I was thinking about an extension to the Shortest Hamiltonian Path (SHP) problem, and I couldn't find a way of solving it. I know it is NP-complete, but I figured I'd ask here for ideas, since I do not want to simply brute force the problem.
The extension is fairly simply: Given an undirected, complete, weighted graph with n vertices, find the shortest hamiltonian path with end vertices v and u.
So, bruteforce would still take O(n!) time, since the remaining n-2 vertices can be visited in (n-2)! ways. I was trying to find a way to maybe solve this slightly faster. My efforts for finding a way to solve this problem in a beneficial manner has so far been fruitless.
Would anyone have an idea how to exploit the knowledge of the end-vertices? Preferably explained alongside some pseudocode. It is required for the solution found to be optimal.
I guess it could be solved by integer programming, since the knowledge of end nodes are fairly limiting, and makes it easy to avoid cycles, but it wouldn't really exploit the composition of the problem.
If you want to find the shortest path to connect all nodes, then you should look at travelling salesman algorithms. I don't exactly see why you approach it as an HSP. The only reason I can think of is that you want to specify your starting cities, but it should be easy to fix that (if you need that I can post it) by changing your graph a bit.
edit: adding how to tweak your graph
Add 1 node (call it E) and only connect it to your starting and ending nodes. A TSP will compute a solution to your problem by connecting all your nodes. As E is only reachable by going from start to E and then to end, the solution (a cycle yes) will contain start - E - end. Then you remove the 2 edges to and from E and you have the optimal solution. Within the TSP, the path from start to end will be optimal.
You can use a metaheuristic algorithm. There are many kinds of them (Local search, constructive search, etc.). One of them could be based on:
For each x belonging to the set of vertices X:
- Find the set of closest vertices to x C.
- Filter the set C in order to include one of them in the path P.
Repeat until all vertices of X have been included in the path P.
End.
Related
I have weighted undirected graph. I need to find spanning tree with minimal possible cost, so that distance between point A and B will be as low as possible.
For example, I have this graph: graph.
Minimal distance between A and B is 2.
Minimal spanning tree would look like this. But that would make distance between A and B = 3.
Right now I am doing this:
Find distance between AB in graph, using BFS.
Find all paths between AB with length from step 1 using DFS.
Generate spanning tree from every path from step 2.
Compare them and get minimal one.
Everything is OK until I got graph with A-B distance = 12.
Second step then take too much time. Is there any faster way of doing this? Thanks.
The fastest/most efficient way to solve this problem is to use Dijkstra's Shortest Path algorithm. This is a greedy algorithm that has the following basic structure:
1-all nodes on the graph start "infinity" distance apart
2-start with your first node (node A in your example) and keep track of each edge weight to get from this node A to each of its neighbors.
3-Choose the shortest current edge and follow it to your next node, let's call it node C for now
4-Now, for each of C's neighbors, compare the current distance (including infinity if applicable) with the sum of A's edge to C and C's shortest edge to the current neighbor. If it is shorter than the current distance, update it to the new distance.
5-Continue this process until all nodes have been visited and you reach the node you were looking for the shortest path to (i.e. B in your example)
This is a pretty efficient way of finding the shortest path between two nodes, with a running time of O(V^2), not O(nlogn) as mentioned above. As you can see, by making this a greedy algorithm, we are constantly choosing the optimal solution based on the available local information, and therefore we will never have to go back and change our decisions.
This also eliminates the need for a BFS and a DFS like in your example. Hope this helps!
While your step two is correct, I think that the problem is that you are doing too many operations.
You are doing both a BFS and a DFS which is going to be very costly. Instead, what I would recommend doing is using one of several different traversal techniques that will minimize in the computational costs.
This is a common problem to find the shortest path, and one of the popular solutions is Dijkstra's algorithm. Here is an article that expounds on this topic. https://www.geeksforgeeks.org/dijkstras-shortest-path-algorithm-greedy-algo-7/
In short, what this algorithm does, is it takes the starting point A, and then it generates the minimal spanning tree until point B is hit, then there is a single path in which A can get to B, and that is the shortest path.
Both this and your algorithm both run in O(nlogn), but in practice this solution can be thought of as running a single BFS instead of both a BFS and a DFS.
I am hoping there is already a straight forward algorithm as a solution to this but not sure what this type of problem is called and therefore where to look for a solution.
