hakuseki

The result actually doesn't depend on any special properties of social graphs. It works for any graph that has a component with non-constant number of edges per node.

They could be overfitting somehow, but I don't see how you get that from the quoted paragraph. Of course they need to use the same subject (i.e. human brain) for training and testing. Otherwise one would need to assume that different people's brains represent visual concepts in the same way.

I'd really like to see the mouse (or other nonhuman animal) version of this experiment.

In the human experiment we can see how the input and output represent the same categories of object. I think it would be really interesting to what extent a mouse's perceptions can be mapped onto human categories such as "fire truck" or "baseball game."

Have you browsed through this list?

https://en.wikipedia.org/wiki/Category:Point-and-click_adven...

A logarithmic representation is great for multiplication, but adding numbers becomes an expensive operation.

The term you're looking for is "simply-connected set."

Simply-connected sets and compact sets are both sometimes informally described as "sets with no holes," but the definition of "holes" is different in both cases.

A compact set contains all of its limiting points (i.e. limits of convergent sequences of points within the set). Therefore it may not have any point-shaped holes. But it may still have a hole shaped like, e.g. an open disc.

To rule out such large holes, you want a simply-connected set, which is a connected set in which any closed path can be continuously shrunk to a point.

This headline contradicts the linked article.

The HN headline (Heavier cars are safer for their drivers, but far deadlier for everyone else) includes the non-driving passengers with "everyone else," implying that heavier cars are deadlier for them.

The first sentence of the abstract in the link states "Heavier vehicles are safer for their own occupants but more hazardous for other vehicles."

It seems to me that the two formal systems can disagree about BB(n) for some n without disagreeing about the state of any given Turing machine at any specific time step. For example, ZFC+CH might non-constructively predict that some machine M halts, while ZFC+notCH might predict that M does not halt. If all machines other than M can be either run to completion or proven not to halt, then the value of BB(n) would be provable in ZFC+notCH but undecidable in ZFC+CH.

I'm not quite grasping how to use this. I tried uploading a photograph and erasing part of it. But instead of painting in the erased portion, it left the erased area blank and replaced my photograph with an entirely new image.

That is not quite the right calculation. To see this, try plugging 0.75 into the same formula to get Stockfish's expected score. The result is about 0.3545. If this were the correct formula, then the two expected scores should sum to 1, but in this case we only get 0.3612.

Instead, you should convert the 0.25 to "odds" form. 0.25 is 1:3 odds, represented by the number 1/3. (1/3)^((3585 - 2864)/200) is about 0.01905 (still in odds form). To convert this back to an expected score you would take 0.01905 / (1 + 0.01905) = 0.0187. So Magnus Carlsen's expected score is 0.0187.

Applying the same method to Stockfish, we have 3:1 odds, which is represented by the number 3. 3^((3585 - 2864)/200) is about 52.48. Converting back to expected score we get 52.48 / (1 + 52.48) = 0.9813. So Stockfish's expected score is 0.9813.

Our sanity check is to add 0.0187 + 0.9813. The result is 1.0, as it should be.

To clarify, my question was basically asking whether the author of the parent comment is satisfied by my comment.

IANAP, but it seems to me that right now we are able to make the statement: "Angular momentum is conserved."

If we were to give different names to different types of angular momentum, we could no longer say that. Instead we would have to say something like: "The sum of Type 1 Angular Momentum and Type 2 Angular Momentum is conserved."

Does that qualify as more complex?

This seems like a point in favor of models like REALM (https://ai.googleblog.com/2020/08/realm-integrating-retrieva...) which could allow for deletion of sensitive information without needing to retrain the model.

> More employees also does not mean faster product development, every developer knows this is often the complete opposite.

Maybe teams are more productive when their membership is stable. Either adding or removing developers mid-project could disrupt development.

Further thoughts:

Alice and Bob must each choose an actual policy for how many confirmations they need before they actually commit to attack. We can number the messages like this:

1. Alice's first message 2. Bob's first message 3. Alice's second message 4. Bob's second message ...

Alice's policy can only commit to attack after receiving some even-indexed message n. Bob can only commit to attack after receiving some odd-indexed message m. So n and m cannot possibly be equal. Therefore the possibility exists that a message fails between n and m, and either Alice or Bob ends up attacking without the other's support.

I guess the idea is that Alice wants to be 100% certain that Bob will attack before she commits her own forces, and Bob similarly wants to be 100% certain that Alice will attack. This level of certainty is not achievable.

Gravity wells are where most of the resources are. For example, Earth outmasses the asteroid belt by a factor of over 1,000.

Yes, and that's why suspending the one car might seem counterintuitive. But that doesn't mean it's a bad solution.

I really couldn't say.

It does occur to me, though, that you are probably talking about the elementwise product, not the dot product. The dot product includes a reduce_sum step and outputs a scalar.

It does sound silly at first, but I think suspending the one car is actually a rather elegant solution on several counts.

1. Easy to implement

2. Doesn't punish fleet scale (compare: suspending the whole fleet means a fleet that's twice as large would be suspended twice as often)

3. Punishes the fleet owner in proportion to the frequency of the error

4. Is exactly correct when the cause is really a hardware problem specific to the car

I really don't find that einsum is less intuitive than reshape. Maybe they both take about the same time to learn the first time. And that's less time than it would take me to struggle through even a single function using reshape ops.

I'm not sure which part you're saying is unbounded, but everything looks bounded to me. A game is played in finite time, on hardware with bounded operations per second, and AFAIK the NES is a deterministic computer, so the solution does not need to involve any algorithms -- it is simply a fixed sequence of button presses.

It's a decidable problem because the duration is bounded (by the best current time).

Øyvind Thorsby also published a similar (somewhat easier) quiz a couple weeks ago:

https://thorsbyprojects.tumblr.com/post/660029925870518272/a...

I really like the world-building that went into this game but I wish it were more challenging.

I think the absence of escape pods is rather telling: they are an unneeded feature, since save-scumming is the default.

What about GPT-f? It's a language model that proved theorems in the metamath formal system.

> there are only around 10,000,000 possible game-states after the first WBBWW

You don't even need to go that far. After WBBW you are already reverting to normal alternating play. Maybe that's around 400,000 positions?

GP doesn't seem to be saying that chess and brick-laying are equally difficult to automate. GP is saying that a belief in brick-laying AI represents no more contempt towards brick-layers than a belief in chess AI represents towards chess players.

Definitely could not have used Kolmogorov complexity, as it is uncomputable.

I do not dispute this. My comment was about the linked paper [1] regarding equivalence of the halting problem and Kolmogorov complexity, not the SAT problem.