I operate by Crocker’s rules.
I try to not make people regret telling me things. So in particular:
- I expect to be safe to ask if your post would give AI labs dangerous ideas.
- If you worry I’ll produce such posts, I’ll try to keep your worry from making them more likely even if I disagree. Not thinking there will be easier if you don’t spell it out in the initial contact.
“I wouldn’t abuse power.” is a lie so old that we evolved to believe it and act like it, in order to convince others, until we’re in power. See Why Does Power Corrupt?.
If they set to 1 they converge in a single backward pass, since they then calculate precisely backprop. Setting to less than that and perhaps mixing up the pass order merely obfuscates and delays this process, but converges because any neuron without incorrect children has nowhere to go but towards correctness. And the entire convergence is for a single input! After which they manually do a gradient step on the weights as usual.
[Preliminary edit: I think this was partly wrong. Replicating...]
It’s neat that you can treat activations and parameters by the same update rule, but then you should actually do it. Every “tick”, replace the input and label and have every neuron update its parameters and data in lockstep, where every neuron can only look at its neighbors. Of course, this only has a chance of working if the inputs and labels come from a continuous stream, as they would if the input were the output of another network. They also notice the possibility of continuous data. And then one could see how its performance degrades as one speeds up the poor brain’s environment :).
: Which has to be in backward order and has to be done once more after the v update line.
Go to the bank and tell them “I need a contract that will pay out money if there is no pandemic.”. (The bank is now rubbing their hands, because this offsets their risk.) Your costs are no longer speculative, and you can safely pass on the cost of the contract to the consumer.
You combine some of the advantages of both approaches, but also some disadvantages:
you need a parking spot
you need to wait for the engine
you need to be where your wagon is (or else have it delivered)
you can be identified both through your wagon and your regular interaction with a centralized service
i was confused about how a comment that reads like this one manages to get to −60, so i’m going to guess that the conflict theory crowd sniffed out that you are their enemy from what you say and from your having created a throwaway account to comment here, and then clicked the “i would like less of this kind of comment” button.
i would like less of this kind of dynamic.
Write clearly. What’s the point of the post? How do the parts of the post argue for its point?
For a moment it seemed like the outpouring of magic might hold, take root in the castle’s stone
Hermione tried and failed to become a ghost.
It was, in fact, mentioned in chapter 39, Pretending to be Wise:
(...) they were just afterimages, burned into the stone of the castle by the death of a wizard, like the silhouettes left on the walls of Hiroshima.
If #4 is true, it is provable:
If #4 is true, changing the color of a single node (except to forbidden colors on edges) cannot change the parity of the trichromatic triangle count, and this would be checkable through a finite case analysis of graphs of size <=7. Given that lemma, we can recolor one corner to red, the remainder of one large edge blue and the remaining nodes green, producing the odd count 1.
#8:
We are looking for a surface [0,1]²->[0,1] whose intersection with the plane x=y does not contain a function of t. It suffices to show that the intersection looks like the letter s, with exactly the endpoints reaching t=0 or t=1. It suffices to show that the intersection can be any continous function of x including the points t=x=y=0 and t=x=y=1. Within each plane of constant x, define the surface as 0 for small t, 1 for large t, and rapidly rising through the plane x=y wherever we want the intersection.
If corrigibility has one central problem, I would call it: How do you say “If A, then prefer B.” instead of “Prefer (if A, then B).”? Compare pytorch’s detach, which permits computation to pass forward, but prevents gradients from propagating backward, by acting as an identity function with derivative 0.
[APPRENTICE] Advanced math or AI alignment. I’m bad at getting homework done and good at grokking things quickly, so the day-to-day should look like pair programming or tutoring.
[MENTOR] See https://www.lesswrong.com/posts/MHqwi8kzwaWD8wEQc/would-you-like-me-to-debug-your-math. The first session has the highest leverage, but if my calendar doesn’t end up booked (there is one slot in the next two weeks booked out of like 50), more time per person makes sense. My specialization is pattern-matching to correctly predict where a piece of math is going if it’s good. When you science that art you get applied category theory.
not having to pay is effectively the same as gaining
No! If you’re going to add/multiply something to your utility function for convenience, you have to do it for every action. When the building is on fire, deciding whether to turn on the sprinklers is a decision on whether to spend T and gain D, so V(TP)-V(FN) needs to be D-T.
The bottom left picture on page 21 in the paper shows that this is not just regularization coming through only after the error on the training set is ironed out: 0 regularization (1/lambda=inf) still shows the effect.
Can we switch to the interpolation regime early if we, before reaching the peak, tell it to keep the loss constant? Aka we are at loss l* and replace the loss function l(theta) with |l(theta)-l*| or (l(theta)-l*)^2.
Somewhere in between, it was called his lich-prophylactery-thingy.
(2) is essentially aiming to take over the world in the name of making it safer, which is not generally considered the kind of thing we should be encouraging lots of people to do.
Wait, you want to do it the hard way? Not only win the AI race with enough head start for safety, but stop right at the finish line and have everyone else stop at the finish line? However would you prevent everyone everywhere from going over? If you manage to find a way, that sounds like taking over the world with extra steps.
Written slightly differently, the reasoning seems sane: Suppose I cross. I must have proven it’s a good idea. Aka I proved that I’m consistent. Aka I’m inconsistent. Aka the bridge blows up. Better not cross.