This appeared on Daily Nous on 13 September 2022, just a couple days after the expression “the present King of England” changed(?) its meaning(?).
The “To φ or not to φ” comics feature on The Daily Nous is done by Tanya Kostocha , Assistant Professor of philosophy at Ashoka University. Russell’s theory of descriptions is long gone, but is still studied for the sake of understanding the variety of refutations and reformulations that succeeded it. Oh, and also for its well-remembered example, “The present King of France is bald” (uttered in 1905, when there was no current King of France).
Andréa – My 94yo mom who was who pointed me to towards being an anglophile, from her interest in Queen Elizabeth from when mom and she were young, cannot understand that the royal line extends beyond Prince William’s children. I keep trying to point out to her that even Princess Margret and her descendants still stand in line for the throne (though with very little chance of getting it).
I tried to explain using the movie “King Ralph” with John Goodman and Peter O’Toole that the line extends beyond the immediate families. Which I have to do by explaining the plot to her as she has never seen it. Basically – the entire main royal family – including Princess Margaret and family and beyond – have been killed at an event (explosion I think). They need the next nearest relative who happens to be an American named Ralph – beer drinking, bowling, blue collar fellow played by Goodman. O’Toole is the servant tasked with getting him “up to snuff” as the new King.
(I won’t give away the rest of the story unless it is asked for.)
“He wrote The Adventures of Humpty Dumpty Junior, a little egg, Humpty Senior’s son.”
Humpty Dumpty had a son? And were no chickens involved, or did they just not speak of same as too facts of lifey for the magazine’s little readers? (“Which came first, the egg or the egg?”)
Maybe you can show her/print out this chart . . .
Shrug: Yes, Humpty Dumpty, an egg, was married to Mrs. Dumpty, another egg, and they had a son, also an egg. How exactly an egg produced an egg was never explained, but that makes sense because it was a children’s magazine, and children’s magazines didn’t give a lot of detail as to where babies come from. I bet though that if they did provide a backstory, a stork would be involved. Which makes sense if you think about it, a stork delivering an egg to a couple of eggs.
Meryl A: Long before “King Ralph”, in 1969 Richard Lester made “The Bed Sitting Room.” War has been declared on Britain, and massive atomic bomb attacks on London result in the shortest war in history. Much of the population is gone, but the remainder continue to carry on as if nothing has happened, living in the bombed-out shells of their apartments, and taking the escalator to the train in the morning, falling off at the top because the rest of the train station is gone.
But even though the Royal Family has been killed along with all the rest of the upper class, the line of succession still works because it includes everybody. The new ruler is Mrs. Ethel Shroake of 393A High Street, Leytonstone.
The words of the National Anthem have been changed to “God save Mrs. Ethel Shroake, Long live Mrs. Ethel Shroake, God save Mrs. Ethel Shroake of 393A High Street, Leytonstone”.
“. . . in 1969 Richard Lester . . .”
He of ‘Hard Day’s Night’ fame . . . and I was just looking at Leo McKern in
‘Rumpole of the Bailey’ . . . feel like I’m in a time machine.
Not just “A Hard Day’s Night”, but also “Superman II” and “Superman III” (the latter with Richard Pryor), and “A Funny Thing Happened On The Way To The Forum” with Zero Mostel and Buster Keaton. And the version of “The Three Musketeers” that had Oliver Reed, Richard Chamberlain, Raquel Welch, Michael York, Frank Finlay and Christopher Lee.
Oliver Reed – I remember him in a PBS (Educational TV or something like that it was known then) bio of Dante Gabriel Rossetti.
So in a circuitous way, I was brought back to our logic discussion today, and how common sense understandings often get in the way of more rigorous logical understanding. I was trying to remember a point made on the TED Radio Hour yesterday, where the speaker used the mnemonic “Preacher, Prosecutor, Politician”, and I couldn’t remember the “Prosecutor” (I knew it was a lawyer of some type, but I couldn’t think of any that began with “p”), but my brain did throw up “Poets, priests, and politicians” …who all have words to thank for their positions; so when their eloquence escapes you; their logic ties you up and Rapes you: the doo doo doo, the da da da, is all I have to say to you! The Doo doo doo; the Da, da, da, they’re meaningless, and all that’s true!
