Scheduling: Minimizing Gaps between Non-Overlapping Time Ranges - python

Using Django to develop a small scheduling web application where people are assigned certain times to meet with their superiors. Employees are stored as models, with a OneToMany relation to a model representing time ranges and day of the week where they are free. For instance:
Bob: (W 9:00, 9:15), (W 9:15, 9:30), ... (W 15:00, 15:20)
Sarah: (Th 9:05, 9:20), (F 9:20, 9:30), ... (Th 16:00, 16:05)
...
Mary: (W 8:55, 9:00), (F 13:00, 13:35), ... etc
My program allows a basic schedule setup, where employers can choose to view the first N possible schedules with the least gaps in between meetings under the condition that they meet all their employees at least once during that week. I am currently generating all possible permutations of meetings, and filtering out schedules where there are overlaps in meeting times. Is there a way to generate the first N schedules out of M possible ones, without going through all M possibilities?
Clarification: We are trying to get the minimum sum of gaps for any given day, summed over all days.

I would use a search algorithm, like A-star, to do this. Each node in the graph represents a person's available time slots and a path from one node to another means that node_a and node_b are in the partial schedule.
Another solution would be to create a graph in which the nodes are each person's availability times and there is a edge from node_a to node_b if the person associated with node_a is not the same as the person associated with node_b. The weight of each node is the amount of time between the time associated with the two nodes.
After creating this graph, you could generate a variant of a minimum spanning tree from the graph. The variant would differ from MSTs in that:
you'll only add a node to the MST if the person associated with the node is not already in the MST.
you finish creating the MST when all persons are in the MST.
The minimum spanning tree generated would represent a single schedule.
To generate other schedules, remove all the edges from the graph which are found in the schedule you just created and then create a new minimum spanning tree from the graph with the removed edges.

In general, scheduling problems are NP-hard, and while I can't figure out a reduction for this problem to prove it such, it's quite similar to a number of other well-known NP-complete problems. There may be a polynomial-time solution for finding the minimum gap for a single day (though I don't know it off hand, either), but I have less hopes for needing to solve it for multiple days. Unfortunately, it's a complicated problem, and there may not be a perfectly elegant answer. (Or, I'm going to kick myself when someone posts one later.)
First off, I'd say that if your dataset is reasonably small and you've been able to compute all possible schedules fairly quickly, you may just want to stick with that solution, as all others will be approximations, and could possibly end up running slower, if the constant factor of their running time is large. (Meaning that it doesn't grow with the size of the dataset, so it will relatively be smaller for a large dataset.)
The simplest approximation would be to just use a greedy heuristic. It will almost assuredly not find the optimum schedules, and may take a long time to find a solution if most of the schedules are overlapping, and there are only a few that are even valid solutions - but I'm going to assume that this is not the case for employee times.
Start with an arbitrary schedule, choosing one timeslot for each employee at random. For each iteration, pick one employee and change his timeslot to the best possible time, with respect to the rest of current schedule. Repeat this process until your satisfied with the result - when it isn't improving quickly enough anymore or has taken too long. You're probably not going to want to repeat until you can't make any more changes that improve the schedule, since this process will likely loop for most data.
It's not a great heuristic, but it should produce some reasonable schedules, and has a lot of room for adjustment. You may want to always try to switch overlapping times first before any others, or you may want to try to flip the employee who currently contributes to the largest gap, or maybe eliminate certain slots that you've already tried. You may want to sometimes allow a move to a less optimal solution in hopes that you're at a local minima and want to get out of it - some randomness can also help with this. Make sure you always keep track of the best solution you've seen so far.
To generate more schedules, the most obvious thing would be to just start the process over with a different random schedule. Or, maybe flip a few arbitrary times from the previous solution you found, and repeat from there.
Edit: This is all fairly related to genetic algorithms, and you may want to use some of the ideas I presented here in a GA.

