private void dfs(Digraph G, int v)
{
marked[v] = true;
low[v] = pre++;
int min = low[v];
stack.Push(v);
foreach (int w in G.Adj(v))
{
if (!marked[w])
{
dfs(G, w);
}
if (low[w] < min)
{
min = low[w];
}
}
if (min < low[v])
{
low[v] = min;
return;
}
int u;
do
{
u = stack.Pop();
id[u] = count;
low[u] = G.V;
} while (u != v);
count++;
}
// check that algorithm computes either the topological order or finds a directed cycle
private void dfs(Digraph G, int v)
{
onStack[v] = true;
marked[v] = true;
foreach (int w in G.Adj(v))
{
// short circuit if directed cycle found
if (cycle != null)
{
return;
}
else if (!marked[w])
{
// found new vertex, so recur
edgeTo[w] = v;
dfs(G, w);
}
else if (onStack[w])
{
// trace back directed cycle
cycle = new LinkedStack <int>();
for (int x = v; x != w; x = edgeTo[x])
{
cycle.Push(x);
}
cycle.Push(w);
cycle.Push(v);
Debug.Assert(check());
}
}
onStack[v] = false;
}
// BFS from multiple sources
private void bfs(Digraph G, IEnumerable <int> sources)
{
LinkedQueue <int> q = new LinkedQueue <int>();
foreach (int s in sources)
{
marked[s] = true;
distTo[s] = 0;
q.Enqueue(s);
}
while (!q.IsEmpty)
{
int v = q.Dequeue();
foreach (int w in G.Adj(v))
{
if (!marked[w])
{
edgeTo[w] = v;
distTo[w] = distTo[v] + 1;
marked[w] = true;
q.Enqueue(w);
}
}
}
}
private void dfs(Digraph G, int v)
{
marked[v] = true;
preorder[v] = pre++;
stack1.Push(v);
stack2.Push(v);
foreach (int w in G.Adj(v))
{
if (!marked[w])
{
dfs(G, w);
}
else if (id[w] == -1)
{
while (preorder[stack2.Peek()] > preorder[w])
{
stack2.Pop();
}
}
}
// found strong component containing v
if (stack2.Peek() == v)
{
stack2.Pop();
int w;
do
{
w = stack1.Pop();
id[w] = count;
} while (w != v);
count++;
}
}
private LinkedStack <int> cycle = null; // Eulerian cycle; null if no such cylce
/// <summary>
/// Computes an Eulerian cycle in the specified digraph, if one exists.
/// </summary>
/// <param name="G">the digraph</param>
///
public DirectedEulerianCycle(Digraph G)
{
// must have at least one edge
if (G.E == 0)
{
return;
}
// necessary condition: indegree(v) = outdegree(v) for each vertex v
// (without this check, DFS might return a path instead of a cycle)
for (int v = 0; v < G.V; v++)
{
if (G.Outdegree(v) != G.Indegree(v))
{
return;
}
}
// create local view of adjacency lists, to iterate one vertex at a time
IEnumerator <int>[] adj = new IEnumerator <int> [G.V];
for (int v = 0; v < G.V; v++)
{
adj[v] = G.Adj(v).GetEnumerator();
}
// initialize stack with any non-isolated vertex
int s = nonIsolatedVertex(G);
LinkedStack <int> stack = new LinkedStack <int>();
stack.Push(s);
// greedily add to putative cycle, depth-first search style
cycle = new LinkedStack <int>();
while (!stack.IsEmpty)
{
int v = stack.Pop();
while (adj[v].MoveNext())
{
stack.Push(v);
v = adj[v].Current;
}
// add vertex with no more leaving edges to cycle
cycle.Push(v);
}
// check if all edges have been used
// (in case there are two or more vertex-disjoint Eulerian cycles)
if (cycle.Count != G.E + 1)
{
cycle = null;
}
Debug.Assert(certifySolution(G));
}
private void dfs(Digraph G, int v)
{
count++;
marked[v] = true;
foreach (int w in G.Adj(v))
{
if (!marked[w])
{
dfs(G, w);
}
}
}
private void dfs(Digraph G, int v)
{
marked[v] = true;
foreach (int w in G.Adj(v))
{
if (!marked[w])
{
edgeTo[w] = v;
dfs(G, w);
}
}
}
// certify that digraph is acyclic
private bool check(Digraph G)
{
// digraph is acyclic
if (HasOrder)
{
// check that ranks are a permutation of 0 to V-1
bool[] found = new bool[G.V];
for (int i = 0; i < G.V; i++)
{