So, they’re meaningless, and all that’s true. Which is to say, nothing is true.
But what does it mean, logically, to say a meaningless phrase is the only thing that’s true? According to Pinocchio’s Hat from above, if Pinocchio only lies, and he has no hats, he cannot say that all his hats a green, because that statement is neither true nor false, and Pinocchio can only make false statements. Colloquially, we’d say this kind of statement is meaningless. Yet Sting assigns it the value “true”, to thus imply that if it is true, then nothing can be true, and we all understand that — even though the logicians were arguing about how to classify those statements, going from they must be false, to they must be out of bounds, and changing the implications depending on where it fell. But we all understand Sting’s sentiments exactly, I think.
OK, granted, Sting is talking about Truth with a capital “t”, which is a different beast from what the logicians are dealing with.
Andréa pointed out in email that this week in Wayno’s blog he brings up Bertrand Russell, and uses this picture as his “pipe pic of the week”:
This week’s blog entry (with his comments on this week’s Bizarro weekday cartoons) is linked here.
As I mentioned somewhere above, you can knock out two of the Pinocchio statements and be left with “no hats” and “one hat”. Those are contradictory statements, so they can’t both be true. For me, since the “one hat” statement causes no problems or ambiguities that I can see (can’t be falsified), I considered it true. That then means that the “no hats” statement must be false. The reasons why aren’t even that important to me.
When I took Graph Theory during my MSCS program, I was a bit unprepared for the number of proofs of various sorts we had to do†. One on an exam was a true or false about the number of nodes for a given graph, and gave a specific number. Some people falsified the statement through counter-example, by showing there was a higher number nodes that fit the criteria.
I started off that way, but then went and asked the professor if it meant exactly X nodes. I figured he’d probably say to use the information given on the exam, but he said “no, it doesn’t meant that”. So I got it right.
The people that got it wrong were unhappy, including one guy who complained loudly in class and demanded the professor admit that he was wrong. That didn’t work.
† I hadn’t done a proof by induction in over 15 years, back in some of the math classes for my BS Physics. We did a lot of them in that class.
Brian, I don’t think that’s the intended answer or really the right approach to the Pinocchio’s Hats puzzle.
But first, let’s agree that this is not done under an earlier age’s understanding of the truth conditions of an “all” statement. If it is of the form “all x’s are P” and it so happens that there are no x’s, the modern understanding is that any such statement counts as TRUE. Not “ill-defined”, not “lacking truth value”; just (vacuously) TRUE. And that holds regardless of our discomfort over noticing that, for instance, “all x’s are not-P” is also TRUE when there are no x’s at all.
Don’t bother arguing that this is not the best system. That rejection may be correct (and Europe’s logicians thought so for a long time), but it’s the only way quantified logic can really work. It’s fine if you disagree, but it’s not something we can settle. And the puzzle was clearly constructed under the vacuous-truth system.
Also, noting that “only one answer is correct” should not be taken as a statement to be factored into the logical evaluations — it is just a reassuring hint from the puzzle setter not to get wrapped up in multiple guesses.
1) Pin always lies. [Any statement he makes we must consider FALSE]
2) Pin says “All hats that belong to Pin are green”.
Suppose Pin has no hats at all. Then condition (2) is TRUE — no, I mean the statement quoted in condition 2 is TRUE. This contradicts condition 1. So we rule out this supposition, that Pin has no hats.
That eliminates answer (C), which was exactly that supposition.
And it shows the truth of (A), which is just the direct negation of that.
If we were using the “only one correct answer” principle we would be done, since we have found one correct answer, A.
But can we, without using “only one correct answer” as a presupposition, rule out B, D, and E?
B cannot be concluded from the premises, because nothing rules out Pin having multiple green hats. Show by counterexample: say he has 3 green hats and none of other colors. This falsifies B, so it cannot be said to logically follow.
It is left to the reader to do the same move for D and E.
When I saw that you were describing a puzzle with colored hats, I thought it was going to be the one I have known and liked the longest, requiring getting into what one character can conclude about other characters’ reasoning.
This may have been from Smullyan, via Gardner when he was first introducing him to his readership in the SciAm column. Or maybe I saw or heard it elsewhere — I think it’s the one I’ve known the longest. I think the original version may have been explained with forehead patches instead of hats; but hats make it easier to set up the right who-can-see-whose-marking conditions.
We also have to make some agreements about all the characters being competent but not super reasoners. Also the oldest versions I recall just said “They are all silent for a while — Then person A announces that she has solved it. What is her answer?” I have introduced the story variant that “The puzzlemaster asks C if they can solve it, and after just a minute C says no. Then the puzzlemster asks B, and after a moment B also says no. Then the puzzlemaster asks C, who does solve it. What is C’s correct solution?”
The puzzlemaster brings in contestants A, B, and C and locks the door. Puzzlemaster shows the contestants that there are a supply of five hats, three white and two black. The contestants now are lined up facing forward in a queue, with A at the back, B in the middle, and C at the front.
The puzzlemaster now puts one hat from the supply on each contestant’s head and puts away the leftover hats. None of the contestants can see their own hat. A can see what color hats B and C are wearing; B can see what C is wearing (but not A); C cannot see any hat.
The puzzlemaster makes these conditions explicit to the contestants, and tells them the goal for each is to determine the color of their own hat.
Now the puzzlemaster asks A “Can you figure out the color of your own hat?”. A answers “No”.
Then the puzzlemaster asks B “Can you figure out the color of your own hat?”. B answers “No”.
Then the puzzlemaster asks C “Can you figure out the color of your own hat?”. C answers “Yes, it has to be ____”. And the puzzlemaster verifies that this is correct.
What was C’s correct conclusion, and the reasoning that got them there?
You’re in maze of twisty little passages, all alike. Eventually you come upon the two doors from “The Lady or the Tiger”; obviously you don’t want to go through the door with the tiger behind it. There are two guards here, one of whom always tells the truth, and the other of whom always lies, but you do not know which is which. You may ask exactly one question and only one question to either one of the guards. What one question do you ask?
(@Mitch: I like you hat puzzle, and I’m confident I know the answer, but I won’t post it yet…)
@larK I won’t try to get it precisely, but as an attempt schematically at your liar puzzle, you should ask either of the guards a question based around “If I were to ask the OTHER guard xxxx, what would he say?”
@ larK – XYZZY !!
Mentioning the truth conditions of an “all” statement reminds me of a very basic difference between mathematical logic and everyday logic. Mathematically, “a or b” is true if a is true or if b is true or if both are true. But when Mom says “You may have pie or cake for dessert” she means you can’t have pie AND cake. I guess she should have said “You may have pie XOR cake.”
Likewise “If pigs have wings, then green is blue” is taken as a nonsensical statement, even though logically it is true because pigs never have wings.
The hat puzzle makes me think of a puzzle I read a long time ago. There was a certain far off city where every resident had a very logical mind. Marriage was taken very seriously; infidelity was considered as one of the worst offenses.
Yet there were 30 unfaithful married women in the city. If a woman was unfaithful, everybody but the woman’s own husband knew she was unfaithful. People were really good at spreading gossip but even better at keeping secrets.
The king of the city had had enough of this infidelity. He declared that he knew there were unfaithful wives and decreed that for the next 30 nights, and those 30 nights only, a man who knew his wife was unfaithful had license to kill his wife in the night, but only on the night of the first day he found out about her infidelity.
29 nights went by with no deaths. Then on the 30th night, 30 unfaithful wives were killed by their husbands. Why did that happen?
If there were only one unfaithful wife, her husband would think “I thought there were no unfaithful wives but the King says there are more than zero. The husband who doesn’t know his wife is unfaithful must be me,” and he would kill his wife on the first night.
But if there were two unfaithful wives, Jones would say “There is only one unfaithful wife and that is Smith’s, and so Smith will kill his wife on the first night,” and Smith would say the same about Jones. But after the night passes with no killing, Smith and Jones will each say “Uh-oh, there are two of us” and kill their wives on the second night.
And so on.
@ larK – In addition to Danny’s “What would the other guy say?” solution, Gardener’s book offers these options: (1) “If I were to ask you whether this door leads to the [lady], would you say ‘yes’?” and (2) “Of the two statements ‘you are a liar’ and ‘this door leads to the [lady]’, is one and only one of them true?”
