All of shinoteki's Comments + Replies

What's the difference between

Boy1Tu/Boy2Tuesday

and

Boy1Tuesday/Boy2Tu

?

0bigjeff5
In Boy1Tu/Boy2Tuesday, the boy referred to as BTu in the original statement is boy 1, in Boy2Tu/Boy1Tuesday the boy referred to in the original statement is boy2. That's why the "born on tuesday" is a red herring, and doesn't add any information. How could it?

LessWrong is sci-fi. Check what's popular. Superintelligent AI, space travel, suspended animation, hyper-advanced nanotech...

It is true that people have written unrealistic books about these things. People also wrote unrealistic books about magicians flying through the air and scrying on each other with crystal balls. Yet we have planes and webcams.

Who is to say there even are concepts that the human mind simply can't grasp? I can't visualize in n-dimensional space, but I can certainly understand the concept

The human mind is finite, and there are in... (read more)

-3BaconServ
Naturally, some of the ideas fiction holds are feasible. In order for your analogy to apply, however, we'd need a comprehensive run-down of how many and which fictional concepts have become feasible to date. I'd love to see some hard analysis across the span of human history. While I believe there is merit in nano-scale technology, I'm not holding my breath for femtoengineering. Nevertheless, if such things were as readily predictable as people seem to think, you have to ask why we don't have the technology already. The answer is that actually expressing our ideas onto physical reality is non-trivial, and by direct consequence, potentially non-viable. I need backing on both of these points. As far as I know, there isn't enough verified neuroscience to determine if our brains are conceptually limited in any way. Primarily because we don't actually know how abstract mental concepts map onto physical neurons. Even ignoring that (contrary to memetic citation) the brain does grow new neural cells and repair itself in adults, even if the number of neurons is finite, the number of and potential for connections between them is astronomical. We simply don't know the maximum conceptual complexity of the human brain. As far as there being infinitely many concepts, "flying car" isn't terribly more complicated than "car" and "flying." Even if something in the far future is given a name other than "car," we can still grasp the concept of "transportation device," paired with any number of accessory concepts like, "cup holder," "flies," "transforms," "teleports," and so on. Maybe it's closer to a "suit" than anything we would currently call a "car;" some sort of "jetpack" or other. I'd need an expansion on "concept" before you could effectively communicate that concept-space is infinite. Countably infinite or uncountably infinite? All the formal math I'm aware of indicates that things like conceptual language are incomputable or give rise to paradoxes or some other such problem t
shinoteki100

Counterexample: P(3^^^...3)(n "^"s) = 1/2^n P(anything else) = 0 This is normalized because the sum of a geometric series with decreasing terms is finite. You might have been thinking of the fact that if a probability distribution on the integers is monotone decreasing (i.e. if P(n)>P(m) then n <m) then P(n) must decrease faster than 1/n. However, a complexity-based distribution will not be monotone because some big numbers are simple while most of them are complex.

Those are the probabilities that both halves of a pair of photons are transmitted, so you can't determine them without the information from both detectors. The distribution at each individual detector doesn't change, it's the correlation between them that changes.

1NickRetallack
Oh. I can imagine a distribution that looks like that. It would have been helpful if he had given us all the numbers. Perhaps he does in this blog post, but I got confused part way through and couldn't make it to the end. Would it look like this?
2Shmi
... And to calculate this correlation one needs to transmit information by classical means, no faster than light.

A' doesn't become A'' by catching up to him, he becomes A'' when he uses his time machine to jump back 3 hours.

There would be three babies for 6 hours, but then the youngest two would use their time machines and disappear into the past.

A'' doesn't cease to exist. A' "ceases to exist" because his time machine sends him back into the past to become A''.

3OnTheOtherHandle
Oh! Alright, thank you. :) So if you go back and do something one hour in the past, then the loop closes an hour later, when the other version of yourself goes back for the same reasons you did, and now once again you are the only "you" at this moment in time. It's not A' that continues on with life leaving A'' off the hook, it is A'' who moves on while A' must go back. That makes much more sense. Edit: This means it is always the oldest Harry that we see, right? The one with all the extra waiting around included in his age? Since all the other Harries are stuck in a six hour loop.

You don't need a time machine to go forward in time - you can just wait. A'' cant leave everything to A' because A' will disappear within three hours when he goes back to become A''. If A' knows A wasn't reminded the A' can't remind A. the other three Harrys use their time turners to go backwards and close the loop. You do need both forward and backward time travel to create a closed loop, but the forward time travel can just be waiting; it doesn't require a machine.

