I'm having a hard time understanding Tarjan's algorithm for articulation points. I'm currently following this tutorial here: https://www.hackerearth.com/practice/algorithms/graphs/articulation-points-and-bridges/tutorial/. What I really can't see, and couldn't see in any other tutorial, is what exactly a "back edge" means. Considering the graph given there, I know 3-1 and 4-2 are back edges, but are 2-1, 3-2, and 4-3 back edges too? Thank you.
...a Back Edge is an edge that connects a vertex to a vertex that is discovered before it's parent.
from your source.
Think about it like this: When you apply a DFS on a graph you fix some path that the algorithm chooses. Now in the given case: 0->1->2->3->4. As in the article mentioned, the source graph contains the edges 4-2 and 3-1. When the DFS reaches 3 it could choose 1 but 1 is already in your path so it is a back edge and therefore, as mentioned in the source, a possible alternative path.
Addressing your second question: Are 2-1, 3-2, and 4-3 back edges too? For a different path they can be. Suppose your DFS chooses 0->1->3->2->4 then 2-1 and 4-3 are back edges.
Consider the following (directed) graph traversal with DFS. Here the colors of the nodes represent the following:
The floral-white nodes are the ones that are yet to be visited
The gray nodes are the nodes that are visited and on stack
The black nodes are the ones that are popped from the stack.
Notice that when the node 13 discovers the node 0 through the edge 13->0 the node 0 is still on the stack. Here, 13->0 is a back edge and it denotes the existence of a cycle (the triangle 0->1->13).
In essence, when you do a DFS, if there are cycles in your graph between nodes A, B and C and you have discovered the edges A-B, later you discover the edge B-C, then, since you have reached node C, you will discover the edge C-A, but you need to ignore this path in your search to avoid infinite loops. So, in your search A-B and B-C were not back edges, but C-A is a back edge, since this edge forms a cycle back to an already visited node.
From article mentioned:
Given a DFS tree of a graph, a Back Edge is an edge that connects a
vertex to a vertex that is discovered before it's parent.
2-1, 3-2, 4-3 are not "Back edge" because they link the vertices with their parents in DFS tree.
Here is the code for a better understand:
#include<bits/stdc++.h>
using namespace std;
struct vertex{
int node;
int start;
int finish;
int color;
int parent;
};
int WHITE=0, BLACK=1, GREY=2;
vector<int> adjList[8];
int num_of_verts = 8;
struct vertex vertices[8];
int t=0;
bool DFS_visit(int u){
bool cycleExists = false;
vertices[u].color=GREY;
t++;
vertices[u].start= t;
for( int i=0; adjList[u][i]!=-1; i++){
if( vertices[adjList[u][i]].color == WHITE ){
if(!cycleExists) cycleExists = DFS_visit(adjList[u][i]);
else DFS_visit(adjList[u][i]);
}
else {
cout << "Cycle detected at edge - ("<<u<<", "<<adjList[u][i]<<")"<<endl;
cycleExists = true;
}
}
vertices[u].color=BLACK;
t++;
vertices[u].finish= t;
return cycleExists;
}
void DFS(){
for(int i=0;i<num_of_verts;i++){
vertices[i].color=WHITE;
vertices[i].parent=NULL;
}
t=0;
for(int i=0;i<num_of_verts;i++){
if(vertices[i].color==WHITE){
cout << "Traversing component "<<i<<"-"<<endl;
bool cycle = DFS_visit(i);
cycle==1? cout<<"Cycle Exists\n\n":cout <<"Cycle does not exist\n\n";
}
}
}
int main(){
adjList[0] = {4, -1};
adjList[1] = {0, 5, -1};
adjList[2] = {1, 5, -1};
adjList[3] = {6, 7, -1};
adjList[4] = {1, -1};
adjList[5] = {-1};
adjList[6] = {2, 5, -1};
adjList[7] = {3, 6, -1};
DFS();
return 0;
}
Related
I'm writing a function which calculates the max path from one node to another in an adjacency list graph after going through the graph in DFS/backtracking manner. The path will not have any cycles, but can have the same nodes within a different path. Ex: A->B->C->D and A-C->B->D is valid while A-B->A->C->D is not. To avoid cycles, I can use a visited set to add nodes once discovered and pop later on.
