Note that a graph is represented as an adjacency list.
I've heard of 2 approaches to find a cycle in a graph:
Keep an array of boolean values to keep track of whether you visited a node before. If you run out of new nodes to go to (without hitting a node you have already been), then just backtrack and try a different branch.
The one from Cormen's CLRS or Skiena: For depth-first search in undirected graphs, there are two types of edges, tree and back. The graph has a cycle if and only if there exists a back edge.
Can somebody explain what are the back edges of a graph and what's the diffirence between the above 2 methods.
Thanks.
Update: Here's the code to detect cycles in both cases. Graph is a simple class that represents all graph-nodes as unique numbers for simplicity, each node has its adjacent neighboring nodes (g.getAdjacentNodes(int)):
public class Graph {
private int[][] nodes; // all nodes; e.g. int[][] nodes = {{1,2,3}, {3,2,1,5,6}...};
public int[] getAdjacentNodes(int v) {
return nodes[v];
}
// number of vertices in a graph
public int vSize() {
return nodes.length;
}
}
Java code to detect cycles in an undirected graph:
public class DFSCycle {
private boolean marked[];
private int s;
private Graph g;
private boolean hasCycle;
// s - starting node
public DFSCycle(Graph g, int s) {
this.g = g;
this.s = s;
marked = new boolean[g.vSize()];
findCycle(g,s,s);
}
public boolean hasCycle() {
return hasCycle;
}
public void findCycle(Graph g, int v, int u) {
marked[v] = true;
for (int w : g.getAdjacentNodes(v)) {
if(!marked[w]) {
marked[w] = true;
findCycle(g,w,v);
} else if (v != u) {
hasCycle = true;
return;
}
}
}
}
Java code to detect cycles in a directed graph:
public class DFSDirectedCycle {
private boolean marked[];
private boolean onStack[];
private int s;
private Graph g;
private boolean hasCycle;
public DFSDirectedCycle(Graph g, int s) {
this.s = s
this.g = g;
marked = new boolean[g.vSize()];
onStack = new boolean[g.vSize()];
findCycle(g,s);
}
public boolean hasCycle() {
return hasCycle;
}
public void findCycle(Graph g, int v) {
marked[v] = true;
onStack[v] = true;
for (int w : g.adjacentNodes(v)) {
if(!marked[w]) {
findCycle(g,w);
} else if (onStack[w]) {
hasCycle = true;
return;
}
}
onStack[v] = false;
}
}
Answering my question:
The graph has a cycle if and only if there exists a back edge. A back edge is an edge that is from a node to itself (selfloop) or one of its ancestor in the tree produced by DFS forming a cycle.
Both approaches above actually mean the same. However, this method can be applied only to undirected graphs.
The reason why this algorithm doesn't work for directed graphs is that in a directed graph 2 different paths to the same vertex don't make a cycle. For example: A-->B, B-->C, A-->C - don't make a cycle whereas in undirected ones: A--B, B--C, C--A does.
Find a cycle in undirected graphs
An undirected graph has a cycle if and only if a depth-first search (DFS) finds an edge that points to an already-visited vertex (a back edge).
Find a cycle in directed graphs
In addition to visited vertices we need to keep track of vertices currently in recursion stack of function for DFS traversal. If we reach a vertex that is already in the recursion stack, then there is a cycle in the tree.
Update: Working code is in the question section above.
For the sake of completion, it is possible to find cycles in a directed graph using DFS (from wikipedia):
L ← Empty list that will contain the sorted nodes
while there are unmarked nodes do
select an unmarked node n
visit(n)
function visit(node n)
if n has a temporary mark then stop (not a DAG)
if n is not marked (i.e. has not been visited yet) then
mark n temporarily
for each node m with an edge from n to m do
visit(m)
mark n permanently
unmark n temporarily
add n to head of L
Here is the code I've written in C based on DFS to find out whether a given undirected graph is connected/cyclic or not. with some sample output at the end. Hope it'll be helpful :)
#include<stdio.h>
#include<stdlib.h>
/****Global Variables****/
int A[20][20],visited[20],count=0,n;
int seq[20],connected=1,acyclic=1;
/****DFS Function Declaration****/
void DFS();
/****DFSearch Function Declaration****/
void DFSearch(int cur);
/****Main Function****/
int main()
{
int i,j;
printf("\nEnter no of Vertices: ");
scanf("%d",&n);
printf("\nEnter the Adjacency Matrix(1/0):\n");
for(i=1;i<=n;i++)
for(j=1;j<=n;j++)
scanf("%d",&A[i][j]);
printf("\nThe Depth First Search Traversal:\n");
DFS();
for(i=1;i<=n;i++)
printf("%c,%d\t",'a'+seq[i]-1,i);
if(connected && acyclic) printf("\n\nIt is a Connected, Acyclic Graph!");
if(!connected && acyclic) printf("\n\nIt is a Not-Connected, Acyclic Graph!");
if(connected && !acyclic) printf("\n\nGraph is a Connected, Cyclic Graph!");
if(!connected && !acyclic) printf("\n\nIt is a Not-Connected, Cyclic Graph!");
printf("\n\n");
return 0;
}
/****DFS Function Definition****/
void DFS()
{
int i;
for(i=1;i<=n;i++)
if(!visited[i])
{
if(i>1) connected=0;
DFSearch(i);
}
}
/****DFSearch Function Definition****/
void DFSearch(int cur)
{
int i,j;
visited[cur]=++count;
seq[count]=cur;
for(i=1;i<count-1;i++)
if(A[cur][seq[i]])
acyclic=0;
for(i=1;i<=n;i++)
if(A[cur][i] && !visited[i])
DFSearch(i);
}
Sample Output:
majid@majid-K53SC:~/Desktop$ gcc BFS.c
majid@majid-K53SC:~/Desktop$ ./a.out
************************************
Enter no of Vertices: 10
Enter the Adjacency Matrix(1/0):
0 0 1 1 1 0 0 0 0 0
0 0 0 0 1 0 0 0 0 0
0 0 0 1 0 1 0 0 0 0
0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0
0 1 0 0 1 0 0 0 0 0
0 0 0 0 0 0 0 1 0 0
0 0 0 0 0 0 0 0 1 0
0 0 0 0 0 0 0 0 0 1
0 0 0 0 0 0 1 0 0 0
The Depdth First Search Traversal:
a,1 c,2 d,3 f,4 b,5 e,6 g,7 h,8 i,9 j,10
It is a Not-Connected, Cyclic Graph!
majid@majid-K53SC:~/Desktop$ ./a.out
************************************
Enter no of Vertices: 4
Enter the Adjacency Matrix(1/0):
0 0 1 1
0 0 1 0
1 1 0 0
0 0 0 1
The Depth First Search Traversal:
a,1 c,2 b,3 d,4
It is a Connected, Acyclic Graph!
majid@majid-K53SC:~/Desktop$ ./a.out
************************************
Enter no of Vertices: 5
Enter the Adjacency Matrix(1/0):
0 0 0 1 0
0 0 0 1 0
0 0 0 0 1
1 1 0 0 0
0 0 1 0 0
The Depth First Search Traversal:
a,1 d,2 b,3 c,4 e,5
It is a Not-Connected, Acyclic Graph!
*/
I think the above code works only for a connected digraph since we start dfs from the source node only, for if the digraph is not connected there may be a cycle in the other component which may go unnoticed!
来源：https://stackoverflow.com/questions/19113189/detecting-cycles-in-a-graph-using-dfs-2-different-approaches-and-whats-the-dif