Is there an algorithm that will allow me to traverse a weighted graph in the following manner?
Start at a specific node
Go through all vertecies in the graph
Do this in the least amount of time (weights are times)
End up at the starting node
Sounds like the Travelling Salesman Problem to me. An NP-hard problem. There is no polynomial time algorithm that will give you the optimum solution. You could use a search heuristic to get a close to optimal solution though.
As Greg Sexton stated before me, it is a classic example of the Travelling Salesman Problem. There are many advanced algorithms about for handling this style of problem, which is best for your particular situation rather depends on the graph. If the number of vertices is high, you will need substantial computational power to get it done within a realistic time frame.
I am not sure, if any efficient algorithm exists, but a brute force approach would surely give you the answer.
In any case, can you give the constraints on the number of vertices/edges.
Related
So I have the following layout:
graph representation
The objective is to collect all the yellow blocks by moving the white ball around. I'm trying to come up with an algorithm that will calculate an efficient path however I'm not too sure where to start.
Initially I thought about path finding algorithms like Djikstra and A* but they don't seem to fit with my goal. I've also thought about hamiltonian paths which is closer to what I want but still doesn't seem to solve the problem.
Any suggestions on what sort of algorithm can be used would be appreciated.
Your problem has a classic name in the litterature, it is the minimum hamiltonian walk problem. Beware not to mistake it with the minimum hamiltonian path problem, its 'cousin', because it is much more famous, and much, much harder (finding a hamiltonian walk can be done in polynomial time, finding a hamiltonian path is NP-complete). The traveling salesman problem is the other name of the minimum hamiltonian path problem (path, not walk).
There are very few ressources on this problem, but nevertheless you can have a look at an article called 'An algorithm for finding a short closed spanning walk in a graph' by Takamizawa, Nishizeki and Saito from 1980. They provide a polynomial algorithm to find such a path.
If the paper is a bit hard to read, or the algorithm too complex to implement, then I'll suggest that you go for the christofides algorithm, because it runs in polynomial time, and is somehow efficient (it is a 2-approximation if I remember well).
Another possible approach would be to go for a greedy algorithm, like a nearest unvisited neigbor algorithm (start somewhere, go to the nearest node that is not in the walk yet, repeat until everyone is in the walk).
Acutally, I think the easiest and maybe best simple solution is to go greedy.
My Problem
Whether there's an efficient algorithm to find a max-weight (or min-weight) k-clique in a complete k-partite graph (a graph in which vertices are adjacent if and only if they belong to different partite sets according to wikipedia)?
More Details about the Terms
Max-weight Clique: Every edge in the graph has a weight. The weight of a clique is the sum of the weights of all edges in the clique. The goal is to find a clique with the maximum weight.
Note that the size of the clique is k which is the largest possible clique size in a complete k-partite graph.
What I have tried
I met this problem during a project. Since I am not a CS person, I am not sure about the complexity etc.
I have googled several related papers but none of them deals with the same problem. I have also programmed a greedy algorithm + simulated annealing to deal with it (the result seems not good). I have also tried something like Dynamic Programming (but it does not seem efficient). So I wonder whether the exact optimal can be computed efficiently. Thanks in advance.
EDIT Since my input can be really large (e.g. the number of vertices in each clique is 2^k), I hope to find a really fast algorithm (e.g. polynomial of k in time) that works out the optimal result. If it's not possible, can we prove some lower bound of the complexity?
Generalized Maximum Clique Problem (GMCP)
I understand that you are looking for the Generalized Maximum/ minimum Clique Problem (GMCP), where finding the clique with maximum score or minimum cost is the optimization problem.
This problem is a NP-Hard problem as shown in Generalized network design problems, so there is currently no polynomial time exact solution to your problem.
Since, there is no known polynomial solution to your problem, you have 2 choices. Reducing the problem size to find the exact solution or to find an estimated solution by relaxing your problem and it leads you to a an estimation to the optimal solution.
