I'm working on implementing a model in Python. As part of this model, I have a set of agents (e.g. humans) that need to visit a set of targets (e.g. places). Each agent has its own initial location (i.e. starting point) and I can calculate the distance from each agent to each target.
What I need at this point is to allocate a first job to each agent in a way that the sum of all travel distances for agents from their starting location to their first job is minimum.
I considered greedy algorithm, but I found examples that proves order of allocation can lead to non-optimal solutions. I also looked into nearest neighbour algorithm in TSP, but all I could find was for one agent (or salesman) not multiple.
Could someone point me to any (non-exhaustive search) algorithm/approach that could be used for this purpose please? Thanks
If the number of agents = number of targets, we end up with a standard assignment problem. This can be solved in different ways:
as an LP (linear programming problem). Technically a MIP but variables are automatically integer-valued, so an LP solver suffices.
as a network problem
or using specialized algorithms.
If, say, the number of locations > number of agents, we still can use an LP/MIP:
min sum((i,j), d(i,j)*x(i,j))
sum(j, x(i,j)) = 1 for all agents i (each agent should be assigned to exactly one location)
sum(i, x(i,j)) <= 1 for all locations j (each location should be assigned to at most one agent)
x(i,j) in {0,1}
For the network approach, we would need to add some dummy nodes.
All these methods are quite fast (this is an easy model). To give you an indication: I solved a random example with 500 agents and 1000 locations as an LP and it took 0.3 seconds.
Related
The "Traveling Salesman Problem" is a problem where a person has to travel between "n" cities - but choose the itinerary such that:
Each city is visited only once
The total distance traveled is minimized
I have heard that if a modern computer were the solve this problem using "brute force" (i.e. an exact solution) - if there are more than 15 cities, the time taken by the computer will exceed a hundred years!
I am interested in understanding "how do we estimate the amount of time it will take for a computer to solve the Traveling Salesman Problem (using "brute force") as the number of cities increase". For instance, from the following reference (https://www.sciencedirect.com/topics/earth-and-planetary-sciences/traveling-salesman-problem):
My Question: Is there some formula we can use to estimate the number of time it will take a computer to solve Traveling Salesman using "brute force"? For example:
N cities = N! paths
Each of these N! paths will require "N" calculations
Thus, N * N calculations would be required for the computer to check all paths and then be certain that the shortest path has been found : If we know the time each calculation takes, perhaps we could estimate the total run time as "time per calculation * N*N! "
But I am not sure if this factors in the time to "store and compare" calculations.
Can someone please explain this?
I have heard that if a modern computer were the solve this problem using "brute force" (i.e. an exact solution) - if there are more than 15 cities, the time taken by the computer will exceed a hundred years!
This is not completely true. While the naive brute-force algorithm runs with a n! complexity. A much better algorithm using dynamic programming runs in O(n^2 2^n). Just to give you an idea, with n=25, n! ≃ 2.4e18 while n^2 2^n ≃ 1e12. The former is too huge to be practicable while the second could be OK although it should take a pretty long time on a PC (one should keep in mind that both algorithm complexities contain an hidden constant variable playing an important role to compute a realistic execution time). I used an optimized dynamic programming solution based on the Held–Karp algorithm to compute the TSP of 20 cities on my machine with a relatively reasonable time (ie. no more than few minutes of computation).
Note that in practice heuristics are used to speed up the computation drastically often at the expense of a sub-optimal solution. Some algorithm can provide a good result in a very short time compared to the previous exact algorithms (polynomial algorithms with a relatively small exponent) with a fixed bound on the quality of the result (for example the distance found cannot be bigger than 2 times the optimal solution). In the end, heuristics can often found very good results in a reasonable time. One simple heuristic is to avoid crossing segments assuming an Euclidean distance is used (AFAIK a solution with crossing segments is always sub-optimal).
My Question: Is there some formula we can use to estimate the number of time it will take a computer to solve Travelling Salesman using "brute force"?
Since the naive algorithm is compute bound and quite simple, you can do such an approximation based on the running-time complexity. But to get a relatively precise approximation of the execution time, you need a calibration since not all processors nor implementations behave the same way. You can assume that the running time is C n! and find the value of C experimentally by measuring the computation time taken by a practical brute-force implementation. Another approach is to theoretically find the value of C based on low-level architectural properties (eg. frequency, number of core used, etc.) of the target processor. The former is much more precise assuming the benchmark is properly done and the number of points is big enough. Moreover, the second method requires a pretty good understanding of the way modern processors work.
