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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 want to do something similar to Appointment scheduling algorithm (N people with N free-busy slots, constraint-satisfaction). But my additional requirement is that I should be able to give 2nd optimal solution, 3rd optimal solution and so on.
Is it possible to achieve this without hitting the performance badly?
There isn't a lot of research (to my knowledge) into finding the solutions, in order, from the most optimal, as most of the time we just care about finding an as efficient as possible solution. So, on the assumption that a better solution might not come to light, I'll give this solution.
To find the most efficient solution, use the accepted answer in the linked question. Copied here for convenience:
Find a maximum matching in a bipartite graph (one set of vertices is the set of people and the other on the set of slots, there is an edge between a person and a slot if the person is available for this slot).
This problem can be solved with the Hopcroft-Karp algorithm.
Complexity O(n5/2) in the worst case, better if the graph is sparse.
Now, in turn, try to remove each edge of the output from the input graph and run the algorithm again.
One of these runs should give you the second-most optimal.
Now, in turn, try to remove each edge of the output from the graph that gave you the second-most optimal and run the algorithm again.
Now the third-most optimal should be among the generated sets.
Now similarly try to remove the edges of the graph of the third-most optimal.
And so on.
Complexity:
O(n5/2) in the worst case for the optimal solution.
O(n7/2) (O(n.n5/2)) in the worst case for each next solution to be generated.
Example:
Say you have edges a,b,c,d,e,f,g.
Let's say the maximum match is a,b,c.
Now you remove a from the input graph and get b,c,d,e,f,g.
Let's say the maximum match of this graph is c,d,e.
Now you remove b from the input graph and get a,c,d,e,f,g.
Let's say the maximum match of this graph is a,d,e.
Now you remove c from the input graph and get a,b,d,e,f,g.
Let's say the maximum match of this graph is a,b,g.
Now either c,d,e, a,d,e or a,b,g will be the second-most optimal (let's say it's a,b,g).
Now try to remove a, b, then g from a,b,d,e,f,g and get the maximum match of each of those 3 graph.
One of these 5 sets (the 6 generated sets excluding the second-most optimal one) should be the third optimal one.
And so on.
Proof:
I'll have to think about that a bit more...
Note:
For example, let's say we have edges a,b,c,d,e with a maximum match of a,b,c.
We remove a and get c,d,e as the maximum match.
We remove b and get c,d,e as the maximum match.
Note that these two are identical, so you shouldn't one as the second-most optimal and another as the third-most optimal.
Although you should keep both around - you need to check the graphs generated from removing c, d and e from both b,c,d,e and a,c,d,e.
Since you will need to check all the edges removed from both when c,d,e is the next-most optimal, this may negatively affect running time a bit.
My apologies if this is a duplicate. I lack the Computer Science knowledge to know what to properly search for.
I need to find a matching algorithm. I've got a series of rooms, and a series of contents-for-rooms. The contents have a minimum size room in which they would fit - so some would happily fit in any room, some would only fit in one or two rooms. I'll also be having a maximum size for some of the rooms, but I assume that this only effects when I determine if the room would be suitable.
Assuming (though - this won't be guaranteed in my actual use) that there is a potential solution, how do I find the optimal allocation, such that each room is only used once and none of the contents are lacking a room?
You problem appears to be a maximum bipartite matching problem. You could think of your problem as an undirected graph G(V,E) where the vertices V are the rooms and contents and the edges E are possible connections between rooms and contents:
The graph is bipartite. If we split the vertices into two sets, rooms and contents, there are no internal edges in each set.
An edge exists in the graph between contents(i) and room(j) if the room is big enough to hold the contents.
A maximum matching produces the maximum number of pairings between vertices in the two sets (i.e. rooms and contents), ensuring that each vertex is only used once. The matching is said to be "perfect" if all vertices are matched. There are a number of algorithms that can be used for such problems, potentially the fastest is the Hopcroft-Karp method.
You could also consider a further optimisation of your problem, in which you try to minimse the total wasted space in the rooms. In this case a "weight" would be associated to the edges defined above based on the difference between the areas of the contents and the rooms.
