The Universe of Discourse

Does someone really have to do the dirty jobs?

Doing the laundry used to be backbreaking toil. Haul the water, chop the wood, light the fire, heat the water, and now you are ready to begin the really tough part of the work. The old saying goes "Wash on Monday", because Monday is the day after your day of rest, and otherwise you won't have the strength to do the washing.

And the saying continues: “Iron on Tuesday, mend on Wednesday”. Routine management of clothing takes half of the six-day work week.

For this reason, washing is the work of last resort for the poorest and most marginal people. Widows are washerwomen. Prisons are laundries. Chinese immigrants run laundries. Anyone with enough money to outsource their laundry does so.

The invention of mechanical washing machines eliminated a great amount of human suffering and toil. Machines do the washing now. Nobody has to break their back scrubbing soiled linens against a washboard.

“Eskimo child with wooden tub and washboard”, c. 1905, by Frank Hamilton Nowell, public domain, via Wikimedia Commons.

But the flip side of that is that there are still poor and marginalized people, who now have to find other work. Mechanical laundry has taken away their jobs. They no longer have to do the backbreaking labor of hand laundry. Now they have the option to starve to death instead.

Is it a net win? I don't know. I'd like to think so. I'd like to free people from the toil of hand laundry without also starving some of them to death. Our present system doesn't seem to be very good at that sort of thing. I'm not sure what a better system would look like.

Anyway, this is on my mind a lot lately because of the recent developments in computer-generated art. I think “well, it's not all bad, because at least now nobody will have to make a living drawing pornographic pictures of other people's furry OCs. Surely that is a slight elevation of the human condition.” On the other hand, some of those people would rather have the money and who am I to deny them that choice?

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The mathematical past is a foreign country

A modern presentation of the Peano axioms looks like this:

1. !!0!! is a natural number
2. If !!n!! is a natural number, then so is the result of appending an !!S!! to the beginning of !!n!!
3. Nothing else is a natural number

This baldly states that zero is a natural number.

I think this is a 20th-century development. In 1889, the natural numbers started at !!1!!, not at !!0!!. Peano's Arithmetices principia, nova methodo exposita (1889) is the source of the Peano axioms and in it Peano starts the natural numbers at !!1!!, not at !!0!!:

There's axiom 1: !!1\in\Bbb N!!. No zero. I think starting at !! 0!! may be a Bourbakism.

In a modern presentation we define addition like this:

$$ \begin{array}{rrl} (i) & a + 0 = & a \\ (ii) & a + Sb = & S(a+b) \end{array} $$

Peano doesn't have zero, so he doesn't need item !!(i)!!. His definition just has !!(ii)!!.

But wait, doesn't his inductive definition need to have a base case? Maybe something like this?

\begin{array}{rrl} (i') & a + 1 = & Sa \\ \end{array}

Nope, Peano has nothing like that. But surely the definition must have a base case? How can Peano get around that?

Well, by modern standards, he cheats!

Peano doesn't have a special notation like !!S!! for successor. Where a modern presentation might write !!Sa!! for the successor of the number !!a!!, Peano writes “!!a + 1!!”.

So his version of !!(ii)!! looks like this:

$$ a + (b + 1) = (a + b) + 1 $$

which is pretty much a symbol-for-symbol translation of !!(ii)!!. But if we try to translate !!(i')!! similarly, it looks like this:

$$ a + 1 = a + 1 $$

That's why Peano didn't include it: to him, it was tautological.

But to modern eyes that last formula is deceptive because it equivocates between the "!!+ 1!!" notation that is being used to represent the successor operation (on the right) and the addition operation that Peano is trying to define (on the left). In a modern presentation, we are careful to distinguish between our formal symbol for a successor, and our definition of the addition operation.

Peano, working pre-Frege and pre-Hilbert, doesn't have the same concept of what this means. To Peano, constructing the successor of a number, and adding a number to the constant !!1!!, are the same operation: the successor operation is just adding !!1!!.

But to us, !!Sa!! and !!a+S0!! are different operations that happen to yield the same value. To us, the successor operation is a purely abstract or formal symbol manipulation (“stick an !!S!! on the front”). The fact that it also has an arithmetic interpretation, related to addition, appears only once we contemplate the theorem $$\forall a. a + S0 = Sa.$$ There is nothing like this in Peano.

It's things like this that make it tricky to read older mathematics books. There are deep philosophical differences about what is being done and why, and they are not usually explicit.

Another example: in the 19th century, the abstract presentation of group theory had not yet been invented. The phrase “group” was understood to be short for “group of permutations”, and the important property was closure, specifically closure under composition of permutations. In a 20th century abstract presentation, the closure property is usually passed over without comment. In a modern view, the notation !!G_1\cup G_2!! is not even meaningful, because groups are not sets and you cannot just mix together two sets of group elements without also specifying how to extend the binary operation, perhaps via a free product or something. In the 19th century, !!G_1\cup G_2!! is perfectly ordinary, because !!G_1!! and !!G_2!! are just sets of permutations. One can then ask whether that set is a group — that is, whether it is closed under composition of permutations — and if not, what is the smallest group that contains it.

It's something like a foreign language of a foreign culture. You can try to translate the words, but the underlying ideas may not be the same.

Addendum 20250326

Simon Tatham reminds me that Peano's equivocation has come up here before. I previously discussed a Math SE post in which OP was confused because Bertrand Russell's presentation of the Peano axioms similarly used the notation “!!+ 1!!” for the successor operation, and did not understand why it was not tautological.

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Baseball on the Moon

We want to adapt baseball to be played on the moon. Is there any way to make it work?

My first impression is: no, for several reasons.

The pitched ball will go a little faster (no air resistance) but breaking balls are impossible (ditto). So the batter will find it easier to get a solid hit. We can't fix this by moving the plate closer to the pitcher's rubber; that would expose both batter and pitcher to unacceptable danger. I think we also can't fix it by making the plate much wider.

Once the batter hits the ball, it will go a long long way, six times as far as a batted ball on Earth. In order for every hit to not be a home run, the outfield fence will have to be about six times as far way, so the outfield will be !!36!! times as large. I don't think the outfielders can move six times as fast to catch up to it. Perhaps if there were 100 outfielders instead of only three?