The algorithm must go through every possible path from the starting node to end node as it is possible paths with the same nodes, but different ordering will be valued differently.
I believe the algorithm may be O(n!) considering everything in the graph may be connected, but I'm not too sure. I'm a bit new to graphs, so I'm having a hard time understanding the exact space/time complexity of things.
I can offer to consider the concrete graph (tree), where vertices are numbers but not letters. The problem, you want to solve, supposes applying the Depth First Search algorithm as you have shown in your requirement.
Assume, we have the graph (tree):
Relatively of this graph, the max path from one vertex to another vertex must be calculated.
Here is the solution of your problem.
#include <iostream>
#include <vector>
using namespace std;
const int maximumSize=10;
vector<int> visited(maximumSize, 0), distances(maximumSize, 0);
vector<int> graph[maximumSize];
int vertices, edges;
void showContentVector1D(vector<int>& input)
{
for(int index=0; index<input.size(); ++index)
{
cout<<input[index]<<", ";
}
return;
}
void createGraph()
{
cin>>vertices>>edges;
int vertex0, vertex1;
for(int edge=1; edge<=edges; ++edge)
{
cin>>vertex0>>vertex1;
graph[vertex0].push_back(vertex1);
graph[vertex1].push_back(vertex0);
}
return;
}
void depthFirstSearch(int currentVertex, int previousVertex)
{
if(visited[currentVertex]==1)
{
return;
}
visited[currentVertex]=1;
distances[currentVertex]=0;
for(int nextVertex : graph[currentVertex])
{
if(nextVertex==previousVertex)
{
continue;
}
depthFirstSearch(nextVertex, currentVertex);
distances[currentVertex]=max(distances[currentVertex], distances[nextVertex]+1);
}
return;
}
void solve()
{
createGraph();
depthFirstSearch(1, 0);
cout<<"distances <- ";
showContentVector1D(distances);
cout<<endl;
return;
}
int main()
{
solve();
return 0;
}
Input:
6 5
1 4
2 4
3 4
4 5
5 6
The first line in the input is 6 5, where 6 is the quantity of vertices in a graph and 5 is the quantity of edges in a graph.
1 4, 2 4, 3 4, 4 5, 5 6 are edges of an provided undirected graph.
Output:
distances <- 0, 3, 0, 0, 2, 1, 0, 0, 0, 0,
The number 3 is assigned into the index 1 which corresponds to the vertex 1 of the provided undirected graph (tree). Also in the concrete indices the distances 2 and 1 are assigned. Those indices are 4 and 5 correspondingly. As you can see the maximum distance is 3 from the vertex 1 to the vertex 6.
I'm trying to implement the list of all possible spanning trees of a graph in order of increasing cost. I'm using the algorithm by Sorensen and Janssens (2005). The graph is initialized as follows:
typedef property<edge_weight_t, int> EdgeWeightProperty;
typedef adjacency_list<vecS, vecS, undirectedS, no_property, EdgeWeightProperty> Graph;
typedef Graph::edge_descriptor Edge;
typedef Graph::vertex_descriptor Vertex;
typedef boost::graph_traits<Graph>::edge_iterator EdgeIterator;
typedef std::pair<EdgeIterator, EdgeIterator> EdgePair;
Graph g;
add_edge(1, 2, 3, g);
add_edge(1, 3, 1, g);
add_edge(1, 4, 2, g);
add_edge(2, 3, 3, g);
add_edge(2, 4, 1, g);
After this it's necessary to find the minimum spanning tree of a graph with some limitations, for instance Edge(2)-(4) shouldn't be in MST and Edge(1)-(2) should be there.
For the edge exclusion it's possible to use remove_edge_if(..) to delete the edge from the graph.
template<typename WMap>
class Remover
{
public:
Remover(const WMap& weights, int threshold)
: m_weights(weights), m_threshold(threshold) {}
template<typename ED>
bool operator()(ED w) const { return m_weights[w] <= m_threshold; }
private:
const WMap& m_weights;
int m_threshold;
};
....
