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Suppose I have a weighted graph with n vertices and the starting point is given. The shortest path is defined as the path with the least sum of weights.
How can I find out the shortest path that passes through m different vertices ? (each vertex can be visited once or more than once. That is, there are exactly m vertices in the set of the vertices that have been visited, but each vertex may have been visited multiple times.)
Note that the number m is given but the specific m vertices are not. (These m vertices are selected by algorithm）
Is it an NP-Hard problem?
We can reduce the Hamiltonian Path Problem (HPP) to your problem: a Hamiltonian Path is a path in a directed unweighted graph that visits each vertex exactly once. To solve an instance of HPP, convert the graph into a weighted graph with weight 1 on every edge, set m to be |V| and solve your problem. This is a polynomial-time reduction, so your problem is NP-hard, since HPP is NP-complete.
It is also NP-complete since it is clearly in NP. So there is also a polynomial-time reduction from your problem to any other NP-complete problem. Algorithms for solving the Travelling Salesman Problem are probably most suited for you: see here for details and examples.
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How can the Floyd-Warshall algorithm be modified to find the shortest path of any negative cost cycle of a directed graph that maintains O(V^3) time complexity?
There is no shortest path in a graph with negative cycle, for every path - one can find a shorter one by traversing the cycle one more time.
If you are referring to Shortest Simple Path (each vertex can be visited at most once) - it cannot be done, unless P=NP, and it most likely isn't.
Assume you have an efficient shortest simple path algorithm A.
Given an instance of the Longest Path Problem and a graph G=(V,E,w), create a new graph G'=(V,E,w') where w'(u,v) = -1*w(u,v). Now invoke A on G', and you got the shortest simple path on G' - which is the longest path on G.
However, since Longest Path is NP-Hard - such a solution is not possible unless P=NP.
tl;dr:
In a graph with negative cycle, there is no such thing as shortest path.
You cannot find a shortest simple path in a graph with negative cycles in O(V^3) time (unless P=NP, and even then it's not sure to be O(V^3)).
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N-numbers , d1,d2,d3..dn are given.
How do we check if it is possible to construct a undirected graph with vertices v1,v2,v3,...vn with degress d1,d2,...dn respectively.
Graph should not contain multiple edges between the same pair of nodes, or "loop" edges
(where both end vertices are the same node).
Also, what is the running time of the algorithm ?
This is what Wikipedia calls the graph realization problem, solvable by the Havel--Hakimi algorithm. Start with a graph having n vertices, v1..vn, and 0 edges. Define the deficit of a vertex vk to be the difference between dk and the current degree of vk. Repeatedly choose the vertex vk with the largest deficit D and connect it to the D other vertices having the D largest deficits. If a vertex would have negative deficit, then the instance is unsolvable. Otherwise, we terminate with a solution. I'll leave the running time as an exercise.
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I know that the problem of checking whether given edge of a weighted digraph belongs to a negative cycle is NP-complete (Finding the minimal subgraph that contains all negative cycles) and Bellman-Ford allows to check a vertex for the same thing in O(|V|*|E|) time. But what if I want to find all vertices belonging to negative cycles? I wonder if it could be done faster than Floyd-Warshall's O(|V|^3).
I don't think Floyd-Warshall does the job of finding these vertices. Using a similar approach as taken in the post you're referring to, it can be shown that finding the set of all vertices that lie on a negative cycle is NP-complete as well.
The related post shows that one may use the algorithm to find the set of all edges that lie on a negative cycle to solve the hamiltonian cycle problem, which means that the former problem is NP-complete.
If we can reduce the problem of finding all edges that lie on a negative cycle to the problem of finding the set of all vertices that lie on a negative cycle, we've shown NP-completeness of the latter problem.
For each edge (u,w) in your weighted digraph, introduce a new auxiliary vertex v, and split (u, w) in two edges (u, v) and (v, w). The weight of (u, w) can be assigned to either (u, v) or (v, w).
Now apply the magic polynomial-time algorithm to find all the vertices that lie on a negative cycle, and take the subset that consists of the auxiliary vertices. Since each auxiliary vertex is associated with an edge, we've solved the problem of finding the minimal subgraph that contains all negative cycles, and we can thus also solve the hamiltonian cycle problem in polynomial time, which implies P = NP. Assuming P != NP, finding all vertices that lie on a negative cycle is NP-complete.
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I have a simple question on DFS and I'm trying to understand how to use it and not how to solve the whole problem. I'm really looking for an explanation and not a solution to my homework.
I'll write down the question first.
"Suppose you have an undirected graph G=(V,E) and let three of its
vertices to be called v1, v2 and v3. Find an algorithm which
determines if these three vertices are part of a clique
(complete graph) (k>=3)"
Now I suppose to use DFS in order to solve it. As far as I understand DFS will let me know if v1, v2 and v3 are in the same strongly connected component. If I'm correct I should also determine if G is also a clique(complete graph).
I read in the internet and I found out that asserting if a graph is clique or not is NP and cannot be solved easily. Am I correct? Am I missing anything? Is there any propery I can use to determine immediately if a graph is comeplete ?
To clarify the confusion about the NP-completeness: checking whether a graph is a clique is not NP-complete; just count the edges and see whether there are n(n-1)/2. What is NP-complete is to find a maximum clique (meaning the subgraph that has the biggest number of vertices and is a clique) or a clique of k vertices in a graph of n vertices (if k is part of the input instead of a fixed number); the latter case is called the clique decision problem.
EDIT: I just realized you asked something regarding strongly connected components as well; that term only applies to directed graphs (i.e. the edges have a direction, which means for two vertices v and w, the edge v->w is not the same as the edge w->v). Cliques are commonly defined on undirected graphs, for which there are only connected components.
If I understood it properly, all you have to check whether these three vertices are connected, i.e., the edges v1-v2, v2-v3 and v3-v1 exists. If they exist, they form a clique of K=3. If at least one of them does not, these three vertices together can not be in a clique of size k>=3.
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I am wring thesis about shortest path algorithms.
And i don't understand one thing...
I have made visualisation of dijkstras algorithm.
1) Is it correct ? Or am i doing something wrong?
2) How would look Bellman-Ford algorithm? As fas as i have looked for difference, i found "Bellman-ford: the basic idea is very similar to Dijkstra's, but instead of selecting the shortest distance neighbour edges, it select all the neighbour edges." But also dijkstra checks all vertexes and all edges, isnt it?
dijkstra assumes that the cost of paths is montonically increasing. that plus the ordered search (using the priority queue) mans that when you first reach a node, you have arrived via the shortest path.
this is not true with negative weights. if you use dijkstra with negative weights then you may find a later path is better than an earlier one (because a negative weight improved the path on a later step).
so in bellman-ford, when you arrive at a node you test to see if the new path is shorter. in contrast, with dijkstra, you can cull nodes
in some (most) cases dijkstra will not explore all complete paths. for example, if G linked only back to C then any path through G would be higher cost that any through C. bellman-ford would still consider all paths through G to F (dijkstra would never look at those because they are of higher cost that going through C). if it does not do this it can't guarantee finding negative loops.
here's an example: the above never calculates the path AGEF. E has already been marked as visited by the time you arrive from G.
I am also thinking the same
Dijkstra's algorithm solves the single-source shortest-path problem when all edges have non-negative weights. It is a greedy algorithm and similar to Prim's algorithm. Algorithm starts at the source vertex, s, it grows a tree, T, that ultimately spans all vertices reachable from S. Vertices are added to T in order of distance i.e., first S, then the vertex closest to S, then the next closest, and so on.