Fielding the ball will be more difficult. Note that even though the vacuum prevents the pitch from breaking, the batted ball can still take unexpected hops off the ground.

Having gotten hold of the ball, the outfielder will then need to throw it back to the infield. They will be able to throw it that far, but they probably won't be able do it accurately enough for the receiving fielder to make the play at the base. More likely the outfielder will throw it wild.

I don't think this can be easily salvaged. People do love home runs, but I don't think they would love this. Games are too long already.

Well, here's a thought. What if instead of four bases, arranged in a !!90!!-foot square, we had, I don't know, eight or ten, maybe !!200!! or !!300!! feet apart? More opportunities for outs on the basepaths, and also the middle bases would not be so far from the outfield. Instead of throwing directly to the infield, the outfielders would have a relay system where one outfielder would throw to another that was farther in, and perhaps one more, before reaching the infield. That might be pretty cool.

I think it's not easy to run fast on the Moon. On the Earth, a runner's feet are pushing against the ground many times each second. On the Moon, the runner is taking big leaps. They may only get in one-sixth as many steps over the same distance, which would give them much less opportunity to convert muscle energy into velocity. (Somewhat countervailing, though: no air resistance.) Runners would have to train specially to be able to leap accurately to the bases. Under standard rules, a runner who overshoots the base will land off the basepaths and be automatically out.

So we might expect to see the runner bounding toward first base. Then one of the thirty or so far-left fielders would get the ball, relay it to the middle-left fielder and then the near-left fielder who would make the throw back to first. The throw would be inaccurate because it has to traverse a very large infield, and the first baseman would have to go chasing after it and pick it up from foul territory. He can't get back to first base quickly enough, but that's okay, the pitcher has bounded over from the mound and is waiting near first base to make the force play. Maybe the runner isn't there yet because one of his leaps was too long and to take another he has to jump high into the air and come down again.

It would work better than Quiddich, anyway.

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Hangeul sign-engraving machine

Last summer I was privileged to visit the glorious Letterpress Museum in Paju Book City, where I spent several hours and took a collection of photos that are probably not of interest to anyone but letterpress geeks, and perhaps not even to them.

Looking back at the photos it's not always clear to me why I took each one. But some of them I can remember. For example, this one:

This is not exactly letterpress. It is a device for engraving lettered signs on thin strips of metal or perhaps plastic. Happily I don't have to spend too much time explaining this because Marcin Wichary has just published an extensively-illustrated article about the Latin-script version. The only thing different about this one is the fonts, which are for writing Korean in Hangeul script rather than English in Latin script.

(Here's my real-quick summary. There is no ink. A stylus goes into the grooves of those brass templates. The stylus is attached with a pantograph to a router bit that rests on the object that the operator wants to engrave. When operator moves the stylus in the template grooves, the router bit follows their motions and engraves matching grooves in the target object. By adjusting the pantograph, one can engrave letters that are larger or smaller than the templates.)

Hangeul has an alphabet of 24 letters, but there's a difficulty in adapting this engraving technique for written Hangeul: The letters aren't written in a simple horizontal row as European languages are. Instead, they are grouped into syllables of two or three letters. For example, consider the consider the Korean word “문어”, pronounced (roughly) "moon-aw". which means “octopus”. This is made up of five letters ㅁㅜㄴㅇㅓ, but as you see they are arranged in two syllables 문 ("moon") and 어 ("aw"). So instead of twenty-four kinds of templates, one for each letter, the Korean set needs one for every possible syllable, and there are thousands of possible syllables.

Unicode gets around this by… sorry, Unicode doesn't get around it, they just allocate eleven thousand codepoints, one for each possible syllable. But for this engraving device, it would be prohibitively expensive to make eleven thousand little templates, then another eleven thousand spares, and impractical to sort and manage them in the shop. Instead there is a clever solution.

Take a look at just one of these templates:

This is not a Hangeul syllable.

Rather, it is five. The upper-right letter in the syllable is the vowel, and the template allows the operator to engrave any of the five vowels

ㅣㅓㅏㅕㅑ

to produce the syllables

잉 엉 앙 영 양

pronounced respectively "ing", "ông", "ang", "yông", and "yang".

Similarly this one can produce six different syllables:

The upper-left part can be used to engrave either of the consonants ㅅ or ㅈ and the upper-right part can be used to engrave any of the vowels ㅣㅓㅏ, to produce the combined set 싱 성 상 징 정 장. I'm not sure why this template doesn't also enable vowels ㅕㅑ as the other one did.

In the picture at top you can see that while the third template can be used to engrave any of the three syllables 송 승 숭 the operator has actually used it to engrave the first of these.

This ingenious mechanism cuts down the required number of templates by perhaps a factor of five, from ten boxes to two.

Addendum 20250325

A great many of the 11,000 Unicode codepoints are for seldom-used syllables that contain four or even five letters, such as 둻. I studied Korean for a while and I think I learned only one with with more than three letters in a syllable: 닭 means “chicken”.

I don't see templates for these syllables in any of my photographs, which probaby accounts for much of the great reduction in templates from the 11,000 possible syllables. But there must have been some way to engrave the syllables with the machine.

Maybe there was a template that had a small four small ㄷsymbols, one in each of the four corners of the template, and another with four ㄹ symbols, and so on? Then the operator could have composed 닭out of bits from four different templates.

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Claude and Merle Miller let me down

Claude

My relationship with Claude has its ups and downs, and I'm still trying to figure out what to use it for and what not. It's great at rewriting my job application cover letters to sound less like an awkward nerd.

Last week I was in the supermarket and decided to try asking it something I've been wondering for a long time:

MJD: What is the song "Mercy Street” about? What is the connection with Anne Sexton? Who is "Mary"?

I thought Claude might do well with this. I had had a conversation with it a while back about Pixies songs, which I was satisfied with. But this time Claude let me down:

Claude: … The "Mary" referenced in the song is likely Mary Gray Staples, who was the name of Sexton's childhood doll. …

(I thought: What? Am I supposed to believe that

Dreaming of the tenderness
The tremble in the hips
Of kissing Mary's lips

is about a doll?)