Numerically, assuming a running time t ≃ C n!, we can say that ln t ≃ ln(C n!) ≃ ln(C) + ln(n!). Based on the Stirling's approximation, we can say that ln t ≃ ln C + n ln n + O(ln n), so ln C ≃ ln t - n ln n - O(ln n). Thus, ln C ≃ ln t - n ln n - O(ln n) and finally, C ≃ exp(ln t - n ln n) (with an O(n) approximation). That being said, the Stirling's approximation may not be precise enough. Using a binary search to numerically compute the inverse gamma function (which is a generalization of the factorial) should give a much better approximation for C.
Each of these N! paths will require "N" calculations
Well, a slightly optimized brute-force algorithm do not need perform N calculation as the partial path length can be precomputed. The last loops just need to read the precomputed sums from a small array that should be stored in the L1 cache (so it take only no more than few cycle of latency to read/store).
I have implemented an algorithm that uses two other algorithms for calculating the shortest path in a graph: Dijkstra and Bellman-Ford. Based on the time complexity of the these algorithms, I can calculate the running time of my implementation, which is easy giving the code.
Now, I want to experimentally verify my calculation. Specifically, I want to plot the running time as a function of the size of the input (I am following the method described here). The problem is that I have two parameters - number of edges and number of vertices.
I have tried to fix one parameter and change the other, but this approach results in two plots - one for varying number of edges and the other for varying number of vertices.
This leads me to my question - how can I determine the order of growth based on two plots? In general, how can one experimentally determine the running time complexity of an algorithm that has more than one parameter?
It's very difficult in general.
The usual way you would experimentally gauge the running time in the single variable case is, insert a counter that increments when your data structure does a fundamental (putatively O(1)) operation, then take data for many different input sizes, and plot it on a log-log plot. That is, log T vs. log N. If the running time is of the form n^k you should see a straight line of slope k, or something approaching this. If the running time is like T(n) = n^{k log n} or something, then you should see a parabola. And if T is exponential in n you should still see exponential growth.
You can only hope to get information about the highest order term when you do this -- the low order terms get filtered out, in the sense of having less and less impact as n gets larger.
In the two variable case, you could try to do a similar approach -- essentially, take 3 dimensional data, do a log-log-log plot, and try to fit a plane to that.
However this will only really work if there's really only one leading term that dominates in most regimes.
Suppose my actual function is T(n, m) = n^4 + n^3 * m^3 + m^4.
When m = O(1), then T(n) = O(n^4).
When n = O(1), then T(n) = O(m^4).
When n = m, then T(n) = O(n^6).
In each of these regimes, "slices" along the plane of possible n,m values, a different one of the terms is the dominant term.
So there's no way to determine the function just from taking some points with fixed m, and some points with fixed n. If you did that, you wouldn't get the right answer for n = m -- you wouldn't be able to discover "middle" leading terms like that.
I would recommend that the best way to predict asymptotic growth when you have lots of variables / complicated data structures, is with a pencil and piece of paper, and do traditional algorithmic analysis. Or possibly, a hybrid approach. Try to break the question of efficiency into different parts -- if you can split the question up into a sum or product of a few different functions, maybe some of them you can determine in the abstract, and some you can estimate experimentally.
Luckily two input parameters is still easy to visualize in a 3D scatter plot (3rd dimension is the measured running time), and you can check if it looks like a plane (in log-log-log scale) or if it is curved. Naturally random variations in measurements plays a role here as well.
In Matlab I typically calculate a least-squares solution to two-variable function like this (just concatenates different powers and combinations of x and y horizontally, .* is an element-wise product):
x = log(parameter_x);
y = log(parameter_y);
% Find a least-squares fit
p = [x.^2, x.*y, y.^2, x, y, ones(length(x),1)] \ log(time)
Then this can be used to estimate running times for larger problem instances, ideally those would be confirmed experimentally to know that the fitted model works.
This approach works also for higher dimensions but gets tedious to generate, maybe there is a more general way to achieve that and this is just a work-around for my lack of knowledge.
I was going to write my own explanation but it wouldn't be any better than this.
Consider a problem whose solution maximizes an objective function.
Problem : From 500 elements, 15 needs to be selected (candidate solution), Value of Objective function depends on the pairwise relationships between the elements in a candidate solution and some more.
The steps for solving such a problem is described here:
1. Generate a set of candidate solutions in guided random manner(population) //not purely random the direction is given to generate the population
2. Evaluating the objective function for current population
3. If the current_best_solution exceeds the global_best_solution, then replace the global_best with current_best
4. Repeat steps 1,2,3 for N (arbitrary number) times
where size of population and N are smaller (approx 50)
After N iterations it returns a candidate solution stored in global_best_solution
Is this the description of a well-known algorithm?
If it is, what is the name of that algorithm or if not under which category these type of algorithms fit?
What you have sounds like you are just fishing. Note that you might as well get rid of steps 3 and 4 since running the loop 100 times would be the same as doing it once with an initial population 100 times as large.
