Chronicles of Harry

Backwards temperature metaphors

Mon, 13 Apr 2026 21:05:51 GMT

Instead of saying "the trail has gone cold", we should say "the trail has gone warm".

We tend to conceptualize recent things as hot, and as cooling off over time. But, cold things also return to room temperature over time. If you've deliberately chilled your drink for instance, then if you wait too long it goes warm, not cold! So, if we restrict ourselves to that level, then either metaphor works. We could speak of recent things as cold and as going warm over time.

Now I'm not saying we should do that in general! But "cold" and "hot" have further meanings that can be used to select between them. Cold is low-entropy, structured, preserves information. Hot is high-entropy, unstructured, destroys information -- "you're not dead until you're warm and dead".

So, a recent trail of information should be thought of as cold, not as hot. And if you wait too long and let it get disrupted by the environment, it hasn't cooled off -- it's warmed up, like ice cream you took too long to eat. Cold preserves; cold is the state one should want, regarding an information trail!

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A breakthrough on integer complexity by Konyagin and Oganesyan!

Mon, 13 Apr 2026 07:04:28 GMT

So joshuazelinsky recently sent me a link to this paper and wow!

This paper contains just two theorems but both are huge advances in integer complexity; one on the upper bound, one on the lower bound.

First the upper bound. Let's review -- what upper bounds are known on integer complexity? If one wants a bound that works for all n, well, there's the naive bound ||n||≤3log₂n, and then there's Josh's improvements on that, and that's it. The empirical maximum of ||n||/(log n) occurs at n=1439, but these bounds aren't good enough to prove that that value is indeed the maximum; they're substantial overestimates. What if you just want bounds that work for all but finitely many n, bounds on the lim sup? Sorry, we don't have any of those that aren't bounds for all n.

But we do know of better results if you just want bounds that work for almost all n, bounds on what in this post I called lim sup ap ||n||/(log n), which are obtained by the averaging method; these bounds are good enough to break the 1439 barrier, but of course they don't bound the actual lim sup. And we know of even better bounds if your idea of "almost all" only requires logarithmic density 1, rather than natural density 1; this is what I've now been denoting lim sup ap* ||n||/(log n), and the bounds come from Steinerberger and Shriver's method. The current best bounds for both of these categories come, to my knowledge, from Kazuyuki Amano's work.

Well. Konyagin and Oganesyan claim that they have shown that the averaging method actually yields an upper bound on lim sup ||n||/(log n)! Indeed, more than that -- you should just go click through to their paper and read their inequality. They have well and truly broken the 1439 barrier, proving that only finitely many numbers can have a complexity that high. Although, of course, their actual inequality comes with an error term, and having asked them, they say the error term is bad enough that their theorem doesn't currently prove that 1439 is the actual maximum. But wow! This is a massive improvement on the state of the art.

But wait, their next theorem is equally impressive. For a long time, nobody's been able to get a nontrivial lower bound on lim sup ||n||/(log n) either. By "nontrivial" here I mean "better than the known liminf, 3/(log 3)". It was an open question whether ||n|| was asymptotic to 3log₃n, and some people thought it was, even though this would mean that ||2^k||=2k would have to fail for large k! Well, Konyagin and Oganesyan say they've now done it -- they've proven a lower bound on lim sup ||n||/(log n) that is larger than the trivial one, showing that ||n|| is not asymptotic to 3log₃n after all.

Except, they're actually claiming something much stronger. Getting a lower bound on the lim sup would mean showing that infinitely many n have ||n||/(log n) above the bound. They say they've shown that in fact, almost all n do.

So this isn't just a lower bound on the lim sup -- it's a lower bound on the lim inf ap! That's basically as good as you could do!

Their proof here is actually based on my work with joshuazelinsky on the defect. At a high level, their approach is to first use our iterative classification theorem -- yes, the original one, they're not using low-defect polynomials -- to establish upper bounds on how many leaders below x have defect in the range (k-1)σ to kσ, where σ is the variable they're using to denote their step size (they pick σ=0.48), with these upper bounds, importantly, being uniform in both x and k. (This is the hard part.) Once they've done that, they apply this to count how many numbers n below a bound x have ||n||<(3+γ)log₃n, where γ is a number they've picked for this to work (they pick γ=0.06), and compute that, oh look, it's o(x). Therefore, almost all numbers have ||n||/(log n) above this bound. Tada!

Josh and I actually worked on a similar idea many years ago (ours didn't require picking a step size below 1; we were looking at defects inbetween the integers k-1 and k, and of course were working based on low-defect polynomials), although I'm unsure if Josh's method would have been good enough to show that it worked for almost all n rather than just infinitely many; if it was, we didn't realize it. But ultimately it didn't work out because we couldn't prove those uniform bounds we needed. But Konyagin and Oganesyan say they've done it!

I do have to wonder about the choice of σ. I would expect larger values of σ to yield better results, so it's surprising to me that they picked it so far below what it could have been. I have asked them about this, however, and they are of the opinion that larger σ would probably not yield much better results. Still, we'll see if anyone manages to do any better with their ideas.

(It's possible they picked σ<½ so that they could start with B_σ and B_2σ both already known, using Josh's and my work classifying numbers with defect less than 1. Of course, my algorithms can be used to compute all numbers with defect less than 2, but maybe they didn't know about this. I've since sent them the output of such a calculation, just in case they can make use of it. Like I said, their opinion was that larger σ wouldn't be much better, but we'll see.)

Now the question becomes, is it all correct? Unfortunately, their arguments are quite analysis-heavy, and so I am not the best person to evaluate them. So right now my answer can only be "I don't know". But I'm hopeful!

-Harry

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What I don't like about the OMNY card

Sat, 21 Mar 2026 19:59:30 GMT

So, New York City recently got rid of the Metro Card (the card readers are still there, so if you have one it's still usable for now, but you can no longer get a new one nor refill your existing one) and replaced it with the OMNY card. The OMNY card has one big obvious advantage: It's much easier to use. Tapping your OMNY card is much easier than the mess that was swiping your Metro Card several times until it finally worked. (They say real New Yorkers could get it consistently on the first try. Well, by that standard I'm no real New Yorker, I suppose.) Also, if you want to get a new OMNY card or refill your existing one, the machines are faster and easier to use, particularly because they support tapping a credit card, whereas previously getting a credit card to register was *also* something that might take multiple tries.

But the OMNY card system has a few problems, ways in which it's distinctly worse than the old Metro Card. First off, when you pay with an OMNY card, the display doesn't show how much you've been charged and how much is is left on the card. Metro Card did this. Why not OMNY?