MJD: Mary Gray Staples is an unusual name for a doll. Are you sure about that?

Claude: You're right to question this - I made an error. Mary Gray was actually Anne Sexton's great-aunt, not the name of her doll. … The doll references in Gabriel's song likely connect to Sexton's imagery in other poems, but Mary specifically refers to her great-aunt.

MJD: I don't remember a doll in Gabriel's song.

Claude: You're absolutely right - I apologize for the error. There isn't actually a reference to a doll in Peter Gabriel's "Mercy Street" lyrics. … The "Mary" referenced in the song is indeed Anne Sexton's great-aunt Mary Gray…

MJD: Are you sure? Wikipedia says that Mary Gray Staples was the birth name of Sexton's mother.

Claude: You're absolutely right, and I apologize for the confusion. …

Claude just flubbed over and over.

I wonder if the grammatical error in “Mary Gray Staples, who was the name of …” is a kind of a tell? Perhaps Claude is fabricating, by stitching together parts of two unrelated sentences that it read somewhere, one with “Mary Gray Staples, who was…” and the other “… was the name of…”? Probably it's not that simple, but the grammatical error is striking.

Anyway, this was very annoying because I tend to remember things like this long past the time when I remember where I heard them. Ten years from now I might remember that Anne Sexton once had a doll with a very weird name.

Merle Miller

A while back I read Merle Miller's book Plain Speaking. It's an edited digest of a series of interviews Miller did with former President Truman in 1962, at his home in Independence, Missouri. The interviews were originally intended to be for a TV series, but when that fell through Miller turned them into a book. In many ways it's a really good book. I enjoyed it a lot, read it at least twice, and a good deal of it stuck in my head.

But I can't recommend it, because it has a terrible flaw. There have been credible accusations that Miller changed some of the things that Truman said, embellished or rephrased many others, that he tarted up Truman's language, and that he made up some conversations entirely.

So now whenever I remember something that I think Truman said, I have to stop and try to remember if it was from Miller. Did Truman really say that it was the worst thing in the world when records were destroyed? I'm sure I read it in Miller, so, uhh… maybe?

Miller recounts a discussion in which Truman says he is pretty sure that President Grant had never read the Constitution. Later, Miller says, he asked Truman if he thought that Nixon had read the Constitution, and reports that Truman's reply was:

I don't know. I don't know. But I'll tell you this. If he has, he doesn't understand it.

Great story! I have often wanted to repeat it. But I don't, because for all I know it never happened.

(I've often thought of this, in years past, and whatever Nixon's faults you could at least wonder what the answer was. Nobody would need to ask this about the current guy, because the answer is so clear.)

Miller, quotes Truman's remarks about Supreme Court Justice Tom Clark, “It isn't so much that he's a bad man. It's just that he's such a dumb son of a bitch.” Did Truman actually say that? Did he just imply it? Did he say anything like it? Uhhh… maybe?

There's a fun anecdote about the White House butler learning to make an Old-fashioned cocktail in the way the Trumans preferred. (The usual recipe involves whiskey, sugar, fresh fruit, and bitters.) After several attempts the butler converged on the Trumans' preferred recipe, of mostly straight bourbon. Hmm, is that something I heard from Merle Miller? I don't remember.

There's a famous story about how Paul Hume, music critic for the Washington Post, savaged an performance of Truman's daughter Margaret, and how Truman sent him an infamous letter, very un-presidential, that supposedly contained the paragraph:

Some day I hope to meet you. When that happens you'll need a new nose, a lot of beef steak for black eyes, and perhaps a supporter below!

Miller reports that he asked Truman about this, and Truman's blunt response: “I said I'd kick his nuts out.” Or so claims Miller, anyway.

I've read Truman's memoirs. Volume I, about the immediate postwar years, is fascinating; Volume II is much less so. They contain many detailed accounts of the intransigence of the Soviets and their foreign minister Vyacheslav Molotov, namesake of the Molotov Cocktail. Probably 95% of what I remember Truman saying is from those memoirs, direct from Truman himself. But some of it must be from Plain Speaking. And I don't know any longer which 5% it is.

As they say, an ice cream sundae with a turd in it isn't 95% ice cream, it's 100% shit. Merle Miller shit in the ice cream sundae of my years of reading of Truman and the Truman administrations.

Now Claude has done the same. And if I let it, Claude will keep doing it to me. Claude caga en la leche.

Addendum

The Truman Library now has the recordings of those interviews available online. I could conceivably listen to them all and find out for myself which things went as Miller said.

So there may yet be a happy ending, thanks to the Wonders of the Internet! I dream of someday going through those interviews and producing an annotated edition of Plain Speaking.

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Reflector grids

Around here, these metal things are commonly found on streetside utility poles, attached maybe a meter off the ground.

Metal reflector

Plastic reflector

When I first noticed one of these I said “I wonder what the holes are for. Maybe to make it more visible? And what do they do with all the leftover rectangles after they've made one?”

I eventually got a better idea: The little metal rectangles are the primary product, and after they have been die-cut out of the metal sheet, there is this waste material left over with all the holes. Instead of throwing it away someone nails it to a utility pole to make the pole easier to see at night. I felt a bit silly that my first idea had been exactly backwards.

I later learned that only the older ones are made of sheet metal. Newer ones are made of some sort of plastic, maybe polyethylene or vinyl or something, about the same thickness. They look pretty much the same. I can only tell them apart by feeling them.

Still I wondered what the little rectangles had been used for. It turns out that the purpose is this:

That's according to an old Philadelphia Inquirer article, Why yellow grids are on some Philly-area utility poles. (Patricia Madej, Aug. 31, 2019.) But I measured them to make sure. They matched.

The answer came as a bit of a surprise to Jay Lipschutz, 73, of Northeast Philly …

His wife, Ruth, he said, had insisted they’re reflectors for drivers to see. She was right.

Jay, my friend, your wife is smarter than you are. Listen to her.

