TL;DR: Short problem description:
I am looking for an efficient algorithm that optimizes how N agents, located in 2D space, can be placed into M shelters by minimizing the distance the agents need to travel.
Each shelter can only hold 1 agent. If N > M (more agents than available shelters), then some agents will not get placed into shelters (all agents are the same).
(Optional simplification: while agents can be freely located in 2D space, shelters are always arranged on a square grid. No agent is located outside of the convex hull of shelters.)
This is all you need to know. However, if you think that this problem has no efficient solution then here is ...
a more specific (and to me most relevant) version of the problem:
There are exactly 9 shelters, arranged on a square grid (with distance d). All N agents are located around the central shelter (in a box of size d*d centered around central shelter). However, in this case, the central shelter is always empty but all other shelters may or may not be available (empty) at the beginning.
For this case, I need an algorithm that solves the problem of arbitrary many agents N (typically N < 9) and arbitrary shelters being available (either all 9, or in the extreme case only the central shelter).
The algorithm should be efficient, since I need to solve many of these problems quickly.
Example:
Here is an example with N=3 agents (black dots) and M=5 available shelters (green dots). The red dots show non-available shelters. ]1 I use letters for shelters and numbers for agents.
What I did so far:
I am sure that this problem has a specific name and has been solved/studied already, but I cannot find its name or any solutions. I need to solve many of those problems fast and I always want the optimal solution (if thats not possible, an almost optimal solution is also sufficient). Here is what I tried/thought of so far:
Brute force: I know that the optimal solution to the problem can be found with brute force by checking all possible options, calculating the total travel distance for each and picking the option with smallest total travel distance. This may involve many computations if M and N are large.
A fast but very non-optimal solution works as follows: for each agent i, calculate the distance to central node E. Starting from the agent i with smallest distance to E, assign i to its closest shelter (in this case: E). Then assign the next agent to its closest shelter, considering that E is now not available anymore, etc, until all agents are assigned or stop if no more free shelters are available. This works, is fast, but of course produces non-optimal results (in the example image: 2->E, 1->B, 3->F, while the optimal solution should be 3->E, 2->F, 1->B)
Another idea I'm working on is to first find the agents that are under the most "pressure", i.e. all of their good options are far away. Starting with the agent under highest pressure, assign it to the closest shelter. Continue for all other agents. However, I am not sure how to properly define "pressure" for this problem, as it likely should be a combination of the distances to the first few shelters. Also, I am not sure that this will lead to the optimal solution, but may result in an almost optimal solution.
I am trying to think of this problem as some sort of weighted permutation, that is, I need to select N shelters and map them to the N agents, but each mapping comes at a cost. I need to minize the total cost, but I have no idea how to do this.
Also, I am thinking of some sort of Simulated Annealing, or some form of push-and-pull algorithm where each shelter is attracting agents, or agents are attracted to shelters based on their distance. While this may sound interesting, I would expect that this is computationally not efficient.
I am happy for any input, especially if this problem already has a proper name and solutions. I am also happy for a simple and fast-to-compute algorithm that achieves an almost optimal solution.
As suggested in the comments (thanks again!), this is indeed answered by this post.
Specifically, this is an assignment problem which gets solved by the Hungarian algorithm, considering agents as workers, shelters as tasks and the cost of worker i doing task j being the Manhatten distance between agent i and shelter j.
The python package munkres implements this algorithm and is very fast for the 9-shelter problem. If there are more shelters than agents, the package handles it automatically. For the case of more agents than shelters I am satisfied with deleting random agents until the number of agents is equal to the number of shelters. Therefore my problem is solved.
Related
Problem: I need to drop (n) employees from office to their homes(co-ordinates available). I have (x) 7-seater & (y) 4-seater cabs available.
I have to design an algorithm to drop all the employees to their homes while travelling minimum distance.
Also, the algorithm must tell me how many 7-seater or/and 4-seater vehicles I must choose so as to travel minimum distance.
eg. If I have 15 employees then the algorithm may tell me to use 1 (7-seater) cab & 2 (4-seater) cab & have the employees in each cab as following:
[(E2, E4, E6, E8), (E1, E3, E5, E7, E9, E10, E12), (E11, E13, E14, E15)]
Approach: I'm thinking of this as a Travelling Salesman Problem with multiple salesmen with an upper limit on number of cities each can travel. Also salesmen do not need to come back to the origin. Ant's colony problem came to my mind, but I can't really choose wisely which algorithm to choose
Requirement: I really need the ALGORITHM. Either TSP or Ant's colony, doesn't matter. I'll welcome opinions, but I really need the ALGORITHM.