Worse, if you don't have sufficient fare, the error message is unhelpful -- it just says "card not accepted". This will make you think there's something wrong with your OMNY card and you need a new one. There isn't! You just need to refill it! This is confusing; the message should say "insufficient fare" like the Metro Card would.

Similarly, free transfers should have a "free transfer" message like Metro Card did, etc -- when I say you should know how much you've been charged, that includes if it's free, and you should know why it's free! OMNY has a feature where the subways become free if you've used them enough in one week. Is there a special message for when you've activated this feature? I bet there isn't, but there definitely should be!

What OMNY card does do is that, if your balance is low, sometimes it will allow you to go a tiny bit into the negatives (not more than like $3). Um, OK, I guess? I don't actually find this a helpful feature. Making it one more ride before I have to refill my OMNY card doesn't really make my life substantially easier (indeed, it throws off my accounting of when I expect to have to do so). What I want is better indication of what's going on when I use the card, not an interest-free loan of $3.

But what really grates is that it seems like, if your balance is negative, OMNY won't necessarily work for things that are supposed to be free. The other day I took a bus, with my card at $0. (Or so I deduced later -- as I've complained, it didn't tell me!) This put me at $-3 (again, or so I deduced later). Then, just a few minutes later, I attempted to take a subway, with what should be a free transfer. It said "card not accepted"! Dude, it's a *free* transfer. You shouldn't need to have money in order to take a *free* transfer. Yes, even if your balance is negative! Because my OMNY card wasn't accepted, I had to pay with my credit card instead, which meant I had to pay for both the bus *and* the subway, which is not how it's supposed to work. I should be able to trust that things will cost what they're supposed to cost, and not that they'll effectively cost double when your card is low on balance. It should be very simple -- if something is free, it should be always permitted, even at negative balance.

(In effect, the "feature" of allowing small negatives cost me $3, because if you couldn't go negative, I wouldn't have hit this bug! I'd have just paid with my credit card for both the bus and the subway, and the free transfer would have worked.)

Finally, there's the odd fact that OMNY doesn't seem to work for PATH like Metro Card did; instead PATH has an apparently identical, yet nonetheless distinct, system called "TAPP". Why don't these work together? I have to assume the reason is political rather than technical, and that probably it'll be fixed at some point, but it's still a problem.

I should figure out who to complain to about this, I guess. :-/

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Snakequake!

Wed, 25 Feb 2026 01:17:30 GMT

Someone else has written about Snake!

You might remember my webpage on the problem, which unfortunately they don't seem to have turned up in their lit review (I guess that's what happens when you don't publish properly :P ).

But these people are studying the same problem and have turned up some cool new results!

One thing to beware of, they numbered their theta graphs differently than I did. Their Θ(2,2,k) is my Θ_1,1,k-1.

But: They showed that if G is bipartite with an odd number of vertices, then the only way it can be winnable is to include some Θ_1,1,k as a spanning subgraph. As a consequence, they derive that determining winnability is NP-hard!

Moreover, they showed that if a winnable graph does not have a Hamilton cycle, it must have girth at most 6, something I certainly never thought of! They also have an example to show that this is tight, but the example seems incorrect, in that it seems to have a Hamilton cycle? Regardless though the bound is tight, as is shown by Θ_2,2,2.

Note that while I didn't put it on the webpage, I also had an inequality about cycles, only mine was about the largest cycle. I've probably written about it here before, but: Let c(G) be the length of the largest cycle, and let k(G) be the size of the largest clique that can be found at the end of a Hamilton path. Then for a winnable graph G with n vertices, k(G)+c(G)≥n, meaning in particular that c(G)≥n/2. But I never looked at the smallest cycle!

Anyway yeah cool results! Obviously I've written to them. :)

-Harry

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Mystery Hunt Roundup 2026

Tue, 24 Feb 2026 06:16:58 GMT

Hey, they finally put up the public version! This post has been a bit delayed while I waited for the public version of Mystery Hunt to go up. Let's see what I still remember.

So I decided to stay home for Mystery Hunt this year and solve remotely. This is because I'd been to the JMM down in DC shortly before. Admittedly, these were two weeks apart, so yeah I probably could have swung it, but, eh. Easier to stay home. Fortunately for me this year wasn't very in-person focused like last year was.

Shaked joined our team this year! He actually went up to Boston for it though.

Do I have more to say about the Hunt overall? Eh I dunno. Maybe I'll just get to the individual rounds/puzzles.
( Cut for spoilers! )
OK I think that's all I have to say, or that I can remember anyway. I think I had more to say at the time, but, oh well, I've forgotten.

-Harry

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Against extra vowels and syllables

Sun, 08 Feb 2026 07:45:52 GMT

People keep needlessly adding vowels to things, and I think it'd be better if they didn't.

The other day my roommate Geoff received a package via the US Postal Service -- the USPS. He decidedly to jokingly pronounce this initialism as if it were a word -- "uspis", he said. (Sorry, no, I'm not going to bother with IPA or anything here.)

Except... "uspis"? No. That should be "usps". One syllable. There isn't a need to add an additional vowel between the "p" and the "s"; not only is it perfectly pronounceable as-is, but that combination of sounds even occurs in such common English words as "wasps". It shouldn't be any sort of phonotactic problem for an English speaker.

I see this sort of thing commonly and it bugs me -- people adding vowels and syllables where they just don't need to. Now I realize that adding vowels requires less thinking -- it's easier to just separate consonants with vowels than think about how to form them into clusters -- and that in many cases there just isn't any reason for people to put in that extra effort, because, they're, y'know, just trying to pronounce something quickly. (Although it seems to me that there are many cases where people add vowels even when no effort is required!) But when you're doing it deliberately for humor value then you'd want to try to optimize that, right? And consonant clusters are funnier than extra vowels! Like, "snunch" is funnier than "sanunnich", right? Sure, an Italian would have trouble with it, but this is English; we're known for our weird consonant clusters, they shouldn't be difficult for us.

Let's consider the exampe of "Fblthp". "Fblthp" is the name of a character from Magic: The Gathering, introduced in 2013 in the flavor text of the card Totally Lost. He's a homunculus, and his name has no vowels on it. People found the flavor text funny and so Fblthp reappeared and eventually got his own card. This also led to a more general pattern in Magic of homunculi receving names with no vowels in their spellings -- Zndrsplt, Vnxwt. But let's focus on Fblthp.