Example and solution for the small problem size
In small k-partite graphs (in our case k is 30 and each partite has 92 nodes), we were able to get the optimal solution in a reasonable time by a heavy branch and bounding algorithm. We have converted the problem into another NP-hard problem (Mixed Integer Programming), reduced number of integer variables, and used IBM Cplex optimizer to find the optimal solution to GMCP.
You can find our project page and paper useful. I can also share the code with you.
How to estimate the solution
One straight forward estimation to this NP-Hard problem is relaxing the Mixed Integer Programming problem and solve it as a linear programming problem. Of course it will give you an estimation of the solution, but still you might get a reasonable answer in practice.
More general problem (Generalized Maximum Multi Clique Problem)
In another work, we solve the Generalized Maximum Multi Clique Problem (GMMCP), where maximizing the score or minimizing the cost of selecting multiple k-cliques in a complete k-partite graph is in interest. You can find the project page by searching for GMMCP Tracking.
The maximum clique problem in a weighted graph in general is intractable. In your case, if the graph contains N nodes, you can enumerate through all possible k-cliques in N ** k time. If k is fixed (don't know if it is), your problem is trivially polynomially solvable, as this is a polynomial in N. I don't believe the problem to be tractable if k is a free parameter because I can't see how the assumption of a k-partite graph would make the problem significantly simpler from the general one.
How hard your problem is in practice depends also on how the weights are distributed. If all the weights are very near to each others, i.e. the difference between "best" and "good" is relatively small, the problem is very hard. If you have wildly different weights on the edges, the problem can be easier, because a greedy algorithm can give you a good "initial" solution, and you can use that and subsequent good solutions to limit your combinatorial search using the well-known branch-and-bound method.
I was reading about dijkstra algorithm and A* star algorithm. I know that the difference is the heuristic used. But what is a heuristic and how this influence the algorithms? Heuristic is just an way to measure the distance? But dijkstra considers the distance too? Sorry, but my question is about heuristic and what it means and why to use them... (I had read about it, but don't understant)
Other question: When should be used each one?
Thank you
In this context, a heuristic is a way of providing the algorithm with some form of extra evaluative information, so that the algorithm can find a 'good enough' solution, without exhaustively searching every possible solution.
Dijkstra's Algorithm does not use a heuristic. It expands outwards from the start node, and examines every node in the graph in order to find the shortest path. While this is accurate, it can be computationally expensive.
By comparison, the A* algorithm uses a distance + cost heuristic, to guide the algorithm in its choice of the next node to explore. This means that the algorithm finds a possible search solution without examining every node on the graph. It is therefore much cheaper to run, but at a loss of complete accuracy. It works because the result is usually close enough to the optimal solution, and is found cheaper than an exhaustive search of the entire graph.
As to when you should use each, it really depends on the application. However, using the A* algorithm requires an admissible heuristic, so this may not be applicable in situations where such information is unavailable to the algorithm.
A heuristic basically means an idea or an intuition! Any strategy that you use to solve a hard problem is a heuristic! In some fields (like combinatorial problems) it refers to the strategy that can help you solve a NP hard problem suboptimally within a polynomial time.
Dijkstra solves the single-source routing problem, i.e. it gives you the cost of going froma single point to any other point in the space.
A* solves a single-source single-target problem. It gives you a minimum distance path from a given point to another given point.
A* is usually faster than Dijkstra provided you give it an admissible heuristic, this is, an estimate of the distance to the goal, that never overestimates said distance. Contrary to a previous answer given here, if the heuristic used is admissible, A* is complete and will give you the optimal answer.
Suppose I have 10 points. I know the distance between each point.
I need to find the shortest possible route passing through all points.
I have tried a couple of algorithms (Dijkstra, Floyd Warshall,...) and they all give me the shortest path between start and end, but they don't make a route with all points on it.
Permutations work fine, but they are too resource-expensive.
What algorithms can you advise me to look into for this problem? Or is there a documented way to do this with the above-mentioned algorithms?
Have a look at travelling salesman problem.
You may want to look into some of the heuristic solutions. They may not be able to give you 100% exact results, but often they can come up with good enough solutions (2 to 3 % away from optimal solutions) in a reasonable amount of time.
