/**
* Reads in a social network from a file, and then repeatedly reads in
* individuals from standard input and prints out their degrees of
* separation.
* Takes three command-line arguments: the name of a file,
* a delimiter, and the name of the distinguished individual.
* Each line in the file contains the name of a vertex, followed by a
* list of the names of the vertices adjacent to that vertex,
* separated by the delimiter.
*
* @param args the command-line arguments
*/
public static void main(String[] args) {
String filename = args[0];
String delimiter = args[1];
String source = args[2];
// StdOut.println("Source: " + source);
SymbolGraph sg = new SymbolGraph(filename, delimiter);
Graph G = sg.graph();
if (!sg.contains(source)) {
StdOut.println(source + " not in database.");
return;
}
int s = sg.indexOf(source);
BreadthFirstPaths bfs = new BreadthFirstPaths(G, s);
while (!StdIn.isEmpty()) {
String sink = StdIn.readLine();
if (sg.contains(sink)) {
int t = sg.indexOf(sink);
if (bfs.hasPathTo(t)) {
for (int v : bfs.pathTo(t)) {
StdOut.println(" " + sg.nameOf(v));
}
}
else {
StdOut.println("Not connected");
}
}
else {
StdOut.println(" Not in database.");
}
}
}
/**************************************************************************
*
* The code below is solely for testing correctness of the data type.
*
**************************************************************************/
// Determines whether a graph has an Eulerian path using necessary
// and sufficient conditions (without computing the path itself):
// - degree(v) is even for every vertex, except for possibly two
// - the graph is connected (ignoring isolated vertices)
// This method is solely for unit testing.
private static boolean satisfiesNecessaryAndSufficientConditions(Graph G) {
if (G.E() == 0) return true;
// Condition 1: degree(v) is even except for possibly two
int oddDegreeVertices = 0;
for (int v = 0; v < G.V(); v++)
if (G.degree(v) % 2 != 0)
oddDegreeVertices++;
if (oddDegreeVertices > 2) return false;
// Condition 2: graph is connected, ignoring isolated vertices
int s = nonIsolatedVertex(G);
BreadthFirstPaths bfs = new BreadthFirstPaths(G, s);
for (int v = 0; v < G.V(); v++)
if (G.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 graph has an Eulerian cycle using necessary
// and sufficient conditions (without computing the cycle itself):
// - at least one edge
// - degree(v) is even for every vertex v
// - the graph is connected (ignoring isolated vertices)
private static boolean satisfiesNecessaryAndSufficientConditions(Graph G) {
// Condition 0: at least 1 edge
if (G.E() == 0) return false;
// Condition 1: degree(v) is even for every vertex
for (int v = 0; v < G.V(); v++)
if (G.degree(v) % 2 != 0)
return false;
// Condition 2: graph is connected, ignoring isolated vertices
int s = nonIsolatedVertex(G);
BreadthFirstPaths bfs = new BreadthFirstPaths(G, s);
for (int v = 0; v < G.V(); v++)
if (G.degree(v) > 0 && !bfs.hasPathTo(v))
return false;
return true;
}
/**
* Unit tests the {@code BreadthFirstPaths} data type.
*
* @param args the command-line arguments
*/
public static void main(String[] args) {
In in = new In(args[0]);
Graph G = new Graph(in);
// StdOut.println(G);
int s = Integer.parseInt(args[1]);
BreadthFirstPaths bfs = new BreadthFirstPaths(G, s);
for (int v = 0; v < G.V(); v++) {
if (bfs.hasPathTo(v)) {
StdOut.printf("%d to %d (%d): ", s, v, bfs.distTo(v));
for (int x : bfs.pathTo(v)) {
if (x == s) StdOut.print(x);
else StdOut.print("-" + x);
}
StdOut.println();
}
else {
StdOut.printf("%d to %d (-): not connected\n", s, v);
}
}
}