People generally tend to pronounce the name "Fblthp" as "fibble-thip" (yes sorry no IPA). And "fibble-thip" is a pretty funny name! But what would be funnier is reading "Fblthp" as it is written, which is totally doable. "Fb" is the onset; "l" is the nucleus; "thp" is the coda. One syllable, no vowels. Yeah not all of those clusters are permitted in English -- but that's part of what makes it funny! And look -- I'm not actually a linguist, y'know? Hell, I'm not even actually a conlanger -- I'm just a guy who reads about linguistics or conlanging sometimes. I'm a monolingual English speaker. There's all *sorts* of clusters that occur in other languages that I can't pronounce. I can't say "Gbagbo" or "Nguyen". Hell, I still can't tell an aspirated consonant from an unaspirated one. But I can say "Fblthp" no problem -- no vowels.

I want to break this down a bit more -- part of the humor is the fact that the name is so alien, right? Not having vowels (or at least none in the spelling) is an alien feature. If you pronounce it as "fibble-thip", you're making it less alien and less humorous. Don't add vowels!

Now for "Zndrsplt" or "Vnwxt" I have no idea how you do that. For those I have to add at least some vowels. Still it bugs me how people add more vowels/syllables than they need to -- like, "Vnwxt" generally gets rendered as three syllables, when it only needs to be two. People pronounce that "wxt" as if it were "wixit", but it could easily be "wixt". Again, this consonant cluster occurs in English words like "betwixt" -- an English speaker should be able to say it no problem! Why are people acting like they can't?

(Another common example is "Cthulhu". Why add a vowel between the "c" and the "th"? Yes, it's an unusual consonant cluster... but that's what makes it sound alien! And like it's not even *that* alien, it occurs in the word "cthonic" for instance. I can say "cth" a hell of a lot easier than I can say "gb" (which I basically just can't), as I noted above. Go with the more alien-sounding pronunciation! I like to draw the syllable break between the "l" and the "h" -- "Cthul-hu". Most people leave the second "h" silent. Bah. What's the point of it then? Anyway I think my way sounds cooler. :P )

Point is, consonant clusters are cool, complex consonant clusters are cooler, people should use more of them in their made-up words; adding extra vowels to break them up is lame and people should do less of that.

-Harry

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I won my race with Josh!

Sun, 11 Jan 2026 08:27:50 GMT

It seems PAMS never actually told me when they published my paper with Andreas, seems it went out (or at least up) in September. And MathReviews has finally indexed it... which means, that's right, I finally have a finite Erdös number! Namely, 4. And joshuazelinsky gets 5. I won the race! :P

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End-of-year book count

Thu, 01 Jan 2026 23:08:09 GMT

We're down to just 120 books left! And the origami supplies are all gone, too...

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Central groupoids are kind of neat but not that great

Sat, 13 Dec 2025 21:32:35 GMT

So, I have recently been reading/thinking a bit about central groupoids. This is a quite misnamed class of object; they're not groupoids, the term comes from an old use of "groupoid" to mean "magma". (Blech.) They're magmas (sets with a binary operation) satisfying the central identity, (ab)(bc)=b.

You may recall my project regarding equational probabilities in algebraic structures. I deliberately limited this to algebraic structures I considered sufficiently interesting. Central groupoids did not initially make the list. But after seeing them mentioned on Terry Tao's blog recently for the second time, I decided to check them out. Turns out, they're a worthy inclusion, and I've added them now! Because they have rather more structure than I expected.

Like, central groupoids can be alternately characterized in terms of "central digraphs" -- directed graphs where, for any x and y, there's a unique vertex z such that x→z→y. Finding this z is the multiplication in your central groupoid. Going the other way, x→y means that y is a right-multiple of x, or equivalently, that x is a left-multiple of y. And then considering the adjacency matrix of such a digraph gives us a third way of looking at things; it's a 0-1 matrix, and the centrality condition is equivalent to saying that its square is the all-ones matrix! Neat.

It turns out that if you have a finite central groupoid, it must have a square number of elements, call it k². This can be seen by considering the degree of the vertices. It turns out that each vertex has the same indegree and the same outdegree, and that these are moreover equal to each other; this constant is k. (If you look at, say, the right-multiples of x, and want to biject them to the left-multiples of y, you simply right-multiply by y! And then this proves the rest.) If you fix some x, look at its right-multiples, and then look at *their* right-multiples (obviously this all works on the left as well), these secondary sets of right-multiples partition the set; this shows that the overall number of elements is k².

There's another significance to k, too. Consider what happens when you square an element x of your central groupoid. Either x²=x, i.e. x is an idempotent, meaning it has a self-loop; or, since (x²)²=x, they form a 2-cycle (this is the only way to have 2-cycles). How many of each are there? Obviously if k is odd there must be at least one idempotent, but that's pretty weak. Well, it turns out there's always exactly k idempotents! The thing is, I haven't seen an algebraic/combinatorial proof of this -- the only proof I've seen goes through the matrix interpretation! See, the number of idempotents is the trace of the adjacency matrix M, and M²=J, the all-ones matrix, whose eigenvalues are k² and a bunch of 0s. So the eigenvalues of M are k and a bunch of 0s, and therefore its trace is k.

Still, I had to wonder -- if there's k idempotents, and k² elements... could it be that each element can be written uniquely as a product of two idempotents? That would certainly explain things and be cool! Unfortunately, it isn't true. It's true of the "standard" central groupoids; these are constructed by taking a set S, and then defining an operation on S×S via (x₁,x₂)(y₁,y₂)=(x₂,y₁). Unfortunately, it's not true in general; there are counterexamples with k=3. (With k=2, there's only the standard one.)

(EDIT 12/14: Worth noting also, if you were hoping for the opposite, that only the standard ones have this property, well, that's also false.)

So, in that way they fail. And they also fail in that -- like inverse-property loops -- despite having enough structure to be "sink-resistant", they still don't have an equational probabilities gap. There's probably lots of equations you could use to demonstrate this, but the one I used was (x(xx))²=x(xx). In a standard central groupoid, x(xx) and (xx)x are always idempotents, but in general this needn't be so; and if you check out the example I wrote up on my page that I linked, you can see that I was able to construct a central groupoid which yielded a probability of 1-n/(n+1)², approaching 1 from below. No gap!

So, those are central groupoids. Kind of neat, but probably not something I'm going to think about any further.

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The density-probability principle is true!

Sun, 07 Dec 2025 04:25:10 GMT

I never properly wrote about this question here -- only obliquely referred to it here with the text "proving it seems annoying" -- but now it has an answer!