// remove edges of weight < 1
Remover< property_map<Graph, edge_weight_t>::type> r(get(edge_weight, g), 1);
remove_edge_if(r, g);
....
std::list < Edge > spanning_treeT;
kruskal_minimum_spanning_tree(g, std::back_inserter(spanning_treeT));
But how should I ensure that one of the edges is always in the spanning tree? I was trying just to add some Edge into output of the Kruskal function, but it didn't work apparently. It yields MST of the graph + added edge:
std::list < Edge > spanning_tree_g2;
Vertex u, v;
EdgePair ep = edges(g2);
u = source(*ep.first, g2);
v = target(*ep.first, g2);
Edge ed = edge(u, v, g2).first;
spanning_tree_g2.push_front(ed);
kruskal_minimum_spanning_tree(g2, std::back_inserter(spanning_tree_g2));
Is it possible to mark the edges in a way that Kruskal algorithm knows what to include and what not to?
I seems you could force the inclusion of a certain edge by splitting this edge and inserting two "artificial" vertices in the middle.
The MST algorithm is already required to produce a tree of edges that covers all vertices.
Because the artifical vertices have been purposely added by you, it's easy to make sure it's never reachable using any other edges.
Before:
------------------[e:w1+w2]------------------
After:
----[e1:w1]---(v1)---[em:0]---(v2)---[e2:w2]----
(where v1 and v2 are the vertices inserted).
After the fact you "collapse" any sequence of (e1,em,e2) or (e2,em,e1) into (e).
You might end up with a tree that reaches v1 and v2 but never traverses em. In that case you can simply drop one of e1 and e2 and replace it with e unconditionally.
The input to the program is the set of edges in the graph. For e.g. consider the following simple directed graph:
a -> b -> c
The set of edges for this graph is
{ (b, c), (a, b) }
So given a directed graph as a set of edges, how do you determine if the directed graph is a tree? If it is a tree, what is the root node of the tree?
First of I'm looking at how will you represent this graph, adjacency list/ adjacency matrix / any thing else? How will utilize the representation that you have chosen to efficiently answer the above questions?
Edit 1:
Some people are mentoning about using DFS for cycle detection but the problem is which node to start the DFS from. Since it is a directed graph we cannot start the DFS from a random node, for e.g. if I started a DFS from vertex 'c' it won't proceed further since there is no back edge to go to any other nodes. The follow up question here should be how do you determine what is the root of this tree.
Here's a fairly direct method. It can be done with either an adjacency matrix or an edge list.
Find the set, R, of nodes that do not appear as the destination of any edge. If R does not have exactly one member, the graph is not a tree.
If R does have exactly one member, r, it is the only possible root.
Mark r.
Starting from r, recursively mark all nodes that can be reached by following edges from source to destination. If any node is already marked, there is a cycle and the graph is not a tree. (This step is the same as a previously posted answer).
If any node is not marked at the end of step 3, the graph is not a tree.
If none of those steps find that the graph is not a tree, the graph is a tree with r as root.
It's difficult to know what is going to be efficient without some information about the numbers of nodes and edges.
There are 3 properties to check if a graph is a tree:
(1) The number of edges in the graph is exactly one less than the number of vertices |E| = |V| - 1
(2) There are no cycles
(3) The graph is connected
I think this example algorithm can work in the cases of a directed graph:
# given a graph and a starting vertex, check if the graph is a tree
def checkTree(G, v):
# |E| = |V| - 1
if edges.size != vertices.size - 1:
return false;
for v in vertices:
visited[v] = false;
hasCycle = explore_and_check_cycles(G, v);
# a tree is acyclic
if hasCycle:
return false;
for v in vertices:
if not visited[v]: # the graph isn't connected
return false;
# otherwise passes all properties for a tree
return true;
# given a Graph and a vertex, explore all reachable vertices from the vertex
# and returns true if there are any cycles
def explore_and_check_cycles(G, v):
visited[v] = true;
for (v, u) in edges:
if not visited[u]:
return explore_and_check_cyles(G, u)
else: # a backedge between two vertices indicates a cycle
return true
return false
Sources:
Algorithms by S. Dasgupta, C.H. Papadimitriou, and U.V. Vazirani
http://www.cs.berkeley.edu/~vazirani/algorithms.html
start from the root, "mark" it and then go to all children and repeat recursively. if you reach a child that is already marked it means that it is not a tree...