If you think of the objective function as a random variable which is a function of random decision variables then what you are doing would e.g. give you something in the 99.9th percentile with very high probability -- but there is no limit to how far the optimum might be from the 99.9th percentile.
To illustrate the difficulty, consider the following sort of Travelling Salesman Problem. Imagine two clusters of points A and B, each of which has 100 points. Within the clusters, each point is arbitrarily close to every other point (e.g. 0.0000001). But -- between the clusters the distance is say 1,000,000. The optimal tour would clearly have length 2,000,000 (+ a negligible amount). A random tour is just a random permutation of those 200 decision points. Getting an optimal or near optimal tour would be akin to shuffling a deck of 200 cards with 100 read and 100 black and having all of the red cards in the deck in a block (counting blocks that "wrap around") -- vanishingly unlikely (It can be calculated as 99 * 100! * 100! / 200! = 1.09 x 10^-57). Even if you generate quadrillions of tours it is overwhelmingly likely that each of those tours would be off by millions. This is a min problem, but it is also easy to come up with max problems where it is vanishingly unlikely that you will get a near-optimal solution by purely random settings of the decision variables.
This is an extreme example, but it is enough to show that purely random fishing for a solution isn't very reliable. It would make more sense to use evolutionary algorithms or other heuristics such as simulated annealing or tabu search.
why do you work with a population if the members of that population do not interact ?
what you have there is random search.
if you add mutation it looks like an Evolution Strategy: https://en.wikipedia.org/wiki/Evolution_strategy
I try to implement k-means as a homework assignment. My exercise sheet gives me following remark regarding empty centers:
During the iterations, if any of the cluster centers has no data points associated with it, replace it with a random data point.
That confuses me a bit, firstly Wikipedia or other sources I read do not mention that at all. I further read about a problem with 'choosing a good k for your data' - how is my algorithm supposed to converge if I start setting new centers for cluster that were empty.
If I ignore empty clusters I converge after 30-40 iterations. Is it wrong to ignore empty clusters?
Check out this example of how empty clusters can happen: http://www.ceng.metu.edu.tr/~tcan/ceng465_f1314/Schedule/KMeansEmpty.html
It basically means either 1) a random tremor in the force, or 2) the number of clusters k is wrong. You should iterate over a few different values for k and pick the best.
If during your iterating you should encounter an empty cluster, place a random data point into that cluster and carry on.
I hope this helped on your homework assignment last year.
Handling empty clusters is not part of the k-means algorithm but might result in better clusters quality. Talking about convergence, it is never exactly but only heuristically guaranteed and hence the criterion for convergence is extended by including a maximum number of iterations.
Regarding the strategy to tackle down this problem, I would say randomly assigning some data point to it is not very clever since we might be affecting the clusters quality since the distance to its currently assigned center is large or small. An heuristic for this case would be to choose the farthest point from the biggest cluster and move that the empty cluster, then do so until there are no empty clusters.
Statement: k-means can lead to
Consider above distribution of data points.
overlapping points mean that the distance between them is del. del tends to 0 meaning you can assume arbitary small enough value eg 0.01 for it.
dash box represents cluster assign
legend in footer represents numberline
N=6 points
k=3 clusters (coloured)
final clusters = 2
blue cluster is orphan and ends up empty.
Empty clusters can be obtained if no points are allocated to a cluster during the assignment step. If this happens, you need to choose a replacement centroid otherwise SSE would be larger than neccessary.
*Choose the point that contributes most to SSE
*Choose a point from the cluster with the highest SSE
*If there are several empty clusters, the above can be repeated several times.
***SSE = Sum of Square Error.
Check this site https://chih-ling-hsu.github.io/2017/09/01/Clustering#
You should not ignore empty clusters but replace it. k-means is an algorithm could only provides you local minimums, and the empty clusters are the local minimums that you don't want.
your program is going to converge even if you replace a point with a random one. Remember that at the beginning of the algorithm, you choose the initial K points randomly. if it can converge, how come K-1 converge points with 1 random point can't? just a couple more iterations are needed.
"Choosing good k for your data" refers to the problem of choosing the right number of clusters. Since the k-means algorithm works with a predetermined number of cluster centers, their number has to be chosen at first. Choosing the wrong number could make it hard to divide the data points into clusters or the clusters could become small and meaningless.
I can't give you an answer on whether it is a bad idea to ignore empty clusters. If you do, you might end up with a smaller number of clusters than you defined at the beginning. This will confuse people who expect k-means to work in a certain way, but it is not necessarily a bad idea.
If you re-locate any empty cluster centers, your algorithm will probably converge anyway if that happens a limited number of times. However, you if you have to relocate too often, it might happen that your algorithm doesn't terminate.