This is a question about automatic sets, so I should probably explain what those are, since a lot of you won't know. A b-automatic set is a set of whole numbers which is given by, you have some DFA on the alphabet {0,...,b-1}, you write numbers in base b and run them through the DFA, and the numbers it accepts are the ones it accepts. (And an automatic set is a b-automatic set for some b. There are also automatic sequences, where instead of just accept/reject, the DFA returns some other output. But we'll stick to sets here.)

Now, there's a number of questions you might have here about the details, but the concept is actually quite robust to small modifications and so precisely how you handle the details in your definition is just not going to matter. For instance, do you write numbers little-endian or big-endian? It doesn't matter, because regular languages are closed under reversal.

For the purposes of this question, though, that particular detail will matter: We're going to assume numbers are fed into the machine little-end first.

So let's say we have a b-automatic set, and it's given by a machine M when the numbers are written from the little end. And let's suppose that M has a particular property: That from every state, it is possible to reach a "sink state" -- a state that only ever loops back to itself. In this case, we can meaningfully ask the question, if we feed into M a uniformly random string of digits, with what probability does it accept? Since with probability 1, it will always hit a sink state eventually, and then we can judge acceptance or rejection by that. Let's call this probability p(M).

The question, then, is: Say S is the automatic set determined by M. Is d(S), the natural density of S, equal to p(M)?

It sure seemed like the answer should be yes, because both can be thought of as plugging random digits into M. But I couldn't prove it! And surprisingly, I couldn't find any results on this in the literature, either. Now I'm no expert on automatic sets, so I could have missed something, but still, I couldn't find it. People said the problem sounded easy -- and then produced proofs that were incorrect. No correct ones.

Well, I asked around, I asked on MathOverflow, and now, a year and a half later, I finally got an answer, due to Alex Mennen. Yes it's true, and it turns out it really wasn't that hard after all! I'm just not much of an analyst. :P

(On MathOverflow, after feedback from Jeffrey Shallit, I ended up reformulating the question a bit; but the version I posted is in fact equivalent to my original version, and also, my original version is what Mennen ended up proving.)

But, the problem as stated above wasn't really what I wanted. Let's throw in some complications.

Complication #1: Multidimensional sets. One can define multidimensional b-automatic sets, subsets of N^k, where the alphabet is {0,...,b-1}^k. Does it still work here?

Complication #2: The set I cared about wasn't actually a b-automatic set at all, but a Fibonacci-automatic set, where the input is in Zeckendorf! (So, some parts of the setup above have to be appropriately modified; the distribution on digit strings is no longer uniform, for instance.) Can we get it to work here?

Complication #3: Actually, it was a 2-dimensional Fibonacci-automatic set.

Fortunately I was able to modify Mennen's proof to handle these complications (going to multiple dimensions is easy; handling Fibonacci required slightly more work). So yay, it's true!

But, there's more we could ask about. Like, what about other numeral systems? Yeah, I don't want to go there. (Handling bijective base-b I'm pretty sure should work the same by the same proof. I expect dual Zeckendorf should also work, but I haven't tried. But those are minor variations.)

And the other question I have to wonder about is... what about the big end? If we feed in numbers from the big end, then it's easy to come up with counterexamples; but maybe it would work if instead of natural density we used logarithmic density? (Recall, logarithmic density isn't really a separate quantity from natural density -- it just exists more often.) The failures I've seen in the big-endian case have always been of the form "the natural density doesn't exist", not "the natural density exists but has the wrong value". I'm guessing it's probably true! But again, I don't really want to be the one to do it.

Jeff says I should write this up as a paper along with Alex Mennen, which, yeah, I guess I should. Kind of annoying, because it'll push other things back. But, oh well, one more thing for the backlog I guess. Still, first I think I might ask around a bit more, see if anyone knows of a result like this already out there. If it's just something I can cite, great! But my guess is that even if someone has done it for base b, they probably wouldn't have done it for Fibonacci...

-Harry

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This is a slime trail

Fri, 07 Nov 2025 20:11:11 GMT

Imagine if someone organized a big festival to celebrate slugs. They could call it SlugFest.

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The numbers rant

Thu, 06 Nov 2025 07:07:15 GMT

The "numbers rant" is someting I've posted various places on the internet in the past, but it seems like I've never posted it here. Figured made sense to get it down here. So, sure. Here we go.

People who haven't seriously studied mathematics might reasonably assume that mathematicians have a unified notion of "number". There are numbers, you can add them, you can subtract them, you can multiply them, you can divide them (except by 0). Sometimes you might restrict to a particular subset of the numbers, like the integers or the real numbers.

The way people learn about numbers in elementary and high school reinforces this. First you learn about whole numbers; then fractions; then negative numbers; at some point you learn there are irrational numbers, implicitly passing from rational numbers to real numbers without learning just what that really means; and then finally you add in complex numbers. And that's all the numbers.

(If you're a bit more into mathematics, you maybe hear that there's some sort of infinities or something, like ℵ₀ and ℵ₁ or something? You don't learn that in school, but you hear about it elsewhere, and you're not sure how it fits into the above picture, but surely it does.)

This picture is wrong, and moreover, I think it encourages some bad ways of thinking. There is no unified notion of "number" in mathematics. Rather, mathematicians use different systems of numbers -- the integers, the rational numbers, the real numbers, the complex numbers -- depending on context. Some of these systems embed into larger ones; the integers embed into the real numbers, for instance. But the integers can also be considered separately from the real numbers.

OK, so what, you might stay. It still sounds like the complex numbers are all the numbers, and sometimes we just focus on subsets of them. Well, they're not.

One way this is false is that you can expand further on the complex numbers -- like, the 2-dimensional complex numbers embed in the 4-dimensional quaternions. But that's sort of a silly way this is false. If that were the only reason this were false, then it would just be the case that the quaternions were all the numbers. But they aren't! And no, not just because you can expand on them to the octonions, etc.

Rather the key point here is that there are distinct number systems that are entirely incompatible with one another. We've already mentioned an example above -- cardinal numbers. Cardinal numbers include (and are therefore compatible with) the whole numbers; but they're incompatible with negative numbers, let alone real numbers or complex numbers. How do you add ℵ₀+½? You don't! This is a nonsense sum, because you're trying to add two numbers from entirely different systems. You have to work within some specified system of numbers; there isn't some overall general thing called "numbers" where you can take any two of them and add them together.