Related

Match the vertices of two identical labelled graphs

I have a rather simple problem to define but I did not find a simple answer so far.
I have two graphs (ie sets of vertices and edges) which are identical. Each of them has independently labelled vertices. Look at the example below:
How can the computer detect, without prior knowledge of it, that 1 is identical to 9, 2 to 10 and so on?
Note that in the case of symmetry, there may be several possible one to one pairings which give complete equivalence, but just finding one of them is sufficient to me.
This is in the context of a Python implementation. Does someone have a pointer towards a simple algorithm publicly available on the Internet? The problem sounds simple but I simply lack the mathematical knowledge to come up to it myself or to find proper keywords to find the information.
EDIT: Note that I also have atom types (ie labels) for each graphs, as well as the full distance matrix for the two graphs to align. However the positions may be similar but not exactly equal.
This is known as the graph isomorphism problem, and probably very hard; although the exactly details of how hard are still subject of research.
(But things look better if you graphs are planar.)
So, after searching for it a bit, I think that I found a solution that works most of the time for moderate computational cost. This is a kind of genetic algorithm which uses a bit of randomness, but it is practical enough for my purposes it seems. I didn't have any aberrant configuration with my samples so far even if it is theoretically possible that this happens.
Here is how I proceeded:
Determine the complete set of 2-paths, 3-paths and 4-paths
Determine vertex types using both atom type and surrounding topology, creating an "identity card" for each vertex
Do the following ten times:
Start with a random candidate set of pairings complying with the allowed vertex types
Evaluate how much of 2-paths, 3-paths and 4-paths correspond between the two pairings by scoring one point for each corresponding vertex (also using the atom type as an additional descriptor)
Evaluate all other shortlisted candidates for a given vertex by permuting the pairings for this candidate with its other positions in the same way
Sort the scores in descending order
For each score, check if the configuration is among the excluded configurations, and if it is not, take it as the new configuration and put it into the excluded configurations.
If the score is perfect (ie all of the 2-paths, 3-paths and 4-paths correspond), then stop the loop and calculate the sum of absolute differences between the distance matrices of the two graphs to pair using the selected pairing, otherwise go back to 4.
Stop this process after it has been done 10 times
Check the difference between distance matrices and take the pairings associated with the minimal sum of absolute differences between the distance matrices.

How do write a code to distribute different weights as evenly as possible among 4 boxes?

I was given a problem in which you are supposed to write a python code that distributes a number of different weights among 4 boxes.
Logically we can't expect a perfect distribution as in case we are given weights like 10, 65, 30, 40, 50 and 60 kilograms, there is no way of grouping those numbers without making one box heavier than another. But we can aim for the most homogenous distribution. ((60),(40,30),(65),(50,10))
I can't even think of an algorithm to complete this task let alone turn it into python code. Any ideas about the subject would be appreciated.
The problem you're describing is similar to the "fair teams" problem, so I'd suggest looking there first.
Because a simple greedy algorithm where weights are added to the lightest box won't work, the most straightforward solution would be a brute force recursive backtracking algorithm that keeps track of the best solution it has found while iterating over all possible combinations.
As stated in #j_random_hacker's response, this is not going to be something easily done. My best idea right now is to find some baseline. I describe a baseline as an object with the largest value since it cannot be subdivided. Using that you can start trying to match the rest of the data to that value which would only take about three iterations to do. The first and second would create a list of every possible combination and then the third can go over that list and compare the different options by taking the average of each group and storing the closest average value to your baseline.
Using your example, 65 is the baseline and since you cannot subdivide it you know that has to be the minimum bound on your data grouping so you would try to match all of the rest of the values to that. It wont be great, but it does give you something to start with.
As j_random_hacker notes, the partition problem is NP-complete. This problem is also NP-complete by a reduction from the 4-partition problem (the article also contains a link to a paper by Garey and Johnson that proves that 4-partition itself is NP-complete).
In particular, given a list to 4-partition, you could feed that list as an input to a function that solves your box distribution problem. If each box had the same weight in it, a 4-partition would exist, otherwise not.
Your best bet would be to create an exponential time algorithm that uses backtracking to iterate over the 4^n possible assignments. Because unless P = NP (highly unlikely), no polynomial time algorithm exists for this problem.

Rising and Falling Edge in multiple signals - PYTHON

This is the global scenario: I'm recording some simple signals from a novel sensor using Python 3.8. I have already filtered signals to have a better representations where let run other algorithms of Data Analysis. Nothing of special.
Following some signals on which I need to run my algorithm:
First Example
Second Example
These signals came out a sensor whose I am working on. My aim is to get the timestamps where signals starting to increase or decrease. I actually need to run this algorithm for only one signal (blue or orange).
I have reported both signals because they have antagonistic behaviour and maybe could be useful to achieve my task.
In other words, these signals are regarded to Foot Flexion Extension (FLE/EXT), then the point where they start to increase represents the point when I start to move my foot. Viceversa, when I move back my foot it results on decreasing signals amplitude.
My job is to identify the FLE/EXT and I tried to examine first derivative but it appears to don't give me any useful information.
I also have tried to use a convolution with a fixed-lenght ones-array by looking for when the successive convulution's average is greater than the current one.
This approach has 2 constraints:
Fixed-lenght array: because when signals represents faster FLE/EXT (then in less temporale distance in x-axis) the window is not enough to catch variation.
Threshold's Criterion for choosing how much has to be the successive average respect to the current one in order to save this iteration for my purpose.
I have stuck here, because I want to use a dynamic threshold approach or something similar which can allow me to exclude any fixed thresholds.
I hope to have a discussion with you for solving my problem. What do you think?
Please, if something is unclear, I am ready to clarify better.
Best regards,
V

Do I need to use a bin packing algorithm, or knapsack?

