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Let G be a directed weighted graph with nodes colored black or white, and all weights non-negative. No other information is specified--no start or terminal vertex.
I need to find a path (not necessarily simple) of minimal weight which alternates colors at least n times. My first thought is to run Kosaraju's algorithm to get the component graph, then find a minimal path between the components. Then you could select nodes with in-degree equal to zero since those will have at least as many color alternations as paths which start at components with in-degree positive. However, that also means that you may have an unnecessarily long path.
I've thought about maybe trying to modify the graph somehow, by perhaps making copies of the graph that black-to-white edges or white-to-black edges point into, or copying or deleting edges, but nothing that I'm brain-storming seems to work.
The comments mention using Dijkstra's algorithm, and in fact there is a way to make this work. If we create an new "root" vertex in the graph, and connect every other vertex to it with a directed edge, we can run a modified Dijkstra's algorithm from the root outwards, terminating when a given path's inversions exceeds n. It is important to note that we must allow revisiting each vertex in the implementation, so the key of each vertex in our priority queue will not be merely node_id, but a tuple (node_id, inversion_count), representing that vertex on its ith visit. In doing so, we implicitly make n copies of each vertex, one per potential visit. Visually, we are effectively making n copies of our graph, and translating the edges between each (black_vertex, white_vertex) pair to connect between the i and i+1th inversion graphs. We run the algorithm until we reach a path with n inversions. Alternatively, we can connect each vertex on the nth inversion graph to a "sink" vertex, and run any conventional path finding algorithm on this graph, unmodified. This will run in O(n(E + Vlog(nV))) time. You could optimize this quite heavily, and also consider using A* instead, with the smallest_inversion_weight * (n - inversion_count) as a heuristic.
Furthermore, another idea hit me regarding using knowledge of the inversion requirement to speedup the search, but I was unable to find a way to implement it without exceeding O(V^2) time. The idea is that you can use an addition-chain (like binary exponentiation) to decompose the shortest n-inversion path into two smaller paths, and rinse and repeat in a divide and conquer fashion. The issue is you would need to construct tables for the shortest i-inversion path from any two vertices, which would be O(V^2) entries per i, and O(V^2logn) overall. To construct each table, for every entry in the preceding table you'd need to append V other paths, so it'd be O(V^3logn) time overall. Maybe someone else will see a way to merge these two ideas into a O((logn)(E + Vlog(Vlogn))) time algorithm or something.
I have been using Dijkstra Algorithm to find the shortest path in the Graph API which is given by the Princeton University Algorithm Part 2, and I have figured out how to find the path with Chebyshev Distance.
Even though Chebyshev can move to any side of the node with the cost of only 1, there is no impact on the Total Cost, but according to the graph, the red circle, why does the path finding line moves zigzag without moving straight?
Will the same thing will repeat if I use A* Algorithm?
If you want to prioritize "straight lines" you should take the direction of previous step into account. One possible way is to create a graph G'(V', E') where V' consists of all neighbour pairs of vertices. For example, vertex v = (v_prev, v_cur) would define a vertex in the path where v_cur is the last vertex of the path and v_prev is the previous vertex. Then on "updating distances" step of the shortest path algorithm you could choose the best distance with the best (non-changing) direction.
Also we can add additional property to the distance equal to the number of changing a direction and find the minimal distance way with minimal number of direction changes.
It shouldn't be straight in particular, according to Dijkstra or A*, as you say it has no impact on the total cost. I'll assume, by the way, that you want to prevent useless zig-zagging in particular, and have no particular preference in general for a move that goes in the same direction as the previous move.
Dijkstra and A* do not have a built-in dislike for "weird paths", they only explicitly care about the cost, implicitly that means they also care about how you handle equal costs. There are a couple of things you can do about that:
Use tie-breaking to make them prefer straight moves whenever two nodes have equal cost (G or F, depending on whether you're doing Dijkstra or A*). This gives some trouble around obstacles because two choices that eventually lead to equal-length paths do not necessarily have the same F score, so they might not get tie-broken. It'll never give you a sub-optimal path though.
Slightly increase your diagonal cost, it doesn't have to be a whole lot, say 10 for straight and 11 for diagonal. This will just avoid any diagonal move that isn't a shortcut. But obviously: if that doesn't match the actual cost, you can now find sub-optimal paths. The bigger the cost difference, the more that will happen. In practice it's relatively rare, and paths have to be long enough (accumulating enough cost-difference that it becomes worth an entire extra move) before it happens.
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 :)
NOTE: Due to the fact that the trip does not end at the same place it started and also the fact that every point can be visited more than once as long as I still visit all of them, this is not really a TSP variant, but I put it due to lack of a better definition of the problem.
So..
Suppose I am going on a hiking trip with n points of interest. These points are all connected by hiking trails. I have a map showing all trails with their distances, giving me a directed graph.
My problem is how to approximate a tour that starts at a point A and visits all n points of interest, while ending the tour anywhere but the point where I started and I want the tour to be as short as possible.
Due to the nature of hiking, I figured this would sadly not be a symmetric problem (or can I convert my asymmetric graph to a symmetric one?), since going from high to low altitude is obviously easier than the other way around.
Also I believe it has to be an algorithm that works for non-metric graphs, where the triangle inequality is not satisfied, since going from a to b to c might be faster than taking a really long and weird road that goes from a to c directly. I did consider if triangle inequality still holds, since there are no restrictions regarding how many times I visit each point, as long as I visit all of them, meaning I would always choose the shortest of two distinct paths from a to c and thus never takr the long and weird road.
I believe my problem is easier than TSP, so those algorithms do not fit this problem. I thought about using a minimum spanning tree, but I have a hard time convincing myself that they can be applied to a non-metric asymmetric directed graph.
What I really want are some pointers as to how I can come up with an approximation algorithm that will find a near optimal tour through all n points
To reduce your problem to asymmetric TSP, introduce a new node u and make arcs of length L from u to A and from all nodes but A to u, where L is very large (large enough that no optimal solution revisits u). Delete u from the tour to obtain a path from A to some other node via all others. Unfortunately this reduction preserves the objective only additively, which make the approximation guarantees worse by a constant factor.
The target of the reduction Evgeny pointed out is non-metric symmetric TSP, so that reduction is not useful to you, because the approximations known all require metric instances. Assuming that the collection of trails forms a planar graph (or is close to it), there is a constant-factor approximation due to Gharan and Saberi, which may unfortunately be rather difficult to implement, and may not give reasonable results in practice. Frieze, Galbiati, and Maffioli give a simple log-factor approximation for general graphs.
If there are a reasonable number of trails, branch and bound might be able to give you an optimal solution. Both G&S and branch and bound require solving the Held-Karp linear program for ATSP, which may be useful in itself for evaluating other approaches. For many symmetric TSP instances that arise in practice, it gives a lower bound on the cost of an optimal solution within 10% of the true value.
You can simplify this problem to a normal TSP problem with n+1 vertexes. To do this, take node 'A' and all the points of interest and compute a shortest path between each pair of these points. You can use the all-pairs shortest path algorithm on the original graph. Or, if n is significantly smaller than the original graph size, use single-source shortest path algorithm for these n+1 vertexes. Also you can set length of all the paths, ending at 'A', to some constant, larger than any other path, which allows to end the trip anywhere (this may be needed only for TSP algorithms, finding a round-trip path).
As a result, you get a complete graph, which is metric, but still asymmetric. All you need now is to solve a normal TSP problem on this graph. If you want to convert this asymmetric graph to a symmetric one, Wikipedia explains how to do it.
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.