComponents)
{
switch (initialVertexStrategy)
{
case InitialVertexStrategy.All:
{
List<int> minList = null;
int minValue = int.MaxValue;
foreach (int vertex in connectedComponent)
{
List<int> temp = ComputeSequence(graph, connectedComponent, candidateStrategy, vertex, out tempValue);
if (tempValue < minValue)
{
minValue = tempValue;
minList = temp;
}
}
sequence.AddRange(minList);
}
break;
case InitialVertexStrategy.BFS:
{
int init = Heuristics.BFS(graph, connectedComponent.Last());
sequence.AddRange(ComputeSequence(graph, connectedComponent, candidateStrategy, init, out tempValue));
}
break;
case InitialVertexStrategy.DoubleBFS:
{
int init = Heuristics.BFS(graph, Heuristics.BFS(graph, connectedComponent.Last()));
sequence.AddRange(ComputeSequence(graph, connectedComponent, candidateStrategy, init, out tempValue));
}
break;
}
}
return new LinearDecomposition(graph, sequence);
}
// Constructor for creating a new representative table. We directly fill the table depending on the given sequence.
public LinearRepresentativeTable(Graph graph, List<int> sequence)
{
// Initialize the table
Table = new Dictionary<BitSet, LinearRepresentativeList>();
int n = graph.Size;
// Save these initial representatives for the cut(empty, V(G))
LinearRepresentativeList reps = new LinearRepresentativeList();
reps.Update(new BitSet(0, n), new BitSet(0, n));
Table[new BitSet(0, n)] = reps;
Table[graph.Vertices] = reps;
FillTable(graph, sequence);
List<int> reverseSequence = new List<int>();
for (int i = sequence.Count - 1; i >= 0; i--)
reverseSequence.Add(sequence[i]);
FillTable(graph, reverseSequence);
}
// Constructs the actual width values
// The idea is the following: If we want to know the optimal width for a certain cut A, thus Width[A], then we can obtain this by
// either taking the max(Cuts[A], the minimum over all widths of Width[A - a]), which is our recurrence relation.
private static void ConstructSequence(Graph graph, BitSet A)
{
// min saves the minimum size of all neighborhoods of the cut [A - a], where a can be any vertex in A
long min = long.MaxValue;
// v will be the optimal choice to leave out in the previous iteration in order to obtain A's full neighborhood
int v = -1;
foreach (int a in A)
{
BitSet previous = A - a;
// If we have not constructed the previous step yet, then go in recursion and do so
if (!Neighborhoods.ContainsKey(previous))
ConstructSequence(graph, previous);
// Save the minimum value
if (Width[previous] < min)
{
min = Width[previous];
v = a;
}
}
// Obtain the neighborhood of v
BitSet nv = graph.OpenNeighborhood(v) * (graph.Vertices - A);
// We save the full set of neighborhood vertices at cut A. It does not matter that v was chosen arbitrarely; we always end up in the same collection of neighboring vertices for the set A
Set<BitSet> un = new Set<BitSet>();
foreach (BitSet _base in Neighborhoods[A - v])
{
un.Add(_base - v); // previous neighbor without v is a possible new neighborhood
un.Add((_base - v) + nv); // previous neighbor without v, unioned with the neighborhood of v is a possible new neighborhood
}
// Save all entries
Neighborhoods[A] = un; // List of all neighbors at cut A
Cuts[A] = Neighborhoods[A].Count; // Dimension at this cut
Width[A] = Math.Max(min, Cuts[A]); // Actual possible width to get to this cut
}
// Counts the number of independent sets in a graph, such that:
// - All vertices in R are included in the independent set, initially empty
// - Some of the vertices in P are included in the independent set, initially all vertices of the graph
// - None of the vertices in X are included in the independent set, initially empty
private static long Compute(Graph graph, BitSet R, BitSet P, BitSet X)
{
// Base case, when P and X are both empty we cannot expand the IS
if (P.IsEmpty && X.IsEmpty)
return 1;
