How to detect if the given graph has a cycle containing all of its nodes? Does the suggested algorithm have any flaws?

Question

I have a connected, non-directed, graph with N nodes and 2N-3 edges. You can consider the graph as it is built onto an existing initial graph, which has 3 nodes and 3 edges. Every node added onto the graph and has 2 connections with the existing nodes in the graph. When all nodes are added to the graph (N-3 nodes added in total), the final graph is constructed.

Originally I'm asked, what is the maximum number of nodes in this graph that can be visited exactly once (except for the initial node), i.e., what is the maximum number of nodes contained in the largest Hamiltonian path of the given graph? (Okay, saying largest Hamiltonian path is not a valid phrase, but considering the question's nature, I need to find a max. number of nodes that are visited once and the trip ends at the initial node. I thought it can be considered as a sub-graph which is Hamiltonian, and consists max. number of nodes, thus largest possible Hamiltonian path).

Since i'm not asked to find a path, I should check if a hamiltonian path exists for given number of nodes first. I know that planar graphs and cycle graphs (C_n) are hamiltonian graphs (I also know Ore's theorem for Hamiltonian graphs, but the graph I will be working on will not be a dense graph with a great probability, thus making Ore's theorem pretty much useless in my case). Therefore I need to find an algorithm for checking if the graph is cycle graph, i.e. does there exist a cycle which contains all the nodes of the given graph.

Since DFS is used for detecting cycles, I thought some minor manipulation to the DFS can help me detect what I am looking for, as in keeping track of explored nodes, and finally checking if the last node visited has a connection to the initial node. Unfortunately I could not succeed with that approach.

Another approach I tried was excluding a node, and then try to reach to its adjacent node starting from its other adjacent node. That algorithm may not give correct results according to the chosen adjacent nodes.

I'm pretty much stuck here. Can you help me think of another algorithm to tell me if the graph is a cycle graph?

Edit

I realized by the help of the comment (thank you for it n.m.):

A cycle graph consists of a single cycle and has N edges and N vertices. If there exist a cycle which contains all the nodes of the given graph, that's a Hamiltonian cycle. – n.m.

that I am actually searching for a Hamiltonian path, which I did not intend to do so:) On a second thought, I think checking the Hamiltonian property of the graph while building it will be more efficient, which is I'm also looking for: time efficiency.

After some thinking, I thought whatever the number of nodes will be, the graph seems to be Hamiltonian due to node addition criteria. The problem is I can't be sure and I can't prove it. Does adding nodes in that fashion, i.e. adding new nodes with 2 edges which connect the added node to the existing nodes, alter the Hamiltonian property of the graph? If it doesn't alter the Hamiltonian property, how so? If it does alter, again, how so? Thanks.

EDIT #2

I, again, realized that building the graph the way I described might alter the Hamiltonian property. Consider an input given as follows:

1 3
2 3
1 5
1 3

these input says that 4th node is connected to node 1 and node 3, 5th to node 2 and node 3 . . .

4th and 7th node are connected to the same nodes, thus lowering the maximum number of nodes that can be visited exactly once, by 1. If i detect these collisions (NOT including an input such as 3 3, which is an example that you suggested since the problem states that the newly added edges are connected to 2 other nodes) and lower the maximum number of nodes, starting from N, I believe I can get the right result.

See, I do not choose the connections, they are given to me and I have to find the max. number of nodes.

I think counting the same connections while building the graph and subtracting the number of same connections from N will give the right result? Can you confirm this or is there a flaw with this algorithm?

Banville

Why would planar graphs be hamiltonian? Any tree is planar, but certainly not hamiltonian. – Shira

might be better suited for math.stackexchange.com – Cislunar

I would say something along these lines: ostermiller.org/find_loop_singly_linked_list.html – Regal

Could also go for cs.stackexchange.com – Regal

A cycle graph consists of a single cycle and has N edges and N vertices. If there exist a cycle which contains all the nodes of the given graph, that's a Hamiltonian cycle. – Brock

Do you mean every node outside the initial triangle is added iteratively, and the k-th node added has exactly two edges to the k-1+3 previous nodes, or that each of the n-3 nodes added has two edges to the initial triangle? Also, could you explain what you mean when you say you will be dealing with "dense" graphs? Usually, dense graphs are defined as having |E| in Θ(|V|²). – Shira

I'm curious why DFS didn't work? Seems like you could sort nodes according to ascending edge count as a heuristic... – Alienor

@G.Bach Yes, they are iteratively added. K-th node has exactly 2 edges to the previous k-1+3 edges. For what i meant about dense graphs, please see this SO question. In this case, I think I should check the hamiltonian property while adding the edges instead of examining whole graph. Isn't it better? – Banville

@Volkanİlbeyli I misread your OP, I thought you said you will be dealing with dense graphs. – Shira

Answer 1

To add some clarification to this thread: finding a Hamiltonian Cycle is NP-complete, which implies that finding a longest cycle is also NP-complete because if we can find a longest cycle in any graph, we can find the Hamiltonian cycle of the subgraph induced by the vertices that lie on that cycle. (See also for example this paper regarding the longest cycle problem)

We can't use Dirac's criterion here: Dirac only tells us minimum degree >= n/2 -> Hamiltonian Cycle, that is the implication in the opposite direction of what we would need. The other way around is definitely wrong: take a cycle over n vertices, every vertex in it has exactly degree 2, no matter the size of the circle, but it has (is) an HC. What you can tell from Dirac is that no Hamiltonian Cycle -> minimum degree < n/2, which is of no use here since we don't know whether our graph has an HC or not, so we can't use the implication (nevertheless every graph we construct according to what OP described will have a vertex of degree 2, namely the last vertex added to the graph, so for arbitrary n, we have minimum degree 2).

The problem is that you can construct both graphs of arbitrary size that have an HC and graphs of arbitrary size that do not have an HC. For the first part: if the original triangle is A,B,C and the vertices added are numbered 1 to k, then connect the 1st added vertex to A and C and the k+1-th vertex to A and the k-th vertex for all k >= 1. The cycle is A,B,C,1,2,...,k,A. For the second part, connect both vertices 1 and 2 to A and B; that graph does not have an HC.

What is also important to note is that the property of having an HC can change from one vertex to the other during construction. You can both create and destroy the HC property when you add a vertex, so you would have to check for it every time you add a vertex. A simple example: take the graph after the 1st vertex was added, and add a second vertex along with edges to the same two vertices of the triangle that the 1st vertex was connected to. This constructs from a graph with an HC a graph without an HC. The other way around: add now a 3rd vertex and connect it to 1 and 2; this builds from a graph without an HC a graph with an HC.
Storing the last known HC during construction doesn't really help you because it may change completely. You could have an HC after the 20th vertex was added, then not have one for k in [21,2000], and have one again for the 2001st vertex added. Most likely the HC you had on 23 vertices will not help you a lot.

If you want to figure out how to solve this problem efficiently, you'll have to find criteria that work for all your graphs that can be checked for efficiently. Otherwise, your problem doesn't appear to me to be simpler than the Hamiltonian Cycle problem is in the general case, so you might be able to adjust one of the algorithms used for that problem to your variant of it.

Shira

I was the +1. I believe there is a way to check HP property while building the graph since its built based on a simple rule, other than the algorithms listed on Wikipedia. Don't you think so? – Banville

I'm not too sure. You can both create and destroy the HP property when you add a vertex, so you would have to check for it every time you add a vertex. Storing the last known HP during construction doesn't really help you because it may change completely. You could have an HP after the 20th vertex was added, then not have one for k in [21,2000], and have one again for the 2001st vertex added. Most likely the HP you had on 23 vertices will not help you a lot. – Shira

The thing is, I am looking for the number of maximum nodes that can be visited when you start from a random node, visit all (possible ones) others, and finish the trip where I started. Therefore, I, sort of, am looking for hamiltonian cycles in sub-graphs. From your example, I would be looking for hamiltonian cycle in the whole graph, which, in fact, i don't. Therefore an algorithm (I think i found one this time, will post it tomorrow after some testing) checking the added nodes and vertices, saving some problematic nodes in a list, and solving the problem this way exists. – Banville

See the first paragraph of my answer; if you can find a subgraph with a Hamiltonian Path (or Cycle) efficiently, then you can do so for the whole graph if it is Hamiltonian. – Shira