The article also tells us that the rectangular leftover is called a “grid reflector”. With a little more research I learned that one manufacturer of grid reflectors is Almetek. They cost $3.50 each. Pricey, for something they would have had to throw away. (Here's the old South Philly Review article that put me on to Almetek.)

What kicked off this article was that I was walking around and I saw this similar reflector grid, which felt to me like it was a bit of a farce, like a teenager sneaking into a bar wearing a fake mustache:

Hey, those aren't holes! When I saw this one I wondered for a moment if I was suffering some sort of mental collapse, or if none of the others had had real holes either. But no, they had, and this one really did have fake holes.

(Also, it has been installed sideways. Normally they are oriented as the two above.)

This isn't the first time I have written about ID numbers on utility poles hereabouts.

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Jonathan Chait

[ Content warning: angry, contemptuous ranting that accomplishes nothing. ]

I didn't really know who Jonathan Chait was until last week when I unfortunately read this essay of his (from February 2016) on “Why Liberals Should Support a Trump Republican Nomination”.

I've said a lot of dumb things in my life but I don't think I've ever been as wrong about anything as Chait was about this. I sure hope I haven't. But if I do ever find out I had been this wrong about something, I would want to retire to a cave or a mountaintop or something.

“Hey, remember Dominus? Whatever happened to him, anyway?”

“Oh, he said he was going away to cleanse himself of error, and might not be back for a long time.”

And yet this guy is still shamelessly writing. And why not? Editors are still buying his essays and maybe people are even still reading them. Why? You'd think that people would look at this essay and say “yeah, that's enough Chait for me, thanks, next time I need an opinion I'll try someone else.” I get it, nobody's right all the time. Whenever you read anyone's essay you're taking a risk, like rolling a die. Sometimes the die rolls high, sometimes it rolls low, and some dice might have higher numbers to begin with. I've usually been well-served by Daniel Dennett's dice, and Robertson Davies'.

But here people have an opportunity to toss a totally unknown die that they haven't tried before but that most likely rolls numbers from 1 to 6, and instead they toss the Jonathan Chait die when they know it has at least one side with a -1000.

A long time ago, I wrote:

I hate trying to predict the future; I don't think I'm good at it and I don't think anyone else is.

I don't think anyone could have predicted the extent of the current fiasco, but I do think it should not have been hard to predict, in 2016, that liberals should not, in fact, have supported a Trump Republican nomination.

Anyone can be wrong, even the wise cannot see all ends. But I think this one was maybe not so hard to see. Chait spends a lot of time comparing Trump with Arnold Schwartzenegger: both nominally conservative, both inexperienced in government, both assholes. I think the part that Chait ignored was that by 2016 — no, scratch that, by 1990 — it was perfectly clear that Trump was a liar, a thief, a racist, and a deadbeat, and that he had no respect for law or truth or ethics or anything other than his own convenience of the moment. (Here are just two examples. More recently, his ridiculous years-long insistence that Barack Obama was born in Kenya. And earlier, his equally ridiculous lies around his destruction of the Bonwit Teller building.)

In that old essay I said:

Most people who try don't seem to revisit their old predictions to see if they were correct, or to learn from their past errors, and the people who listen to them never do this.

I looked around a little to see if Jonathan Chait had written an essay titled “I was wrong, I was so, so wrong, I just couldn't have been wronger” but I didn't find one and I also didn't find any recent essays that said anything like “here's why I think this new essay is more reliable than that embarrassing Trump one I wrote for The New Yorker in 2016.”

I don't understand how Chait still has a job after writing this essay. Why isn't he selling shoes? How does a writer come back from this? Isn't there some charitable society for the protection for the public that could pay to have someone follow Chait around, quoting out loud from this essay, as a warning to everyone he meets for the rest of his life?

It least now I've been warned. Now when I read “Jonathan Chait said recently…” I'll remember: “Oh, you mean Jonathan ‘Why Liberals Should Support a Trump Republican Nomination’ Chait! Thanks, I'll pass.”

[Other articles in category /politics] permanent link

Claude chokes on graph theory

Having had some pleasant surprises from Claude, I thought I'd see if it could do math. It couldn't. Apparently some LLMs can sometimes solve Math Olympiad problems, but Claude didn't come close.

First I asked something simple as a warmup:

MJD: What is the largest number that is less than 1000?

I had tried this on ChatGPT a couple of years back, with tragic results:

ChatGPT: The largest number that is less than 1000 is 999.

But it should have quit while it was ahead, because its response continued:

ChatGPT: Any number that is less than 1000 will have three digits, with the first digit being 9, the second digit being 9, and the third digit being any number from 0 to 8.

and then when I questioned it further it drove off the end of the pier:

ChatGPT: Any number with four or more digits can be less than 1000, depending on the specific digits that are used. For example, the number 9991 is a four-digit number that is less than 1000.

Claude, whatever its faults, at least knew when to shut up:

MJD: What is the largest number that is less than 1000?

Claude: 999

I then asked it “What if it doesn't have to be an integer?” and it didn't do so well, but that's actually a rather tricky question, not what I want to talk about today. This article is about a less tricky question.

I have omitted some tedious parts, and formatted the mathematics to be more readable. The complete, unedited transcript can be viewed here.

I started by setting up context:

MJD: Let's say that the cube graph !!Q_n!! has as vertices the set !!\{0,1\}^n!! and two vertices are connected with an edge if and only if their coordinates differ in exactly one position. This is a standard construction. Are you familiar with it?

Claude asserted that it was familiar with this family of graphs. ((Wikipedia on cube graphs.) The basic examples, !!Q_0!! through !!Q_3!!, look like this:

Each graph consists of two copies of the previous graph, with new edges added between the corresponding vertices in the copies.

Then I got to the real question:

Now let's say that the Keane number for a graph !!G!! is the largest number !!k!! such that:

1. !!G!! can be partitioned into !!k!! connected subgraphs, and
2. each pair of subgraphs is adjacent, in the sense that for each pair !!H!!, !!K!! there is an edge of !!G!! with one vertex in !!H!! and one in !!K!!.

If you understand so far, tell me the Keane numbers of the cube graphs !!Q_0, Q_1!!, and !!Q_2!!.

Here are the maximal partitions for those three graphs:

The Keane number of !!Q_0!! is !!1!! because it has only one vertex. For !!Q_1!! we can put each of the two vertices into a separate part to get two parts. For !!Q_2!! we can get three parts as above. But there is no partition of !!Q_2!! into four parts that satisfies the second condition, because two of the parts would have to comprise the upper-left and lower-right vertices, and would not be connected by an edge.

Claude got this initial question right. So far so good.

Then I asked Claude for the Keane number of !!Q_3!! and this it was unable to produce.

The correct number is !!4!!. There are several essentially different partitions of !!Q_3!! into four parts, each of which touches the other three, which proves that the number is at least !!4!!. Here are two examples:

In addition to these there are also partitions into parts of sizes !!1+1+2+4!!, and sizes !!1+1+3+3!!.

On the other hand, more than !!4!! parts is impossible, and the proof is one sentence long: !!Q_3!! has only !!8!! vertices, so any partition into !!5!! or more parts must have a part of size !!1!!, and this part can't be adjacent to the other four parts, because a single vertex has only three outgoing edges. I would expect a bright middle-schooler to figure this out in at most a few minutes.

At first, Claude got the right number, but with completely bogus reasoning. To avoid belaboring the obvious, I'll highlight the errors but I won't discuss them at length.

Claude:

• Let's find the maximum possible !!k!!:
  
  • One valid partition with !!k=4!! exists:
    
    • Take two opposite vertices (e.g., !!000!! and !!111!!)
    • Make these two single-vertex subgraphs
    • Split the remaining !!6!! vertices into two paths of !!3!! vertices each
    • Each subgraph is connected
    • Each pair of subgraphs will have an edge between them

The diagrams I showed earlier display some of the partitions that show !!k≥4!!, but the one Claude gave here does not, because two of its parts (!!D!! and !!B!!) are not adjacent:

Claude:

• Why can't !!k!! be !!5!! or more?
  • If we had !!5!! or more subgraphs, we would need !!{5 \choose2} = 10!! edges between pairs
  • !!Q_3!! only has !!12!! edges total
  • Some edges would need to be reused for multiple pairs
  • But each edge can only connect one pair of subgraphs
  • Therefore !!5!! or more subgraphs is impossible

Okay, Claude, I agree we would need ten edges for the connections between the parts, but we have !!12!!, so why do you say that “some edges would need to be reused”? It may be correct, but it does not follow.

Claude:
Therefore, the Keane number of !!Q_3!! is !!4!!.

Would you like me to verify any part of this reasoning?

So, right answer, but seemingly by luck, since there were serious errors in reasoning, in both the !!k≥4!! part and also in the !!k< 5!! part. I decided to ignore the second one.

MJD: You said two of the subgraphs are !!H=\{000\}!! and !!K=\{111\}!!. You said each pair of subgraps will have an edge between them. What edge connects !!H!! and !!K!!?

Claude: Ah, I made a mistake.

The vertices !!000!! and !!111!! differ in all three positions, so they are not adjacent in !!Q_3!! (they need to differ in exactly one position to have an edge between them). This means that making them single-vertex subgraphs would violate condition (2) since these subgraphs would not be adjacent.

I need to rethink my claimed partition. Would you like me to try again to determine the Keane number of !!Q_3!!?

MJD: Yes.

Claude: [finds a specific example showing !!k≥3!!, and remarks that !!k=4!! “seems difficult”]
Let me work on verifying whether !!k=4!! is possible or !!k=3!! is maximum. Would you like me to continue?

MJD: Please go on.

Claude: Let me try to prove !!k=3!! is maximum in !!Q_3!!.

Suppose we had !!k=4!! subgraphs. Then:

• We need !!{4 \choose 2} = 6!! edges between different subgraphs
  • Each subgraph needs to be connected:
    
    • A subgraph with !!3!! vertices needs at least !!2!! edges
    • A subgraph with !!2!! vertices needs !!1!! edge
    • A subgraph with !!1!! vertex needs !!0!! edges
  • There are !!12!! edges total in !!Q_3!!
  • If we have !!4!! subgraphs:
    • First, !!6!! edges must go between different subgraphs
    • That leaves only !!6!! edges for internal connectivity
    • But with !!8!! vertices split into !!4!! subgraphs:
      • If any subgraph has !!3+!! vertices, it needs !!2+!! edges internally
      • If any has !!4+!! vertices, it needs !!3+!! edges internally
      • With !!8!! vertices total divided into !!4!! parts, at least one part must have !!3+!! vertices
      • After using !!2+!! edges for that part and !!6!! edges between parts, we don't have enough edges left to make all other parts connected

Therefore !!k=4!! is impossible, and the Keane number of !!Q_3!! is !!3!!.

Claude seems completely undirected here. Some of the things it says are correct facts, but it can't link the facts together into actual reasoning, except by accident. Sometimes it utters A followed by B where A implies B, and it seems exciting, but just as often B is a nonsequitur.

This is not that different from how ChatGPT was two years ago. It can spout text, and often it can spout enough plausible, grammatically-correct text to suggest that it knows things and is thinking about them. Claude's text-spouting prowess is much more impressive than ChatGPT's was two years ago. But it is still just a text-spouter.

I went on, trying to detect actual thinking.

MJD: What if there were !!4!! parts of !!2!! vertices each?

Claude's response here correctly applied its earlier analysis: four parts of size !!2!! would use up four edges for internal connectivity, leaving !!8!! for external connections, and we only need !!6!!.

Claude: I was too quick to conclude 4 was impossible. Let me try to find an explicit partition into 4 parts of 2 vertices each...
Would you like me to try to find such a partition, or would you first like me to verify that this kind of partition would definitely satisfy the adjacency requirement (condition 2)?

MJD: You keep going until you think you ha e the right answer.

Claude: Let me try to construct a partition of !!Q_3!! into !!4!! parts of !!2!! vertices each: …

This time Claude tried this partition:

It noticed that two of the four parts were not adjacent, and gave up without searching further.

