/**
* Returns a random rooted-out DAG on <tt>V</tt> vertices and <tt>E</tt> edges. The DAG returned
* is not chosen uniformly at random among all such DAGs.
*
* @param V the number of vertices
* @param E the number of edges
* @return a random rooted-out DAG on <tt>V</tt> vertices and <tt>E</tt> edges
*/
public static Digraph rootedOutDAG(int V, int E) {
if (E > (long) V * (V - 1) / 2) throw new IllegalArgumentException("Too many edges");
if (E < V - 1) throw new IllegalArgumentException("Too few edges");
Digraph G = new Digraph(V);
SET<Edge> set = new SET<Edge>();
// fix a topological order
int[] vertices = new int[V];
for (int i = 0; i < V; i++) vertices[i] = i;
StdRandom.shuffle(vertices);
// one edge pointing from each vertex, other than the root = vertices[V-1]
for (int v = 0; v < V - 1; v++) {
int w = StdRandom.uniform(v + 1, V);
Edge e = new Edge(v, w);
set.add(e);
G.addEdge(vertices[w], vertices[v]);
}
while (G.E() < E) {
int v = StdRandom.uniform(V);
int w = StdRandom.uniform(V);
Edge e = new Edge(v, w);
if ((v < w) && !set.contains(e)) {
set.add(e);
G.addEdge(vertices[w], vertices[v]);
}
}
return G;
}
public static void main(String[] args) {
In in1 = new In(Inputs.DIGRAPH_SMALL);
In in2 = new In(Inputs.DIGRAPH_SMALL);
edu.princeton.cs.algs4.Digraph sedgewicks = new edu.princeton.cs.algs4.Digraph(in1);
Digraph digraph = new Digraph(in2);
in1.close();
in2.close();
StdOut.printf("Sedgewicks - v: %d, e: %d\n", sedgewicks.V(), sedgewicks.E());
for (int i = 0; i < sedgewicks.V(); i++) {
for (int w : sedgewicks.adj(i)) {
StdOut.printf("%d -> %d\n", i, w);
}
}
StdOut.println();
StdOut.printf("My - v: %d, e: %d\n", digraph.V(), digraph.E());
for (int i = 0; i < digraph.V(); i++) {
for (int w : digraph.adj(i)) {
StdOut.printf("%d -> %d\n", i, w);
}
}
StdOut.println();
edu.princeton.cs.algs4.Digraph sedwickDi = sedgewicks.reverse();
StdOut.printf("Sedgewicks reverse - v: %d, e: %d\n", sedwickDi.V(), sedwickDi.E());
for (int i = 0; i < sedwickDi.V(); i++) {
for (int w : sedwickDi.adj(i)) {
StdOut.printf("%d -> %d\n", i, w);
}
}
StdOut.println();
Digraph myReverse = digraph.reverse();
StdOut.printf("My reverse - v: %d, e: %d\n", myReverse.V(), myReverse.E());
for (int i = 0; i < myReverse.V(); i++) {
for (int w : myReverse.adj(i)) {
StdOut.printf("%d -> %d\n", i, w);
}
}
}
/** Copy constructor. */
public Digraph(Digraph G) {
this(G.V());
this.E = G.E();
for (int v = 0; v < G.V(); v++) {
// reverse so that adjacency list is in same order as original
Stack<Integer> reverse = new Stack<Integer>();
for (int w : G.adj[v]) {
reverse.push(w);
}
for (int w : reverse) {
adj[v].add(w);
}
}
}
/**
* Computes an Eulerian cycle in the specified digraph, if one exists.
*
* @param G the digraph
*/
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
Iterator<Integer>[] adj = (Iterator<Integer>[]) new Iterator[G.V()];
for (int v = 0; v < G.V(); v++) adj[v] = G.adj(v).iterator();
// initialize stack with any non-isolated vertex
int s = nonIsolatedVertex(G);
Stack<Integer> stack = new Stack<Integer>();
stack.push(s);
// greedily add to putative cycle, depth-first search style
cycle = new Stack<Integer>();
while (!stack.isEmpty()) {
int v = stack.pop();
while (adj[v].hasNext()) {
stack.push(v);
v = adj[v].next();
}
// 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.size() != G.E() + 1) cycle = null;
assert certifySolution(G);
}
/**
* Returns a random simple digraph containing <tt>V</tt> vertices and <tt>E</tt> edges.
*
* @param V the number of vertices
* @param E the number of vertices
* @return a random simple digraph on <tt>V</tt> vertices, containing a total of <tt>E</tt> edges
* @throws IllegalArgumentException if no such simple digraph exists
*/
public static Digraph simple(int V, int E) {
if (E > (long) V * (V - 1)) throw new IllegalArgumentException("Too many edges");
if (E < 0) throw new IllegalArgumentException("Too few edges");
Digraph G = new Digraph(V);
SET<Edge> set = new SET<Edge>();
while (G.E() < E) {
int v = StdRandom.uniform(V);
int w = StdRandom.uniform(V);
Edge e = new Edge(v, w);
if ((v != w) && !set.contains(e)) {
set.add(e);
G.addEdge(v, w);
}
}
return G;
}
/**
* Returns a random simple DAG containing <tt>V</tt> vertices and <tt>E</tt> edges. Note: it is
* not uniformly selected at random among all such DAGs.
*
* @param V the number of vertices
* @param E the number of vertices
* @return a random simple DAG on <tt>V</tt> vertices, containing a total of <tt>E</tt> edges
* @throws IllegalArgumentException if no such simple DAG exists
*/
public static Digraph dag(int V, int E) {
if (E > (long) V * (V - 1) / 2) throw new IllegalArgumentException("Too many edges");
if (E < 0) throw new IllegalArgumentException("Too few edges");
Digraph G = new Digraph(V);
SET<Edge> set = new SET<Edge>();
int[] vertices = new int[V];
for (int i = 0; i < V; i++) vertices[i] = i;
StdRandom.shuffle(vertices);
while (G.E() < E) {
int v = StdRandom.uniform(V);
int w = StdRandom.uniform(V);
Edge e = new Edge(v, w);
if ((v < w) && !set.contains(e)) {
set.add(e);
G.addEdge(vertices[v], vertices[w]);
}
}
return G;
}
// check that solution is correct
private boolean certifySolution(Digraph G) {
// internal consistency check
if (hasEulerianCycle() == (cycle() == null)) return false;
// hashEulerianCycle() returns correct value
if (hasEulerianCycle() != hasEulerianCycle(G)) return false;
// nothing else to check if no Eulerian cycle
if (cycle == null) return true;
// check that cycle() uses correct number of edges
if (cycle.size() != G.E() + 1) return false;
// check that cycle() is a directed cycle of G
// TODO
return true;
}
// 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 boolean 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++) for (int w : 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;
}