0OnTheOtherHandle
I think I get it, but I'm still a bit confused, because both A' and A'' are moving forward at the same rate, which means since A'' started off older, A' will never really "catch up to" and become A'', because A'' continues to age. A'' is still three hours older than A', right, forever and ever? To consider a weird example, what about a six hour old baby going back in time to witness her own birth? Once the fetus comes out, wouldn't there just be two babies, one six hours older than the other? Since they're both there and they're both experiencing time at a normal forward rate of one second per second, can't they just both grow up like siblings? If the baby that was just born waited an hour and went back to witness her own birth, she would see her six hour older version there watching her get born, and she would also see the newborn come out, and then there'd be three babies, age 0, age six hours, and age twelve hours, right? How exactly would the "witnessing your own birth" thing play out with time travel? I think your explanation implies that there will never be multiple copies running around for any length of time, but why does A'' cease to exist once A' ages three hours? A'' has also aged three hours and become someone else in the meantime, right?
1NoriMori1992
…You know that paper goes on to assert that the two problems are meaningfully different, such that it's rational to both one-box in Newcomb's Problem and chew gum in Solomon's Problem, right?

The nice part about modal agents is that there are simple tools for finding the fixed points without having to search through proofs; in fact, Mihaly and Marcello wrote up a computer program to deduce the outcome of the source-code-swap Prisoner's Dilemma between any two (reasonably simple) modal agents. These tools also made it much easier to prove general theorems about such agents.

Would it be possible to make this program publicly available? I'm curious about how certain modal agents play against each other, but struggling to caculate it manually.

If you can prove a contradiction, defect.

Should this be "If you can prove that you will cooperate, defect"? As it is, I don't see how this prevents cooperation with Cooperatebot, unless the agent uses an inconsistent system for proofs.

2Will_Sawin
It kills the Lobian argument, I believe, since this implication "if there's a proof that you cooperate, then cooperate " is no longer true. Instead, here's a Lobian argument for defection: Suppose there is a proof that you defect. Then either there is a proof of contradiction, or there is no proof that your move is the same as your opponent's. Either way, you defect.

It's true that if you can prove that your opponent will cooperate counterfactual-if you cooperate and defect counterfacual-if you defect, then you should cooperate. But we don't yet have a good formalization of logical counterfactuals, and the reasoning that cooperates with cooperatebot just uses material-if instead of conterfactual-if.

4Will_Sawin
We have Ambient Decision Theory, which is a pretty good formalization of logical counterfactuals.

If there is a feasible psuedorandomness generator that is computationally indistinguishable from randomness, then randomness is indeed not necessary. However, the existence of such a pseudorandomness generator is still an open problem.

5Eliezer Yudkowsky
What? No it's not. There are no pseudo-random generators truly ultimately indistinguishable in principle from the 'branch both ways' operation in quantum mechanics, the computations all have much lower Kolmogorov complexity after running for a while. There are plenty of cryptographically strong pseudo-random number generators which could serve any possible role a cognitive algorithm could possibly demand for a source of bits guaranteed not to be expectedly correlated with other bits playing some functional role, especially if we add entropy from a classical thermal noise source, the oracular knowledge of which would violate the second law of thermodynamics. This is not an open problem. There is nothing left to be confused about.

Constructively, (not ((not A) and (not B))) is weaker than (A or B). While you could call the former "A or B", you then have to come up with a new name for the latter.

0prase
I haven't been suggesting using (A or B) as a name for (not ((not A) and (not B))) in constructive logic where they aren't equivalent. Rather, I have been suggesting using classical logic (where the above sentences are equivalent) with a constructivist interpretation, i.e. not making difference between "true" and "theorem". But since it is possible for (A or B) to be a theorem and simultaneously for both A and B to be non-theorems, logical "or" would not have the same interpretation, namely it wouldn't match the common language "or" (for when we say "A or B is true", we mean that indeed one of them must be true).

The Metamorphosis of Prime Intellect. The chapters aren't in chronological order; the bootstrapping and power leveling happen in chapters two and four.

0shiftedShapes
I had read most of this many years ago...not sure why I didn't finish then but glad I did this time. Thanks for the link. Also I think the answer is to contine to evolve and constantly push against ones limits rather than outsourcing the work andliving a life Iif leisure as the majority did after the change.

No. To get the 1/3 probability you have to assume that she would be just as likely to say what she says if she had 1 boy as if she had 2 (and that she wouldn't say it if she had none). In your scenario she's only half as likely to say what she says if she has one boy as if she has two boys, because if she only has one there's a 50% chance it's the one she's just given birth to.