Claude: This makes me think !!3!! really is the maximum.

If Claude were a human mathematician, this would be a serious error. Finding one invalid partition proves nothing at all.

Claude: My previous argument was wrong, but the conclusion might be right. Would you like me to try to prove more carefully that !!4!! is impossible?

There is no proof that !!4!! is impossible, and I thought it would be unenlightening to watch Claude flounder around with it. But I wanted to see what would happen if I asked it to prove a false claim that should be easier because its premise is stronger:

MJD: Try a weaker conjecture: prove more carefully the four parts of size 2 is impossible.

Claude: Okay, let me prove that we cannot partition !!Q_3!! into four parts of size 2 each.

1. Suppose we had such a partition !!H_1, H_2, H_3, H_4!!, where each !!H_i!! contains exactly two vertices.
2. For any such !!H_i!!, its two vertices must be adjacent (to be connected). This means they differ in exactly one coordinate.
3. Therefore, each !!H_i!! must be of one of these forms:
   • Two vertices differing in 1st coordinate (like !!000,100!!)
   • Two vertices differing in 2nd coordinate (like !!000,010!!)
   • Two vertices differing in 3rd coordinate (like !!000,001!!)
4. Consider any vertex !!v!! in !!Q_3!!. It has exactly 3 neighbors:
   • One differing in 1st coordinate
   • One differing in 2nd coordinate
   • One differing in 3rd coordinate
5. Therefore, if !!v!! is in !!H_i!!:
   • One of its neighbors must also be in !!H_i!!
   • That leaves only two edges from !!v!! that could connect to other parts
   • So !!v!! can only help !!H_i!! be adjacent to at most two other parts
6. But each !!H_i!! needs to be adjacent to all three other parts

This contradiction proves no such partition exists.

It's tempting to look at this and say that Claude was almost right. It produced 16 lines and at least 15 of them, on their own, were correct. But it's less impressive than it might first appear. Again Claude displays the pattern of spouting text, some of which is correct, and some of which is related. But that is all I can say in its favor. Most of its statements are boilerplate. Sections 2–4 can be deleted with nothing lost. Claude has buried the crux of the argument, and its error, in section 5.

MJD: In step 5 I see that if !!v!! is in !!H_i!!, it can only help !!H_i!! be adjacent to two other parts. But !!H_i!! has size 2, so it contains a vertex other than !!v!!. If !!v!! connects !!H_i!! to two other parts, couldn't the other vertex of !!H_i!! connect !!H_i!! to a third part?

Claude: Ah, you're absolutely right! My proof attempt failed. Let me back up and try a different approach to the original question of whether !!Q_3!! can be partitioned into 4 parts that satisfy both conditions.

…

This time Claude did find a correct partition into four parts, showing that !!k≥4!!.

MJD: How did you come up with that example partition?

Claude: I should be honest - I didn't find it systematically. I was trying different arrangements somewhat randomly until I found one that worked. …

I don't think there is any sense in which this is true, but at this point I hadn't yet internalized that Claude's descriptions of its own internal processes are text-spouting just like the rest of its output. In any case, I ignored this and asked it to analyze its own earlier mistake:

MJD: Earlier you purported to prove that !!k=4!! was impossible, saying that !!6!! edges would be needed to connect the !!4!! parts, and that this did not leave enough edges to connect the parts internally. You didn't analyze this carefully, so you missed the solution where all four parts have !!2!! vertices each. Do that analysis more carefully than you did last time and explain why that counting method does not rule out a partition of !!Q_3!! into !!2+2+2+2!!.

Claude: (details elided)

So this counting argument shows:

• We need 10 edges
• We have 12 edges
• Therefore the counting argument does not rule out a !!2+2+2+2!! partition

My earlier analysis erred by not being explicit about the numbers. I just had a vague sense that we needed "too many" edges, without actually counting them. This shows the importance of precise counting rather than relying on intuition about what seems like "too many".

Claude got the counting part right, although I think the final paragraph is just spouting, especially the claim “I just had a vague sense that…”, which should not be taken seriously.

[ Digression: This reminds me of a section in Daniel Dennett's Consciousness Explained in which he discusses the perils of asking humans about their internal processes. The resulting answers, he says, may provide interesting informaiton about what people think is going on in their heads, but we should be very wary about ascribing any accuracy or insight to these descriptions. Dennett makes an analogy with an anthropologist who asks a forest tribe about their forest god. The tribespeople agree that the forest god is eight feet tall, he wears a panther skin, and so on. And while this might be folklorically interesting, we should be very reluctant to conclude from this testimony that there is actually an eight-foot-tall fur-clad god in the forest somewhere. We should be similarly reluctant to ascribe real existence to Claude's descriptions of its “vague senses” or other internal processes suggested by its remarks like “Ah, let me think...” or “Ah, you're absolutely right!”. Claude has even less access to its own mental states (such as they are) than a human would. ]

As I pointed out earlier in this article, there are several essentially different solutions to the problem of partitioning !!Q_3!! into four parts. Claude has found one of them, the one I showed above on the left. I wanted to see if it could find another:

MJD: Are there any other partitions that are not ruled out by that argument? Or is 2+2+2+2 the only possibility?

Claude: Let me think about what other partitions could satisfy the edge-counting constraint.

If we have !!4!! parts, need !!6!! edges between parts, and have !!12!! total edges, then we can use up to !!6!! edges for internal connectivity of parts.

Possible partitions of !!8!! vertices into !!4!! parts could be:

• !!2+2+2+2!! (needs !!4!! internal edges)
• !!3+2+2+1!! (needs !!3!! internal edges)
• !!4+2+1+1!! (needs !!4!! internal edges)
• !!3+3+1+1!! (needs !!4!! internal edges)
• !!5+1+1+1!! (needs !!6!! internal edges)

So from just edge counting:

(Claude concludes that none of these is ruled out by edge-counting.)

So the edge-counting argument alone doesn't rule out any of these distributions of part sizes. We'd need additional arguments to determine which are actually possible in !!Q_3!!.

