Section 4 of this paper states that it is easy to see that A* is equivalent to dijkstra used on a modified graph where edge weights add the potential from the outgoing node and subtract the potential from the ingoing node. I can see that the priority queue would have the same order between the two algorithms since the additional terms telescope away, but I don't think that constitute a proof of the equivalence. Can someone enlighten me?
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I have a complete, weighted, undirected graph. The edge weights are the cost of a connection between two nodes, so the minimum spanning tree is the subset of the edges with the lowest total cost such that the graph remains connected.
The MST must be connected at all times, but unfortunately the connections aren't very reliable, so I would like to add redundancy to this graph/network.
Is it possible to compute a subset of edges such that the total edge cost is minimised and edge-connectivity is over a certain minimum?
I can see how it would be possible by bruteforcing, but I was looking for something more practical. I haven't been able to find much about this problem, I think mainly because I don't posses the vocabulary necessary to search.
My current idea is:
Compute the MST
While the it is still below a certain connectivity
Find a node most below that connectivity
Activate that node's edge with the lowest weight
The reason I don't find all the nodes below a certain connectivity all at once is because activating an edge may give another one enough connectivity.
I'm pretty sure this does not yield 100% provably optimal networks, because with this method, it is possible to over-connect nodes (e.g. you activate k edges for a node, then another node activates more shared edges, making some of those k redundant). I hope that makes sense.
Any tips would be much appreciated!
The Wikipedia article on edge connected graphs ends with, A related problem: finding the minimum k-edge-connected spanning subgraph of G (that is: select as few as possible edges in G that your selection is k-edge-connected) is NP-hard for k >= 2. They then cite a 1979 paper that shows it.
Therefore I'd suggest taking a greedy approach, and tip-toeing away.
Assume we're given a graph on a 2D-plane with n nodes and edge between each pair of nodes, having a weight equal to a euclidean distance. The initial problem is to find MST of this graph and it's quite clear how to solve that using Prim's or Kruskal's algorithm.
Now let's say we have k extra nodes, which we can place in any integer point on our 2D-plane. The problem is to find locations for these nodes so as new graph has the smallest possible MST, if it is not necessary to use all of these extra nodes.
It is obviously impossible to find the exact solution (in poly-time), but the goal is to find the best approximate one (which can be found within 1 sec). Maybe you can come up with some hints of the most efficient way of going throw possible solutions, or provide with some articles, where the similar problem is covered.
It is very interesting problem which you are working on. You have many options to attack this problem. The best known heuristics in such situation are - Genetic Algorithms, Particle Swarm Optimization, Differential Evolution and many others of this kind.
What is nice for such kind of heuristics is that you can limit their execution to a certain amount of time (let say 1 second). If it was my task to do I would try first Genetic Algorithms.
You could try with a greedy algorithm, try the longest edges in the MST, potentially these could give the largest savings.
Select the longest edge, now get the potential edge from each vertex that are closed in angle to the chosen one, from each side.
from these select the best Steiner point.
Fix the MST ...
repeat until 1 sec is gone.
The challenge is what to do if one of the vertexes is itself a Steiner point.
I have questions about an optimal algorithm problem on a weighted graph. I am given an edgelist with weights, a list with savepoints, a starte- and end- node and the max distance for a step.
The output should be a list of savepoints, which are accessible in one step from starting- and end- node.
I thought of some kind of dijkstra's algorithm from each point of the list of savepoints.
I'm not sure if that's a good idea, since if I have many savepoints I calculate a lot of paths multiple times. Every idea/help is welcome!
Thank you very much in advance!
You have to have the condition that a weight cannot be negative, otherwise the problem becomes very intractable. Otherwise it's just a breadth first search, with marking the distance for every visited node. So you don't revisit a node is a previous move has visited it earlier at lower cost.
You keep a priority queue of all active nodes, so you are checking the lowest cost node each time. The priority queue is in fact the hardest part to get right. If you check the A* algorithm for my binary image library https://github.com/MalcolmMcLean/binaryimagelibrary you can take the priority queue for there. A* over a maze is very similar to shortest path over a graph, but you don't have a heuristic because you must have the exact shortest path, and instead of 4 / 8 edges per tile, you have nodes with arbitrary numbers of connections.
