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Open thread, September 18 - September 24, 2017

2 Post author: Thomas 18 September 2017 08:30AM
If it's worth saying, but not worth its own post, then it goes here.

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Comments (26)

Comment author: Brillyant 19 September 2017 05:45:08PM 2 points [-]

Fix RECENT ON RATIONALITY BLOGS.

Comment author: Lumifer 18 September 2017 04:00:11PM 1 point [-]

The IgNobels for 2017 are out.

I think LW should re-focus on more important issues under discussion in peer-reviewed science, e.g. "Never Smile at a Crocodile: Betting on Electronic Gaming Machines is Intensified by Reptile-Induced Arousal" (link)

Comment author: MrMind 19 September 2017 07:54:18AM 0 points [-]

Wonderful as always!

Comment author: Bound_up 24 September 2017 12:38:42AM 0 points [-]

I'm thinking of going to Australia as recommended here: http://lesswrong.com/lw/43m/optimal_employment/

It looks like the program run by a fellower LWer at http://ozworkvisa.com/ is gone now?

Does anyone know if there are still people who've put together a fast track for going to work in Australia?

Comment author: Elo 24 September 2017 12:52:05AM 1 point [-]

I am in Australia and I don't know of a program. Send me a message and I can see if I can help.

Comment author: wMattDodd 20 September 2017 07:29:08PM *  0 points [-]

I've finally been able to put words to some things I've been pondering for awhile, and a Google search on the most sensible terms (to me) for these things turned up nothing. Looking to see if there's already a body of writing on these topics using different terms, and my ignorance of such would lead to me just re-inventing the wheel in my ponderings. If these are NOT discussed topics for some reason, I'll post my thoughts because I think they could be critically important to the development of Friendly AI.

implicit utility function ('survive' is an implicit utility function because regardless of what your explicit utility function is, you can't progress it if you're dead)

conflicted utility function (a utility function that requires your death for optimal value is conflicted, as in the famous Pig That Wants to be Eaten)

dynamic utility function (a static utility function is a major effectiveness handicap, probably a fatal one on a long enough time scale)

meta utility function (a utility function that takes the existence of itself into account)

Comment author: J_Thomas_Moros 20 September 2017 07:56:28PM 1 point [-]

What you label "implicit utility function" sounds like instrumental goals to me. Some of that is also covered under Basic AI Drives.

I'm not familiar with the pig that wants to be eaten, but I'm not sure I would describe that as a conflicted utility function. If one has a utility function that places maximum utility on an outcome that requires their death, then there is no conflict, that is the optimal choice. Though I think human's who think they have such a utility function are usually mistaken, but that is a much more involved discussion.

Not sure what the point of a dynamic utility function is. Your values really shouldn't change. I feel like you may be focused on instrumental goals that can and should change and thinking those are part of the utility function when they are not.

Comment author: Brillyant 20 September 2017 05:35:52PM 0 points [-]

Anyone following the role American football may play in long term brain injuries? Subconcussive hits to the head accumulating to cause problems?

Anyone have thoughts?

Comment author: IlyaShpitser 20 September 2017 05:38:09PM 1 point [-]

Yes, US football and boxing are very bad for the brain. Plenty of evidence.

Comment author: Brillyant 21 September 2017 01:29:54PM 0 points [-]

Plenty of evidence.

Any that you find particularly clear and compelling?

Comment author: Thomas 18 September 2017 08:33:13AM 0 points [-]

A problem to ponder this week.

Comment author: gjm 18 September 2017 02:11:38PM *  0 points [-]

Well, it seems that there is a "crossword" of size 270343. That's in decimal; in binary the same approach gives you 37156667.

Comment author: cousin_it 18 September 2017 11:01:15AM *  0 points [-]

Nice exercise, thank you! With the right algorithm, even a slow language like Python can find an 8x8 square of primes in less than a minute:

4 6 3 3 3 3 3 7

6 1 0 0 0 0 0 1

3 0 0 0 0 0 0 1

3 0 0 0 0 0 0 1

3 0 0 0 0 0 0 1

3 0 0 0 0 0 0 1

3 0 0 0 0 0 0 1

7 1 1 1 1 1 1 1

Here's my code if anyone's curious. The idea is simple, precompute all suffixes of primes and then fill the square by backtracking from the bottom right corner.

Comment author: Thomas 18 September 2017 11:33:07AM 0 points [-]

Congratulation! It's essential that you don't tell the algorithm, at least for now. You have an extra solution, where every horizontal has its equal vertical. Which is perfectly okay, but I wonder if that is the property of your algorithm?

Comment author: cousin_it 18 September 2017 11:41:43AM *  1 point [-]

No, it gives plenty of non-symmetric solutions as well. Here's one:

8 1 8 9 9 1 4 3

6 5 4 0 4 9 0 3

6 0 2 7 2 9 5 1

8 2 7 8 7 4 1 3

3 9 2 9 3 8 6 1

2 4 7 8 9 7 6 7

2 2 0 6 6 3 3 7

9 9 7 3 7 9 3 3

Comment author: Thomas 18 September 2017 12:21:10PM 0 points [-]

Very well. What do you think, are there arbitrary large squares possible or not?