It is in some ways similar to the traveling salesman problem but I think it should be much simpler.
The main difference to the problem is there are limited connections (3 to 6 per city) between cities.
The path does not need to return to starting, only that it visits each city just once. Also the connections are all the same length so the full path length will always be the same (NOT A SHORTEST DISTANCE PROBLEM). There are 84 cites and therefore the final path will always be 87 units long.
Basically I am looking for any solution that can from a random start, get to all cities just once. I am hoping for a "random" solution that won't look orderly.
Any suggestions on what this type of problem is called and where I might find an algorithm.
Thanks.
You are looking for a Hamiltonian Path. Unfortunately, this problem is NP-Complete, although the fact that the vertices in your graph have a limited degree with help its tractability. You can find more information about solving this problem on the linked Wikipedia page or in this answer.
I have a undirected graph with about 100 nodes and about 200 edges. One node is labelled 'start', one is 'end', and there's about a dozen labelled 'mustpass'.
I need to find the shortest path through this graph that starts at 'start', ends at 'end', and passes through all of the 'mustpass' nodes (in any order).
( http://3e.org/local/maize-graph.png / http://3e.org/local/maize-graph.dot.txt is the graph in question - it represents a corn maze in Lancaster, PA)
Everyone else comparing this to the Travelling Salesman Problem probably hasn't read your question carefully. In TSP, the objective is to find the shortest cycle that visits all the vertices (a Hamiltonian cycle) -- it corresponds to having every node labelled 'mustpass'.
In your case, given that you have only about a dozen labelled 'mustpass', and given that 12! is rather small (479001600), you can simply try all permutations of only the 'mustpass' nodes, and look at the shortest path from 'start' to 'end' that visits the 'mustpass' nodes in that order -- it will simply be the concatenation of the shortest paths between every two consecutive nodes in that list.
In other words, first find the shortest distance between each pair of vertices (you can use Dijkstra's algorithm or others, but with those small numbers (100 nodes), even the simplest-to-code Floyd-Warshall algorithm will run in time). Then, once you have this in a table, try all permutations of your 'mustpass' nodes, and the rest.
Something like this:
//Precomputation: Find all pairs shortest paths, e.g. using Floyd-Warshall
n = number of nodes
for i=1 to n: for j=1 to n: d[i][j]=INF
for k=1 to n:
for i=1 to n:
for j=1 to n:
d[i][j] = min(d[i][j], d[i][k] + d[k][j])
//That *really* gives the shortest distance between every pair of nodes! :-)
//Now try all permutations
shortest = INF
for each permutation a[1],a[2],...a[k] of the 'mustpass' nodes:
shortest = min(shortest, d['start'][a[1]]+d[a[1]][a[2]]+...+d[a[k]]['end'])
print shortest
(Of course that's not real code, and if you want the actual path you'll have to keep track of which permutation gives the shortest distance, and also what the all-pairs shortest paths are, but you get the idea.)
It will run in at most a few seconds on any reasonable language :)
[If you have n nodes and k 'mustpass' nodes, its running time is O(n3) for the Floyd-Warshall part, and O(k!n) for the all permutations part, and 100^3+(12!)(100) is practically peanuts unless you have some really restrictive constraints.]
run Djikstra's Algorithm to find the shortest paths between all of the critical nodes (start, end, and must-pass), then a depth-first traversal should tell you the shortest path through the resulting subgraph that touches all of the nodes start ... mustpasses ... end
This is two problems... Steven Lowe pointed this out, but didn't give enough respect to the second half of the problem.
You should first discover the shortest paths between all of your critical nodes (start, end, mustpass). Once these paths are discovered, you can construct a simplified graph, where each edge in the new graph is a path from one critical node to another in the original graph. There are many pathfinding algorithms that you can use to find the shortest path here.
Once you have this new graph, though, you have exactly the Traveling Salesperson problem (well, almost... No need to return to your starting point). Any of the posts concerning this, mentioned above, will apply.
Actually, the problem you posted is similar to the traveling salesman, but I think closer to a simple pathfinding problem. Rather than needing to visit each and every node, you simply need to visit a particular set of nodes in the shortest time (distance) possible.
The reason for this is that, unlike the traveling salesman problem, a corn maze will not allow you to travel directly from any one point to any other point on the map without needing to pass through other nodes to get there.