For example, we have a graph consisting of vertices (cities) and edges (roads) and each edge(road) has a particular cost, find the minimal cost to visit all cities ATLEAST ONCE. Cost is the sum of the edge costs of the edges traversed.
The part "ATLEAST ONCE" caught me. In a TSP we can visit a node only once according to Wiki. Consider the graph,
A-B 11
A-C 5
B-C 2
B-E 4
C-E 3
C-D 20
D-E 100
In a TSP, The cyclic path would be A-B-E-D-C-A cost- 140 (or) A-C-D-E-B-A cost- 140. Where as from my problem description we can visit each vertex ATLEAST ONCE so we can have a cyclic path A-C-D-C-E-B-A cost- 63 which is << a TSP. This is where I had a problem. Any specific algorithm here? I'm pretty sure TSP wont work well here.
Pointers or pseudo code will be very helpful.
For each pair of nodes, you can apply the shortest path algorithm and calculate the shortest distance. This will be the new cost matrix for each pair.
Now it is reduced to Travelling Salesman Problem.
Then you can apply TSP solving technique.
Given that you are allowing a vertex to be visited multiple times, this effectively turns your incomplete graph into a complete graph (all vertices connected), which is what TSP requires. Solving your problem in the general case is exactly the same as solving the metric TSP. The good news is that this is a heavily researched topic. The bad news is that you aren't able to sidestep the TSP - since your problem is identical to a form of the TSP.
As pointed out by others, you complete the graph by computing the shortest cost between each pair of vertices and adding those edges where missing. You also need to replace any existing direct edge for which you've found a lower indirect path cost so that you have a Metric TSP. You can store with the new synthetic edges their actual paths (through intermediate vertices) so you can recover those for your final answer, or you can recompute those paths as needed upon receiving the result of the TSP.
Now you can solve this as a TSP. However, solving TSP optimally is too expensive in the general case, so you'll likely want to use an approximate solution algorithm. A variety of these (e.g. Christofides algorithm, Lin–Kernighan heuristic) are available which make differing tradeoffs between guaranteed levels of optimality and performance of the algorithm.
If you actually don't care about completing the cycle, and just want a minimum path that visits all vertices, starting and ending at any vertex, this is a somewhat different problem. For this, read my answer here: https://stackoverflow.com/a/33601043/5237297
I am looking for a way to perform a topological sorting on a given directed unweighted graph, that contains cycles. The result should not only contain the ordering of vertices, but also the set of edges, that are violated by the given ordering. This set of edges shall be minimal.
As my input graph is potentially large, I cannot use an exponential time algorithm. If it's impossible to compute an optimal solution in polynomial time, what heuristic would be reasonable for the given problem?
Eades, Lin, and Smyth proposed A fast and effective heuristic for the feedback arc set problem. The original article is behind a paywall, but a free copy is available from here.
There’s an algorithm for topological sorting that builds the vertex order by selecting a vertex with no incoming arcs, recursing on the graph minus the vertex, and prepending that vertex to the order. (I’m describing the algorithm recursively, but you don’t have to implement it that way.) The Eades–Lin–Smyth algorithm looks also for vertices with no outgoing arcs and appends them. Of course, it can happen that all vertices have incoming and outgoing arcs. In this case, select the vertex with the highest differential between incoming and outgoing. There is undoubtedly room for experimentation here.
The algorithms with provable worst-case behavior are based on linear programming and graph cuts. These are neat, but the guarantees are less than ideal (log^2 n or log n log log n times as many arcs as needed), and I suspect that efficient implementations would be quite a project.
Inspired by Arnauds answer and other interesting topological sorting algorithms have I created the cyclic-toposort project and published it on github. cyclic_toposort does exactly what you desire in that it quickly sorts a directed cyclic graph providing the minimal amount of violating edges. It optionally also provides the maximum groupings of nodes that are on the same topological level (and can therefore be activated at the same time) if desired.
If the problem is still relevant to you then I would be happy if you check out my project and let me know what you think!
This project was useful to my own neural network topology research, so I had to create something like this anyway. I am answering your question this late in case anyone else stumbles upon this thread in search for the same question.