You would then seek a maximum weight maximum matching.
You could solve this as an http://en.wikipedia.org/wiki/Assignment_problem. You don't have matching numbers of things to be matched, but you can make up things for whichever side runs short first. If you make the cost for the made up things the same for every possible match, the minimum cost answer for a solution with made up things will also be the minimum cost answer for a solution without made up things which produces only a partial match, because the contribution of the made-up things to the cost is the same no matter how they are assigned.
(of course there may be a faster way to solve your specific problem - for instance if you only have one thing to match on one of the sides, just try it in every possible location).
In a tower defense game, you have an NxM grid with a start, a finish, and a number of walls.
Enemies take the shortest path from start to finish without passing through any walls (they aren't usually constrained to the grid, but for simplicity's sake let's say they are. In either case, they can't move through diagonal "holes")
The problem (for this question at least) is to place up to K additional walls to maximize the path the enemies have to take. For example, for K=14
My intuition tells me this problem is NP-hard if (as I'm hoping to do) we generalize this to include waypoints that must be visited before moving to the finish, and possibly also without waypoints.
But, are there any decent heuristics out there for near-optimal solutions?
[Edit] I have posted a related question here.
I present a greedy approach and it's maybe close to the optimal (but I couldn't find approximation factor). Idea is simple, we should block the cells which are in critical places of the Maze. These places can help to measure the connectivity of maze. We can consider the vertex connectivity and we find minimum vertex cut which disconnects the start and final: (s,f). After that we remove some critical cells.
To turn it to the graph, take dual of maze. Find minimum (s,f) vertex cut on this graph. Then we examine each vertex in this cut. We remove a vertex its deletion increases the length of all s,f paths or if it is in the minimum length path from s to f. After eliminating a vertex, recursively repeat the above process for k time.
But there is an issue with this, this is when we remove a vertex which cuts any path from s to f. To prevent this we can weight cutting node as high as possible, means first compute minimum (s,f) cut, if cut result is just one node, make it weighted and set a high weight like n^3 to that vertex, now again compute the minimum s,f cut, single cutting vertex in previous calculation doesn't belong to new cut because of waiting.
But if there is just one path between s,f (after some iterations) we can't improve it. In this case we can use normal greedy algorithms like removing node from a one of a shortest path from s to f which doesn't belong to any cut. after that we can deal with minimum vertex cut.
The algorithm running time in each step is:
min-cut + path finding for all nodes in min-cut
O(min cut) + O(n^2)*O(number of nodes in min-cut)
And because number of nodes in min cut can not be greater than O(n^2) in very pessimistic situation the algorithm is O(kn^4), but normally it shouldn't take more than O(kn^3), because normally min-cut algorithm dominates path finding, also normally path finding doesn't takes O(n^2).
I guess the greedy choice is a good start point for simulated annealing type algorithms.
P.S: minimum vertex cut is similar to minimum edge cut, and similar approach like max-flow/min-cut can be applied on minimum vertex cut, just assume each vertex as two vertex, one Vi, one Vo, means input and outputs, also converting undirected graph to directed one is not hard.
it can be easily shown (proof let as an exercise to the reader) that it is enough to search for the solution so that every one of the K blockades is put on the current minimum-length route. Note that if there are multiple minimal-length routes then all of them have to be considered. The reason is that if you don't put any of the remaining blockades on the current minimum-length route then it does not change; hence you can put the first available blockade on it immediately during search. This speeds up even a brute-force search.
But there are more optimizations. You can also always decide that you put the next blockade so that it becomes the FIRST blockade on the current minimum-length route, i.e. you work so that if you place the blockade on the 10th square on the route, then you mark the squares 1..9 as "permanently open" until you backtrack. This saves again an exponential number of squares to search for during backtracking search.
You can then apply heuristics to cut down the search space or to reorder it, e.g. first try those blockade placements that increase the length of the current minimum-length route the most. You can then run the backtracking algorithm for a limited amount of real-time and pick the best solution found thus far.