There are many more systems of numbers; for instance, there are the p-adic numbers, which is not one system but a family of systems, one for every prime p (the 2-adic numbers, the 3-adic numbers, etc). The p-adic numbers (for some fixed p) can be thought of as base-p expansions, except that, whereas in the real numbers, base-p expansions can go finitely far to the left and infinitely far to the right, p-adic numbers can go infinitely far to the left and finitely far to the right. (Note that, as mentioned above, different values of p yield different, incompatible systems; this is in contrast to the real numbers, where, regardless of what base you write it in, it's still the real numbers.) How do you add a p-adic number to a real number? You don't! These are incompatible systems. Now, the p-adic numbers always contain the rational numbers; so if the p-adic number and the real number you're adding are both rational numbers, then sure you could perform the addition within the rational numbers; but there is no more general system that contains both the p-adics and the reals and allows you to add one to the other without special additional knowledge, they're not compatible.

So, there is no unified mathematical notion of "number". Or as I sometimes put it, there's no such thing as a number.

There's something else that bothers me about the way number systems are taught, beyond just that it's wrong, though. It's that if you present things as if complex numbers are all the numbers, it puts the focus on complex numbers as the most fundamental type of numbers. It's like, complex numbers, that's the real meaning of numbers, why would we focus on silly particular subsets like the integers? Presenting the complex numbers as "all the numbers" devalues number theory. It suggests complex numbers as the most fundamental thing, when of course, in reality, whole numbers are the most fundamental thing; other number systems are built on top of the whole numbers, and much of what they do is in a sense just reflecting on and elucidating properties of the whole numbers.

(It's also worth noting that it's important to pick the correct number system for your application. Many people use the real numbers without thinking about it because they're not aware of other possibilities. To be fair, the real numbers are indeed very useful and frequently the right answer, but it's still worth thinking about.)

Annoyingly, there's no formal mathematical definition of a "number system" -- most things we might consider "number systems" are rings or at least semirings, but some fail these criteria (ordinals with their usual arithmetic are just a near-semiring). And of course plenty of rings and semirings are of things we wouldn't normally consider "numbers". Still -- numbers come in systems; different systems are used for different purposes, so pick one that matches your purpose; and there's no such thing as "all the numbers".

That's the numbers rant!

-Harry

PS: If you want a semi-trollish take, one could argue that p-adic numbers are numbers, but complex numbers aren't, because only systems lacking nontrivial automorphisms should be deemed systems of numbers. :P (Because a number should have some sort of absolute meaning, and if there's a nontrivial automorphism, then it doesn't.)

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This has happened before

Mon, 06 Oct 2025 06:58:47 GMT

But that time it involved ping pong

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In a 1-dimensional universe, there'd be no natural units

Sat, 06 Sep 2025 03:26:41 GMT

If you're unfamiliar with the concept, the idea of natural units is that you can create a "natural" system of units, determined by physical constants, by declaring that certain dimensioned physical constants should be equal to 1. E.g., when doing relativity, one often sets c=1, causing length and time to have the same dimensions; the idea is that by extending this convention to other physical constants, you fix enough things that you get a natural length, a natural time, etc.

The best known of these is Planck units, and when I say "natural units", I really mean "Planck units and slight variants", because the other systems of natural units seem much less natural. Planck units work by setting c=1 (from relativity), ℏ=1 (from quantum physics), and G=1 (from general relativity). Personally, I'd prefer setting κ=1, where κ=8πG/c⁴ is Einstein's gravitational constant, since that's the constant that actually appears in Einstein's equation (presumably more fundamental than Newton's law of gravity!), but G is the version that's caught on (partly because Planck units predate general relativity![0]).

Setting these three constants to 1 gets you natural units of length, time, and mass. You can then also include temperature by setting k_B=1 (from thermodynamics), and include electric charge by setting... oogh, this one's tricky. Planck's choice was to set the Coulomb constant to 1, but the naturality of that choice is questionable; OTOH, it's not really clear what the correct choice should be; there's several good candidates (vacuum permittivity, elementary charge). Well -- I'm not going to discuss temperature or electricity, so let's ignore that.

Anyway, the key to the whole system is the fact that c, ℏ, and κ have dimensions that are linearly independent (under multiplication), so you can in fact get the isolated dimensions of length, time, and mass out of them. But... what if we changed the number of spatial dimensions[1] of the universe?[2] The dimensions of κ are curvature/density (or energy density, whatever, we're setting c=1). The dimensions of curvature are inverse area, 1/L². But the dimensions of density are M/L^d, where d is the number of spatial dimensions. Which means that, in a 1-dimensional universe, κ and ℏ would (up to length/time conversions, since we're setting c=1) have the same dimensions as each other, making it impossible to extract natural units from them!

But wait, there's a way to rescue this! Remember the electricity mess above? If you're not getting natural units from ℏ and G, this changes from a problem to a solution -- you could set both the elementary charge to 1, and either the Coulomb constant or ε₀ to 1, depending on your preference!

Except... this doesn't work either! Once you set c=1 and e=1, you find that ℏ and the Coulomb constant have the same dimensions. You'd need G here to make things work, but with only 1 spatial dimension, G doesn't help you.

So: In a world with only one spatial dimension, there would be no natural units.

(Yeah OK the systems of natural units based around the electron or proton masses would still work fine but c'mon those clearly aren't that natural, I did say I was only considering slight variants of Planck units. :P )

-Harry

[0]Actually, IINM, Planck units also predate the switch to ℏ from h=2πℏ, but that change got made... the G to κ change never became popular, partly because I think switching from G to κ never got that popular in other contexts either. The version using κ instead of G is sometimes known as "reduced Planck units". So, yeah.
[1]Yes, this post uses the word "dimension" in two different senses a bunch. Don't get confused!
[2]I'm assuming there's always exactly 1 time dimension; if we allow changing that, then what I really mean is "two total dimensions", rather than "1 spatial dimension".

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NYC subway challenge

Wed, 13 Aug 2025 08:25:58 GMT

Inspired by seeing this old article on Hacker News, here's a challenge for you: In the (current) NYC subway system, find two stops A and B such that to get from A to B, you need to take 4 trains. (This isn't 1980 anymore; I'm pretty certain there aren't any pairs requiring 5 trains.) We can set out several difficulty levels:

1. Ultra-easy version: There just has to be some time of day / day of the week at which it works.
2. Easy version: It has to work on the normal subway schedule (meaning, weekday, not late-night, not rush hour, no peak-direction-only services, no other special stuff, the Times square / Bryant Park transfer is open).
3. Full version: It has to work no matter the time of day or day of the week.

We assume that the subways are in fact running as scheduled -- no reroutes or service reductions. You have to stay inside the subway system, inside the fare gates; out-of-system transfers are not allowed even if they're free, and if you want to turn around, you have to do it at a station where that's possible.

