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Does anyone else have the problem that stuff written in the morning comes out fine, but stuff written late at night comes out horrible when you reread it in the morning? Is there anything I can do about that?
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?
I am in Australia and I don't know of a program. Send me a message and I can see if I can help.
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)
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.
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?
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.
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?
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
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.
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.
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.
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 :-)
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?
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.
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.)
There are 143 primes between 100 and 999. We can, therefore, make 2,924,207 3x3 different squares with 3 horizontal primes. 50,621 of them have all three vertical numbers prime. About 1.7%.
There are 1061 primes between 1000 and 9999. We can, therefore, make 1,267,247,769,841 4x4 different squares with 4 horizontal primes. 406,721,511 of them have all four vertical numbers prime. About 0.032%.
I strongly suspect that this goes to 0, quite rapidly.
How many Sudokus can you get with 9 digit primes horizontally and vertically?
Not a single one. Which is quite obvious when you consider that you can't have a 2, 4, 6, or 8 in the bottom row. But you have to, to have a Sudoku, by the definition.
It's a bit analogous situation here.
I'm sure you're right that the fraction of all-horizontals-prime grids that have all verticals prime tends to 0 as the size increases. But the number of such grids increases rapidly too.
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.
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.
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.