I think not. Even in binary notations NxN and above, probably don't exist for an N, large enough.

Comment author: cousin_it 18 September 2017 01:13:22PM *  1 point [-]

I'm pretty sure arbitrarily large squares are possible. Here's an argument that assumes primes behave like random numbers, which is often okay to assume. By the prime number theorem, the chance that an N-digit number is prime is proportional to 1/N. So the chance that N^2 random digits arranged in a square will form 2N primes (N rows and N columns) is about 1/N^(2N). But the number of ways to select N^2 digits is 10^(N^2) which easily overwhelms that.

Comment author: Thomas 18 September 2017 03:02:08PM *  0 points [-]

The bottom and the rightmost prime can both have only odd digits without 5. The probability for each prime to fit there is then only (2/5)^N times that. We can't see them as independent random numbers.

Comment author: cousin_it 18 September 2017 03:19:28PM *  3 points [-]

If you're pointing out that my argument isn't rigorous, I know. It can be overcome by some kind of non-random conspiracy among primes. But it needs to be a hell of a strong conspiracy, much stronger than what you mention. Even if the whole square had to consist of only 1 3 7 9, you'd still have 4^(N^2) possible squares, and 1/N^(2N) of them would still be a huge number.

Example, just for fun:

9 7 9 7 7 9 9 1

1 7 9 9 7 1 3 1

7 9 3 9 7 3 9 9

3 3 1 7 9 1 9 7

7 3 3 1 9 7 3 1

7 7 9 1 7 1 3 9

3 1 1 7 9 1 1 9

3 9 3 3 3 3 1 1

Heck, I can even make these:

1 1 1 9 1 9 9 1

9 1 1 1 1 1 9 9

1 9 1 9 9 1 1 9

1 1 1 1 1 9 1 1

1 9 1 1 9 9 1 1

9 1 9 1 1 1 1 9

9 9 1 1 9 1 9 1

1 9 1 9 9 1 1 9

Bottom line, primes are much more common than you think :-)

Comment author: cousin_it 20 September 2017 12:12:58PM *  0 points [-]

Here's another fun argument. The question boils down to "how common are primes?" And the answer is, very common. We can define a subset of positive integers as "small" if the sum of their reciprocals converges, and "large" otherwise. For example, the set of all positive integers is large (because the harmonic series diverges), and the complement of any small set is large. Well, it's possible to prove that the set of all primes is large, while the set of all numbers not containing some digit (say, 7) is small. So once you go far enough from zero, there are way more primes than there are numbers not containing 7. Now it doesn't sound as surprising that you can make squares out of primes, does it?

Comment author: Thomas 22 September 2017 10:44:28AM 0 points [-]

Say, that we have N-1 lines, with N-1 primes. Each N digits. What we now need is an N digit prime number to put it below.

Its most significant digit may be 1, 3, 7 or 9. Otherwise, the leftmost vertical number wouldn't be prime. If the sum of all N-1 other rightmost digits is X, then:

If X mod 3 = 0, then just 1 and 7 are possible, otherwise the leftmost vertical would be divisible by 3. If X mod 3 = 1, then 1, 3, 7 and 9 are possible. If X mod 3 = 2, then just 3 and 9 are possible, otherwise the leftmost vertical would be divisible by 3.

The probability is (1/3)*(((1+2+1)/5))=4/15 that the first digit fits. (4/15)^N, that all N digit fit.

Actually, we must consider the probability of divisibility by 11, which is roughly 1/11, which further reduces 4/15 per number to 40/165. And with 7 ... and so on.

For the divisibility with 3, we render out not only one permutation of N-1 primes but all of them. For the divisibilty with 11, some of them.

It's quite complicated.

Comment author: gjm 22 September 2017 05:25:13PM 0 points [-]

It is indeed quite complicated. But if you handwavily estimate the results of all that complexity -- the probabilities of divisibility by various things -- then the estimate you get is the one cousin_it gave earlier, because the Prime Number Theorem is what you get when you estimate the density of prime numbers by treating divisibility-by-a-prime as a random event. (Which for many purposes works very well.)

Comment author: Thomas 20 September 2017 07:52:41PM *  0 points [-]

Interesting line of inferring... I am quite aware how dense primes are, but that might not be enough.

I have counted all these 4x4 (decimal) crossprimes. There are 913,425,530 of them if leading zeros are allowed. But only 406,721,511 without leading zeros.

if leading zeros ARE allowed, then there are certainly arbitrary large crossprimes out there. But if leading zeros aren't allowed, I am not that sure. I have no proof, of course.

Comment author: username2 18 September 2017 12:00:38PM 0 points [-]

This one is also attractive in that primes are not repeated.