For "Choosing good k for your data", Andrew Ng gives the example of a tee shirt manufacturer looking at potential customer measurements and doing k-means to decide if you want to offer S/M/L (k=3) or 2XS/XS/S/M/L/XL/2XL (k=7). Sometimes the decision is driven by the data (k=7 gives empty clusters) and sometimes by business considerations (manufacturing costs are less with only three sizes, or marketing says customers want more choices).
Set a variable to track the farthest distanced point and its cluster based on the distance measure used.
After the allocation step for all the points, check the number of datapoints in each cluster.
If any is 0, as is the case for this question, split the biggest cluster obtained and split further into 2 sub-clusters.
Replace the selected cluster with these sub-clusters.
I hope the issue is fixed now.. Random assignment will affect the clustering structure of the already obtained clustering.
I am using simulated annealing to solve an NP-complete resource scheduling problem. For each candidate ordering of the tasks I compute several different costs (or energy values). Some examples are (though the specifics are probably irrelevant to the question):
global_finish_time: The total number of days that the schedule spans.
split_cost: The number of days by which each task is delayed due to interruptions by other tasks (this is meant to discourage interruption of a task once it has started).
deadline_cost: The sum of the squared number of days by which each missed deadline is overdue.
The traditional acceptance probability function looks like this (in Python):
def acceptance_probability(old_cost, new_cost, temperature):
if new_cost < old_cost:
return 1.0
else:
return math.exp((old_cost - new_cost) / temperature)
So far I have combined my first two costs into one by simply adding them, so that I can feed the result into acceptance_probability. But what I would really want is for deadline_cost to always take precedence over global_finish_time, and for global_finish_time to take precedence over split_cost.
So my question to Stack Overflow is: how can I design an acceptance probability function that takes multiple energies into account but always considers the first energy to be more important than the second energy, and so on? In other words, I would like to pass in old_cost and new_cost as tuples of several costs and return a sensible value .
Edit: After a few days of experimenting with the proposed solutions I have concluded that the only way that works well enough for me is Mike Dunlavey's suggestion, even though this creates many other difficulties with cost components that have different units. I am practically forced to compare apples with oranges.
So, I put some effort into "normalizing" the values. First, deadline_cost is a sum of squares, so it grows exponentially while the other components grow linearly. To address this I use the square root to get a similar growth rate. Second, I developed a function that computes a linear combination of the costs, but auto-adjusts the coefficients according to the highest cost component seen so far.
For example, if the tuple of highest costs is (A, B, C) and the input cost vector is (x, y, z), the linear combination is BCx + Cy + z. That way, no matter how high z gets it will never be more important than an x value of 1.
This creates "jaggies" in the cost function as new maximum costs are discovered. For example, if C goes up then BCx and Cy will both be higher for a given (x, y, z) input and so will differences between costs. A higher cost difference means that the acceptance probability will drop, as if the temperature was suddenly lowered an extra step. In practice though this is not a problem because the maximum costs are updated only a few times in the beginning and do not change later. I believe this could even be theoretically proven to converge to a correct result since we know that the cost will converge toward a lower value.
One thing that still has me somewhat confused is what happens when the maximum costs are 1.0 and lower, say 0.5. With a maximum vector of (0.5, 0.5, 0.5) this would give the linear combination 0.5*0.5*x + 0.5*y + z, i.e. the order of precedence is suddenly reversed. I suppose the best way to deal with it is to use the maximum vector to scale all values to given ranges, so that the coefficients can always be the same (say, 100x + 10y + z). But I haven't tried that yet.
mbeckish is right.
Could you make a linear combination of the different energies, and adjust the coefficients?
Possibly log-transforming them in and out?
I've done some MCMC using Metropolis-Hastings. In that case I'm defining the (non-normalized) log-likelihood of a particular state (given its priors), and I find that a way to clarify my thinking about what I want.
I would take a hint from multi-objective evolutionary algorithm (MOEA) and have it transition if all of the objectives simultaneously pass with the acceptance_probability function you gave. This will have the effect of exploring the Pareto front much like the standard simulated annealing explores plateaus of same-energy solutions.
However, this does give up on the idea of having the first one take priority.
You will probably have to tweak your parameters, such as giving it a higher initial temperature.
I would consider something along the lines of:
If (new deadline_cost > old deadline_cost)
return (calculate probability)
else if (new global finish time > old global finish time)
return (calculate probability)
else if (new split cost > old split cost)
return (calculate probability)
else
return (1.0)
Of course each of the three places you calculate the probability could use a different function.
It depends on what you mean by "takes precedence".
For example, what if the deadline_cost goes down by 0.001, but the global_finish_time cost goes up by 10000? Do you return 1.0, because the deadline_cost decreased, and that takes precedence over anything else?
This seems like it is a judgment call that only you can make, unless you can provide enough background information on the project so that others can suggest their own informed judgment call.