The readers’ letter rejects all of these “logical” solutions: “It is a sad commentary on the rise of logic that it leads to the decay of the art of lying…. the proposed solution [implies]… that liars can be made dupes of their own principles,… whenever lying takes the form of slavish adherence to arbitrary rules. No [true liar] could be expected to display the scrupulous consistency required … nor would any liar capable of such acumen be so easily outwitted.”
“We therefore propose as the most general solution the following question or its moral equivalent: ‘Did you know that [the lady] is serving free beer?’“
Mark in Boston, the “infidelity” puzzle strikes me as a version of the “Unexpected Hanging” or as Martin Gardner recast it,”The Unexpected Egg”.
If anyone’s inclined to disagree about empty “for all” propositions being true, the reason to prefer this interpretation is that it’s the limit as the set becomes empty.
This follows from “for all” being the transitive closure of “and” (that is, “for all X, P(X)” can be expanded to “X1 and X2 and X3 and …”, at least for finite sets).
So if you have a bag of statements X, all the statements in X are true if you take one (X1) out and it’s true, and all the rest of the statements in X are true. So when you have one statement (Xn) left, the condition is that it’s true and all none of the statements left in the bag are true. So you want to treat the empty bag as containing all true statements so that this case doesn’t need to become a special case.
For “there exists”, which is the transitive closure of “or”, it’s the other way around; that is, “there exists a statement in the bag such that XYZ” is false when the bag is empty. This is both necessary to maintain standard equivalences and also fortunately consistent with the general understanding of what existence means.
The reason it may seemsweird is that in natural language when we say things like “all hats are green” we normally actually mean something like “some hats exist, and in general any hat you find will be green… though there might be exceptions if there are a lot of hats”. We don’t speak in first-order logic and it’s not natural to try.
Andréa – Thank you. She understands through William’s children, but not that the rest of the family, cousins, etc follow beyond that – she says that they are too far away in the line (including Edward and Anne are too far away). Best also sometimes to just agree with her these days. I am sure we will discuss it again on another weekend.
Here is a solution to the black/white hats puzzle I posed previously. (Not to be confused with the green hats puzzle, about satisfaction of an empty “for all”.) With luck, you could get to it by the link https://cidu.info/2022/09/15/russell-on-denoting/#comment-118077 . Or by slogging thru the comments in the “Russell on Denoting?” thread to the timestamp “SEPTEMBER 24, 2022 AT 7:11 PM”. All of this is trying to avoid a big paste job here…
So, taking the conditions and question as read, let’s proceed by steps.
What could Contestant A be seeing, consistent with his declaration that he cannot conclude the color of his own hat? Well, one thing we can observe is that A cannot be seeing two Black hats (one on B and one on C) — for if he saw that, and knows there are only two black hats in the original supply, he must have a white hat.
Now consider what B (middle position) could be seeing on C’s head, and what could be concluded from that. Suppose B saw a Black hat on C. Then B couold frame a hypothtical —
“Suppose I have on a Black hat. Then A would be seeing my Black hat and C’s Black hat, and by Step 1 A would have concluded that he has White hat. But he does not announce that conclusion. So the supposition is wrong, and I do not have Black. So I can respond that I have concluded I have a White hat.”
But B did not have that response. So the failed hypothetical was “Suppose B saw a Black hat on C.”. Now C can follow this reasoning, and conclude that his own hat is White.
It’s like the US Presidential succession. Even though some are far down the line, they are still in line. Someone in the line goes into seclusion as the “designated survivor” during the State of the Union speech.
Brian, I don’t think that’s the intended answer or really the right approach to the Pinocchio’s Hats puzzle.
I missed this first part back when, and only noticed it while going through email. I think I didn’t clearly state that I meant the “at least one hat” and not “exactly one hat” and did not provide a letter answer, so you probably didn’t see that we actually agreed.
If the vice president of the United States dies, who’s president?
If the plane crashes right on the border of Czech Republic and Slovakia, where are the survivors buried?
@ Mitch – Usually in their own home towns, but it may take years (or even decades) for that to happen.
P.S. This isn’t the right opportunity for the classic question about a rooster laying an egg exactly on the apex of a roof (“Will it roll down to the east or to the west?”), so instead I will repeat the following linguistic riddle:
“What comes between ‘fear‘ and ‘sex‘?“
How many legs does a cow have, if you call her tail a leg?
Ah! Now we’re getting somewhere! So how many green hats may Pinocchio claim owning when you call having no hats ‘true’?
