问题
I'm back with another similar question. I am currently working on a Java program that will check if a graph is 2-colorable, i.e. if it contains no odd cycles (cycles of odd number length). The entire algorithm is supposed to run in O(V+E) time (V being all vertices and E being all edges in the graph). My current algorithm does a Depth First Search, recording all vertices in the path it takes, then looks for a back edge, and then records between which vertices the edge is between. Next it traces a path from one end of the back edge until it hits the other vertex on the other end of the edge, thus retracing the cycle that the back edge completes.
I was under the impression that this kind of traversing could be done in O(V+E) time for all cycles that exist in my graph, but I must be missing something, because my algorithm is running for a ridiculously long time for very large graphs (10k nodes, no idea how many edges).
Is my algorithm completely wrong? And if so, can anyone point me in the right direction for a better way to record these cycles or possibly tell if they have odd numbers of vertices? Thanks for any and all help you guys can give. Code is below if you need it.
Addition: Sorry I forgot, if the graph is not 2-colorable, I need to provide an odd cycle that proves that it is not.
package algorithms311;
import java.util.*;
import java.io.*;
public class CS311 {
public static LinkedList[] DFSIter(Vertex[] v) {
LinkedList[] VOandBE = new LinkedList[2];
VOandBE[0] = new LinkedList();
VOandBE[1] = new LinkedList();
Stack stack = new Stack();
stack.push(v[0]);
v[0].setColor("gray");
while(!stack.empty()) {
Vertex u = (Vertex) stack.peek();
LinkedList adjList = u.getAdjList();
VOandBE[0].add(u.getId());
boolean allVisited = true;
for(int i = 0; i < adjList.size(); i++) {
if(v[(Integer)adjList.get(i)].getColor().equals("white")) {
allVisited = false;
break;
}
else if(v[(Integer)adjList.get(i)].getColor().equals("gray") && u.getPrev() != (Integer)adjList.get(i)) {
int[] edge = new int[2]; //pair of vertices
edge[0] = u.getId(); //from u
edge[1] = (Integer)adjList.get(i); //to v
VOandBE[1].add(edge);
}
}
if(allVisited) {
u.setColor("black");
stack.pop();
}
else {
for(int i = 0; i < adjList.size(); i++) {
if(v[(Integer)adjList.get(i)].getColor().equals("white")) {
stack.push(v[(Integer)adjList.get(i)]);
v[(Integer)adjList.get(i)].setColor("gray");
v[(Integer)adjList.get(i)].setPrev(u.getId());
break;
}
}
}
}
return VOandBE;
}
public static void checkForTwoColor(String g) { //input is a graph formatted as assigned
String graph = g;
try {
// --Read First Line of Input File
// --Find Number of Vertices
FileReader file1 = new FileReader("W:\\Documents\\NetBeansProjects\\algorithms311\\src\\algorithms311\\" + graph);
BufferedReader bReaderNumEdges = new BufferedReader(file1);
String numVertS = bReaderNumEdges.readLine();
int numVert = Integer.parseInt(numVertS);
System.out.println(numVert + " vertices");
// --Make Vertices
Vertex vertex[] = new Vertex[numVert];
for(int k = 0; k <= numVert - 1; k++) {
vertex[k] = new Vertex(k);
}
// --Adj Lists
FileReader file2 = new FileReader("W:\\Documents\\NetBeansProjects\\algorithms311\\src\\algorithms311\\" + graph);
BufferedReader bReaderEdges = new BufferedReader(file2);
bReaderEdges.readLine(); //skip first line, that's how many vertices there are
String edge;
while((edge = bReaderEdges.readLine()) != null) {
StringTokenizer ST = new StringTokenizer(edge);
int vArr[] = new int[2];
for(int j = 0; ST.hasMoreTokens(); j++) {
vArr[j] = Integer.parseInt(ST.nextToken());
}
vertex[vArr[0]-1].addAdj(vArr[1]-1);
vertex[vArr[1]-1].addAdj(vArr[0]-1);
}
LinkedList[] l = new LinkedList[2];
l = DFSIter(vertex);//DFS(vertex);
System.out.println(l[0]);
for(int i = 0; i < l[1].size(); i++) {
int[] j = (int[])l[1].get(i);
System.out.print(" [" + j[0] + ", " + j[1] + "] ");
}
LinkedList oddCycle = new LinkedList();
boolean is2Colorable = true;
//System.out.println("iterate through list of back edges");
for(int i = 0; i < l[1].size(); i++) { //iterate through the list of back edges
//System.out.println(i);
