What is the best algorithm to use to solve a maze, which is ideally a graph, while gathering the most number of coins in a fixed number of steps?
Each edge has a distance and each node has a certain number of coins(0 - n coins)
The total fixed number of steps is given as input and it is guaranteed that there is a solution which solves the maze after these steps.
Thanks
Sorry, you will not find an efficient solution for this problem, it is NP-hard, which means if you want a precise solution, it will have exponential complexity. This can be shown by reducing the Knapsack problem to it.
Assume you have an instance of Knapsack with n items, set w_0...w_n of weights and set v_0..._v_n of values, and capacity W, we can build a graph where each vertex corresponds to a value, plus 2 additional vertices s and e for start and end. Create an edge from s to each vertex v_i with weight w_i/2, and also an edge from v_i to the vertex e with corresponding weight w_i/2. Now find the path from s to e, limited to length W. This path will not visit vertex v_i more then once since after the coin is collected no reason to return to that node. Also, to visit any v_i and getting to e it will spend exactly w_i steps, whether if going from s or returning from e. The solution guarantees that the limit of steps is not exceeded and the combination of values is maximal. So by just picking all vertices v_i visited we have a solution to the optimization version of Knapsack which is NP-hard.
Now it is not all that bad. The problem has a solution, it is just inefficient. For a small maze, it may still work using brute force, i.e. start with a vertex and try to go in any direction recursively. If exceeded maximum number of steps - abort. If reached destination - return the path. Then of all the returned paths select the one that yields maximum coins.
Additionally, if you do not need a precise solution - you may try to give an approximate solution. For example, you can use Dijkstra algorithm to produce shortest path tree. So now you have a shortest path from source to each vertex. Pick the path from source to destination, and try to improve it iteratively. In each iteration pick a vertex on the path, look at all its neighbors and see it going through this neighbor gives you better value while still under steps constraint. This will not necessarily give you the optimal solution, but will likely produce some good solutions.
And then you can try other optimizations, like "pick a coin and run". Look at your path and try to improve it by going from some vertex to its neighbor and return immediately. It may be useful if your maze has a dead end, so that naturally you would not go there to reach destination, but you do have enough steps to go there collect some coins and return.
I hope this helps.
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.
Most of the time when implementing a pathfinding algorithm such as A*, we seek to minimize the travel cost along the path. We could also seek to find the optimal path with the fewest number of turns. This could be done by, instead of having a grid of location states, having a grid of location-direction states. For any given location in the old grid, we would have 4 states in that spot representing that location moving left, right, up, or down. That is, if you were expanding to a node above you, you would actually be adding the 'up' state of that node to the priority queue, since we've found the quickest route to this node when going UP. If you were going that direction anyway, we wouldnt add anything to the weight. However, if we had to turn from the current node to get to the expanded node, we would add a small epsilon to the weight such that two shortest paths in distance would not be equal in cost if their number of turns differed. As long as epsilon is << cost of moving between nodes, its still the shortest path.
I now pose a similar problem, but with relaxed constraints. I no longer wish to find the shortest path, not even a path with the fewest turns. My only goal is to find a path of ANY length with numTurns <= n. To clarify, the goal of this algorithm would be to answer the question:
"Does there exist a path P from locations A to B such that there are fewer than or equal to n turns?"
I'm asking whether using some sort of greedy algorithm here would be helpful, since I do not require minimum distance nor turns. The problem is, if I'm NOT finding the minimum, the algorithm may search through more squares on the board. That is, normally a shortest path algorithm searches the least number of squares it has to, which is key for performance.
Are there any techniques that come to mind that would provide an efficient way (better or same as A*) to find such a path? Again, A* with fewest turns provides the "optimal" solution for distance and #turns. But for my problem, "optimal" is the fastest way the function can return whether there is a path of <=n turns between A and B. Note that there can be obstacles in the path, but other than that, moving from one square to another is the same cost (unless turning, as mentioned above).
I've been brainstorming, but I can not think of anything other than A* with the turn states . It might not be possible to do better than this, but I thought there may be a clever exploitation of my relaxed conditions. I've even considered using just numTurns as the cost of moving on the board, but that could waste a lot of time searching dead paths. Thanks very much!
Edit: Final clarification - Path does not have to have least number of turns, just <= n. Path does not have to be a shortest path, it can be a huge path if it only has n turns. The goal is for this function to execute quickly, I don't even need to record the path. I just need to know whether there exists one. Thanks :)
I'm searching for an algorithm to find pairs of adjacent nodes on a hexagonal (honeycomb) graph that minimizes a cost function.
each node is connected to three adjacent nodes
each node "i" should be paired with exactly one neighbor node "j".
each pair of nodes defines a cost function
c = pairCost( i, j )
The total cost is then computed as
totalCost = 1/2 sum_{i=1:N} ( pairCost(i, pair(i) ) )
Where pair(i) returns the index of the node that "i" is paired with. (The sum is divided by two because the sum counts each node twice). My question is, how do I find node pairs that minimize the totalCost?
The linked image should make it clearer what a solution would look like (thick red line indicates a pairing):
Some further notes:
I don't really care about the outmost nodes
My cost function is something like || v(i) - v(j) || (distance between vectors associated with the nodes)
I'm guessing the problem might be NP-hard, but I don't really need the truly optimal solution, a good one would suffice.
Naive algos tend to get nodes that are "locked in", i.e. all their neighbors are taken.
Note: I'm not familiar with the usual nomenclature in this field (is it graph theory?). If you could help with that, then maybe that could enable me to search for a solution in the literature.
This is an instance of the maximum weight matching problem in a general graph - of course you'll have to negate your weights to make it a minimum weight matching problem. Edmonds's paths, trees and flowers algorithm (Wikipedia link) solves this for you (there is also a public Python implementation). The naive implementation is O(n4) for n vertices, but it can be pushed down to O(n1/2m) for n vertices and m edges using the algorithm of Micali and Vazirani (sorry, couldn't find a PDF for that).
This seems related to the minimum edge cover problem, with the additional constraint that there can only be one edge per node, and that you're trying to minimize the cost rather than the number of edges. Maybe you can find some answers by searching for that phrase.
Failing that, your problem can be phrased as an integer linear programming problem, which is NP-complete, which means that you might get dreadful performance for even medium-sized problems. (This does not necessarily mean that the problem itself is NP-complete, though.)
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.