This is a cost minimization problem, not a travelling salesman problem. It is related to TSP in the sense that TSP is a very specific cost minimization problem.
The solution consists of three steps:
Generate a list of employee drop-off points (nodes)
Create distinct paths that do not intersect, nor branch. These will be your routes and help prevent wasteful route overlaps. Use cost(path) = distance(furthest node and origin) + taxi_cost(nodes) + sum(distance between nodes) to compare paths and/or brute-force all potential networks. Networks are layouts of paths. DO NOT BRANCH THE PATHS!!
Total distance is a line of defense against waste ensuring that routes are not too long.
Sum of distances helps the algorithm converge on neighbourhoods where many employees live (when possible).
Because this variation of the coin problem allows imperfect solutions, it reduces to a variant of the Knapsack Problem. The utility of each taxi is capacity. If you also wish to choose the cheapest way to transport your employees, utility(taxi) = capacity/cost. From this our simplest solution is to be greedy; who cares about empty space? If you really care about filling up taxis perfectly (as opposed to cost efficiently), you'll need a much more complex solution. You only specify the least distance as your metric (with each additional taxi multiplying cost). I assume this is a proxy to say 'I don't want to pay too much'.
Therefore: taxi_cost(nodes) = math.floor(amount(nodes)/max(utility(taxis)+1). This equation selects the cheapest, roomiest taxi, and figures out how many of them are required to fully service the route.
Be sure to calculate the cost of each network you examine as sum(cost(path))
Once you've found the cheapest network to service, for each path in the chosen network:
make a list of employees travelling to the furthest node
fill the preferred taxi with those employees
repeat with the next furthest node until you have a full taxi, then add the filled taxi to the list. If you run out of employees, you've finished assigning taxis to the route. (The benefit of furthest-first selection is that you can ask employees in unfilled taxis to walk if that part of the route is within blocks of the office).
The algorithm above is not perfect, but it will have many desirable tendencies.
routes will be as short as possible and cover the greatest possible area (by not looping or branching)
routes will tend to service neighbourhoods, rather than trying to overlap responsibilities. This part of the algorithm isn't optimal, but is effective. This makes it really easy to remove service routes without needing to recalculate the transportation network.
the taxis chosen will be cost-efficient, helping to avoid paying more than necessary.
routes will use as few taxis as possible, taking into account the relative cost of upgrading to roomier ones with higher capacity
because the taxis travelling furthest will be full, it has less of an impact on your employee's ability to get to work if you decide to cancel service to emptier taxis.
Every step closer to perfection costs you many times more than the previous step, so diminished returns are acceptable if the solution provide desirable features. Although the algorithm makes some potentially sub-optimal tradeoffs, they come with huge value; your network of taxi routes becomes much easier to modify.
If you'd like to make an optimal solution, the Knapsack Problem, Coin Problem, and Change-making Problem help determine the cost of taxis and routes.
Spanning Trees are the most effective way to determine routes. Center the spanning tree at the office and calculate the cost of each branch as the maximum distance from the office. Try to keep each branch servicing areas with high density to make it easier to add and remove taxi routes.
Studying pathfinding can help you learn how to determine good cost functions so that you can numerically compare different potential paths. Remember that your network consists of a set of paths, but will require its own cost function so that you can compare different layouts.
I've written an in-depth guide to pathfinding for this answer. Pathfinding articles are few and just don't go into enough depth for a lot of problem spaces. A good cost function can get you a nearly perfect solution if you have multiple priorities. Unfortunately, good cost functions are domain specific so you will need to identify them yourself. Feel free to message me if you aren't sure how to make a path with certain traits and I'll help you figure out a good cost function.
It's a constraint satisfaction problem, not really a TSP. If you're looking for a project that may be able to help you, you could look into cspdb, which is what I wrote some time ago:
https://github.com/gtoonstra/cspdb
You'd be using a database in the backend that maintains the state and write a couple of scripts in it's own grammar that manipulates that state. A couple of examples are included to solve nqueens and classroom scheduling with multiple constraints.