I would actually recommend A* pathfinding as a technique to consider. You set this up by deciding which nodes have access to which other nodes directly, and what the "cost" of each hop from a particular node is. In this case, it looks like each "hop" could be of equal cost, since your nodes seem relatively closely spaced. A* can use this information to find the lowest cost path between any two points. Since you need to get from point A to point B and visit about 12 inbetween, even a brute force approach using pathfinding wouldn't hurt at all.
Just an alternative to consider. It does look remarkably like the traveling salesman problem, and those are good papers to read up on, but look closer and you'll see that its only overcomplicating things. ^_^ This coming from the mind of a video game programmer who's dealt with these kinds of things before.
This is not a TSP problem and not NP-hard because the original question does not require that must-pass nodes are visited only once. This makes the answer much, much simpler to just brute-force after compiling a list of shortest paths between all must-pass nodes via Dijkstra's algorithm. There may be a better way to go but a simple one would be to simply work a binary tree backwards. Imagine a list of nodes [start,a,b,c,end]. Sum the simple distances [start->a->b->c->end] this is your new target distance to beat. Now try [start->a->c->b->end] and if that's better set that as the target (and remember that it came from that pattern of nodes). Work backwards over the permutations:
[start->a->b->c->end]
[start->a->c->b->end]
[start->b->a->c->end]
[start->b->c->a->end]
[start->c->a->b->end]
[start->c->b->a->end]
One of those will be shortest.
(where are the 'visited multiple times' nodes, if any? They're just hidden in the shortest-path initialization step. The shortest path between a and b may contain c or even the end point. You don't need to care)
Andrew Top has the right idea:
1) Djikstra's Algorithm
2) Some TSP heuristic.
I recommend the Lin-Kernighan heuristic: it's one of the best known for any NP Complete problem. The only other thing to remember is that after you expanded out the graph again after step 2, you may have loops in your expanded path, so you should go around short-circuiting those (look at the degree of vertices along your path).
I'm actually not sure how good this solution will be relative to the optimum. There are probably some pathological cases to do with short circuiting. After all, this problem looks a LOT like Steiner Tree: http://en.wikipedia.org/wiki/Steiner_tree and you definitely can't approximate Steiner Tree by just contracting your graph and running Kruskal's for example.
Considering the amount of nodes and edges is relatively finite, you can probably calculate every possible path and take the shortest one.
Generally this known as the travelling salesman problem, and has a non-deterministic polynomial runtime, no matter what the algorithm you use.
http://en.wikipedia.org/wiki/Traveling_salesman_problem
The question talks about must-pass in ANY order. I have been trying to search for a solution about the defined order of must-pass nodes. I found my answer but since no question on StackOverflow had a similar question I'm posting here to let maximum people benefit from it.
If the order or must-pass is defined then you could run dijkstra's algorithm multiple times. For instance let's assume you have to start from s pass through k1, k2 and k3 (in respective order) and stop at e. Then what you could do is run dijkstra's algorithm between each consecutive pair of nodes. The cost and path would be given by:
dijkstras(s, k1) + dijkstras(k1, k2) + dijkstras(k2, k3) + dijkstras(k3, 3)
How about using brute force on the dozen 'must visit' nodes. You can cover all the possible combinations of 12 nodes easily enough, and this leaves you with an optimal circuit you can follow to cover them.
Now your problem is simplified to one of finding optimal routes from the start node to the circuit, which you then follow around until you've covered them, and then find the route from that to the end.
Final path is composed of :
start -> path to circuit* -> circuit of must visit nodes -> path to end* -> end
You find the paths I marked with * like this
Do an A* search from the start node to every point on the circuit
for each of these do an A* search from the next and previous node on the circuit to the end (because you can follow the circuit round in either direction)
What you end up with is a lot of search paths, and you can choose the one with the lowest cost.
There's lots of room for optimization by caching the searches, but I think this will generate good solutions.
It doesn't go anywhere near looking for an optimal solution though, because that could involve leaving the must visit circuit within the search.
One thing that is not mentioned anywhere, is whether it is ok for the same vertex to be visited more than once in the path. Most of the answers here assume that it's ok to visit the same edge multiple times, but my take given the question (a path should not visit the same vertex more than once!) is that it is not ok to visit the same vertex twice.