found[Rank(i)] = true;
}
for (int i = 0; i < G.V; i++)
{
if (!found[i])
{
Console.Error.WriteLine("No vertex with rank " + i);
return(false);
}
}
// check that ranks provide a valid topological order
for (int v = 0; v < G.V; v++)
{
foreach (int w in G.Adj(v))
{
if (Rank(v) > Rank(w))
{
Console.Error.WriteLine("{0}-{1}: rank({2}) = {3}, rank({4}) = {5}\n",
v, w, v, Rank(v), w, Rank(w));
return(false);
}
}
}
// check that order() is consistent with rank()
int r = 0;
foreach (int v in Order())
{
if (Rank(v) != r)
{
Console.Error.WriteLine("Order() and Rank() inconsistent");
return(false);
}
r++;
}
}
return(true);
}
// run DFS in digraph G from vertex v and compute preorder/postorder
private void dfs(Digraph G, int v)
{
marked[v] = true;
pre[v] = preCounter++;
preorder.Enqueue(v);
foreach (int w in G.Adj(v))
{
if (!marked[w])
{
dfs(G, w);
}
}
postorder.Enqueue(v);
post[v] = postCounter++;
}
private int[] rank; // rank[v] = order where vertex v appers in order
/// <summary>
/// Determines whether the digraph <c>G</c> has a topological order and, if so,
/// finds such a topological order.
/// </summary>
/// <param name="G">the digraph</param>
///
public TopologicalX(Digraph G)
{
// indegrees of remaining vertices
int[] indegree = new int[G.V];
for (int v = 0; v < G.V; v++)
{
indegree[v] = G.Indegree(v);
}
// initialize
rank = new int[G.V];
order = new LinkedQueue <int>();
int count = 0;
// initialize queue to contain all vertices with indegree = 0
LinkedQueue <int> queue = new LinkedQueue <int>();
for (int v = 0; v < G.V; v++)
{
if (indegree[v] == 0)
{
queue.Enqueue(v);
}
}
for (int j = 0; !queue.IsEmpty; j++)
{
int v = queue.Dequeue();
order.Enqueue(v);
rank[v] = count++;
foreach (int w in G.Adj(v))
{
indegree[w]--;
if (indegree[w] == 0)
{
queue.Enqueue(w);
}
}
}
// there is a directed cycle in subgraph of vertices with indegree >= 1.
if (count != G.V)
{
order = null;
}
Debug.Assert(check(G));
}
/**************************************************************************
*
* The code below is solely for testing correctness of the data type.
*
**************************************************************************/
// Determines whether a digraph has an Eulerian path using necessary
// and sufficient conditions (without computing the path itself):
// - indegree(v) = outdegree(v) for every vertex,
// except one vertex v may have outdegree(v) = indegree(v) + 1
// (and one vertex v may have indegree(v) = outdegree(v) + 1)
// - the graph is connected, when viewed as an undirected graph
// (ignoring isolated vertices)
// This method is solely for unit testing.
private static bool hasEulerianPath(Digraph G)
{
if (G.E == 0)
{
return(true);
}
// Condition 1: indegree(v) == outdegree(v) for every vertex,
// except one vertex may have outdegree(v) = indegree(v) + 1
int deficit = 0;
for (int v = 0; v < G.V; v++)
{
if (G.Outdegree(v) > G.Indegree(v))
{
deficit += (G.Outdegree(v) - G.Indegree(v));
}
}
if (deficit > 1)
{
return(false);
}
// Condition 2: graph is connected, ignoring isolated vertices
Graph H = new Graph(G.V);
for (int v = 0; v < G.V; v++)
{
foreach (int w in G.Adj(v))
{
H.AddEdge(v, w);
}
}
// check that all non-isolated vertices are connected
int s = nonIsolatedVertex(G);
BreadthFirstPaths bfs = new BreadthFirstPaths(H, s);
for (int v = 0; v < G.V; v++)
{
if (H.Degree(v) > 0 && !bfs.HasPathTo(v))
{
return(false);
}
}
return(true);
}
/**************************************************************************
*
* The code below is solely for testing correctness of the data type.