Note: By no means the most efficient way, but conceptually useful. Sometimes you want efficiency, sometimes you want an alternate viewpoint for pedagogic reasons. This is most certainly the later.
Algorithm: Starting with an adjacency matrix A of size n. Take the matrix power A**n. If the matrix is zero for every entry you know that it is at least a collection of trees (a forest). If you can show that it is connected, then it must be a tree. See a Nilpotent matrix. for more information.
To find the root node, we assume that you've shown the graph is a connected tree. Let k be the the number of times you have to raise the power A**k before the matrix becomes all zero. Take the transpose to the (k-1) power A.T ** (k-1). The only non-zero entry must be the root.
Analysis: A rough worse case analysis shows that it is bounded above by O(n^4), three for the matrix multiplication at most n times. You can do better by diagonalizing the matrix which should bring it down to O(n^3). Considering that this problem can be done in O(n), O(1) time/space this is only a useful exercise in logic and understanding of the problem.
For a directed graph, the underlying undirected graph will be a tree if the undirected graph is acyclic and fully connected. The same property holds good if for the directed graph, each vertex has in-degree=1 except one having in-degree=0.
If adjacency-list representation also supports in-degree property of each vertex, then we can apply the above rule easily. Otherwise, we should apply a tweaked DFS to find no-loops for underlying undirected graph as well as that |E|=|V|-1.
Following is the code I have written for this. Feel free to suggest optimisations.
import java.util.*;
import java.lang.*;
import java.io.*;
class Graph
{
private static int V;
private static int adj[][];
static void initializeGraph(int n)
{
V=n+1;
adj = new int[V][V];
for(int i=0;i<V;i++)
{
for(int j=0;j<V ;j++)
adj[i][j]= 0;
}
}
static int isTree(int edges[][],int n)
{
initializeGraph(n);
for(int i=0;i< edges.length;i++)
{
addDirectedEdge(edges[i][0],edges[i][1]);
}
int root = findRoot();
if(root == -1)
return -1;
boolean visited[] = new boolean[V];
boolean isTree = isTree(root, visited, -1);
boolean isConnected = isConnected(visited);
System.out.println("isTree= "+ isTree + " isConnected= "+ isConnected);
if(isTree && isConnected)
return root;
else
return -1;
}
static boolean isTree(int node, boolean visited[], int parent)
{
// System.out.println("node =" +node +" parent" +parent);
visited[node] = true;int j;
for(j =1;j<V;j++)
{
// System.out.println("node =" +node + " j=" +j+ "parent" + parent);
if(adj[node][j]==1)
{
if(visited[j])
{
// System.out.println("returning false for j="+ j);
return false;
}
else
{ //visit all adjacent vertices
boolean child = isTree(j, visited, node);
if(!child)
{
// System.out.println("returning false for j="+ j + " node=" +node);
return false;
}
}
}
}
if(j==V)
return true;
else
return false;
}
static int findRoot()
{
int root =-1, j=0,i;
int count =0;
for(j=1;j<V ;j++)
{
count=0;
for(i=1 ;i<V;i++)
{
if(adj[i][j]==1)
count++;
}
// System.out.println("j"+j +" count="+count);
if(count==0)
{
// System.out.println(j);
return j;
}
}
return -1;
}
static void addDirectedEdge(int s, int d)
{
// System.out.println("s="+ s+"d="+d);
adj[s][d]=1;
}
static boolean isConnected(boolean visited[])
{
for(int i=1; i<V;i++)
{
if(!visited[i])
return false;
}
return true;
}
public static void main (String[] args) throws java.lang.Exception
{
int edges[][]= {{2,3},{2,4},{3,1},{3,5},{3,7},{4,6}, {2,8}, {8,9}};
int n=9;
int root = isTree(edges,n);
System.out.println("root is:" + root);
int edges2[][]= {{2,3},{2,4},{3,1},{3,5},{3,7},{4,6}, {2,8}, {6,3}};
int n2=8;
root = isTree(edges2,n2);
System.out.println("root is:" + root);
}
}
I'm learning about pathfinding from a book, but I have spotted some weird statement.