From a list d destinations you can make the array of pairwise travel costs c. where c[a,b] is the travel cost from a to b...
Now you have a start point p. add to the array c2 values for p to each point in d.
So you now have the concept of groups.
You can look at this as a greedy algorithm. Given the list c2 you can take the cheapest option given your state.
Your state is the vector all you cab vectors (the costs of getting from where ever they are to where ever they could go next) .* the assignment vector where k == 0 for k in the. You find the minimum option given your state (considering adding another onerous to a cab of 4 costs the difference between the 4 person cab and the 7 person cab and adding a person to a zero person cab or adding a new cab also has a cost. Once all your people have been assigned to their cabs you have an answer.
The Idea of a greedy algorithm is most often characterized by the backpack problem but it can also be implemented for statistical methods such as feature selection.
Like #Aaron3468 this approach is not perfect and does not guarantee the best solution.
If you want the best possible solutions you can iterate through all the combinations but this becomes impractical quickly.
From my point of view your algorithm should solve 2 problems: the number of cars of each type and the shortest distance (how you number your employees depends on you or you should give more details). Sorry I'm a using a phone and I don't have have all features of the site.
For the number of cars you can use below algorithm. To solve issues related to distances, you should give more info about the paths and their lengths. A graph algorithm may then be combined with this to do the trick. Here 11=7+4.
Begin
Integer temp:= n/11
Integer rem:= n mod 11
If rem=0
x:=temp
y:=temp
Else if rem<=4
x:=temp
y:=temp+1
Else if rem<=7
x:=temp+1
y:=temp
Else
x:=temp+1
y:=temp+1
Endif
End
I'm trying to come up with a fast and reasonably optimal algorithm to solve the following TSP/hamiltonian-path-like problem:
A delivery vehicle has a number of pickups and dropoffs it needs to
perform:
For each delivery, the pickup needs to come before the
dropoff.
The vehicle is quite small and the packages vary in size.
The total carriage cannot exceed some upper bound (e.g. 1 cubic
metre). Each delivery has a deadline.
The planner can run mid-route, so the vehicle will begin with a number of jobs already picked up and some capacity already taken up.
A near-optimal solution should minimise the total cost (for simplicity, distance) between each waypoint. If a solution does not exist because of the time constraints, I need to find a solution that has the fewest number of late deliveries. Some illustrations of an example problem and a non-optimal, but valid solution:
I am currently using a greedy best first search with backtracking bounded to 100 branches. If it fails to find a solution with on-time deliveries, I randomly generate as many as I can in one second (the most computational time I can spare) and pick the one with the fewest number of late deliveries. I have looked into linear programming but can't get my head around it - plus I would think it would be inappropriate given it needs to be run very frequently. I've also tried algorithms that require mutating the tour, but the issue is mutating a tour nearly always makes it invalid due to capacity constraints and precedence. Can anyone think of a better heuristic approach to solving this problem? Many thanks!
Safe Moves
Here are some ideas for safely mutating an existing feasible solution:
Any two consecutive stops can always be swapped if they are both pickups, or both deliveries. This is obviously true for the "both deliveries" case; for the "both pickups" case: if you had room to pick up A, then pick up B without delivering anything in between, then you have room to pick up B first, then pick up A. (In fact a more general rule is possible: In any pure-delivery or pure-pickup sequence of consecutive stops, the stops can be rearranged arbitrarily. But enumerating all the possibilities might become prohibitive for long sequences, and you should be able to get most of the benefit by considering just pairs.)
A pickup of A can be swapped with any later delivery of something else B, provided that A's original pickup comes after B was picked up, and A's own delivery comes after B's original delivery. In the special case where the pickup of A is immediately followed by the delivery of B, they can always be swapped.
If there is a delivery of an item of size d followed by a pickup of an item of size p, then they can be swapped provided that there is enough extra room: specifically, provided that f >= p, where f is the free space available before the delivery. (We already know that f + d >= p, otherwise the original schedule wouldn't be feasible -- this is a hint to look for small deliveries to apply this rule to.)
If you are starting from purely randomly generated schedules, then simply trying all possible moves, greedily choosing the best, applying it and then repeating until no more moves yield an improvement should give you a big quality boost!