(Note: I'm not a subway expert, so I'm relying on the maps (including their legends and info) to be essentially accurate here. I'm aware that there's some more info that's not on the maps, but so far, I've yet to see anything relevant to the challenge, so I'm just going to hope there's truly nothing!)

Can you find how to do it?
( Answers in the cut )

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Nonstandard subnational flag emoji: An answer!

Wed, 13 Aug 2025 07:26:37 GMT

So, remember this?

Well, I finally found out where it's coming from! What's going on here is that these flags are included in my system's emoji font, which is from Google's Noto Emoji family. And the version of this font included with Debian and Ubuntu, at least, includes these nonstandard flags. Which ones? Well, the files for them are right here! And yup, it's exactly the ones I said.

So yay, that's that question answered!

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Horseshoe crabs!

Wed, 25 Jun 2025 06:39:07 GMT

So, last summer, I learned from HK about the horeshoe crabs in New York City. She had -- I think, I forget the exact story -- been out exploring the distant parts of the city, and had come across Broad Channel, and she saw the horseshoe crabs on the beach there, and she gathered some people to go see them again. So she and I and some other friends went out to see them.

See, around May and June, it's horsehsoe crab mating season. I had no idea that horseshoe crabs even lived in the waters around here, but they do. And during the mating season they come up into the shallows to mate. But, you can't see them at just any time. My understanding is, they come out mostly when the tide is highest. You want to go at high tide, but not just at high tide, you want to go at high tide around a spring tide.

Anyway, last year's trip out to Broad Channel and we saw lots of horseshoe crabs. Although, at first, we made the mistake of heading north to the wildlife refuge, rather than south to the beach. The wildlife refuge did not have horseshoe crabs. What did have was lots and lots of mosquitoes. We all got really bitten before deciding to turn back and head to the beach instead.

I wanted to try the same thing again this year -- but maybe we didn't need to go all the way out to Broad Channel? The city has a webpage listing where horseshoe crabs can be seen. None of these locations are all that convenient, but Kaiser Park, in Coney Island, seemed like the most overall accessible.

So a few friends and I tried it out. Unfortunately, we made a bit of an error in timing. Yes, we went near a spring tide... but I hadn't checked the detailed tide charts, and it was actually near low tide when we went. Oops. We ended up mostly only seeing dead horseshoe crabs -- although we did find one live one, buried in the sand. Once the tide came in, it came out, and we got to see it swimming around! Still, just one is a bit disappointing. We did see a really cool bird which I have since learned was a yellow-crowned night heron!

We (well, not the same group) tried again two weeks later, same location. In this case we knew it would be around low tide, but, well, Kate insisted it be then (and then didn't show up...). The result was basically the same. A bunch of dead horseshoe crabs, only one living one. Well -- I spotted one living one buried in the sand before the others arrived, but once they showed up and the tide had come in a bit, I couldn't find it anymore. I guess it swam away? We did not see any herons this time, which was a bit disappointing, as Kippy (a birder) had come hoping to see one.

We did see lots of fiddler crabs! They're difficult to see, because in addition to being small, you can only see them from some distance -- get anywhere near them and they'll scuttle back into their holes (their holes covered that section of the beach). But move away and they'll come out and wave their claws up and down!

Anyway, I decided to make one final attempt this past Sunday. This time, the tides should be right. (And for whatever reason, this spring tide was projected to be especially high, higher than the other recent ones. I wonder why?) HK suggested we try Calvert Vaux Park, across the creek from Coney Island rather than on it, although, like Kate, she then didn't show up! But this time it worked -- we did find horseshoe crabs! And also an egret, and a fiddler crab that wasn't so good at hiding, and also lots of cats? By the entrance near 27th Ave I encountered 3 cats. Huh.

The park seems to consist of 3 peninsulas (do all 3 count as the same park? I think the south one is separate), but the three seem pretty split off from one another -- I think they all have separate entrances. We didn't try the northern peninsula because we just kind of missed it, but it looks like there's a beach there we could have tried. We did try the cove that lies between the northern and central peninsulas, since there was a beach there, but it was a dump, and also had no horseshoe crabs. (It also required taking some paths through the brush that were not the easiest.)

Ultimately the place we found horseshoe crabs was one of the beaches on the south side of the central peninsula. This is basically where we found all the animals, except the cats! There weren't as many as at Broad Channel last year -- possibly still an issue of timing, we were a bit after high tide and a few days before the highest tides -- but still a good number. One of them was tagged! So, yes, I reported it afterward.

We didn't try the southern peninsula (which I think is actually not part of the park). We were kind of tired from the heat at that point, and I'm not sure that the southern peninsula has any beaches, or how to get to them if so.

June is ending now so I think that'll be it for this year. But I'll go look for them again next year, I figure, and that time I'll be more considerate of just when high tide is...

(Meanwhile, the goats arrive on July 12th!)

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The great enbookening

Sat, 24 May 2025 05:23:07 GMT

Huh, I never posted about this, did I? So Kappa finally moved in back in January, and with her, many boxes of books. This month she finally unpacked most of those boxes. (There's still two remaining, to be handled at some undetermined future date.) There had been preparations earlier -- she and Geoff had gotten new bookshelves to replace Geoff's old bookshelves in the main room, that had been kind of falling apart for years. One of the old bookshelves was thrown out, the other I ended up taking into my room.

But now finally it was happening, the unpacking of the books. The idea was that Kappa and Geoff's book collections would be merged, with the new books going onto what had been Geoff's bookshelves -- whether the ones in the main room or the ones in their room. But -- as was anticipated -- in merging the collections, many duplicates were found. Where do you think these went?

That's right. They went on the giveaway shelves.

Now, which shelves those are had to change. The sheer number of books unpacked was such that Geoff and Kappa ended up taking over all the shelves along the west wall -- my books got kicked out to the bookshelf that was now in my room. So did my board games on the bookshelf in the corner. Now the west wall is all Geoff and Kappa's books, and the east wall is all books being given away. (Due to the change, I decided to finally label the shelves of free books.)

It wasn't just duplicates from between Geoff and Kappa that got put on the giveaway shelves, though. On unpacking her books, Kappa found various ones that she actually just didn't want anymore, so those ended up on the giveaway shelves too. Meanwhile, in preparation for the unpacking, Geoff had had to actually organize his book collection... and in doing so, turned up some duplicates purely within it! (And some books he decided he didn't want, either.) So that's more for the free shelves!