MJD: Why didn't you consider !!4+2+2+1!!?

Claude: Ah, you're right - I should have considered !!4+2+2+1!!! For !!4+2+2+1!!:

• The part of size !!4!! needs at least !!3!! internal edges to be connected
• The parts of size !!2!! each need !!1!! internal edge
• So that's !!5!! internal edges needed
• Plus !!6!! edges needed between parts
• Total needed = !!11!! edges
• We have !!12!! edges available

So this case is also not ruled out by edge counting. I apologize for missing it in my list. Would you like me to check if there are any other distributions I missed?

Oh, such failure! Failures that I have highlighted are clearly false statements, But the failure here is much worse. Again, almost everything Claude said was correct! And yet deeply wrong!

A bright middle-schooler could have missed this also. It was a bit of a trick question. Not because it's ambiguous, but because problem with !!4+2+2+1!! is not in line with the direction that the discussion was taking at that point. There is nothing wrong with !!4+2+2+1!! from an edge-counting point of view!

But I think a human mathematician, even a very young one, would have been more likely to answer the question I actually asked, which was “Why didn't you consider !!4+2+2+1!!?”, and from there might have quickly come to the useful conclusion that it doesn't need to be considered. Claude didn't do anything like that.

MJD: !!4+2+2+1!! doesn't work. Think about it more.

Claude: Let me think more carefully about a potential !!4+2+2+1!! partition.

1. Consider the part !!H!! of size !!4!!:

   • To be connected, it needs at least !!3!! edges internally
   • One vertex in !!H!! must connect to all !!3!! other vertices in !!H!!
   • But in !!Q_3!!, each vertex has only !!3!! edges total
   • If one vertex in !!H!! uses all its edges internally to connect to the other !!3!! vertices in !!H!!, it has no edges left to connect to other parts
   • But every part needs to be adjacent to every other part

2. Therefore !!4+2+2+1!! is impossible, not because of edge counting, but because:

   • Either !!H!! isn't connected (if we don't use enough internal edges)
   • Or !!H!! can't connect to all other parts (if we use too many internal edges)

This shows why just counting edges isn't enough - we also need to think about the structure of !!Q_3!! and how the edges can be arranged.

Terrible. This is so muddled I wasn't even sure what to highlight as the errors. Yes, if one vertex in !!H!! uses all its edges internally to connect to the other !!3!! vertices in !!H!!, it has no edges left to connect to other parts. But the other three vertices in !!H!! have two edges each, so !!H!! could easily connect to the other three parts.

And, of course, Claude has still missed the main point, which is that one needn't consider !!4+2+2+1!!, because !!4+2+2+1= 9!! and !!Q_3!! has only !!8!! vertices.

MJD: There is a much simpler argument why 4+2+2+1 won't work.

Claude: Ah, let me think...

In a !!4+2+2+1!! partition:

• The single vertex in the part of size !!1!! has degree !!3!! in !!Q_3!!
• But it needs to connect to all three other parts!
• It needs all three of its edges just to be adjacent to the other parts
• So it can't connect to all three other parts AND connect to three different vertices

This is a much simpler contradiction. I was making it too complicated by thinking about the part of size !!4!! when the part of size !!1!! immediately shows it's impossible.

MJD: How many vertices are.in !!Q_3!!?

Claude: !!Q_3!! has !!8!! vertices (all binary strings of length !!3!!):
!!\{000, 001, 010, 011, 100, 101, 110, 111\}!!

And now I see an even simpler argument why !!4+2+2+1!! is impossible: !!4 + 2 + 2 + 1 = 9!! vertices, but !!Q_3!! only has !!8!! vertices!
I was doing complex analysis of connectivity and edges when I should have just checked that the claimed partition added up to the right number of vertices. This is a good reminder to check the most basic facts first.

Finally! I would love to know what really happened here. What caused Claude to emit a sentence connecting !!4 + 2 + 2 + 1 = 9!! with !!Q_8!! having only !!8!! vertices?

MJD: So remind me, did you decide that !!k(Q_3)!! is !!3!!, or !!4!!?

I wasn't sure it would remember this, by which I really mean that I wasn't sure it would be able to pick out from its previous mass of statements which one was correct. But whether by luck or not, it did answer the question correctly:

Claude: Earlier I found a valid partition of Q_3 into 4 parts:

!!H_1 = \{000, 001\} !! !!H_2 = \{100, 110\}!! !!H_3 = \{010, 011\}!! !!H_4 = \{101, 111\}!!

I verified that:

• Each part is connected
• Each part is adjacent to every other part

Therefore !!k(Q_3) ≥ 4!!.

I haven't yet proven that !!k(Q_3) ≤ 4!!. Would you like me to try to prove that !!5!! parts is impossible?

MJD: Please go ahead.

I found Claude's attempt at this interesting, but not in a good way. After reading it I shrugged and quit the conversation.

Claude:
Let me prove that !!5!! parts is impossible in !!Q_3!!.

1. Five parts would require !!{5 \choose 2} = 10!! different edges just to connect all pairs of parts to each other.
2. Each such edge must be distinct because a single edge can only connect one pair of parts.
3. Each vertex in !!Q_3!! has degree !!3!!. With !!8!! vertices, the total number of edges is !!(8 × 3)/2 = 12!!.
4. Out of these !!12!! edges:
   • !!10!! must be used to connect different parts to each other
   • That leaves only !!2!! edges for internal connectivity of parts
   • But with !!8!! vertices split into !!5!! parts, some part must have size !!≥ 2!!
   • Any part of size !!≥ 2!! must use at least one edge internally to be connected
   • If one part uses one edge internally, that leaves only one edge for all other parts
   • But if any other part has size !!≥ 2!!, it also needs an internal edge
   • With !!8!! vertices in !!5!! parts, at least two parts must have size !!≥ 2!!
   • Therefore !!2!! edges is not enough for internal connectivity

Thus !!k(Q_3) = 4!!.

Claude says that with !!8!! vertices in !!5!! parts, at least two parts must have size !!2!! or more. This is wrong, because you could have !!4+1+1+1+1!!.