I believe we can reduce the contained maximum manifold problem to boolean satisifiability and show NP-completeness through any dependency on this subproblem. Because of this, the algorithms spinning_plate provided are reasonable as heuristics, precomputing and machine learning is reasonable, and the trick becomes finding the best heuristic solution if we wish to blunder forward here.
Consider a board like the following:
..S........
#.#..#..###
...........
...........
..........F
This has many of the problems that cause greedy and gate-bound solutions to fail. If we look at that second row:
#.#..#..###
Our logic gates are, in 0-based 2D array ordered as [row][column]:
[1][4], [1][5], [1][6], [1][7], [1][8]
We can re-render this as an equation to satisfy the block:
if ([1][9] AND ([1][10] AND [1][11]) AND ([1][12] AND [1][13]):
traversal_cost = INFINITY; longest = False # Infinity does not qualify
Excepting infinity as an unsatisfiable case, we backtrack and rerender this as:
if ([1][14] AND ([1][15] AND [1][16]) AND [1][17]:
traversal_cost = 6; longest = True
And our hidden boolean relationship falls amongst all of these gates. You can also show that geometric proofs can't fractalize recursively, because we can always create a wall that's exactly N-1 width or height long, and this represents a critical part of the solution in all cases (therefore, divide and conquer won't help you).
Furthermore, because perturbations across different rows are significant:
..S........
#.#........
...#..#....
.......#..#
..........F
We can show that, without a complete set of computable geometric identities, the complete search space reduces itself to N-SAT.
By extension, we can also show that this is trivial to verify and non-polynomial to solve as the number of gates approaches infinity. Unsurprisingly, this is why tower defense games remain so fun for humans to play. Obviously, a more rigorous proof is desirable, but this is a skeletal start.
Do note that you can significantly reduce the n term in your n-choose-k relation. Because we can recursively show that each perturbation must lie on the critical path, and because the critical path is always computable in O(V+E) time (with a few optimizations to speed things up for each perturbation), you can significantly reduce your search space at a cost of a breadth-first search for each additional tower added to the board.
Because we may tolerably assume O(n^k) for a deterministic solution, a heuristical approach is reasonable. My advice thus falls somewhere between spinning_plate's answer and Soravux's, with an eye towards machine learning techniques applicable to the problem.
The 0th solution: Use a tolerable but suboptimal AI, in which spinning_plate provided two usable algorithms. Indeed, these approximate how many naive players approach the game, and this should be sufficient for simple play, albeit with a high degree of exploitability.
The 1st-order solution: Use a database. Given the problem formulation, you haven't quite demonstrated the need to compute the optimal solution on the fly. Therefore, if we relax the constraint of approaching a random board with no information, we can simply precompute the optimum for all K tolerable for each board. Obviously, this only works for a small number of boards: with V! potential board states for each configuration, we cannot tolerably precompute all optimums as V becomes very large.
The 2nd-order solution: Use a machine-learning step. Promote each step as you close a gap that results in a very high traversal cost, running until your algorithm converges or no more optimal solution can be found than greedy. A plethora of algorithms are applicable here, so I recommend chasing the classics and the literature for selecting the correct one that works within the constraints of your program.
The best heuristic may be a simple heat map generated by a locally state-aware, recursive depth-first traversal, sorting the results by most to least commonly traversed after the O(V^2) traversal. Proceeding through this output greedily identifies all bottlenecks, and doing so without making pathing impossible is entirely possible (checking this is O(V+E)).
Putting it all together, I'd try an intersection of these approaches, combining the heat map and critical path identities. I'd assume there's enough here to come up with a good, functional geometric proof that satisfies all of the constraints of the problem.
At the risk of stating the obvious, here's one algorithm
1) Find the shortest path
2) Test blocking everything node on that path and see which one results in the longest path
3) Repeat K times
Naively, this will take O(K*(V+ E log E)^2) but you could with some little work improve 2 by only recalculating partial paths.
As you mention, simply trying to break the path is difficult because if most breaks simply add a length of 1 (or 2), its hard to find the choke points that lead to big gains.
If you take the minimum vertex cut between the start and the end, you will find the choke points for the entire graph. One possible algorithm is this
1) Find the shortest path
2) Find the min-cut of the whole graph
3) Find the maximal contiguous node set that intersects one point on the path, block those.