Of course, putting books on the free shelves doesn't actually immediately save any space -- it only does so once someone takes them. Maybe once enough books are taken, I'll be able to move my own books back out into the main room. But that'll take a while.

Prior to the deduplication, the number of books on the giveaway shelves sat at 137. After, it sat at... I don't know, I didn't even try counting. See, among the books being given away due to inter-spouse duplication was a nearly-complete set of Discworld books. These got placed on the top shelf of the tall bookcase behind the chair, double-stacked. I didn't want to deal with the inconvenience of counting that, so I didn't.

Now of course, these new books are going rather faster than the old ones. After all, the books still on the shelves before this happened were subject to adverse selection; of the books we were giving away, these were the 137 that people wanted the least. By contrast, the bulk of the ones being newly added were because they were books that two distinct people had bought! (Also, it's a lot of fantasy and science fiction, whereas the 137 that were there prior were heavy on abstruse philosophy.)

In particular, those Discworld books lasted on the shelves less than a week before Zvi and Laura took the collection in its entirety. After that, counting became practical. The new total -- noting that when Zvi and Laura took the Discworld books, that same night some other people took other books, so this shouldn't be taken as "the grand total minus Discworld" -- was 222.

Two weeks later we're down to 212. And two of the books taken have been ones that were already on the shelves prior to this -- perhaps having better books to give away encourages people to explore the other ones more thoroughly? We'll see how low it goes!

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An order-theoretic interpretation of Jacobsthal multiplication that makes associativity obvious

Mon, 28 Apr 2025 01:46:20 GMT

So, here's something that had bugged me for a long time. Jacobsthal multiplication is associative, but the proof of this is unsatisfying. Normally the way you prove an operation is associative is that you show that a multiary version of it makes sense; you don't just literally show that a(bc)=(ab)c, you show that abc makes sense on its own as a ternary thing (or with even more operands) and that a(bc) and (ab)c are both equal to this.

But for Jacobsthal multiplication, the only proof I knew of its associativity was to show that a×(b×c) was equal to (a×b)×c; I wasn't able to find an interpretation of a×b×c on its own.

Now, some years ago, Paolo Lipparini found an order-theoretic interpretation of the Jacobsthal product, but it wasn't one that led to an immediate proof of associativity. But based on an observation of Isa Vialard, I think I finally have one! It's a little awkward, but of course it is; it's Jacobsthal multiplication, after all.

Let's say A and B are WPOs, and let's consider the lexicographic product A·B (note: B here is the coordinate that's compared first). Then o(A)o(B) ≤ o(A·B) ≤ o(A)×o(B).

But, actually, we can say more. Suppose B has k maximal elements, so that o(B) has finite part greater than or equal to k. Then, Vialard noted, we have o(A·B) = o(A)·(o(B)-k) + o(B)×k.

In particular, if o(B) has finite part equal to its number of maximal elements, then o(A·B)=o(A)×o(B). (In fact, as long as A has multiple distinct powers of ω in its Cantor normal form, the converse also holds.)

Btw, the finite part of o(B) is equal to the number of elements of B that have only finitely many elements above them. We could call such elements "almost maximal". So B satisfies this condition iff every almost maximal element is maximal.

Anyway, let's call a WPO that satisfies this condition "flat-topped". Note that flat-topped WPOs are closed under disjoint union, under Cartesian product, and, yes, under lexicographic product! So we can interpret α×β as the type of A·B for any flat-topped A,B with o(A)=α and o(B)=β; and we can do similarly with α×β×γ, etc. So this does it! We've interpreted multiary × and proved associativity nicely. Hooray!

EDIT 5/5: Actually, on writing to Lipparini about this, he pointed out to me that another observation of Vialard provides an even easier answer, namely, without needing the flat-topped condition, one has w(A·B)=w(A)×w(B), where w is width. I hadn't noticed this because, well, I've never really paid a lot of attention to widths, tbh.

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While/else (with a better name: maybe while/coda?) should be a standard construct

Wed, 16 Apr 2025 19:44:27 GMT

I'm trying to just write down more old thoughts, and you know what, I haven't written this one down.

The Python programming language has an interesting construct I haven't seen elsewhere: while/else. A while block with an else block following. Now this is terribly misnamed; based on the name you'd think that, like, the else block executes if the while block ends up being skipped, like if/else, but that's not what it is. The way it actually works that the else block executes after the while block *unless* you broke out of the while block.

This is misnamed -- it should be called something like while/coda or while/then -- but I think this is a nice enough construct that it should be standard! Yes, obviously just like labeled breaks this can be replicated by setting a flag or whatever (and often is), but it would still be nicer to just have it directly.

In particular, it really bugs me that Rust in particular doesn't have it, and here's why. Rust has this neat feature where any block can return a value (indeed, *does* return a value -- if it doesn't seem to, it's returning a value of unit type), and in a loop block, when you break, you can specify a value, and then the loop returns a value! That's pretty neat!

But, while you can do this from a loop block, you can't do this (with a non-unit type) from a while block! Why not? Because if the while block concludes naturally, what value would it return? The only sensible thing is to say that the while block returns unit type -- since it must return a consistent type -- but that means it can't return a non-unit value, you can't break out of it with a value. But if you could add a coda to a while block, this wouldn't be a problem anymore!

(Obviously all of this also implies to other types of looping blocks that naturally terminate.)

This lack of consistency really bugs me, and it's something that people have repeatedly suggested adding, but it's never been added because they haven't come up with a syntax for it that is both clear and not too breaking. (Apparently breaking out of a loop with a value was a somewhat late addition to Rust, which is why this wasn't gotten right at the beginning.) Boo! Anyway yeah it bugs me.

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My one comment about the Switch 2

Wed, 02 Apr 2025 21:06:57 GMT

Why didn't they call it the Super Switch??

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I think I may have finally come up with a way to explain to people what the Freemasons are

Wed, 02 Apr 2025 06:16:14 GMT

This is a problem that has come up several times recently, where I have mentioned the Freemasons to someone, and it turned out they'd never heard of them, and I had to try to explain it, only to find that it's hard to actually explain! But I think I've got it: The Freemasons are an organization about pretending they're about something. :P

...OK, sure, they're not exactly an organization per se, but still...

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On the significance of Eliezer Yudkowsky

Sun, 09 Mar 2025 08:36:42 GMT

There's a question that I occasionally see asked on the internet -- what's the big deal about Eliezer Yudkowsky? Oh, he suggested that artificial intelligence might be dangerous? That's not original! People have been talking about robot uprisings for as long as they've been talking about robots. Why is this Yudkowsky guy getting credit for it?