For completeness, I will include most of the code, but feel free to skip to the second part (Search())
template <class graph_type >
class Graph_SearchDijkstra
{
private:
//create typedefs for the node and edge types used by the graph
typedef typename graph_type::EdgeType Edge;
typedef typename graph_type::NodeType Node;
private:
const graph_type & m_Graph;
//this vector contains the edges that comprise the shortest path tree -
//a directed sub-tree of the graph that encapsulates the best paths from
//every node on the SPT to the source node.
std::vector<const Edge*> m_ShortestPathTree;
//this is indexed into by node index and holds the total cost of the best
//path found so far to the given node. For example, m_CostToThisNode[5]
//will hold the total cost of all the edges that comprise the best path
//to node 5 found so far in the search (if node 5 is present and has
//been visited of course).
std::vector<double> m_CostToThisNode;
//this is an indexed (by node) vector of "parent" edges leading to nodes
//connected to the SPT but that have not been added to the SPT yet.
std::vector<const Edge*> m_SearchFrontier;
int m_iSource;
int m_iTarget;
void Search();
public:
Graph_SearchDijkstra(const graph_type& graph,
int source,
int target = -1):m_Graph(graph),
m_ShortestPathTree(graph.NumNodes()),
m_SearchFrontier(graph.NumNodes()),
m_CostToThisNode(graph.NumNodes()),
m_iSource(source),
m_iTarget(target)
{
Search();
}
//returns the vector of edges defining the SPT. If a target is given
//in the constructor, then this will be the SPT comprising all the nodes
//examined before the target is found, else it will contain all the nodes
//in the graph.
std::vector<const Edge*> GetAllPaths()const;
//returns a vector of node indexes comprising the shortest path
//from the source to the target. It calculates the path by working
//backward through the SPT from the target node.
std::list<int> GetPathToTarget()const;
//returns the total cost to the target
double GetCostToTarget()const;
};
Search():
template <class graph_type>
void Graph_SearchDijkstra<graph_type>::Search()
{
//create an indexed priority queue that sorts smallest to largest
//(front to back). Note that the maximum number of elements the iPQ
//may contain is NumNodes(). This is because no node can be represented
// on the queue more than once.
IndexedPriorityQLow<double> pq(m_CostToThisNode, m_Graph.NumNodes());
//put the source node on the queue
pq.insert(m_iSource);
//while the queue is not empty
while(!pq.empty())
{
//get the lowest cost node from the queue. Don't forget, the return value
//is a *node index*, not the node itself. This node is the node not already
//on the SPT that is the closest to the source node
int NextClosestNode = pq.Pop();
//move this edge from the search frontier to the shortest path tree
m_ShortestPathTree[NextClosestNode] = m_SearchFrontier[NextClosestNode];
//if the target has been found exit
if (NextClosestNode == m_iTarget) return;
//now to relax the edges. For each edge connected to the next closest node
graph_type::ConstEdgeIterator ConstEdgeItr(m_Graph, NextClosestNode);
for (const Edge* pE=ConstEdgeItr.begin();
!ConstEdgeItr.end();
pE=ConstEdgeItr.next())
{
//the total cost to the node this edge points to is the cost to the
//current node plus the cost of the edge connecting them.
double NewCost = m_CostToThisNode[NextClosestNode] + pE->Cost();
//if this edge has never been on the frontier make a note of the cost
//to reach the node it points to, then add the edge to the frontier
//and the destination node to the PQ.
if (m_SearchFrontier[pE->To()] == 0)
{
m_CostToThisNode[pE->To()] = NewCost;
pq.insert(pE->To());
m_SearchFrontier[pE->To()] = pE;
}
//else test to see if the cost to reach the destination node via the
//current node is cheaper than the cheapest cost found so far. If
//this path is cheaper we assign the new cost to the destination
//node, update its entry in the PQ to reflect the change, and add the
//edge to the frontier
else if ( (NewCost < m_CostToThisNode[pE->To()]) &&
(m_ShortestPathTree[pE->To()] == 0) )
{
m_CostToThisNode[pE->To()] = NewCost;
//because the cost is less than it was previously, the PQ must be
//resorted to account for this.