Scoring Solutions
It's very useful to have a way to score a solution, so that they can be ordered. The nice thing about a score is that it's easy to incorporate levels of importance: just as the first digit of a two-digit number is more important than the second digit, you can design the score so that more important things (e.g. deadline violations) receive a much greater weight than less important things (e.g. total travel time or distance). I would suggest something like 1000 * num_deadline_violations + total_travel_time. (This assumes of course that total_travel_time is in units that will stay beneath 1000.) We would then try to minimise this.
Managing Solutions
Instead of taking one solution and trying all the above possible moves on it, I would instead suggest using a pool of k solutions (say, k = 10000) stored in a min-heap. This allows you to extract the best solution in the pool in O(log k) time, and to insert new solutions in the same time.
You could initially populate the pool with randomly generated feasible solutions; then on each step, you would extract the best solution in the pool, try all possible moves on it to generate child solutions, and insert any child solutions that are better than their parent back into the pool. Whenever the pool doubles in size, pull out the first (i.e. best) k solutions and make a new min-heap with them, discarding the old one. (Performing this step after the heap grows to a constant multiple of its original size like this has the nice property of leaving the amortised time complexity unchanged.)
It can happen that some move on solution X produces a child solution Y that is already in the pool. This wastes memory, which is unfortunate, but one nice property of the min-heap approach is that you can at least handle these duplicates cheaply when they arrive at the front of the heap: all duplicates will have identical scores, so they will all appear consecutively when extracting solutions from the top of the heap. Thus to avoid having duplicate solutions generate duplicate children "down through the generations", it suffices to check that the new top of the heap is different from the just-extracted solution, and keep extracting and discarding solutions until this holds.
A note on keeping worse solutions: It might seem that it could be worthwhile keeping child solutions even if they are slightly worse than their parent, and indeed this may be useful (or even necessary to find the absolute optimal solution), but doing so has a nasty consequence: it means that it's possible to cycle from one solution to its child and back again (or possibly a longer cycle). This wastes CPU time on solutions we have already visited.
You are basically combining the Knapsack Problem with the Travelling Salesman Problem.
Your main problem here seems to be actually the Knapsack Problem, rather then the Travelling Salesman Problem, since it has the one hard restriction (maximum delivery volume). Maybe try to combine the solutions for the Knapsack Problem with the Travelling Salesman.
If you really only have one second max for calculations a greedy algorithm with backtracking might actually be one of the best solutions that you can get.
Massively edited this question to make it easier to understand.
Given an environment with arbitrary dimensions and arbitrary positioning of an arbitrary number of obstacles, I have an agent exploring the environment with a limited range of sight (obstacles don't block sight). It can move in the four cardinal directions of NSEW, one cell at a time, and the graph is unweighted (each step has a cost of 1). Linked below is a map representing the agent's (yellow guy) current belief of the environment at the instant of planning. Time does not pass in the simulation while the agent is planning.
http://imagizer.imageshack.us/a/img913/9274/qRsazT.jpg
What exploration algorithm can I use to maximise the cost-efficiency of utility, given that revisiting cells are allowed? Each cell holds a utility value. Ideally, I would seek to maximise the sum of utility of all cells SEEN (not visited) divided by the path length, although if that is too complex for any suitable algorithm then the number of cells seen will suffice. There is a maximum path length but it is generally in the hundreds or higher. (The actual test environments used on my agent are at least 4x bigger, although theoretically there is no upper bound on the dimensions that can be set, and the maximum path length would thus increase accordingly)
I consider BFS and DFS to be intractable, A* to be non-optimal given a lack of suitable heuristics, and Dijkstra's inappropriate in generating a single unbroken path. Is there any algorithm you can think of? Also, I need help with loop detection, as I've never done that before since allowing revisitations is my first time.
One approach I have considered is to reduce the map into a spanning tree, except that instead of defining it as a tree that connects all cells, it is defined as a tree that can see all cells. My approach would result in the following:
http://imagizer.imageshack.us/a/img910/3050/HGu40d.jpg
In the resultant tree, the agent can go from a node to any adjacent nodes that are 0-1 turn away at intersections. This is as far as my thinking has gotten right now. A solution generated using this tree may not be optimal, but it should at least be near-optimal with much fewer cells being processed by the algorithm, so if that would make the algorithm more likely to be tractable, then I guess that is an acceptable trade-off. I'm still stuck with thinking how exactly to generate a path for this however.