-1JohnBonaccorsi
Although I don't see what you're getting at, shinoteki, I appreciate your replying. Maybe you didn't notice; but about half an hour after I posted my comment to which you replied, I posted a comment with a different scenario, which involves no reference to birth order. (That is not to say I see that birth order bears on this.) I will certainly appreciate a reply, from you or from anyone else, to the said latter comment, whose time-stamp is 02 December 2012 06:51:25PM.

Correspondence of beliefs to reality being desirable is no closer to being a tautology than financial institutes being on the side of rivers, undercover spies digging tunnels in the ground, or spectacles being drinking vessels.

-3Logos01
If I had said something that meant something loosely correlated to this, your point would be valid. Instead, what I said was: "I value being right because it is right; I disvalue being wrong because it is wrong."

It is hard to tell whether anyone took this seriously - but it seems that an isomorphic argument 'proves' that computer programs will crash - since "almost any" computer program crashes. The “AGI Apocalypse Argument” as stated thus appears to be rather silly.

I don't see why this makes the argument seem silly. It seems to me that the isomorphic argument is correct, and that computer programs do crash.

5timtyler
Some computer programs crash - just as some possible superintelligences would kill alll humans. However, the behavior of a computer program chosen at random tells you very little about how an actual real-world computer program will behave - since computer programs are typically produced by selection processes performed by intelligent agents. The "for almost any goals" argument is bunk.

He's not talking about impossibility

I know Owen was not talking about impossibility, I brought up impossibility to show that what you thought Owen meant could not be true.

both of which involve moving faster than light.

Moving from B to A slower than the speed of light does not involve moving faster than light.

It shouldn't. Moving from B to A slower than light is possible*, moving from A to B faster than light isn't, and you can't change whether something is possible by changing reference frames.

*(Under special relativity without tachyons)

0PhilGoetz
What I'm trying to get at is, What does a physicist mean when she says she saw X move from A to B faster than light? The measurement is made from a single point; say A. So the physicist is at A, sees X leave at time tX, sends a photon to B at time t0, and gets a photon back from B at time t1, which shows X at B at some time tB. I'm tempted to set tB = (t0+t1)/2, but I don't think relativity lets me do that, except within a particular reference frame. "X travelled faster than light" only means that tX < t1. The FTL interpretation is t0 < tX < tB < t1: The photon left at t0, then X left at tX, and both met at B at time tB, X travelling faster than light. Is there a mundane interpretation under which tB < tX < t1? The photon left A at t0, met X at B at tB, causing X to travel back to A and arrive there at tX. The answer appears to be No, because X would need to travel faster than light on the return trip. And this also explains that Owen's original answer was correct: You can say that X travelled from A to B faster than light, or from B to A faster than light.
0PhilGoetz
My question is whether he meant to say * moving from A to B faster than the speed of light in one reference frame is equivalent to moving from B to A faster than the speed of light in another reference frame or * moving from A to B faster than the speed of light in one reference frame is equivalent to moving from B to A slower than the speed of light in another reference frame both of which involve moving faster than light.

A piece of code that runs a piece of source code does not thereby become that piece of source code.

4DuncanS
That may well be right - a cliquebot may well decide I'm not in its clique, and defect. And I, running its source code, will decide that my counterparty IS in the clique, and cooperate. Not a good outcome. Clearly I need to modify my algorithm so that I run the other bot's code without bothering to analyse it - and have it play against not its actual opponent, but me, running it.

It's not clear to me that ZFC without regularity, replacement, infinity, choice, power set or foundation with a totally ordered field with the LUB property does allow you to talk about most things you want to do with the reals : without replacement or powerset you can't prove that cartesian products exist, so there doesn't seem to be any way of talking about the plane or higher-dimensional spaces as sets. If you add powerset back in you can carry out the Hartogs number construction to get a least uncountable ordinal

1JoshuaZ
Hmm, that's a good point. Lack of cartesian products is annoying. We don't however need the full power set axiom to get them. We can simply have an axiom that states that cartesian products exist. Or even weaker do the following (ad hoc axioms) with a new property of being Cartesian: 1. The cartesian product of any two Cartesian sets exist. 2. Any subset of R is Cartesian. 3. The cartesian product of two Cartesian sets is Cartesian. 4. If A and B are Cartesian then A union B, A intersect B, and A\B are all Cartesian. That should be enough and is a lot weaker than general power set I think.

You don't need Z, third-order arithmetic is sufficient. Every set of ordinals is well-ordered by the usual ordering of ordinals.

0[anonymous]
Only if you accept excluded middle.

It is basically the main point of the definition of ordinals that for any property of ordinals , there is a first ordinal with that property. There are, however, foundational theories without uncountable ordinals , for instance Nik Weaver's Mathematical Conceptualism.