And I think this question is interesting for what it reveals about how much things have changed since, oh, 20 years ago, let's say. Yes, people have been talking about robot uprisings for as long as they've been talking about robots. But the question you have to ask is, who was saying artificial intelligence might be dangerous? Because back then, the general assumption among... technical people, science fans, transhumanists, etc -- the general assumption among that crowd was that of course artificial intelligence would be a good thing, just like technology generally is a good thing; and that if you thought otherwise, if you thought actually it might be dangerous, you were a backwards luddite. (This is roughly the view expressed now by the "e/acc" crowd.) Moreover, the arguments for it being dangerous were largely based on anthropomorphization. Yes, there were people talking about the dangers of artificial intelligence -- but the sort of person who might go work on artificial intelligence, or become any sort of expert on it, would never listen to that sort of person, and with good reason; they generally weren't worth listening to. They really were often anti-technology luddites, and their arguments were generally pretty bad.

(Frankly, I'm not even the best person to be relating this -- someone older, who had an adult's view of all this, would be better. I was prompted to finally write this post due to a conversation with an acquaintance today who occasionally attends OBNYC, but who also attends New York skeptics' meetings, and mentioned how the latter is a very different, substantially older crowd. I like to point out sometimes that a lot of Yudkowsky's writing is fairly continuous with older writings on rationality by people like Feynman, Sagan, etc; but this is only somewhat true, as evidenced by this split!)

So the significance of Eliezer Yudkowsky -- I mean, certainly not the only significance, likely not the main signifiance, but the significance for the purposes of this question -- isn't that he proposed that artificial intelligence could be dangerous; it's that he A. convinced of the risks of artificial intelligence the sort of people who were inclined to dismiss such, and B. did so by pointing out that the arguments for AI being safe are in fact based on anthropomorphization and rebutting these arguments in detail.

But it seems that people just coming to this conversation now often don't realize all this! They don't think there's anything usual about technical people, transhumanists, etc., considering AI to be dangerous -- they think of it as continuous with earlier arguments about robot uprisings, rather than considering those earlier arguments as something such people would have dismissed.

Somewhat similarly -- and this is something Yudkowsky himself has often remarked upon -- you get people who've never bothered to actually read Yudkowsky calling him an anti-progress luddite, which is of course not remotely correct. Indeed, the fact that he was not a luddite, and had credibility as a transhumanist, lent credibility to his eventual turn against artificial intelligence, made it influential rather than something reflexively dismissed! The thing (well, one of the things, but the relevant thing) that was unusual about Yudkowsky (but, thanks to him, is not unusual anymore) was his stance of being pro-technology and pro-progress but against the development of artificial intelligence specifically. But some people who comment on things based only on impressions assume that it still must be the case that anyone opposed to artificial intelligence must be a luddite. This is basically the opposite way one might miss Yudkowsky's significance; instead of failing to realize that the past was not like the present, one could fail to realize that the present is no longer like the past. (Well, OK, that's not quite right, because these people aren't so much assuming that the present is like the past, as they are assuming that the present is like the obvious thing you expect. It isn't!)

(Things are also different from 15 years ago in that it seems like transhumanism as it used to exist seems to be much reduced, because discussion of artificial intelligence has largely subsumed it all! Not entirely, but to a pretty good extent.)

Anyway, yeah, some context for those confused about such things...

-Harry

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Where's Calvin when you need him

Sun, 02 Feb 2025 20:43:57 GMT

Have you all read about the geology on Venus? How there's no real cracks in the crust, no distinct tectonic plates, just one big ("stagnant") lid? How this doesn't allow heat to escape slowly like it does on Earth? How it's hypothesized that every half a billion years or so, the trapped heat finally breaks out in the form of giant volcanic eruptions all over the entire planet, where possibly the planetary entire surface just kind of melts, erasing all record of what came before?

Yeah. You know what they call this kind of worldwide catastrophe?

A global resurfacing event.

That really needs a better name.

ADDENDUM: OK, I'll kick things off and suggest "omnivolcano apocalypse". Or, if we want to to tone things down just a little, perhaps "omnivolcanic catastrophe"?

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Mystery Hunt Roundup 2025!

Thu, 30 Jan 2025 04:56:59 GMT

Yes solutions have been up for a while but I've been busy! Anyway, Mystery Hunt this year!

I once again went up to Boston for Mystery Hunt this year. Still on Plant.

The Hunt had an interesting structure this year. The first round had five different metapuzzles -- four ordinary metas and one supermeta, although I believe it was designed to be solvable without solving all of the other metas first -- and solving each one of the five unlocked a different subsequent round. Each of these later rounds had their own metapuzzle or metapuzzles, and solving the metapuzzles for all rounds unlocked the runaround. We made it to all five of the later rounds, but didn't solve any of their metas.

The other way the structure of the hunt was unusual was that, rather than solves unlocking specific puzzles, instead they separately "discovered" puzzles and gave you keys. You see, when you discovered a puzzle, it would still be locked -- they'd give you a name and a minimal description that they considered spoiler-free -- and you had to spend a key to unlock it. The idea was that this would allow teams to focus their efforts on the sort of puzzle that they prefer. I think this was a positive effect, but there was also a negative one. You know how clues and free solves require strong team leadership to make sure they get used? I think keys have a similar sort of thing going on. Without strong leadership a team might just not spend their keys and not unlock enough puzzles!

Also, one of the subsequent rounds this year was a fish round, and, I have mixed feelings about that. I mean, it's nice to have some easy puzzles, but they're so long, and it's kind of exhausting to solve so many puzzles and still not unlock the meta. If you want to have a fish round, maybe break it up into several rounds, rather than making it one long one?

The other reason I'm maybe a bit cool on fish rounds is that if there's always easy puzzles available, it becomes less appealing to sit down with the harder ones and stare at them and try lots of things. It's distracting! At least it's not like last year where, AIUI, the fish round barely contributed to progression... here at least it was a necessary part of making it to the runaround. But, yeah, if you want to have easier puzzles, maybe just mix them in with the other puzzles as an occasional "here have an easy one" rather than sticking them all in one round to make sure one's always available?

I'm actually a little unsure I helped all that much this year! I mean I definitely contributed to the team but I feel like I was often so distracted... well, I guess I'll go over the individual puzzles and see!

(Also: Li-Mei was on the writing team this year! Hi Li-Mei! :) )
( Cut for spoilers )
OK and I think that's all I have to say about this year!

Next year I think I might go to the JMM because it'll be in Washington DC, so I'll probably stay home for Mystery Hunt and solve remotely. But we'll see...

-Harry

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