tsp | - Defund the Police | Frontend Framework library

I'm trying to solve a classic Traveling Salesman Problem (TSP). I'm using Google OR Tools default TSP wrapper code. I want some arcs to be forbidden, in the sense that perhaps there is no path between node 10 and node 12. In this instance, I set the value to a large number, like 10^6, but the solver still uses that arc.

I verified that this particular instance has a feasible solution, in the sense that when I run the same code twice or extend the time limit of the solver from 30s to 60s, the solver found a solution that does not use this arc.

The question I have: is there a way to specify a hard constraint that this can't be used, instead of setting the arc cost to ∞? I imagine that in a classic TSP Linear Program, a constraint can be set on the binary decision variable.

I am trying to figure out why getBBox() for tspan element of a svg does not return the dimension.

To demonstrate this with an example, if I run BBox on both tsp1 and rect1, it returns the correct dimension for rect1 but not for tsp1

I have a large pandas dataframe with this structure:

I am trying to run a javaagent in a modular application, but I cannot make it work from command line. I have created the smallest repository possible:

I am trying to implement TSP using kruskal's algorithm such that when inserting, the insertion goes to the beginning or the end. Meaning there is only chain of nodes and no node is connected to more than 2 nodes.

Here is the pseudocode for above illustrated idea. I want to implement End(u) and End(v).But I can't seem to find how these End(v) and End(u) would be implemented. Can anyone help? here is my code:

I am trying to implement stacking orders

While the most optimal solution would be to consider picking up orders from nearby restaurants that have similar food prep time AND nearby delivery locations.

I'd like to start off with something slightly easier - that is stacking orders only from the SAME restaurants that have similar food prep time to multiple deliver points. (Deliveroo example: https://riders.deliveroo.com.sg/en/tech-round-up-stacking-orders)

The scale is about 200k orders hourly and 5000 riders, 5000 restaurants. Run time is important here ~ ideally less than 1 minute (on demand service)

What I have in mind is this:

(1) Collect orders per few minutes interval and sort all orders by their prep time O(nlogn)

(2) group orders by restaurant O(n)

(3) Starting from the order that has the smallest remaining prep time, look for any orders in the same restaurant within the time window (let's say 3-5 mins), if exists group them as a stack. O(1).

• locations of delivery points are not considered here to reduce computations - most delivery points are within 3km in a given zone.
• not so interested in global optima for computational time. Picking the order that has the smallest remaining prep time is to avoid considering all combinations for rider - orders permutation matching.

(4) Run simulated annealing for semi-optimal TSP for vehicle routing. (ex. Pickup order A, B, C from the same restaurants -> deliver C to A to B)

I understand for multiple PICKUP and Dropoff the problem would translate into VRPTW - a hard problem to solve in real-time.

A somewhat easier problem - single Pickup and multiple Drop off would there be any better way than what I have in mind right now?

Many thanks in advance.

The following Prolog program defines a predicate sorted/2 for sorting by permutation (permutation sort) in ascending order a list passed in first argument, which results in the list passed in second argument:

I have an assignment to use a greedy approach to satisfy TSP. The problem has 33708 cities. because I had a lot of helpful tools for this from the previous assignment, I decided to reuse that approach and precompute the distances.

so that is barely more than half a billion entries (33708 choose 2), each comfortably fitting in a float32. The x and y coordinates, likewise, are numbers $|n| < 10000 $ with no more than 4 decimal places.

My python for the same was: