I have some interesting question that I would like to get your help:
Let's say I have a graph (Data structure) with n(n-1)/2 edges, which means, completed graph.
How many possible different DFS trees can I get from one DFS scan from some random element in the graph?
Your question is interesting. I believe you are talking about complete graph with n vertices and n(n-1)/2 edges in between them.
If we begin depth first search (DFS) from any vertex, it will end up visiting exactly n vertices. In DFS, we keep track of visited vertices so that we will not visit them once they are visited and hence outgoing option reduces as long as DFS progresses. We can summarize this as:
There are total n options to choose first vertex.
There are total n-1 options to choose second vertex as 1 vertex is already visited.
There are total n-2 options to choose third vertex as 2 vertices are already visited.
There are total n-3 options to choose third vertex as 3 vertices are already visited.
And so on . . .
There is only 1 option to choose nth vertex.
Hence, different possible DFS trees that we can get from DFS in such graph is :
Total ways = n*(n-1)*(n-2)*(n-3)*....*1
= n! ( n factorial )
In fact it's not a tree but a list of nodes of the given complete graph. Thus, the question is: How many permutations of n nodes of the graph? Apparently, the answer is n!.
Related
I know that there are a ton of questions out there about the time complexity of BFS which is : O(V+E)
However I still struggle to understand why is the time complexity O(V+E) and not O(V*E)
I know that O(V+E) stands for O(max[V,E]) and my only guess is that it has something to do with the density of the graph and not with the algorithm itself unlike say Merge Sort where it's time complexity is always O(n*logn).
Examples I've thought of are :
A Directed Graph with |E| = |V|-1 and yeah the time complexity will be O(V)
A Directed Graph with |E| = |V|*|V-1| and the complexity would in fact be O(|E|) = O(|V|*|V|) as each vertex has an outgoing edge to every other vertex besides itself
Am I in the right direction? Any insight would be really helpful.
Your "examples of thought" illustrate that the complexity is not O(V*E), but O(E). True, E can be a large number in comparison with V, but it doesn't matter when you say the complexity is O(E).
When the graph is connected, then you can always say it is O(E). The reason to include V in the time complexity, is to cover for the graphs that have many more vertices than edges (and thus are disconnected): the BFS algorithm will not only have to visit all edges, but also all vertices, including those that have no edges, just to detect that they don't have edges. And so we must say O(V+E).
The complexity comes off easily if you walk through the algorithm. Let Q be the FIFO queue where initially it contains the source node. BFS basically does the following
while Q not empty
pop u from Q
for each adjacency v of u
if v is not marked
mark v
push v into Q
Since each node is added once and removed once then the while loop is done O(V) times. Also each time we pop u we perform |adj[u]| operations where |adj[u]| is the number of
adjacencies of u.
Therefore the total complexity is Sum (1+|adj[u]|) over all V which is O(V+E) since the sum of adjacencies is O(E) (2E for undirected graph and E for a directed one)
Consider a situation when you have a tree, maybe even with cycles, you start search from the root and your target is the last leaf of your tree. In this case you will traverse all the edges before you get into your destination.
E.g.
0 - 1
1 - 2
0 - 2
0 - 3
In this scenario you will check 4 edges before you actually find a node #3.
It depends on how the adjacency list is implemented. A properly implemented adjacency list is a list/array of vertices with a list of related edges attached to each vertex entry.
The key is that the edge entries point directly to their corresponding vertex array/list entry, they never have to search through the vertex array/list for a matching entry, they can just look it up directly. This insures that the total number of edge accesses is 2E and the total number of vertex accesses is V+2E. This makes the total time O(E+V).
In improperly implemented adjacency lists, the vertex array/list is not directly indexed, so to go from an edge entry to a vertex entry you have to search through the vertex list which is O(V), which means that the total time is O(E*V).
Given an unoriented tree with weightless edges with N vertices and N-1 edges and a number K find K nodes so that every node from a tree is within S distance of at least one of the K nodes. Also, S has to be the smallest possible S, so that if there were S' < S at least one node would be unreachable in S' steps.
I tried solving this problem, however, I feel that my supposed solution is not very fast.
