What is a method of making another graph featuring vertices that can be only gotten to with an edge length of l from every vertex V in the original unweighted (assume edges of length 1) and undirected graph G=(V,E). I came up with a solution that just searches through each branch from each V using depth-first search on each vertex until I found all the vertices of path length l from each vertex. This gives a runtime of O(V^(l+1)) so of course, this is not the optimal solution. Can anyone help me find a better solution with a better asymptotic runtime?
You could use the Floyd-Warshall algorithm that uses a matrix representation (as #Hammar suggested) but finishes in O(V^3) regardless of l. Instead of l matrix exponentiations, you determine all distances by sequentially inserting nodes and determining the effect on the shortest paths.
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Given a undirected weighted graph with n vertices and m edges. How do I construct an algorithm which takes at most O((n+m) log(n+m)) for finding the vertex which it's min shortest path distance to a set of vertices S \in V is maximized?
I know I can loop through all vertices and using dijkstra's algorithm find the shortest path to each of the vertices in S but that will surely take much more than O((n+m) log(n+m)).
I am assuming from your ideas (possible use of dijkstra's shortest path algorithm) that weight of every edge is greater or equal than zero.
A clean solution to your problem is to create a new vertex v and add zero weight edges between v and every vertex in S. Run dijkstra's algorithm from v and the answer you look for is the farthest vertex from v.
You can prove by yourself that solution of your problem is equal to the one calculated on the new graph as it is straighforward. Also, it only run dijkstra's algorithm once so it fits on your time complexity constraints.
I have directed Graph G(V,E) with weight function w. so that weight of each (u,v) is a positive value. I need to find the most lightweight circle in the graph that vertex k' is part of it.
I've also given an algorithm i can use which can find the most lightweight path for a graph with positives weights ( i can use it only once).
I thought about creating a sub graph G' where all vertices and edges that are strongly connected components. find the graph which k' is part of it. then find for the most lightweight adjacent edge from k' to some v of vertices. from that v i can run the algorithm given and find the lightweight path then add the weight of the vertex missing ( (k',v) ).
is that seems correct ? I'm in the beginning of this course and I feel i'm not there yet.
It is a single-source shortest-path problem, where you exclude k->k self-loop as a solution, and find a longer path from k to k. The trick is always expand the shortest path thread.
Given this definition, you can start Googling...
I can't imagine why you called your source vertex k'. Anyway...
Add a new vetrex w that has the same outgoing edges as k'.
Then use Dijkstra's algorithm to find the shortest path from w to k'.
Substitute k' for w in the path, and you have the smallest cycle including k'.
Very interesting problem. I am assuming that there are no negative values in the graph, or otherwise the following solutions requires first normalizing the vertices such that the negative values become at least 0. First method (trivial) is to detect all cycles starting from the target vertex (k). Then compute the weight of all those cycles and take the minimum. The second method is to run Dijkstra algorithm (again watch out negative weights) from the target node (k). Then iterate over all incoming edges of (k), and select the source node that has the minimum Dijkstra value. Now the lightest cycle includes the single path (formed by Dijkstra traversal) from (k) to the chosen node + the bridge to come back to (k). I hope that helps :)
Hi I am struggling with this question. It is the following:
Devise an algorithm to find the lowest-weight cycle(i.e. of all cycles in the graph, the one with the smallest sum of edge weights), in a weighted, directed graph G = (V,E). Briefly justify the runtime and space complexity. Assume all edges are non-negative. It should run in O(|V||E|log|V|) time.
Hint: Use multiple calls to Dijkstra's algorithm.
I have seen solutions that use Floyd-Warshall but I was wondering how we would do this using Dijkstra's and how to do it within the time constraint given.
I have a few points of confusion:
Firstly, how do we even know how many cycles are in the graph and how
to check those?
Also, why is it |E||V|log|V|? By my understanding you should traverse
through all the vertices therefore making it |V|log|V|.
This is for my personal learning so if anyone has an example they could use, it would greatly help me! I am not really looking for pseudo-code - just a general algorithm to understand how using the shortest path from one node to all nodes is going to help us solve this problem.
Thank you!
Call Dijkstra's algorithm from each vertex to find the shortest path to itself, if one exists. The shortest path from any vertex to itself is the smallest cycle.
Dijkstra's algorithm takes O(|E| log |V|), so total time is O(|V||E| log |V|).
Note that this time can be worse than Floyd-Warshall, because there can be O(|V|^2) edges in the graph.
Call Dijkstra's algorithm |V| times, using each vertex in V as the start vertex. Store the results in a matrix, where dist(u,v) is the shortest path length from u to v.
For each pair of vertices (u,v), dist(u,v) is the shortest parth from u to v and dist(v,u) is the shortest path from v to u. Therefore, dist(u,v) + dist(v, u) is the length of the lowest weight cycle containing u and v. For each pair of vertices, computer this value and take the minimum. This is the lowest weigtht cycle.