This is obviously Travelling Salesman problem. Specifically for N=10, you can either try the O(N!) naive algorithm, or using Dynamic Programming, you can reduce this to O(n^2 2^n), by trading space.
Beyond that, since this is an NP-hard problem, you can only hope for an approximation or heuristic, given the usual caveats.
As others have mentioned, this is an instance of the TSP. I think Concord, developed at Georgia Tech is the current state-of-the-art solver. It can handle upwards of 10,000 points within a few seconds. It also has an API that's easy to work with.
I think this is what you're looking for, actually:
Floyd Warshall
In computer science, the Floyd–Warshall algorithm (sometimes known as
the WFI Algorithm[clarification needed], Roy–Floyd algorithm or just
Floyd's algorithm) is a graph analysis algorithm for finding shortest
paths in a weighted graph (with positive or negative edge weights). A
single execution of the algorithm will find the lengths (summed
weights) of the shortest paths between all pairs of vertices though it
does not return details of the paths themselves
In the "Path reconstruction" subsection it explains the data structure you'll need to store the "paths" (actually you just store the next node to go to and then trivially reconstruct whichever path is required as needed).
I want to solve the following problem:
I have a DAG which contains cities and jobs between them that needs to be done. The jobs are for trucks which can load a definied limit. The more the truck is loaded the better is the tour. Some jobs are for loading something in and some are for loading defined things out. You can always drive from city a to b even if there is no job to be done between them.
The last restriction is that I always need to start in city a and return to a because there is the home of the trucks :)
I first thought of Dijkstra's shortest path algorithm. I could easly turn that into longest path calculation. My problem in mind is now that all these algorithms are for calculating a shortest or longest path from vertex a to b, but I need it from a returning to a - in a circle.
Has some one some kicks for my mind?
Thanks for your feedback!
Marco
This crazy combination of knapsack and travelling salesman is surely NP-hard.
Virtually everywhere, when you want to load your agent with optimal job schedule, or when you want to build a route through all vertexes in the graph, or when you feel that you need to look for a "longest path*", you most likely run into an NP-complete or an NP-hard problem.
That means, that there is no known fast and exact solution to the problem, i.e. the one that runs in a polynomial time.
So you have to create approximations and implement non-optimal algorithms based on your particular conditions. What time loss is acceptable? Are there additional patterns the trucks can drive? Do you know more about the graph (e.g. is the area divided into distant dense regions)? Answer these questions and you'll find a non-strict heuristics that satisfies your customers.
*yes, searching for longest paths is not as easy as you think. Longest path problem is NP-complete, given that your graph is not acyclic.
You're trying to find the smallest possible way to get everything done? This reminds me of a max-flow/min-cut problem. You might be able to approximate the best answer by:
Connect all terminal nodes to a final end node.
Run one of the various maximum flow algorithms to find the max flow between a and end.
Return to city a. Update the graph to reflect what you just did. Repeat until all jobs are done.
The idea is that you get the most bang for every trip. Each trip after the 1st will be less efficient and less efficient, but that's to be expected.
Note: This only works because you have a DAG. Travelling salesman wouldn't be NP-Complete on a DAG, either, and it will likely be impossible to even hit all nodes on the graph. Re-reading your problem, it seems like you don't have a DAG, since you can return to city a - is that true?
You can adjust the traveling sales man problem dynamic programming algorithm to do what you want.
The classic approach says that you want to maximize the optimum function from all cities but you can take in consideration, at each step also the possibility of returning home.
And like Pavel mentioned, this problem is definitively NP-hard. Do you have some upper bounds for the number of cities or maximum number of objects that can be loaded in a truck?
PS: Do you want the BEST solution (maximum profit - might not be realistic in terms of processing time) or you accept some approximation?
Isn't this a Transportation problem?
Depending on the trucks number and starting points, you could add a fake transporations or add costs in order to satisfy your restrictions.
I'd also ask about truck restrictions:
are they all based in the same city?
do you have a fixed number of them?
and what you win if you use less then
you have?
is there a cycle time restriction?