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).
Related
Given an undirected weighted graph (or a single connected component of a larger disjoint graph) which typically will contain numerous odd and even cycles, I am searching for algorithms to remove the smallest possible number of edges necessary in order to produce one or more bipartite subgraphs. Are there any standard algorithms in the literature such as exist for minimum cut, etc.?
The problem I am trying to solve looks like this in the real world:
Presentations of about 1 hour each are given to students about different subjects in one or two time blocks. Students can sign up for at least one presentation of their choice, or two, or three (3rd choice is an alternative in case one of the others isn't going to be presented). They have to be all different choices. If there are less than three sign-ups for a given presentation, it will not be given. If there are 18 or more, it will be given twice in both blocks. I have to schedule the presentations such that the maximum number of sign-ups are satisfied.
Scheduling is trivial in the following cases:
Sign-ups for only one presentation can always be satisfied if the presentation is given (i.e. sign-ups >= 3);
Sign-ups for two given presentations are always satisfiable if at least one of them is given twice.
First, all sign-ups are aggregated to determine which ones are given once and which are given twice. If a student has signed up for a presentation with too few other sign-ups, the alternative presentation is chosen if it will also be given.
At the end of the day, I am left with an undirected weighted graph where the vertices are the presentations and the edges represent students who have signed up for that combination of presentations, each of which is only presented once. The weight corresponds to the number of sign-ups for the unique combination of presentations (thus avoiding parallel edges).
If the number of vertices, or presentations, is around 20 or less, I have come up with a brute force solution which finishes in acceptable time. However, each additional vertex will double the runtime of that solution. After 28 or so, it rapidly becomes unmanageable.
This year we had 37 presentations, thirty of which were only given once and thus ended up in the graph. What I am trying right now for larger graphs is the following:
Find all discrete components and solve each component individually;
For each component, remove leaf nodes and bridge edges recursively;
Generate a spanning tree (I am using Kruskal's algorithm which works very well), saving the removed edges;
Generate the fundamental cycle set by adding one removed edge back into the tree at a time and stripping off the rest of the tree;
Using the Gibbs-Welch algorithm, I generate the complete set of all elemental cycles starting with the fundamental set obtained in step 4;
Count the number of odd and even cycles to which each edge belongs;
Create a priority queue of edges (ordering discussed below) and remove each edge successively from its connected component until the resulting component is bipartite.
I cannot find an ordering of the priority queue for which I can prove that the result would be as acceptable as a solution obtained using the brute force method (it is probably NP-hard). However, I am trying something along these lines:
a. If the edge belongs only to odd cycles, remove it first;
b. If the edge belongs to more odd than even cycles, remove it before any other edges which belong to more even cycles than odd;
c. Edges with the smallest weight should be removed first.
If an edge belongs to both an odd and an even cycle, removing it would leave a larger odd cycle behind. That is why I am ordering them like that. Obviously, the larger the number of odd cycles to which an edge belongs, the higher the priority, but only if less even cycles are affected.
There are additional criteria which exist but need to be considered outside of the graph problem; for example, removing an edge effectively removes one of the sign-ups for one of the presentations, so an eye has to be kept on not letting the number of sign-ups get too small.
(EDIT: there is also the possibility of splitting presentations into two blocks which have almost enough sign-ups, e.g. 15-16 instead of 18. But this means that whoever is giving the presentation would have to do it twice, so it is a trade-off.)
Thanks in advance for any suggestions!
This problem is equivalent to the NP-hard weighted max cut problem, which asks for a partition of the vertices into two parts such that the maximum number of edges go between the parts.
I think the easiest way to solve a problem size such as you have would be to formulate it as a quadratic integer program and then apply an off the shelf solver. The formulation looks like
maximize (1/2) sum_{ij} w_{ij} (1 - y_i y_j)
subject to
y_i in {±1} for all i
where w_ij is the weight of the undirected edge ij if present else zero (so the corresponding variable and its constraint can be omitted).
I have a graph of many hundred nodes that are mainly connected with each other. I can do processing on entire graph but it really takes a lot of time, so I would like to divide it to smaller sub-graphs of approximately similar size.
With other words. I have a collection of aerial images and I do pairwise image matching on all of them. As a result I get a set of matches for each pair (pixel from first image matched with pixel on second image). Number of matches is considered as weight of this (undirected) edge. These edges then form a graph mentioned above.
I'm not so familiar with graph theory (as it is a very broad topic). What is the best algorithm for this job?
Thank you.
Edit:
This problem has a perfect analogy which I think is easier to understand. Imagine you have a set of people and their connections/friendships, like I social network. Each friendship has a numeric value/weight representing how good friends they are. So in a large group of people I want to get k most interconnected sub-groups .
Unfortunately, the problem you're describing is almost certainly NP-hard. From a graph perspective, you have a graph where each edge has a weight on it. You're trying to split the graph into relatively equal pieces while cutting the lowest total cost of edges cut. This problem is called the maximum k-cut problem and is NP-hard. If you introduce the constraint that you also want to try to make the pieces roughly even in size, you have the balanced k-cut problem, which is also NP-hard.
The good news is that there are nice approximation algorithms for these problems, so if you're looking for solutions that are just "good enough," then you can probably find a library somewhere that implements them. There are also other techniques like spectral clustering which work well in practice and are really fast, but which don't have any guarantees on how well they'll do.
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.
I have an undirected graph. One edge in that graph is special. I want to find all other edges that are part of a even cycle containing the first edge.
I don't need to enumerate all the cycles, that would be inherently NP I think. I just need to know, for each each edge, whether it satisfies the conditions above.
A brute force search works of course but is too slow, and I'm struggling to come up with anything better. Any help appreciated.
I think we have an answer (I must credit my colleague with the idea). Essentially his idea is to do a flood fill algorithm through the space of even cycles. This works because if you have a large even cycle formed by merging two smaller cycles then the smaller cycles must have been both even or both odd. Similarly merging an odd and even cycle always forms a larger odd cycle.
This is a practical option only because I can imagine pathological cases consisting of alternating even and odd cycles. In this case we would never find two adjacent even cycles and so the algorithm would be slow. But I'm confident that such cases don't arise in real chemistry. At least in chemistry as it's currently known, 30 years ago we'd never heard of fullerenes.
If your graph has a small node degree, you might consider using a different graph representation:
Let three atoms u,v,w and two chemical bonds e=(u,v) and k=(v,w). A typical way of representing such data is to store u,v,w as nodes and e,k as edges in a graph.
However, one may represent e and k as nodes in the graph, having edges like f=(e,k) where f represents a 2-step link from u to w, f=(e,k) or f=(u,v,w). Running any algorithm to find cycles on such a graph will return all even cycles on the original graph.
Of course, this is efficient only if the original graph has a small node degree. When a user performs an edit, you can easily edit accordingly the alternative representation.
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.