So a brute force approach would still apply, but you'd have to remove vertices already used when you attempt to calculate each subset of the path.
I have a very simple pathfinding task- a board game played on an 8x8 grid, with each square either being passable or not. What I'm looking for is an algorithm which will give me the best n paths to get from some square A to square B (assuming there are any).
I've been looking at A*, but as far as I can see, there's no clear way to extend it to find more than one path.
So, what's critical is that the paths it gives are actually the shortest n paths, that it doesn't miss any. Efficiency is also very important. Could anybody suggest an algorithm that would be appropriate, or point me in the right direction?
Dijkstra's is a good algorithm for most situations like these, but since you're on an 8x8 grid, I'm going to assume that all the distances between each cell are both equal and static. In this case, a BFS (Breadth First Search) should suit you well.
Given the small size of the board a breadth-first exhaustive search is something you should be considering. 8 x 8 means only 64 squares, x8 moves (or 4 if you don't permit diagonals) and the total search is pretty small.
Dijkstra's works well to find the single shortest path. To find the second, third ... nth shortest paths, you'd need to use an extension to Dijksta's algorithm. Once a shortest path from N1, N2, N3 ... Nx is found, clone all of the intermediate nodes on that path to create nodes N2' through Nx-1'. Clone all of the entering edges on the shortest path as well except for (N1,N2') and remove edge (Nx-1,Nx). Relax all the edges into nodes on the cloned path which now represents the second fastest way of getting to the nodes on the shortest path from the previous iteration.
Check out k-shortest paths, an open-source implementation that also includes some references.
I'm looking for a general idea (and maybe some code example or at least pseudocode)
Now, this is from a problem that someone gave me, or rather showed me, I don't have to solve it, but I did most of the questions anyway, the problem that I'm having is this:
Let's say you have a directed weighted graph with the following nodes:
AB5, BC4, CD8, DC8, DE6, AD5, CE2, EB3, AE7
and the question is:
how many different routes from C to C with a distance of less than x. (say, 10, 20, 30, 40)
The answer of different trips is: CDC, CEBC, CEBCDC, CDCEBC, CDEBC, CEBCEBC, CEBCEBCEBC.
The main problem I'm having with it is that when I do DFS or BFS, my implementation first chooses the node and marks it as visited therefore I'm only able to find 2 paths which are CDC and CEBC and then my algorithm quits. If I don't mark it as visited then on the next iteration (or recursive call) it will choose the same node and not next available route, so I have to always mark them as visited however by doing that how can I get for example CEBCEBCEBC, which is pretty much bouncing between nodes.
I've looked at all the different algorithms books that I have at home and while every algorithm describes how to do DFS, BFS and find shortest paths (all the good stufF), none show how to iterate indefinitively and stop only when one reaches certain weight of the graph or hits certain vertex number of times.
So why not just keep branching and branching; at each node you will evaluate two things; has this particular path exceeded the weight limit (if so, terminate the branch) and is this node where I started (in which case log my path history to an 'acceptable solutions' list); then make new branches which each take a step in each possible direction.
You should not mark nodes as visited; as MikeB points out, CDCDC is a valid solution and yet it revisits D.
I'd do it lke this:
Start with two lists of paths:
Solutions (empty) and
ActivePaths (containing one path, "C").
While ActivePaths is not empty,
Take a path out of ActivePaths (suppose it's "CD"[8]).
If its distance is not over the limit,
see where you are by looking at the last node in the path ("D").
If you're at "C", add a copy of this path to Solutions.
Now for each possible next destination ("C", "E")
make a copy of this path, ("CD"[8])
append the destination, ("CDC"[8])
add the weight, ("CDC"[16])
and put it in ActivePaths
Discard the path.
Whether this turns out to be a DFS, a BFS or something else depends on where in ActivePaths you insert and remove paths.
No offense, but this is pretty simple and you're talking about consulting a lot of books for the answer. I'd suggest playing around with the simple examples until they become more obvious.
In fact you have two different problems:
Find all distinct cycles from C to C, we will call them C_1, C_2, ..., C_n (done with a DFS)
Each C_i has a weight w_i, then you want every combination of cycles with a total weight less than N. This is a combinatorial problem (and seems to be easily solvable with dynamic programming).