I am having problem with shortest path in directed weighted graph. I know Dijkstra, BFS, DFS. However, I have a set of vertices S for starting points and a set of vertices E to end. S and E doesn't overlap. So how can I find the set of edges with minimal sum of edge weight? The edge set doesn't have to include all vertices in S, but have to reach all vertices in E. Should I start with Dijkstra on all permutation of {Si, Ei} and optimize or I miss any important algorithm I should know? Or even I am over-thinking....
If I understand you correctly, you want to find the tree of minimal weight in the graph that contains all the vertices of E and at least one vertex from S.
The problem is called general Steiner tree, and it is NP-hard. So the best you can probably hope for is an exponential-time algorithm or some kind of approximation (the minimum spanning tree of the whole graph comes to mind, maybe after removing some unneeded subtrees).
There is a simple DP solution that works in O(2^n * (n + m)): Let f(S) be the cost of the minimum tree in the graph that spans all the nodes in S. It can be shown that there is such a tree T such that the weight of T \ {x} is f(S \ {x}) for some x, so the transition can be done in O(n + m).
I have an unweighted, connected graph. I want to find a connected subgraph that definitely includes a certain set of nodes, and as few extras as possible. How could this be accomplished?
Just in case, I'll restate the question using more precise language. Let G(V,E) be an unweighted, undirected, connected graph. Let N be some subset of V. What's the best way to find the smallest connected subgraph G'(V',E') of G(V,E) such that N is a subset of V'?
Approximations are fine.
This is exactly the well-known NP-hard Steiner Tree problem. Without more details on what your instances look like, it's hard to give advice on an appropriate algorithm.
I can't think of an efficient algorithm to find the optimal solution, but assuming that your input graph is dense, the following might work well enough:
Convert your input graph G(V, E) to a weighted graph G'(N, D), where N is the subset of vertices you want to cover and D is distances (path lengths) between corresponding vertices in the original graph. This will "collapse" all vertices you don't need into edges.
Compute the minimum spanning tree for G'.
"Expand" the minimum spanning tree by the following procedure: for every edge d in the minimum spanning tree, take the corresponding path in graph G and add all vertices (including endpoints) on the path to the result set V' and all edges in the path to the result set E'.
This algorithm is easy to trip up to give suboptimal solutions. Example case: equilateral triangle where there are vertices at the corners, in midpoints of sides and in the middle of the triangle, and edges along the sides and from the corners to the middle of the triangle. To cover the corners it's enough to pick the single middle point of the triangle, but this algorithm might choose the sides. Nonetheless, if the graph is dense, it should work OK.
The easiest solutions will be the following:
a) based on mst:
- initially, all nodes of V are in V'
- build a minimum spanning tree of the graph G(V,E) - call it T.
- loop: for every leaf v in T that is not in N, delete v from V'.
- repeat loop until all leaves in T are in N.
b) another solution is the following - based on shortest paths tree.
- pick any node in N, call it v, let v be a root of a tree T = {v}.
- remove v from N.
loop:
1) select the shortest path from any node in T and any node in N. the shortest path p: {v, ... , u} where v is in T and u is in N.
2) every node in p is added to V'.
3) every node in p and in N is deleted from N.
--- repeat loop until N is empty.
At the beginning of the algorithm: compute all shortest paths in G using any known efficient algorithm.
Personally, I used this algorithm in one of my papers, but it is more suitable for distributed enviroments.
Let N be the set of nodes that we need to interconnect. We want to build a minimum connected dominating set of the graph G, and we want to give priority for nodes in N.
We give each node u a unique identifier id(u). We let w(u) = 0 if u is in N, otherwise w(1).
We create pair (w(u), id(u)) for each node u.
each node u builds a multiset relay node. That is, a set M(u) of 1-hop neigbhors such that each 2-hop neighbor is a neighbor to at least one node in M(u). [the minimum M(u), the better is the solution].
u is in V' if and only if:
u has the smallest pair (w(u), id(u)) among all its neighbors.
or u is selected in the M(v), where v is a 1-hop neighbor of u with the smallest (w(u),id(u)).
-- the trick when you execute this algorithm in a centralized manner is to be efficient in computing 2-hop neighbors. The best I could get from O(n^3) is to O(n^2.37) by matrix multiplication.
-- I really wish to know what is the approximation ration of this last solution.
I like this reference for heuristics of steiner tree:
The Steiner tree problem, Hwang Frank ; Richards Dana 1955- Winter Pawel 1952
You could try to do the following:
Creating a minimal vertex-cover for the desired nodes N.
Collapse these, possibly unconnected, sub-graphs into "large" nodes. That is, for each sub-graph, remove it from the graph, and replace it with a new node. Call this set of nodes N'.
Do a minimal vertex-cover of the nodes in N'.
"Unpack" the nodes in N'.
Not sure whether or not it gives you an approximation within some specific bound or so. You could perhaps even trick the algorithm to make some really stupid decisions.
As already pointed out, this is the Steiner tree problem in graphs. However, an important detail is that all edges should have weight 1. Because |V'| = |E'| + 1 for any Steiner tree (V',E'), this achieves exactly what you want.
For solving it, I would suggest the following Steiner tree solver (to be transparent: I am one of the developers):
https://scipjack.zib.de/
For graphs with a few thousand edges, you will usually get an optimal solution in less than 0.1 seconds.