Your problem is very similar to a canonical Reinforcement Learning (RL) problem, the Grid World. I would formalize it as a standard Markov Decision Process (MDP) and use any RL algorithm to solve it.
The formalization would be:
States s: your NxM discrete grid.
Actions a: UP, DOWN, LEFT, RIGHT.
Reward r: the value of the cells that the agent can see from the destination cell s', i.e. r(s,a,s') = sum(value(seen(s')).
Transition function: P(s' | s, a) = 1 if s' is not out of the boundaries or a black cell, 0 otherwise.
Since you are interested in the average reward, the discount factor is 1 and you have to normalize the cumulative reward by the number of steps. You also said that each step has cost one, so you could subtract 1 to the immediate reward rat each time step, but this would not add anything since you will already average by the number of steps.
Since the problem is discrete the policy could be a simple softmax (or Gibbs) distribution.
As solving algorithm you can use Q-learning, which guarantees the optimality of the solution provided a sufficient number of samples. However, if your grid is too big (and you said that there is no limit) I would suggest policy search algorithms, like policy gradient or relative entropy (although they guarantee convergence only to local optima). You can find something about Q-learning basically everywhere on the Internet. For a recent survey on policy search I suggest this.
The cool thing about these approaches is that they encode the exploration in the policy (e.g., the temperature in a softmax policy, the variance in a Gaussian distribution) and will try to maximize the cumulative long term reward as described by your MDP. So usually you initialize your policy with a high exploration (e.g., a complete random policy) and by trial and error the algorithm will make it deterministic and converge to the optimal one (however, sometimes also a stochastic policy is optimal).
The main difference between all the RL algorithms is how they perform the update of the policy at each iteration and manage the tradeoff exploration-exploitation (how much should I explore VS how much should I exploit the information I already have).
As suggested by Demplo, you could also use Genetic Algorithms (GA), but they are usually slower and require more tuning (elitism, crossover, mutation...).
I have also tried some policy search algorithms on your problem and they seems to work well, although I initialized the grid randomly and do not know the exact optimal solution. If you provide some additional details (a test grid, the max number of steps and if the initial position is fixed or random) I can test them more precisely.
I'm dealing with a war game. I have a list of my bases B(x,y) from which I can send attacks on the enemy (they have bases between my own bases). Each base B can attack at a range R (the same radius for all bases). How can I find my bases to be able to attack as many enemy bases as possible, but use a minimum number of my bases?
I've reduced the problem to finding the minimum number of bases (and their coordinates) required to cover the largest area possible. I wonder if there is a better way than looking at all the possible combinations and because the number of bases could reach thousands.
Example: If the attack radius is 10 and I have five bases in a square and its center: (0,0), (10,0), (10,10), (0,10), (5,5) then the answer is that only the first four would be needed because all the area covered by the one in the center is already covered by the others.
Note 1 The solution must be single-threaded.
Note 2 The solution doesn't have to be perfect if that means a big gain in speed. The number of bases reaches thousands and this needs to use as little time as possible. I would consider running time greater than 100 ms for 10,000 bases in Python on a modern computer unacceptable, so I was thinking maybe I could start by eliminating the obvious, like if there are multiple bases within R/10 distance of each other, simply eliminate all except for one (whichever).
If I understand you correctly, the enemy bases and your bases are given as well as the (constant) attack radius. I.e. if you select one of your bases, you know exactly which of the enemy bases get attacked due to the selection.
The first step would be to eliminate those enemy cities from the problem which can not be attacked by any of your bases. Then, selecting all of your bases guarantees attacking all attackable enemy bases, so there is solution that attacks as many enemy bases as possible.
Under all those solutions you are looking for the one that uses the minimum number of your bases. This problem is equivalent to the https://en.wikipedia.org/wiki/Set_cover_problem, which is unfortunately NP-hard. You can apply all known solution methods such as Integer Linear Programming or the already mentioned greedy algorithm / metaheuristics.
If your problem instance is large and runtime is the primary concern, greedy is probably the way to go. For example you could always add that particular base of yours to the selection which adds the highest number of enemy bases that can be attacked which were previously not under attack by your already selected bases.
