All of Oscar_Cunningham's Comments + Replies

I got a good answer here: https://stats.stackexchange.com/q/579642/5751

Or is that still too closely tied to the explore-exploit paradigm?

Right. The setup for my problem is the same as the 'bernoulli bandit', but I only care about the information and not the reward. All I see on that page is about exploration-exploitation.

What's the term for statistical problems that are like exploration-exploitation, but without the exploitation? I tried searching for 'exploration' but that wasn't it.

In particular, suppose I have a bunch of machines which each succeed or fail independently with a probability that is fixed separately for each machine. And suppose I can pick machines to sample to see if they succeed or fail. How do I sample them if I want to become 99% certain that I've found the best machine, while using the fewest number of samples?

The difference with exploration-exploitat... (read more)

The problem is that this measures their amount of knowledge about the questions as well as their calibration.

My model would be as follows. For a fixed source of questions, each person has a distribution describing how much they know about the questions. It describes how likely it is that a given question is one they should say p on. Each person also has a calibration function f, such that when they should say p they instead say f(p). Then by assigning priors over the spaces of these distributions and calibration functions, and applying Bayes' rule we get a... (read more)

I was expecting a post about tetraethyllead.

I gave some more examples here: https://www.reddit.com/r/math/comments/9gyh3a/if_you_know_the_factors_of_an_integer_x_can_you/e68u5x4/

You just have to carefully do the algebra to get an inductive argument. The fact that the last digit is 5 is used directly.

Suppose n is a number that ends in 5, and such that the last N digits stay the same when you square it. We want to prove that the last N+1 digits stay the same when you square n^2.

We can write n = m*10^N + p, where p has N digits, and so n^2 = m^2*10^(2N) + 2mp*10^N + p^2. Note that since 2p ends in 0, the term 2mp*10^N actually divides by 10^(N+1). Then since the two larger terms divide by 10^N+1, n^2 agrees with p^2 on its last N+1 d... (read more)

If we start with 5 and start squaring we get 5, 25, 625, 390625, 152587890625.... Note how some of the end digits are staying the same each time. If we continue this process we get a number ...8212890625 which is a solution of x^2 = x. We get another solution by subtracting this from 1 to get ...1888119476.

if not, then we'll essentially need another way to define determinant for projective modules because that's equivalent to defining an alternating map?

There's a lot of cases in mathematics where two notions can be stated in terms of each other, but it doesn't tell us which order to define things in.

The only other thought I have is that I have to use the fact that $W$ is projective and finitely generated. This is equivalent to $W$ being dualisable. So the definition is likely to use $W^{\ast }$ somewhere.

I'm curious about this. I can see a reasonable way to define ${\Lambda }^W\left(V\right)$ in terms of sheaves of modules over $Spec\left(R\right)$: Over each connected component, $W$ has some constant dimension $n$, so we just let ${\Lambda }^W\left(V\right)$ be ${\Lambda }^n\left(V\right)$ over that component.

If we call this construction $F(W,V)$ then the construction I'm thinking of is $F(W,V)\otimes F(W,W{)}^{\ast }$. Note that $F(W,W)$ is locally $1$-dimensional, so my construction is locally isomorphic to yours but globally twisted. It depends on $W$ via more than just its... (read more)

I agree that 'credence' and 'frequency' are different things. But round here the word 'probability' does refer to credence rather than frequency. This isn't a mistake; it's just the way we're using words.

Another thing to say is if $A\left(0\right)=I$ then

$\frac{d}{dx}det\left(A\right(x\left)\right){|}_{x=0}=tr\left(A^{\prime }\right(0\left)\right)$.

My intuition for $\exp$ is that it tells you how an infinitesimal change accumulates over finite time (think compound interest). So the above expression is equivalent to $det(I+\epsilon A)=1+\epsilon tr\left(A\right)+O\left({\epsilon }^2\right)$. Thus we should think 'If I perturb the identity matrix, then the amount by which the unit cube grows is proportional to the extent to which each vector is being stretched in the direction it was already pointing'.

Thank you for that intuition into the trace! That also helps make sense of $det\left(\exp \right(A\left)\right)=\exp \left(tr\right(A\left)\right)$.

This is close to one thing I've been thinking about myself. The determinant is well defined for endomorphisms on finitely-generated projective modules over any ring. But the 'top exterior power' definition doesn't work there because such things do not have a dimension. There are two ways I've seen for nevertheless defining the determinant.