Here's the problem statement:
I have m chocolate bars, of integer length, and n children who
want integer amounts of chocolate. Where the total chocolate needs of
the children are less than or equal to the total amount of chocolate
you have. You need to write an algorithm that distributes chocolate to
the children by making the least number of cuts to the bars.
For example, for M = {1,3,7}, N = {1,3,4}, the least number of cuts would be 1.
I don't have any formal experience with algorithms, could anyone give me any hints on what I should start reading to tackle this problem in an efficient way?
This task can be reduced to solving several knapsack problems. To solve them, the principle of greedy search is usually used, and the number of cuts is the criterion of the search.
The first obvious step of the algorithm is checking the balance.
The second step is to arrange the arrays of bars and chocolate needs, which will simplify further calculations. This implements the principle of greedy search.
The third obvious step is to find and use all the bars, the sizes of which coincide with the needs.
The next step is to find and use all combinations of the bars what satisfy the needs. This task requires a "greedy" search in descending order of needs, which continues in the further calculations. This criterion is not optimal, but it allows to form a basic solution.
If not all the children have received chocolate, then the cuts become obvious. The search should be done according to the descending sizes of the tiles. First, one should check all possibilities to give the cut tiles to two children at once, then the same, but if one existing tile is used, etc.
   
After that there is an obvious variant "one cut - one need", allowing to form the base variant. But if there remain computational resources, they can be used first to calculate the options of the type "two slits - three needs", etc.
Further optimization consists in returning back to the steps and calculation for the following variants.

Appropriate encoding using Particle Swarm Optimization

The Problem
I've been doing a bit of research on Particle Swarm Optimization, so I said I'd put it to the test.
The problem I'm trying to solve is the Balanced Partition Problem - or reduced simply to the Subset Sum Problem (where the sum is half of all the numbers).
It seems the generic formula for updating velocities for particles is
but I won't go into too much detail for this question.
Since there's no PSO attempt online for the Subset Sum Problem, I looked at the Travelling Salesman Problem instead.
They're approach for updating velocities involved taking sets of visited towns, subtracting one from another and doing some manipulation on that.
I saw no relation between that and the formula above.
My Approach
So I scrapped the formula and tried my own approach to the Subset Sum Problem.
I basically used gbest and pbest to determine the probability of removing or adding a particular element to the subset.
i.e - if my problem space is [1,2,3,4,5] (target is 7 or 8), and my current particle (subset) has [1,None,3,None,None], and the gbest is [None,2,3,None,None] then there is a higher probability of keeping 3, adding 2 and removing 1, based on gbest
I can post code but don't think it's necessary, you get the idea (I'm using python btw - hence None).
So basically, this worked to an extent, I got decent solutions out but it was very slow on larger data sets and values.
My Question
Am I encoding the problem and updating the particle "velocities" in a smart way?
Is there a way to determine if this will converge correctly?
Is there a resource I can use to learn how to create convergent "update" formulas for specific problem spaces?
Thanks a lot in advance!
Encoding
Yes, you're encoding this correctly: each of your bit-maps (that's effectively what your 5-element lists are) is a particle.
Concept
Your conceptual problem with the equation is because your problem space is a discrete lattice graph, which doesn't lend itself immediately to the update step. For instance, if you want to get a finer granularity by adjusting your learning rate, you'd generally reduce it by some small factor (say, 3). In this space, what does it mean to take steps only 1/3 as large? That's why you have problems.
The main possibility I see is to create 3x as many particles, but then have the transition probabilities all divided by 3. This still doesn't satisfy very well, but it does simulate the process somewhat decently.
Discrete Steps
If you have a very large graph, where a high velocity could give you dozens of transitions in one step, you can utilize a smoother distance (loss or error) function to guide your model. With something this small, where you have no more than 5 steps between any two positions, it's hard to work with such a concept.
Instead, you utilize an error function based on the estimated distance to the solution. The easy one is to subtract the particle's total from the nearer of 7 or 8. A harder one is to estimate distance based on that difference and the particle elements "in play".
Proof of Convergence
Yes, there is a way to do it, but it requires some functional analysis. In general, you want to demonstrate that the error function is convex over the particle space. In other words, you'd have to prove that your error function is a reliable distance metric, at least as far as relative placement goes (i.e. prove that a lower error does imply you're closer to a solution).
Creating update formulae
No, this is a heuristic field, based on shape of the problem space as defined by the particle coordinates, the error function, and the movement characteristics.
Extra recommendation
Your current allowable transitions are "add" and "delete" element.
Include "swap elements" to this: trade one present member for an absent one. This will allow the trivial error function to define a convex space for you, and you'll converge in very little time.

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