Hum the solution depends on your needs. If you need real time answer, maybe a greedy algorithm could provide good solution.
Other solution could be using meta-heuristic with constraint time(http://en.wikipedia.org/wiki/Metaheuristic). I probably would use genetic algorithm to search a solution for this problem under a limited time.
If interested I can provide a toy example of implementation in Python.
EDIT :
When you have to provide solution quickly a greedy algorithm is often better. But in your case I doubt. Particularity of many greedy algorithm is that you need to start from scratch each time you try to compute a new result.
Speaking again of genetic algorithm, you could for example each time you have to take a decision restart the search process from its last result. In fact you could probably let him turning has a subprocess and each 100ms take the better solution computed during the last loop.
If not too greedy in computing resource, this solution would provide better results than greedy one on the long run as the solution will probably need to be adapted to the changes of the situation but many element will stay unchanged. Just be aware that initializing a meta-search with the solution of a greedy algorithm is anyway a good idea!
Let's say I have N taxis, and N customers waiting to be picked up by the taxis. The initial positions of both customers and taxis are random/arbitrary.
Now I want to assign each taxi to exactly one customer.
The customers are all stationary, and the taxis all move at identical speed. For simplicity, let's assume there are no obstacles, and the taxis can move in straight lines to assigned customers.
I now want to minimize the time until the last customer enters his/her taxi.
Is there a standard algorithm to solve this? I have tens of thousands of taxis/customers. Solution doesn't have to be optimal, just ‘good’.
The problem can almost be modelled as the standard “Assignment Problem”, solvable using the Hungarian algorithm (the Kuhn–Munkres algorithm or Munkres assignment algorithm). However, I want to minimize the cost of the costliest assignment, not minimize the sum of costs of the assignments.
Since you mentioned Hungarian Algorithm, I guess one thing you could do is using some different measure of distance rather than the euclidean distance and then run t Hungarian Algorithm on it. For example, instead of using
d = sqrt((x0 - x1) ^ 2 + (y1 - y0) ^ 2)
use
d = ((x0 - x1) ^ 2 + (y1 - y0) ^ 2) ^ 10
that could cause the algorithm to penalize big numbers heavily, which could constrain the length of the max distance.
EDIT: This paper "Geometry Helps in Bottleneck Matching and Related
Problems" may contains a better algorithm. However, I am still in the process of reading it.
I'm not sure that the Hungarian algorithm will work for your problem here. According to the link, it runs in n ^ 3 time. Plugging in 25,000 as n would yield 25,000 ^ 3 = 15,625,000,000,000. That could take quite a while to run.
Since the solution does not need to be optimal, you might consider using simulated annealing or possibly a genetic algorithm instead. Either of these should be much faster and still produce close to optimal solutions.
If using a genetic algorithm, the fitness function can be designed to minimize the longest period of time that an individual would need to wait. But, you would have to be careful because if that is the sole criteria, then the solution won't work too well for cases when there is just one cab that is closest to the passenger that is furthest away. So, the fitness function would need to take into account the other waiting times as well. One idea to solve this would be to run the model iteratively and remove the longest cab trip (both cab & person) after each iteration. But, doing that for all 10,000+ cabs/people could be expensive time wise.
I don't think any cab owner or manager would even consider minimizing the waiting time for the last customer entering his cab over minimizing the sum of the waiting time for all cabs - simply because they make more money overall when minimizing the sum of the waiting times. At least Louie DePalma would never do that... So, I suspect that the real problem you have has little or nothing to do with cabs...
A "good" algorithm that would solve your problem is a Greedy Algorithm. Since taxis and people have a position, these positions can be related to a "central" spot. Sort the taxis and people needing to get picked up in order (in relation to the "centre"). Then start assigning taxis, in order, to pick up people in order. This greedy rule will ensure taxis closest to the centre will pick up people closest to the centre and taxis farthest away pick up people farthest away.
A better way might be to use Dynamic Programming however, I am not sure nor have the time to invest. A good tutorial for Dynamic Programming can be found here
For an optimal solution: construct a weighted bipartite graph with a vertex for each taxi and customer and an edge from each taxi to each customer whose weight is the travel time. Scan the edges in order of nondecreasing weight, maintaining a maximum matching of the subgraph containing the edges scanned so far. Stop when the matching is perfect.