• View the module as a sheaf of modules over the spectrum of the ring. Then the dimension is constant on each connected component, so you can take the top exterior power on each and then glue them back together.
• Use the fact

I think the determinant is more mathematically fundamental than the concept of volume. It just seems the other way around because we use volumes in every day life.

https://www.lesswrong.com/posts/AAqTP6Q5aeWnoAYr4/?commentId=WJ5hegYjp98C4hcRt

I don't dispute what you say. I just suggest that the confusing term "in the worst case" be replaced by the more accurate phrase "supposing that the environment is an adversarial superintelligence who can perfectly read all of your mind except bits designated 'random'".

In this case P is the cumulative distribution function, so it has to approach 1 at infinity, rather than the area under the curve being 1. An example would be 1/(1+exp(-x)).

A simple way to do this is for ROB to output the pair of integers {n, n+1} with probability K((K-1)/(K+1))^|n|, where K is some large number. Then even if you know ROB's strategy the best probability you have of winning is 1/2 + 1/(2K).

If you sample an event N times the variance in your estimate of its probability is about 1/N. So if we pick K >> √N then our probability of success will be statistically indistinguishable from 1/2.

The only difficulty is implementing code to sample from a geometric distribution with a parameter so close to 1.

The original problem doesn't say that ROB has access to your algorithm, or that ROB wants you to lose.

Note that Eliezer believes the opposite: https://www.lesswrong.com/posts/GYuKqAL95eaWTDje5/worse-than-random.

Right, I should have chosen a more Bayesian way to say it, like 'suceeds with probability greater than 1/2’.

Yes, I'm a big fan of the Entropic Uncertainty Principle. One thing to note about it is that the definition of entropy only uses the measure space structure of the reals, whereas the definition of variance also uses the metric on the reals as well. So Heisenberg's principle uses more structure to say less stuff. And it's not like the extra structure is merely redundant either. You can say useful stuff using the metric structure, like Hardy's Uncertainty Principle. So Heisenberg's version is taking useful information and then just throwing it away.

I'd almos... (read more)

The link still works for me. Perhaps you must first become a member of that discord? Invite link: https://discord.gg/nZ9JV5Be (valid for 7 days)

The weird thing is that there are two metrics involved: information can propagate through a nonempty universe at 1 cell per generation in the sense of the l_infinity metric, but it can only propagate into empty space at 1/2 a cell per generation in the sense of the l_1 metric.

https://en.wikipedia.org/wiki/Norm_(mathematics)#p-norm

You're probably right, but I can think of the following points.

Its rule is more complicated than Life's, so its worse as an example of emergent complexity from simple rules (which was Conway's original motivation).

It's also a harder location to demonstrate self replication. Any self replicator in Critters would have to be fed with some food source.

Yeah, although probably you'd want to include a 'buffer' at the edge of the region to protect the entity from gliders thrown out from the surroundings. A 1,000,000 cell thick border filled randomly with blocks at 0.1% density would do the job.

https://en.wikipedia.org/wiki/Critters_(cellular_automaton)

This is very much a heuristic, but good enough in this case.

Suppose we want to know how many times we expect to see a pattern with n cells in a random field of area A. Ignoring edge effects, there are A different offsets at which the pattern could appear. Each of these has a 1/2^n chance of being the pattern. So we expect at least one copy of the pattern if n < log_2(A).

In this case the area is (10^60)^2, so we expect patterns of size up to 398.631. In other words, we expect the ash to contain any pattern you can fit in a 20 by 20 box.

The glider moves at c/4 diagonally, while the c/2 ships move horizontally. A c/2 ship moving right and then down will reach its destination at the same time the c/4 glider does. In fact, gliders travel at the empty space speed limit.

Most glider guns in random ash will immediately be destroyed by the chaos they cause. Those that don't will eventually reach an eater which will neutralise them. But yes, such things could pose a nasty surprise for any AI trying to clean up the ash. When it removes the eater it will suddenly have a glider stream coming towards it! But this doesn't prove it's impossible to clear up the ash.

Making a 'ship sensor' is tricky. If it collides with something unexpected it will create more chaos that you'll have to clear up.

This sounds like you're treating the area as empty space, whereas the OP specifies that it's filled randomly outside the area where our AI starts.

My understanding was that we just want to succeed with high probability. The vast majority of configurations will not contain enemy AIs.