But highlighting that wrong statement doesn't get at why this answer is so bad. It's bad because the conclusion that appears next is a complete nonsequitur.

The argument can be fixed up. I would put it like this:

• As Claude observed, we have only !!2!! edges available for internal connections
• A part with !!n!! vertices requires at least !!n-1!! edges for internal connections
• Let the sizes of the parts be !!n_1\dots n_5!!. Since !!n_1+\dots +n_5 = 8!!, we need at least !!(n_1-1)+\dots + (n_5-1) = 8-5 = 3!! edges for internal connections
• But we have only !!2!!.

It's true that !!2!! edges is not enough for internal connectivity. But in my opinion Claude didn't come close to saying why.

Back in the early part of the 20th century, we thought that chess was a suitable measure of intelligence. Surely a machine that could play chess would have to be intelligent, we thought. Then we built chess-playing computers and discovered that no, chess was easier than we thought. We are in a similar place again. Surely a machine that could hold a coherent, grammatical conversation on any topic would have to be intelligent. Then we built Claude and discovered that no, holding a conversation was easier than we thought.

Still by the standards of ten years ago this is stunning. Claude may not be able to think but it can definitely talk and this puts it on the level of most politicians, Directors of Human Resources, and telephone santizers. It will be fun to try this again next year and see whether it has improved.

The complete chat is available here.

Addendum

20250301

Many thanks to Jacob Vosmaer for his helpful discussion of how to improve this article.

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Furniture Mecca

[ Content warning: shitpost ]

It's that time of year again! Furniture Mecca is having their annual sale.

This year the sale will run until Friday the 28th. At closing time on that day any remaining furniture will be hurriedly moved to the owner's other furniture store, Furniture Medina.

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Leopards, faces, etc.

I expect we in the United States are about to see a wave of domestic terrorism unprecendented since the 1870s. In the wake of the Civil War, white Southerners used systematic terrorism to continue white supremacy. If a black person became too prosperous, masked thugs would come in the night to burn down their house. If a white person was seen helping a black one, the thugs would arrive, and might let them off lightly for a first offense, and administer only a severe beating, or a tar-and-feathering. If a black man voted, masked thugs would come to murder him perhaps by night, or perhaps in broad daylight and publicly. Blacks in the reconstruction South were met at polling places by armed mobs.

Local law enforcement ignored these lawless acts, and in many cases the terrorists were the local law enforcement: sheriffs, police, judges. The terrorism continued for decades, and the terrorists were restrained, to the extent they were restrained, only by federal enforcement of the anti-Klan acts.

A few weeks ago I hoped Trump might forget about the imprisoned January 6 rioters. Trump discards anyone for whom he has no use, I thought, and he has no more use for them. I was wrong. His pardon of hundreds of rioters sends a clear signal, to his followers and to his enemies, that political terrorism is now supported or at least condoned by the Federal executive branch. The federal executive will not enforce antiterrorism laws unless the terrorists are politically opposed to Trump.

Don't count on anyone to restrain Trump. If a judge rules the wrong way, they may be assaulted by masked thugs. If a congressperson becomes troublesome, their house may burn down in the night. If a newspaper reporter writes an article critical of Trump, masked thugs may kill them, perhaps gun them down in the street.

Nothing will be done. The FBI will shrug. Trump will call it fake news or will blame immigrants, Muslims, or Antifa.

And if you were one of the people cheering for Luigi Mangione last month, remember that that's what you were cheering for, a country where it's okay to gun down people in the street, as long as you hate them enough.

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Think speak, now speak!

Katara is now in her sixth semester in college and can speak Mandarin. I am so proud!

For class she recently wrote a talk (in Mandarin) about Hua Guofeng, the often overlooked second chairman of the Chinese Communist Party. She videoed herself giving the talk, and posted it to YouTube. This somehow attracted over 700 views, and comments from a number of strangers, most of which were in Chinese. Some even offered suggestions — only minor suggestions, which she found very gratifying.

One comment, however, expressed irritation. Google translates it as:

You, a foreigner, don’t need to comment on the Chinese people’s affairs.

Shortly afterward though, there came a defense, which began with this delightful phrase:

想讲就讲

An idiomatic translation is "You can talk if you want!"

A character-by-character translation is:

• 想 - think
• 讲 - speak
• 就 - right now
• 讲 - speak

which I just love. If anyone is looking for a name for their new Chinese-language-themed blog, I think this would be a great choice.

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David McShane's mural with 18 Franks, revealed

Yesterday I offered Gentle Readers a chance to identify the 18 famous Franks in David McShane's mural.

1. Benjamin Franklin
2. Frankie Avalon
3. Detail of ornament from the Pennsylvania Academy of the Fine Arts building, designed by architect Frank Furness
4. A frankfurter
5. Aretha Franklin

I did not recognize the PAFA architectural detail myself, I had to find out from the Mural Arts website. I have sometimes looked for this detail on the PAFA building, but I have never found it.

6. Barney Frank
7. Frankenstein's monster
8. Franklin Delano Roosevelt
9. Frank Zappa

10. Frank Lloyd Wright
11. Major Frank Burns, as played by Larry Linville
12. Frank “Tug” McGraw

13. French 20-centime coin, representing the Franc
14. Frank Oz
15. Frank Perdue

I think these pictures might be so old that they predate the European currency union.

16. Frank Sinatra
17. Frank Morgan, in his role as the Wizard from The Wizard of Oz. (The book was written by L. Frank Baum.)
18. St. Francis of Assisi, with bird friend. The model is the artist's brother Frank.

Not depicted: Frank Rizzo, who is burning in Hell.

I was certain that Tim Curry was there somewhere, in his role as Dr. Frank-N-Furter, but if he ever was I can't find any evidence of it. I even emailed the muralist, who confirmed that Frank-N-Furter had never been there. Still, he is in all our hearts, forever.

The mural was restored in 2015, at which time two more figures were added:

19. Pope Francis, who had visited the city that year, and
20. Frank Sherlock, noted poet and longtime employee at Dirty Frank's

My pictures are at least that old.

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