pq.ChangePriority(pE->To());
m_SearchFrontier[pE->To()] = pE;
}
}
}
}
What I don't get is this part:
//else test to see if the cost to reach the destination node via the
//current node is cheaper than the cheapest cost found so far. If
//this path is cheaper we assign the new cost to the destination
//node, update its entry in the PQ to reflect the change, and add the
//edge to the frontier
else if ( (NewCost < m_CostToThisNode[pE->To()]) &&
(m_ShortestPathTree[pE->To()] == 0) )
if the new cost is lower than the cost already found, then why do we also test if the node has not already been added to the SPT? This seems to beat the purpose of the check?
FYI, in m_ShortestPathTree[pE->To()] == 0 the container is a vector that has a pointer to an edge (or NULL) for each index (the index represents a node)
Imagine the following graph:
S --5-- A --2-- F
\ /
-3 -4
\ /
B
And you want to go from S to F. First, let me tell you the Dijkstra's algorithm assumes there are no loops in the graph with a negative weight. In my example, this loop is S -> B -> A -> S or simpler yet, S -> B -> S
If you have such a loop, you can infinitely loop in it, and your cost to F gets lower and lower. That is why this is not acceptable by Dijkstra's algorithm.
Now, how do you check that? It is as the code you posted does. Every time you want to update the weight of a node, besides checking whether it gets smaller weight, you check if it's not in the accepted list. If it is, then you must have had a negative weight loop. Otherwise, how can you end up going forward and reaching an already accepted node with smaller weight?
Let's follow the algorithm on the example graph (nodes in [] are accepted):
Without the if in question:
Starting Weights: S(0), A(inf), B(inf), F(inf)
- Accept S
New weights: [S(0)], A(5), B(-3), F(inf)
- Accept B
New weights: [S(-3)], A(-7), [B(-3)], F(inf)
- Accept A
New weights: [S(-3)], [A(-7)], [B(-11)], F(-5)
- Accept B again
New weights: [S(-14)], [A(-18)], [B(-11)], F(-5)
- Accept A again
... infinite loop
With the if in question:
Starting Weights: S(0), A(inf), B(inf), F(inf)
- Accept S
New weights: [S(0)], A(5), B(-3), F(inf)
- Accept B (doesn't change S)
New weights: [S(0)], A(-7), [B(-3)], F(inf)
- Accept A (doesn't change S or B
New weights: [S(0)], [A(-7)], [B(-3)], F(-5)
- Accept F
How to find all chordless cycles in an undirected graph?
For example, given the graph
0 --- 1
| | \
| | \
4 --- 3 - 2
the algorithm should return 1-2-3 and 0-1-3-4, but never 0-1-2-3-4.
(Note: [1] This question is not the same as small cycle finding in a planar graph because the graph is not necessarily planar. [2] I have read the paper Generating all cycles, chordless cycles, and Hamiltonian cycles with the principle of exclusion but I don't understand what they're doing :). [3] I have tried CYPATH but the program only gives the count, algorithm EnumChordlessPath in readme.txt has significant typos, and the C code is a mess. [4] I am not trying to find an arbitrary set of fundametal cycles. Cycle basis can have chords.)
Assign numbers to nodes from 1 to n.
Pick the node number 1. Call it 'A'.
Enumerate pairs of links coming out of 'A'.
Pick one. Let's call the adjacent nodes 'B' and 'C' with B less than C.
If B and C are connected, then output the cycle ABC, return to step 3 and pick a different pair.
If B and C are not connected:
Enumerate all nodes connected to B. Suppose it's connected to D, E, and F. Create a list of vectors CABD, CABE, CABF. For each of these:
if the last node is connected to any internal node except C and B, discard the vector
if the last node is connected to C, output and discard
if it's not connected to either, create a new list of vectors, appending all nodes to which the last node is connected.
