hermanschaaf

Thanks! Would love to hear your thoughts if you get a chance to play it. I'm currently working on adding a few more towers and challenges. After that maybe I'll deep-dive into the "best" automated scheduling algorithm for all this :)

Hah, I thought a lot about this problem while developing Sky Lobby, a mobile game (iOS / Android) about controlling and automating elevators.

In the game, some of the elevators are automated, and I wanted to choose an algorithm that best aligned with what players would expect an elevator to do. I ended up going with something close to LOOK, which (as the article states) is mostly what people expect. Except in case of ambiguity, I had it prioritise floors that have been waiting longer, to improve the p90, which is important in the game.

But now, add in these challenges:

- double-deck cabs (where attached cabs cover two floors at once)

- transfer floors between shafts

- express shafts

and the best (or at least, most expected) algorithm becomes much less obvious! I'll leave it as an exercise for the reader to figure out the best one.

Luckily I was in charge of a game, not a real elevator system, so I just had to find a heuristic that was good-enough, and could tweak the rules to let players manually override the elevator's plan if they don't like where it's going. This seems to have satisfied most players.

That's a great point, and yes, probably. Perhaps society is pushing in two opposing directions, the tech-enhanced and the purist / artisan. Whether one will win or they will be able to live peacefully side by side is hard to say

Love the simple landing page! How big is the free tier? The pricing slider seems to show that you pay $39 for anything more than 0 requests.

Thanks for the great article, you've inspired me to take another shot at making a habit of learning through spaced repetition!

I had a question: how much time do you typically spend on this activity in a day? Do you have tips for how to adjust based on the time you have available?

Here is another one I thought of the other day.

Q: Given V (the number of vertices), E (the number of edges) and L (a list of the edges) for a connected undirected graph, determine whether the graph is a tree.

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A: It can be done in O(1) by: return V == E + 1

The number of vertices in a tree is always one more than the number of edges. We also know that the graph is connected, so it has to be a tree.

Out of interest, how do you find such auctions in the UK?