My solution:
set x=1
find nodes which are x distance from every node
let the node which has the most nodes in its distance be one of the K nodes.
recompute for every node whilst not counting already covered nodes.
do this till I find K number of K nodes. Then if every node is covered we are done else increase x.
This problem is called p-center, and you can find several papers online about it such as this. It is indeed NP for general graphs, but polynomial on trees, both weighted and unweighted.
For me it looks like a clustering problem. Try it with the k-Means (wikipedia) algorithm where k equals to your K. Since you have a tree and all vertices are connected, you can use as distance measurement the distance/number of edges between your vertices.
When the algorithm converts you get the K nodes which should be found. Then you can determine S by iterating through all k clusters. There you calculate the maximum distance for every node in the cluster to the center node. And the overall max should be S.
Update: But actually I see that the k-means algorithm does not produce a global optimum, so this algorithm wouldn't also produce the best result ...
You say N nodes and N-1 vertices so your graph is a tree. You are actually looking for a connected K-subset of nodes minimizing the longest edge.
A polynomial algorithm may be:
Sort all your edges increasing distance.
Then loop on edges:
if none of the 2 nodes are in a group, create a new group.
else if one node is in 1 existing goup, add the other to the group
else both nodes are in 2 different groups, then fuse the groups
When a group reach K, break the loop and you have your connected K-subset.
Nevertheless, you have to note that your group can contain more than K nodes. You can imagine the problem of having 4 nodes, closed two by two. There would be no exact 3-subset solution of your problem.
I got this question from a data structure and algorithm textbook saying
A simple undirected graph is complete if it contains an edge between every pair of distinct vertices. A star graph is a tree of n nodes with one node having vertex degree n-1 and the other n-1 having vertex degree 1.
(a) Draw a complete undirected graph with 6 vertices.
(b) Show that applying breath-first algorithm on the undirected graph in (a) will produce a star graph.
I know how the BFS works using queues and I can provide a result of the traversal. What I'm confused about is on part (b) how can I show that applying BFS on an undirected graph will produce a star graph?
In a, there is n * (n - 1) / 2 edges in total.
It means that for every two nodes, there is an edge between them.
if applying BFS on (a) using a queue, Steps as follow:
1.) you pick up a random node, which is the root of the graph.
2.) you travel from the root node, and put all the nodes having edges with it to the queue. In addition, you have a boolean array to mark who is already processed.
in 2.), all the nodes except the root will be put into the queue.
At last, the root node have N - 1 edges, others have only one edge, and this edge is with the root
What is the maximum and minimum number of edges to be considered in krushkal's algorithm with an example for both cases.
What I thought was since the Krushkal's algorithm is for finding minimum spanning tree the maximum number of edges is (V-1) where V is the number of vertices. Adding one more edge would result in a cycle in the graph. How can we obtain at a minimum value ?
Kruskal's algorithm stops when you've added V - 1 edges to your MST, so this is the minimum that have to be considered. This happens when the lowest value V - 1 edges of your graph do not form a cycle, and they will be added one after the other by the algorithm, after which it will stop.
For example, consider a complete graph with edges with cost 1, which is minimum in the graph, between node 1 and every other node. Make all the other edges have cost 2.
The worst case is when you must inspect every edge (of which there are O(V^2)) until you finally select V - 1. This means that you have to force a lot of cycles to be created before the last edge is added.
Consider a complete graph again. Have the V - 2 edges between node 1 and V - 2 nodes have cost 1, which is minimum in the graph. These will be selected first. Now let node k be the one that is not part of a selected edge, so that is left out of the graph. Have the edge between node k and node 1 have the largest cost. This will cause it to be inspected and added to the MST last, forcing the algorithm to inspect all O(V^2) edges before building the MST.
Remember the Kruskal's algorithm processes edges in increasing order of their cost, rejecting edges that would form a cycle if added to the MST we are building.
A tree of N vertices always has N-1 edges. Consequently you have to consider at least N-1 edges during Kruskal's algorithm. An example may be a graph which is a tree.
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 (Cn) 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?
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(2N) 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).
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.
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.