4) Wash, rinse, repeat
3) is the big part and why this algorithm may perform badly, too. You could also try
the smallest node set that connects with other existing blocks.
finding all groupings of contiguous verticies in the vertex cut, testing each of them for the longest path a la the first algorithm
The last one is what might be most promising
If you find a min vertex cut on the whole graph, you're going to find the choke points for the whole graph.
Here is a thought. In your grid, group adjacent walls into islands and treat every island as a graph node. Distance between nodes is the minimal number of walls that is needed to connect them (to block the enemy).
In that case you can start maximizing the path length by blocking the most cheap arcs.
I have no idea if this would work, because you could make new islands using your points. but it could help work out where to put walls.
I suggest using a modified breadth first search with a K-length priority queue tracking the best K paths between each island.
i would, for every island of connected walls, pretend that it is a light. (a special light that can only send out horizontal and vertical rays of light)
Use ray-tracing to see which other islands the light can hit
say Island1 (i1) hits i2,i3,i4,i5 but doesn't hit i6,i7..
then you would have line(i1,i2), line(i1,i3), line(i1,i4) and line(i1,i5)
Mark the distance of all grid points to be infinity. Set the start point as 0.
Now use breadth first search from the start. Every grid point, mark the distance of that grid point to be the minimum distance of its neighbors.
But.. here is the catch..
every time you get to a grid-point that is on a line() between two islands, Instead of recording the distance as the minimum of its neighbors, you need to make it a priority queue of length K. And record the K shortest paths to that line() from any of the other line()s
This priority queque then stays the same until you get to the next line(), where it aggregates all priority ques going into that point.
You haven't showed the need for this algorithm to be realtime, but I may be wrong about this premice. You could then precalculate the block positions.
If you can do this beforehand and then simply make the AI build the maze rock by rock as if it was a kind of tree, you could use genetic algorithms to ease up your need for heuristics. You would need to load any kind of genetic algorithm framework, start with a population of non-movable blocks (your map) and randomly-placed movable blocks (blocks that the AI would place). Then, you evolve the population by making crossovers and transmutations over movable blocks and then evaluate the individuals by giving more reward to the longest path calculated. You would then simply have to write a resource efficient path-calculator without the need of having heuristics in your code. In your last generation of your evolution, you would take the highest-ranking individual, which would be your solution, thus your desired block pattern for this map.
Genetic algorithms are proven to take you, under ideal situation, to a local maxima (or minima) in reasonable time, which may be impossible to reach with analytic solutions on a sufficiently large data set (ie. big enough map in your situation).
You haven't stated the language in which you are going to develop this algorithm, so I can't propose frameworks that may perfectly suit your needs.
Note that if your map is dynamic, meaning that the map may change over tower defense iterations, you may want to avoid this technique since it may be too intensive to re-evolve an entire new population every wave.
I'm not at all an algorithms expert, but looking at the grid makes me wonder if Conway's game of life might somehow be useful for this. With a reasonable initial seed and well-chosen rules about birth and death of towers, you could try many seeds and subsequent generations thereof in a short period of time.
You already have a measure of fitness in the length of the creeps' path, so you could pick the best one accordingly. I don't know how well (if at all) it would approximate the best path, but it would be an interesting thing to use in a solution.
I am aware that Dijkstra's algorithm can find the minimum distance between two nodes (or in case of a metro - stations). My question though concerns finding the minimum number of transfers between two stations. Moreover, out of all the minimum transfer paths I want the one with the shortest time.
Now in order to find a minimum-transfer path I utilize a specialized BFS applied to metro lines, but it does not guarantee that the path found is the shortest among all other minimum-transfer paths.
I was thinking that perhaps modifying Dijkstra's algorithm might help - by heuristically adding weight (time) for each transfer, such that it would deter the algorithm from making transfer to a different line. But in this case I would need to find the transfer weights empirically.
Addition to the question:
I have been recommended to add a "penalty" to each time the algorithm wants to transfer to a different subway line. Here I explain some of my concerns about that.