Repeat until you run out of vectors.
Repeat steps 3-5 with all pairs.
Remove node 1 and all links that lead to it. Pick the next node and go back to step 2.
Edit: and you can do away with one nested loop.
This seems to work at the first sight, there may be bugs, but you should get the idea:
void chordless_cycles(int* adjacency, int dim)
{
for(int i=0; i<dim-2; i++)
{
for(int j=i+1; j<dim-1; j++)
{
if(!adjacency[i+j*dim])
continue;
list<vector<int> > candidates;
for(int k=j+1; k<dim; k++)
{
if(!adjacency[i+k*dim])
continue;
if(adjacency[j+k*dim])
{
cout << i+1 << " " << j+1 << " " << k+1 << endl;
continue;
}
vector<int> v;
v.resize(3);
v[0]=j;
v[1]=i;
v[2]=k;
candidates.push_back(v);
}
while(!candidates.empty())
{
vector<int> v = candidates.front();
candidates.pop_front();
int k = v.back();
for(int m=i+1; m<dim; m++)
{
if(find(v.begin(), v.end(), m) != v.end())
continue;
if(!adjacency[m+k*dim])
continue;
bool chord = false;
int n;
for(n=1; n<v.size()-1; n++)
if(adjacency[m+v[n]*dim])
chord = true;
if(chord)
continue;
if(adjacency[m+j*dim])
{
for(n=0; n<v.size(); n++)
cout<<v[n]+1<<" ";
cout<<m+1<<endl;
continue;
}
vector<int> w = v;
w.push_back(m);
candidates.push_back(w);
}
}
}
}
}
#aioobe has a point. Just find all the cycles and then exclude the non-chordless ones. This may be too inefficient, but the search space can be pruned along the way to reduce the inefficiencies. Here is a general algorithm:
void printChordlessCycles( ChordlessCycle path) {
System.out.println( path.toString() );
for( Node n : path.lastNode().neighbors() ) {
if( path.canAdd( n) ) {
path.add( n);
printChordlessCycles( path);
path.remove( n);
}
}
}
Graph g = loadGraph(...);
ChordlessCycle p = new ChordlessCycle();
for( Node n : g.getNodes()) {
p.add(n);
printChordlessCycles( p);
p.remove( n);
}
class ChordlessCycle {
private CountedSet<Node> connected_nodes;
private List<Node> path;
...
public void add( Node n) {
for( Node neighbor : n.getNeighbors() ) {
connected_nodes.increment( neighbor);
}
path.add( n);
}
public void remove( Node n) {
for( Node neighbor : n.getNeighbors() ) {
connected_nodes.decrement( neighbor);
}
path.remove( n);
}
public boolean canAdd( Node n) {
return (connected_nodes.getCount( n) == 0);
}
}
Just a thought:
Let's say you are enumerating cycles on your example graph and you are starting from node 0.
If you do a breadth-first search for each given edge, e.g. 0 - 1, you reach a fork at 1. Then the cycles that reach 0 again first are chordless, and the rest are not and can be eliminated... at least I think this is the case.
Could you use an approach like this? Or is there a counterexample?
How about this. First, reduce the problem to finding all chordless cycles that pass through a given vertex A. Once you've found all of those, you can remove A from the graph, and repeat with another point until there's nothing left.
And how to find all the chordless cycles that pass through vertex A? Reduce this to finding all chordless paths from B to A, given a list of permitted vertices, and search either breadth-first or depth-first. Note that when iterating over the vertices reachable (in one step) from B, when you choose one of them you must remove all of the others from the list of permitted vertices (take special care when B=A, so as not to eliminate three-edge paths).
Find all cycles.
Definition of a chordless cycle is a set of points in which a subset cycle of those points don't exist. So, once you have all cycles problem is simply to eliminate cycles which do have a subset cycle.
For efficiency, for each cycle you find, loop through all existing cycles and verify that it is not a subset of another cycle or vice versa, and if so, eliminate the larger cycle.
Beyond that, only difficulty is figuring out how to write an algorithm that determines if a set is a subset of another.