this is a problem I've got from my class (I assure you it's not homework). I'm still pondering about it until now. You will receive a graph with at most 25 nodes and 25 edges. Additionally, each node will have degree of at most 3. The task is to find the longest path in this graph. However, you won't only receive 1 graph, but 15,000 graphs, and you'll need to find the longest path in all of them in 1 second. Could anyone please give me a solution (or better yet, just a hint) to this problem? Thank you very much!
Info:- Nodes can be revisited, the only constraints are the edges.- The graphs are given by the edges. So the first line states how many nodes and edges there are, and the lines after that represent the edges, each edge by a pair of integers.- The edges are unweighted.- The only answer required is the length of the path, not the path itself.- This might be important: the graph isn't necessarily connected.
After I saw that "nodes can be revisited", I realised that this is in some ways a trick question. To satisfy those seemingly unbelievable time constraints, what you actually need here is not an algorithm for constructing such a path (usually called a trail, BTW, if vertices can be reused), but rather, for each connected component of the graph, a way of quickly detecting whether all or nearly all edges in that component can be included in a single trail.
So here is my hint: Did you know that there are seven bridges in Königsberg? ;-)
That might seem cryptic, but I think some quick searching around will point you in the right direction, and you will soon find a way to quickly detect whether all edges in a component can form part of the same trail. (You'll need to do some more thinking to figure out how many edges can be included when the answer to the above question is "no".)
Related
Let's say we have a maze, which has a width of W and a height of H. In this maze there are multiple people and multiple towers. The people are the sources (S) and the towers (D) are the destinations. It should be known that we have an omniscient view of the maze. My question is then this:
If I want to find the shortest path between any of the different SD combinations, how do I go about this?
At first, I can think of a naive solution that involves breaking this down into SD different OSOD operations, the problem is that this is quite time-consuming.
The second option would be to break it down into S different OSMD operations. But this I suspect is still too time inefficient for what I'm looking for.
I need an algorithm that can perform the pathfinding in O(WH) time. I've not been able to find anything that gives me the shortest path in O(WH) time and which is MSMD based. Hopefully, someone can point me in the right direction.
Imagine a graph that includes the maze as well as start and end vertexes outside the maze, with an edge from the start vertex to every S, and an edge from every D to the end vertex.
Now breadth-first search (since there are no weights) to find the shortest path from the single start to the single end.
I say 'imagine' that graph, because you don't actually have to build it. This ends up being a simple breadth-first search with minor modifications -- you start with all the S nodes in the root level, and stop when you reach any D.
I am working on an atmospheric simulation for a video game, and a problem I have stumbled into is that I need a cheap (in processing time) way to determine if a graph of nodes in a rectangular grid (each node is connected to up to four neighbours, NSEW) would become partitioned if I removed a particular node.
I have tried searching for ways of detecting if a graph is partitioned but so far I have not found anything that suits my problem. I have not taken advanced math courses and only have basic knowledge of graph theory so it is possible that I just have not been searching with the right terms.
If possible, it would be very very desirable to avoid having to search through the whole graph.
You can find articulation points using a modified depth first search - see http://en.wikipedia.org/wiki/Biconnected_component. An articulation point of a graph is a node that, if removed, disconnects the graph. Every graph can be split up at the articulation points into biconnected components. If you are lucky, you just need to know whether a point is an articulation point. If not, perhaps splitting the graph up into a tree of biconnected components and analysing them will help.
I'm stuck on a graph problem. I'm working with OpenSG and am trying to find paths between two points in a building.
I constructed this graph:
I use it to find the shortest path between two points in a building:
This is how I find the path:
I add the start/target node in the graph
I add edges from start/target node to all nodes that are reachable (this is done with
IntersectAction in OSG)
I use the A*-algorithm to find the
shortest path
I want to minimize the graph now. It doesn't matter if the paths between two points get a little longer i just want to elimate the redundancy that i have now. For example all the light green nodes could be deleted. There is no point in the room that doesnt "see" the door, so there is no need for the nodes. It should look something like this:
So I have an algorithm that does more or less what I want, but I just thought that this must be a well-known problem. It's not a minimal vertex cover, because for example if in the minimal vertex cover a door-node is not included I won't find a way between the two rooms.