*
**************************************************************************/
// Determines whether a digraph has an Eulerian cycle using necessary
// and sufficient conditions (without computing the cycle itself):
// - at least one edge
// - indegree(v) = outdegree(v) for every vertex
// - the graph is connected, when viewed as an undirected graph
// (ignoring isolated vertices)
private static bool hasEulerianCycle(Digraph G)
{
// Condition 0: at least 1 edge
if (G.E == 0)
{
return(false);
}
// Condition 1: indegree(v) == outdegree(v) for every vertex
for (int v = 0; v < G.V; v++)
{
if (G.Outdegree(v) != G.Indegree(v))
{
return(false);
}
}
// Condition 2: graph is connected, ignoring isolated vertices
Graph H = new Graph(G.V);
for (int v = 0; v < G.V; v++)
{
foreach (int w in G.Adj(v))
{
H.AddEdge(v, w);
}
}
// check that all non-isolated vertices are conneted
int s = nonIsolatedVertex(G);
BreadthFirstPaths bfs = new BreadthFirstPaths(H, s);
for (int v = 0; v < G.V; v++)
{
if (H.Degree(v) > 0 && !bfs.HasPathTo(v))
{
return(false);
}
}
return(true);
}
public static void MainTest(string[] args)
{
string filename = args[0];
string delimiter = args[1];
SymbolDigraph sg = new SymbolDigraph(filename, delimiter);
Digraph G = sg.G;
string t = Console.ReadLine();
while (t != null)
{
if (sg.Contains(t))
{
foreach (int v in G.Adj(sg.Index(t)))
{
Console.WriteLine(" " + sg.Name(v));
}
}
t = Console.ReadLine();
}
}
private bool[] marked; // marked[v] = is there an s->v path?
/// <summary>
/// Computes the vertices reachable from the source vertex <c>s</c> in the
/// digraph <c>G</c>.</summary>
/// <param name="G">the digraph</param>
/// <param name="s">the source vertex</param>
///
public NonrecursiveDirectedDFS(Digraph G, int s)
{
marked = new bool[G.V];
// to be able to iterate over each adjacency list, keeping track of which
// vertex in each adjacency list needs to be explored next
IEnumerator <int>[] adj = new IEnumerator <int> [G.V];
for (int v = 0; v < G.V; v++)
{
adj[v] = G.Adj(v).GetEnumerator();
}
// depth-first search using an explicit stack
LinkedStack <int> stack = new LinkedStack <int>();
marked[s] = true;
stack.Push(s);
while (!stack.IsEmpty)
{
int v = stack.Peek();
if (adj[v].MoveNext())
{
int w = adj[v].Current;
// Console.Write("check {0}\n", w);
if (!marked[w])
{
// discovered vertex w for the first time
marked[w] = true;
// edgeTo[w] = v;
stack.Push(w);
// Console.Write("dfs({0})\n", w);
}
}
else
{
// Console.Write("{0} done\n", v);
stack.Pop();
}
}
}
public static void MainTest(string[] args)
{
int V = int.Parse(args[0]);
int E = int.Parse(args[1]);
// Eulerian cycle
Digraph G1 = DigraphGenerator.EulerianCycle(V, E);
DirectedEulerianCycle.UnitTest(G1, "Eulerian cycle");
// Eulerian path
Digraph G2 = DigraphGenerator.EulerianPath(V, E);
DirectedEulerianCycle.UnitTest(G2, "Eulerian path");
// empty digraph
Digraph G3 = new Digraph(V);
DirectedEulerianCycle.UnitTest(G3, "empty digraph");
// self loop
Digraph G4 = new Digraph(V);
int v4 = StdRandom.Uniform(V);
G4.AddEdge(v4, v4);
DirectedEulerianCycle.UnitTest(G4, "single self loop");
// union of two disjoint cycles
Digraph H1 = DigraphGenerator.EulerianCycle(V / 2, E / 2);
Digraph H2 = DigraphGenerator.EulerianCycle(V - V / 2, E - E / 2);
int[] perm = new int[V];
for (int i = 0; i < V; i++)
{
perm[i] = i;
}
StdRandom.Shuffle(perm);
Digraph G5 = new Digraph(V);