@Volkanİlbeyli Just to make this clear: I'm not saying that what you ask for can't be done efficiently, maybe it can; but it can't be done efficiently using any of the algorithms any of the answers give. If you want to solve this, you will have to find properties of the graphs that you operate on that make your problem simpler than the general Hamilton Path/Cycle problem. – Shira

Answer 2

What we have in this problem is a connected, non-directed graph with N nodes and 2N-3 edges. Consider the graph given below,

            A
           / \
          B _ C
         ( )  
          D 

The Graph does not have a Hamiltonian Cycle. But the Graph is constructed conforming to your rules of adding nodes. So searching for a Hamiltonian Cycle may not give you the solution. More over even if it is possible Hamiltonian Cycle detection is an NP-Complete problem with O(2^N) complexity. So the approach may not be ideal.

What I suggest is to use a modified version of Floyd's Cycle Finding algorithm (Also called the Tortoise and Hare Algorithm).

The modified algorithm is,

1. Initialize a List CYC_LIST to ∅.

2. Add the root node to the list CYC_LIST and set it as unvisited. 

3. Call the function Floyd() twice with the unvisited node in the list CYC_LIST for each of the two edges. Mark the node as visited.

4. Add all the previously unvisited vertices traversed by the Tortoise pointer to the list CYC_LIST.

5. Repeat steps 3 and 4 until no more unvisited nodes remains in the list.

6. If the list CYC_LIST contains N nodes, then the Graph contains a Cycle involving all the nodes.

The algorithm calls Floyd's Cycle Finding Algorithm a maximum of 2N times. Floyd's Cycle Finding algorithm takes a linear time ( O(N) ). So the complexity of the modied algorithm is O(N²) which is much better than the exponential time taken by the Hamiltonian Cycle based approach.

One possible problem with this approach is that it will detect closed paths along with cycles unless stricter checking criteria are implemented.

Reply to Edit #2

Consider the Graph given below,

            A------------\
           / \            \
          B _ C            \
          |\ /|             \
          | D |             F
          \   /            /
           \ /            / 
            E------------/

According to your algorithm this graph does not have a cycle containing all the nodes. But there is a cycle in this graph containing all the nodes.

A-B-D-C-E-F-A

So I think there is some flaw with your approach. But suppose if your algorithm is correct, it is far better than my approach. Since mine takes O(n²) time, where as yours takes just O(n).

Crossbreed

I think you misunderstood my algorithm. See, my algorithm first assumes you have N nodes in the hamiltonian cycle, and then starts to build up the graph (reads the input, actually, nothing is built up). It simply compares the inputs, and if it sees a collision, it decreases the # of nodes in the hamiltonian cycle. Thus, according to the data structure, it may take O(n²) time for completing the task. Now consider your example: the input file should be like this for that graph: 2 2 (B B) 2 3 (D C) However, i said new nodes are connected to 2 other nodes, meaning that... – Banville

the input file may NOT contain a line such as 2 2 (B B) in your case. Therefore, correcting the algorithm in the OP, eliminating the lines such as 2 2 in the input, my algorithm seems to work right, doesn't it? – Banville

This answer seems wrong as well; if I understand you correctly you suggest that you can detect a cycle covering all vertices of the graph in O(n^2), which would mean that the Hamiltonian Path problem is solvable in polynomial time. Anyway, Floyd's algorithm doesn't detect longest cycles (which would imply detecting Hamiltonian Paths if such are present) but only cycles. – Shira

@G. Bach: No I was not suggesting that P=NP :) . In this case there are some restrictions on adding new vertices in the graph. Only for such graphs my approach will have any effect. More over I have mentioned that it might fail if there is a closed path. – Crossbreed

Answer 3

Below I have added three extra nodes (3,4,5) in the original graph and it does seem like I can keep adding new nodes indefinitely while keeping the property of Hamiltonian cycle. For the below graph the cycle would be 0-1-3-5-4-2-0

  1---3---5
 / \ / \ /
0---2---4

As there were no extra restrictions about how you can add a new node with two edges, I think by construction you can have a graph that holds the property of hamiltonian cycle.

Lundberg

The thing is, I will not build the graph, the graph will be built according to given input. Therefore it can be as you described as well as something different. So I have to check while building it. I'll update the question, can you see it? – Banville