I compared with various Graph-Problems but I couldnt find a real match...
Its very late (6:20am) and I should go to bed and maybe it seems a little bit confusing or maybe its really obvious. Thanks for any kind of input on the problem.
I guess I found the answer myself. If anyone is interested. It is not the normal Vertex Cover but the "Minimum Connected Vertex Cover Problem" (CVCP). There is a nice paper about it:
Vertex covers and connected vertex covers in 3-connected graphs
Here's an interesting problem/riddle a friend asked me recently:
You are painting the lines of a tennis court. You have a machine to paint straight lines, but once it is turned on, you can't stop it until the job is done. Clearly you will need to go over some lines twice. What is the smallest amount of extra paint you have to use, in feet?
The dimensions of a court:
The sum of all lines is 480ft. I happen to know that the answer is 63ft extra (543ft total), but I can't help but wonder what the best algorithm is to solve this.
It seems similar to the traveling salesman problem, where each line on the court is represented by a vertex in a graph and junctions of court-lines translate to edges. (Otherwise, if lines were edges and corners were vertices, it would require a path that goes through all edges, and I don't know of any algorithms for that). Maybe you need to be more clever about how junctions of lines are represented and I have some ideas about that, but nothing has really worked yet.
I think the problem is small enough, though, that it can be brute-force-checked with all paths through the line graph. How would you code that?
An undirected graph has an Eulerian trail if and only if at most two vertices have odd degree, and if all of its vertices with nonzero degree belong to a single connected component. （ Found from http://en.wikipedia.org/wiki/Eulerian_path ）
When we get a non-Eulerian graph, we can change it to an Eulerian by adding edges to the odd degree vertices.
So, this problem is turned to: find the lowest cost to turn the graph to a Eulerian.
The algorithm should be:
list all vertices with odd degree , suggest it is a list_vodd[];
find the shortest edge between the vertices in list_vodd[], get two vertices of the edge: pa, pb;
add an edge between pa,pb ( which means this edge should be paint twice );
delete pa,pb from list_vodd[];
goto 2 until there are only two vertices in list_vodd[].
use any existing algorithm to find out a Eulerian router in the new graph with the added edges.
I'm a little late to the game here, but I came across this when I was trying to find an algorithm to determine the shortest path to hike every trail in a state park. Here's a sketch that explains why the answer is 63
As Richard mentioned, this is a Chinese Postman problem. Since the problem didn't specify that we have to start and end in the same location we can use a semi-eulerian graph, which is why all the nodes have an even number, except the start and end points which share a common edge.
Here is a very nice video that explains how to graph and solve this type of problem.
https://www.youtube.com/watch?v=spaUY8PlyYA
I have a directed graph with negative edge weights. The graph is modified by the program and sometimes will form negative cycles. When that happens, shortest path algorithms (Bellman-ford/Johnson/Floyd-Warshall) would detect the existence of such negative cycle and fail, but no other useful information is produced.
I would like to identify what edge causes the negative cycle and disallow such modifications in the graph. Can someone help me with a pointer?
Thanks,
Paul
Paul, If you're about to add an edge (source, destination, weight), and you know the distance from destination to source, then you're creating a negative cycle if and only if new weight + old distance is negative.
On the other hand, if you've just got a graph, the bellman-ford algorithm detects negative cycles and can exhibit one when it finds one. You just need to either find an implementation that does that (rather than just failing), or write one yourself. It's not a difficult algorithm and there's lots of pseudocode on the web.
(It's probably a couple of days consultancy work if you want one custom-written for you. I do this sort of thing for a living and would be happy to.)
I'm not sure exactly what you need. I don't know, but I'd imagine that there's an on-line version of Bellman-Ford that keeps its distances up to date cheaply as new edges come in, and will scream if you add a bad one.