I have put off this problem for a few days and got back to it today. After looking at the problem again it looks like doing Dijkstra algorithm on stations and figuring out where the transfer occurs is hard, it's not as obvious as one might think.
Here's an example:
If here I have a partial graph (just 4 stations) and their metro lines: A (red), B (red, blue), C (red), D (blue). Let station A be the source.
And the connections are :
---- D(blue) - B (blue, red) - A (red) - C (red) -----
If I follow the Dijkstra algorithm: initially I place A into the queue, then dequeue A in the 1st iteration and look at its neighbors :
B and C, I update their distances according to the weights A-B and A-C. Now even though B connects two lines, at this point I don't know
if I need to make a transfer at B, so I do not add the "penalty" for a transfer.
Let's say that the distance between A-B < A-C, which causes on the next iteration for B to be dequeued. Its neighbor is D and only at this
point I see that the transfer had to be made at B. But B has already been processed (dequeued). S
So I am not sure how this "delay" in determining the need for transfer would affect the integrity of the algorithm.
Any thoughts?
You can make each of your weights a pair: (# of transfers, time). You can add these weights in the obvious way, and compare them in lexicographic order (compare # of transfers first, use time as the tiebreaker).
Of course, as others have mentioned, using K * (# of transfers) + time for some large enough K produces the same effect as long as you know the maximum time apriori and you don't run out of bits in your weight storage.
I'm going to be describing my solution using the A* Algorithm, which I consider to be an extension (and an improvement -- please don't shoot me) of Dijkstra's Algorithm that is easier to intuitively understand. The basics of it goes like this:
Add the starting path to the priority queue, weighted by distance-so-far + minimum distance to goal
Every iteration, take the lowest weighted path and explode it into every path that is one step from it (discarding paths that wrap around themselves) and put it back into the queue. Stop if you find a path that ends in the goal.
Instead of making your weight simply distance-so-far + minimum-distance-to-goal, you could use two weights: Stops and Distance/Time, compared this way:
Basically, to compare:
Compare stops first, and report this comparison if possible (i.e., if they aren't the same)
If stops are equal, compare distance traveled
And sort your queue this way.
If you've ever played Mario Party, think of stops as Stars and distance as Coins. In the middle of the game, a person with two stars and ten coins is going to be above someone with one star and fifty coins.
Doing this guarantees that the first node you take out of your priority queue will be the level that has the least amount of stops possible.
You have the right idea, but you don't really need to find the transfer weights empirically -- you just have to ensure that the weight for a single transfer is greater than the weight for the longest possible travel time. You should be pretty safe if you give a transfer a weight equivalent to, say, a year of travel time.
As Amadan noted in a comment, it's all about creating right graph. I'll just describe it in more details.
Consider two vertexes (stations) to have edge if they are on a single line. With this graph (and weights 1) you will find minimum number of transitions with Dijkstra.
Now, lets assume that maximum travel time is always less 10000 (use your constant). Then, weight of edge AB (A and B are on one line) is a time_to_travel_between(A, B) + 10000.
Running Dijkstra on such graph will guarantee that minimal number of transitions is used and minimum time is reached in the second place.
update on comment
Let's "prove" it. There're two solution: with 2 transfers and 40 minutes travel time and with 3 transfers and 25 minutes travel time. In first case you travel on 3 lines, so path weight will be 3*10000 + 40. In second: 4*10000 + 25. First solution will be chosen.
I had the same problem as you, until now. I was using Dijkstra. The penalties for transfers is a very good idea indeed and I've been using it for a while now. The main problem is that you cannot use it directly in the weight as you first you have to identify the transfer. And I didn't want to modify the algorithm.
So what I'be been doing, is that each time and you find a transfer, delete the node, add it with the penalty weight and rerun the graph.
But this way I found out that Dijkstra wont work. And this is where I tried Floyd-Warshall which au contraire to Dijkstra compares all possible paths through the graph between each pair of vertices.
It helped me with my problem switching to Floyd-Warshall. Hope it helps you as well.
Its easier to code and lot more easier to implement.