long count = 0;
int pivot = -1;
int min = int.MaxValue;
// Procedure to find a pivot
// The idea is that in any maximal IS, either vertex u or a neighbor of u is included (else we could expand by adding u to the IS)
// We find the u with the smallest neighborhood, so that we will keep the number of branches low
foreach (int u in (P + X))
{
int size = (P * graph.OpenNeighborhood(u)).Count;
if (size < min)
{
min = size;
pivot = u;
}
}
// There should always be a pivot after the selection procedure, else P and X should both have been empty
if (pivot == -1)
throw new Exception("Pivot has not been selected");
// Foreach vertex v in the set containing the legal choices of the the closed neighborhood of the pivot,
// we include each choice in the IS and compute how many maximal IS will include v by going into recursion
foreach (int v in (P * graph.ClosedNeighborhood(pivot)))
{
count += Compute(graph, R + v, P - graph.ClosedNeighborhood(v), X - graph.OpenNeighborhood(v));
P.Remove(v);
X.Add(v);
}
return count;
}
/*************************/
// Trivial cases
/*************************/
public static int TrivialCases(Graph graph, BitSet left, BitSet right)
{
int chosen = -1;
// Check if any vertex in right belongs to one of the trivial cases, if yes then we can add this vertex directly to the sequence
foreach (int v in right)
{
// Trivial case 1. If the neighbors of a vertex v in right are all contained in left, then select v
// What this means is that v has an empty neighborhood, thus it will not add anything to the boolean-dimension
if ((graph.OpenNeighborhood(v) - left).IsEmpty)
{
chosen = v;
break;
}
bool stop = false;
// 2. If there are two vertices, v in right and u in left, such that N(v) * right == (N(u)\v) * right,
// then v is selected as our next vertex
// What this means is that all neighbors of v are already 'represented' by u, thus making the choice for v will not add anything to the dimension
foreach (int u in left)
{
BitSet nv = graph.OpenNeighborhood(v) * right; // N(v) * right
BitSet nu = (graph.OpenNeighborhood(u) - v) * right; // (N(u)\v) * right
if (nv.Equals(nu))
{
chosen = v;
stop = true;
break;
}
}
if (stop) break;
}
return chosen;
}
/*************************/
// Candidate strategy
/*************************/
public static BitSet Candidates(Graph graph, BitSet left, BitSet right, CandidateStrategy candidateStrategy)
{
BitSet nl = graph.Neighborhood(left) * right;
return candidateStrategy == CandidateStrategy.All ?
right.Copy() : (nl + graph.Neighborhood(nl)) * right;
}
// Retreives the actual ordering of a certain ordering that will not validate our bound
private static List<int> RetrieveSequence(Graph graph, BitSet A)
{
List<int> sequence = new List<int>();
// We know that the value saved at Width[graph.Vertices] will be the best value that we can get
// Furthermore, we know that there is a possibility to remove a vertex and get a smaller or equal Width value, in other words Width[A - a] <= Width[A]
// This way we navigate through the table and find a sequence that never will exceed the bound that is set, and we know for sure
// that such a path exists.
foreach (int a in A)
{
BitSet previous = A - a;
if (Width[previous] <= Width[graph.Vertices])
{
sequence.Add(a);
sequence.AddRange(RetrieveSequence(graph, previous));
break;
}
}
return sequence;
}
public static bool Connected(Graph graph)
{
return Connected(graph, graph.Vertices);
}
// Implementation of Algorithm 13 of the PhD thesis by Sadia Sharmin
// The LeastCutValue selection method picks the next vertex based on a greedy choice, namely what is the vertex that will have the least
// influence on the boolean dimension at this point.
// While this generally gives great results, the runtime is very high because we construct multiple bipartite graphs during every iteration.