int[] q = (int[])(l[1].get(i)); // q = pair of vertices that make up a back edge
int u = q[0]; // edge (u,v)
int v = q[1];
LinkedList cycle = new LinkedList();
if(l[0].indexOf(u) < l[0].indexOf(v)) { //check if u is before v
for(int z = l[0].indexOf(u); z <= l[0].indexOf(v); z++) { //if it is, look for u first; from u to v
cycle.add(l[0].get(z));
}
}
else if(l[0].indexOf(v) < l[0].indexOf(u)) {
for(int z = l[0].indexOf(v); z <= l[0].indexOf(u); z++) { //if it is, look for u first; from u to v
cycle.add(l[0].get(z));
}
}
if((cycle.size() & 1) != 0) { //if it has an odd cycle, print out the cyclic nodes or write them to a file
is2Colorable = false;
oddCycle = cycle;
break;
}
}
if(!is2Colorable) {
System.out.println("Graph is not 2-colorable, odd cycle exists");
if(oddCycle.size() <= 50) {
System.out.println(oddCycle);
}
else {
try {
BufferedWriter outFile = new BufferedWriter(new FileWriter("W:\\Documents\\NetBeansProjects\\algorithms311\\src\\algorithms311\\" + graph + "OddCycle.txt"));
String cyc = oddCycle.toString();
outFile.write(cyc);
outFile.close();
}
catch (IOException e) {
System.out.println("Could not write file");
}
}
}
}
catch (IOException e) {
System.out.println("Could not open file");
}
System.out.println("Done!");
}
public static void main(String[] args) {
//checkForTwoColor("smallgraph1");
//checkForTwoColor("smallgraph2");
//checkForTwoColor("smallgraph3");
//checkForTwoColor("smallgraph4");
checkForTwoColor("smallgraph5");
//checkForTwoColor("largegraph1");
}
}
Vertex class
package algorithms311;
import java.util.*;
public class Vertex implements Comparable {
public int id;
public LinkedList adjVert = new LinkedList();
public String color = "white";
public int dTime;
public int fTime;
public int prev;
public boolean visited = false;
public Vertex(int idnum) {
id = idnum;
}
public int getId() {
return id;
}
public int compareTo(Object obj) {
Vertex vert = (Vertex) obj;
return id-vert.getId();
}
@Override public String toString(){
return "Vertex # " + id;
}
public void setColor(String newColor) {
color = newColor;
}
public String getColor() {
return color;
}
public void setDTime(int d) {
dTime = d;
}
public void setFTime(int f) {
fTime = f;
}
public int getDTime() {
return dTime;
}
public int getFTime() {
return fTime;
}
public void setPrev(int v) {
prev = v;
}
public int getPrev() {
return prev;
}
public LinkedList getAdjList() {
return adjVert;
}
public void addAdj(int a) { //adds a vertex id to this vertex's adj list
adjVert.add(a);
}
public void visited() {
visited = true;
}
public boolean wasVisited() {
return visited;
}
}
回答1:
I was under the impression that this kind of traversing could be done in O(V+E) time for all cycles that exist in my graph
There may be much more cycles than O(V+E) in a graph. If you iterate all of them, you will run long.
Back to your original idea, you could just try to implement a straightforward algorithm to color graph in two colors (mark an arbitrary node as black, all neighbors in white, all their neighbors in black, etc; that would be a breadth-first search). That is indeed done in O(V+E) time. If you succeed, then graph is 2-colorable. If you fail, it's not.
Edit: If you need a cycle that proves graph is not 2-colorable, just record for each node the vertex you traversed into it from. When you happen to traverse from black vertex A to black vertex B (thus needing to color black B into white and proving your graph is not 2-colorable), you get the cycle by looking back to parents:
X -> Y -> Z -> U -> V -> P -> Q -> A
\-> D -> E -> B
Then, A-B-E-D-V-P-Q (the paths up to their common ancestor) is the cycle you needed.
Note that in this version you don't have to check all cycles, you just output a first cycle, where back-edge in the tree has both vertexes colored in the same color.
回答2:
you are describing a bipartite graph. a bipartite graph is 2 colorable and it contains no odd length cycles. You can use BFS to prove that a graph is bipartite or not. Hope this helps.
来源：https://stackoverflow.com/questions/1838934/checking-for-odd-cycles-in-an-undirected-graph