for (int v = 0; v < H1.V; v++)
{
foreach (int w in H1.Adj(v))
{
G5.AddEdge(perm[v], perm[w]);
}
}
for (int v = 0; v < H2.V; v++)
{
foreach (int w in H2.Adj(v))
{
G5.AddEdge(perm[V / 2 + v], perm[V / 2 + w]);
}
}
DirectedEulerianCycle.UnitTest(G5, "Union of two disjoint cycles");
// random digraph
Digraph G6 = DigraphGenerator.Simple(V, E);
DirectedEulerianCycle.UnitTest(G6, "simple digraph");
// 4-vertex digraph - no data file
//Digraph G7 = new Digraph(new TextInput("eulerianD.txt"));
//DirectedEulerianCycle.UnitTest(G7, "4-vertex Eulerian digraph");
}
private LinkedStack <int> cycle; // the directed cycle; null if digraph is acyclic
/// <summary>
/// Determines whether the digraph <c>G</c> has a directed cycle and, if so,
/// finds such a cycle.</summary>
/// <param name="G">the digraph</param>
///
public DirectedCycleX(Digraph G)
{
// indegrees of remaining vertices
int[] indegree = new int[G.V];
for (int v = 0; v < G.V; v++)
{
indegree[v] = G.Indegree(v);
}
// initialize queue to contain all vertices with indegree = 0
LinkedQueue <int> queue = new LinkedQueue <int>();
for (int v = 0; v < G.V; v++)
{
if (indegree[v] == 0)
{
queue.Enqueue(v);
}
}
for (int j = 0; !queue.IsEmpty; j++)
{
int v = queue.Dequeue();
foreach (int w in G.Adj(v))
{
indegree[w]--;
if (indegree[w] == 0)
{
queue.Enqueue(w);
}
}
}
// there is a directed cycle in subgraph of vertices with indegree >= 1.
int[] edgeTo = new int[G.V];
int root = -1; // any vertex with indegree >= -1
for (int v = 0; v < G.V; v++)
{
if (indegree[v] == 0)
{
continue;
}
else
{
root = v;
}
foreach (int w in G.Adj(v))
{
if (indegree[w] > 0)
{
edgeTo[w] = v;
}
}
}
if (root != -1)
{
// find any vertex on cycle
bool[] visited = new bool[G.V];
while (!visited[root])
{
visited[root] = true;
root = edgeTo[root];
}
// extract cycle
cycle = new LinkedStack <int>();
int v = root;
do
{
cycle.Push(v);
v = edgeTo[v];
} while (v != root);
cycle.Push(root);
}
Debug.Assert(check());
}
private LinkedStack <int> path = null; // Eulerian path; null if no suh path
/// <summary>
/// Computes an Eulerian path in the specified digraph, if one exists.</summary>
/// <param name="G">the digraph</param>
///
public DirectedEulerianPath(Digraph G)
{
// find vertex from which to start potential Eulerian path:
// a vertex v with outdegree(v) > indegree(v) if it exits;
// otherwise a vertex with outdegree(v) > 0
int deficit = 0;
int s = nonIsolatedVertex(G);
for (int v = 0; v < G.V; v++)
{
if (G.Outdegree(v) > G.Indegree(v))
{
deficit += (G.Outdegree(v) - G.Indegree(v));
s = v;
}
}
// digraph can't have an Eulerian path
// (this condition is needed)
if (deficit > 1)
{
return;
}
// special case for digraph with zero edges (has a degenerate Eulerian path)
if (s == -1)
{
s = 0;
}
// create local view of adjacency lists, to iterate one vertex at a time
IEnumerator <int>[] adj = new IEnumerator <int> [G.V];
for (int v = 0; v < G.V; v++)
{
adj[v] = G.Adj(v).GetEnumerator();
}
// greedily add to cycle, depth-first search style
LinkedStack <int> stack = new LinkedStack <int>();
stack.Push(s);
path = new LinkedStack <int>();
while (!stack.IsEmpty)
{
int v = stack.Pop();
while (adj[v].MoveNext())
{
stack.Push(v);
v = adj[v].Current;
}
// push vertex with no more available edges to path
path.Push(v);
}
// check if all edges have been used
if (path.Count != G.E + 1)
{
path = null;
}
Debug.Assert(check(G));
}
