Open thread, August 7 - August 13, 2017

Why is anonymous posting not welcome?

That seems rather obvious to me.

Kind of...

Like, part of being 'highly skilled' as a programmer is being able to work with other people. I mean, I get what you are saying, but working with assholes is part of the devs tool bag, or he hasn't been a dev very long.

I wrote a little program to attack this question for a smaller number of primes. The results don't encourage me to think that there's a "principled" answer in general. I ran it for (as it happens) the first 1002 primes, up to 7933. The visibility counts fluctuate wildly; it looks as if there may be a tendency for "typical" visibility counts to decrease, but the largest is 256 for the 943rd prime (7451) which not coincidentally has a large gap before it and smallish gaps immediately after.

It seems plausible that the winner with a billion primes might be the 981,765,348th prime, which is preceded by a gap of length 38 (the largest in the first billion primes), but I don't know and I wouldn't bet on it. With 1200 primes you might think the winner would be at position 1183, after the first ever gap of size 14 -- but in fact that gap is immediately followed by another of size 12, and the prime after that does better even though it can't see its next-but-one neighbour, and both are handily beaten by the 943rd prime which sees lots of others above as well as below.

It's still feeling to me as if any solution to this is going to involve more brute force than insight. Thomas, would you like to tell us whether you know of a solution that doesn't involve a lot of calculation? (Since presumably any solution will at least need to know something about all the first billion primes, maybe I need to be less vague. If the solution looked like "the winning prime is the prime p_i for which p_{i+1}-p_{i-1} is greatest" or something similarly simple, I would not consider it to be mostly brute force.)

Initial handwaving:

Super-crudely the n'th prime number is about n log n. If this were exact then each tower would see all the others, because the function x -> x log x is convex. In practice there are little "random" fluctuations which make a difference. It's possible that the answer to the question depends critically on those random fluctuations and can be found only by brute force...

Isn't this literally asking for the largest increasing prime gap sequence between 1 and 1bil? Probably some number theorist knows this.

The intuitive answer seems to me to be: the last one. It's the tallest, so it witness exactly one billion towers. Am I misinterpreting something?

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P.S. If this worries you that AGI is closer than you expected do not watch Jeff dean's overview lecture of DL research at Google.

The overview lecture doesn't really get me worried. It basically means that we are at the point where we can use machine learning to solve well-defined problems with plenty of training data. At the moment that seems to require a human machine learning expert and recent Google experiments suggest that they are confident to develop an API that can do this without machine learning experts being involved.

At a recent LW discussion someone told me that this kind of research doesn't even count as an attempt to develop AGI.

I think I don't know the solution, and if so it's impossible for me to guess what he thinks if he's right :)

But maybe he's thinking of something vague like CIRL, or hierarchical self-supervised learning with generation, etc. But I think he's thinking of some kind of recurrent network. So maybe he has some clever idea for unsupervised credit assignment?

You can set PB predictions to be private. Of course, this doesn't guarantee privacy since there are so many ways to hack websites and PB is not the best maintained codebase nor has it ever been audited... You could encrypt your private predictions, which would offer security but also reminders+scoring.

I don't know of any offline CLI versions but the core functionality is pretty trivial so you could hack one up easily.

For mobile, there's LW Predictions on Android.