private static int LeastCutValue(Graph graph, BitSet left, BitSet right)
{
// Minimum boolean dimension that we can have for our next cut
long minBoolDim = long.MaxValue;
// Vertex that we will choose
int chosen = -1;
// Foreach candidate vertex we construct a bipartite graph and count the number of minimal independent sets in this bipartite graph
//This number is equal to the boolean-dimension at this cut
foreach (int v in right)
{
// Construct the bipartite graph
BipartiteGraph bg = new BipartiteGraph(graph, left + v, right - v);
// Count the number of MIS in the bipartite graph
long boolDim = CC_MIS.Compute(bg);
// Possibly update the minimum value found so far
if (boolDim < minBoolDim)
{
chosen = v;
minBoolDim = boolDim;
}
}
return chosen;
}
/*************************/
// Constructor
/*************************/
public Decomposition(Graph graph, Tree tree)
{
Tree = tree;
Graph = graph;
}
public static List<BitSet> ConnectedComponents(Graph graph)
{
return ConnectedComponents(graph, graph.Vertices);
}
// Basic constructor
public LinearDecomposition(Graph graph, List<int> sequence)
{
Sequence = sequence;
Graph = graph;
}
// Initial call
public static long Compute(Graph graph)
{
return Compute(graph, new BitSet(0, graph.Size), graph.Vertices, new BitSet(0, graph.Size));
}
// Constructs the actual width values
private static void ConstructTree(Graph graph, BitSet A)
{
long min = A.Count == 1 ? 0 : long.MaxValue;
int n = graph.Size;
int v = -1; // v is the vertex that if we remove it from A, we have the smallest number of neighbors
BitSet optimal = new BitSet(0, n);
Set<BitSet> subsets = new Set<BitSet>(new BitSet(0, n));
foreach (int a in A)
{
Set<BitSet> newSubsets = new Set<BitSet>();
foreach (BitSet j in subsets)
{
BitSet subset = j + a;
BitSet inverse = A - subset;
if (subset.Equals(A)) continue; // only consider strict subsets
if (!Width.ContainsKey(subset))
ConstructTree(graph, subset);
if (!Width.ContainsKey(inverse))
ConstructTree(graph, inverse);
newSubsets.Add(subset); // add this for the next iteration
long max = Math.Max(Width[subset], Width[inverse]); // either S or A\S will be the bottleneck
if (max < min)
{
min = max;
optimal = subset; // it doesn't matter if we take j + a or A - (j + a), since when retrieving the tree we split them anyway
if (inverse.Count == 1)
v = inverse.First();
}
}
subsets.AddRange(newSubsets);
}
v = v == -1 ? A.First() : v;
BitSet nv = graph.OpenNeighborhood(v) * (graph.Vertices - (A - v));
Set<BitSet> un = new Set<BitSet>();
foreach (BitSet _base in Neighborhoods[A - v])
{
un.Add(_base - v); // previous neighbor without v is a possible new neighborhood
un.Add((_base - v) + nv); // previous neighbor without v, unioned with the neighborhood of v is a possible new neighborhood
}
Neighborhoods[A] = un;
Cuts[A] = Neighborhoods[A].Count;
Width[A] = Math.Max(min, Cuts[A]); // Actual possible width to get to this cut
OptimalChild[A] = optimal;
}
// Parses a file of DGF format, from which we can build a graph
public static Graph ParseGraph(string filename)
{
StreamReader sr = new StreamReader (filename);
string line;
int nVertices = 0;
// First we parse the number of vertices
while ((line = sr.ReadLine ()) != null)
{
// Skip everything until we find the first useful line
if (!line.StartsWith ("p ")) continue;
// First line always reads 'p edges {n} {e}'
string[] def = line.Split ();
nVertices = int.Parse (def[2]); // number of vertices
break;
}
// Initialize the graph
Graph graph = new Graph (nVertices);
Dictionary<int, int> identifiers = new Dictionary <int, int>();
int counter = 0;
bool renamed = false;
while ((line = sr.ReadLine ()) != null)
{
// Skip comments
if (line.StartsWith ("c ")) continue;
// If a line starts with an n it means we do not work with integers [1,...,n], but we use an arbitrary set of integers
// This means we need to keep track of which edges we should create internally, since we use [0,...,n)
if (line.StartsWith ("n "))
{
renamed = true;
string[] node = line.Split();
int id = int.Parse(node[1]); // actual identifier
identifiers[id] = counter++; // counter will be the internal identifier
}
// Parsing of the edges
if (line.StartsWith ("e "))
{
string[] edge = line.Split(' ');
int i = int.Parse(edge[1]);
int j = int.Parse(edge[2]);
// If there were arbitrary numbers used, look them up to find what we will use internally
if (renamed)
{
i = identifiers[i];
j = identifiers[j];
}
// If no other identifiers are given, we have only have to an element
// DGF files use the range [1...n], while internally we use [0...n), thus a simple minus 1 will suffice
else
{
i--;
j--;
}
// Create the edge
graph.Connect (i, j);
}
}
sr.Close ();
return graph;
}
// Retreives the actual ordering of a certain ordering that will not validate our bound
private static Tree RetrieveTree(Graph graph)
{
Tree tree = new Tree();
Queue<BitSet> queue = new Queue<BitSet>();
queue.Enqueue(graph.Vertices);
while (queue.Count != 0)
{
BitSet A = queue.Dequeue();
tree.Insert(A);
if (OptimalChild[A].IsEmpty) continue;
queue.Enqueue(OptimalChild[A]);
queue.Enqueue(A - OptimalChild[A]);
}
return tree;
}