Shrug offers: “How many legs does a cow have, if you call her tail a leg?”
Phrased with dog instead of cow, this was attributed to Abe Lincoln, with the answer “Four. Calling a tail a leg doesn’t make it one.”
Speaking of quantification (as we have been elsewhere in this same thread), there is potential for confusing understanding of the quantifiers in another saying attributed to Lincoln: “You can fool some of the people all the time, and you can fool all of the people some of the time; but you cannot fool all the people all the time.” Both the first and second parts have their own ambiguity of quantifier scoping. But it makes no real difference to the point.
Kilby, you almost fell for the deceptive wording! The survivors do not need to be buried! (Not yet anyway, as you notes.)
For your addendum riddle, I can’t think of an English word that would match “fünf”; maybe just “fun”?
On the Lincoln quantification ambiguities:
Clause 1, You can fool some of the people all of the time
Cl 1 Repr 1 is like “Some people are so gullible, they can always be fooled”
Cl 1 Repr 2 is like “There are always going to be some people you can fool at the moment”
Clause 2, You can fool all of the people some of the time
Cl 2 Repr 1 is like “Everybody is susceptible to getting fooled, nobody is immune”
Cl 2 Repr 2 is like “Once in a while you can score a total 100% fooled-them triumph!”
@ Mitch – The answer can be delivered in either language, so both “Fünf” or “Five” would be acceptable. It depends on the audience.
Those puzzlers who are still reading this thread might be interested in the website “Gathering 4 Gardner”, which was mentioned today in the Post’s “Style Invitational”.
So I’m working my way through Gardner’s book mentioned earlier in this thread. The first thing that strikes me is the level of intellectual discourse for a popular, general population book! (I have a copy from 1959, the 11th reprint.) I feel like a dunce! He introduces a problem with a casual, of course everyone knows this, and states a problem that I feel thrilled I can generally solve, though I’m sweating, but that is just for introduction! Then he gets to the real problem, and generally it requires a clever insight just to see that a really clever insight is required to actually solve the problem. Any time there’s a question involving how few steps needed to solve the problem (how may weighings, how many questions asked), you can be sure that the eventual answer will be 1, if not zero — an extreme and extremely unlikely answer looking at the problem. (Yeah, it looks like I’d have to weigh at least 5 of those stacks of coins, but I guess if I’m clever, I could reduce that to 4, and if I’m really clever, I see a way to reduce it to 3 weighings… but the solution actually only involves one weighing…) I’d like to say I was going along at a C- level, but honestly, I’m below 50%, so I’m flunking royally!
The other thing that strikes me is how many famous names are casually mentioned as having contributed to or worked on a particular problem; a lot of them possibly even before they were generally famous, but others who might have already been famous, but messed around on this silly, trivial (and I can’t solve) thing in their earlier days. R. P. Feynman, A. Turing, John Nash, Claude Shannon are just a few off the top of my head I remember coming up.
larK, which Gardner book do you mean? I think there were several mentioned. If the one you’re reading is not the first in the series, there may be an explanation for the “everybody knows this one” effect, if he is referring to one he treated in an earlier column / collected in an earlier volume.
You indicated earlier that you generally knew how to approach the hat puzzle I posted. Did you see my solution? And does it fit with what your approach was to be?
Here is one of those coin-weighing problems; and an answer. Is this among the ones he presents early?
You are given a bunch of commemorative coins — actually there are 27 of them. [Yes, the choice of that number is a hint.] You are told that due to a manufacturing accident, one of them is heavier than all the rest, and your task is to identify that incorrect one. You have a two-pan balance which only shows if one side or the other is heavier, but does not provide actual weights. How many comparison-weighings will you need to identify the one incorrect coin?
Answer: Divide them into three piles of nine coins each. Place two of the piles on the balance pans, nine on either side. If the balance swings, take the heavier bunch into the next weighing; if the two sides are in balance, take the bunch which was not weighed to move into the next step.
Now you have nine coins, including the one heavier one. For your second weighing, divide the nine into three groups of three, and weigh two of them against each other. As before, if one side swings down, keep that group for the next step; if the two sides balance, keep the unweighed group of this step to go into the next step.
Now you have three coins. For the third weighing, put one coin on each pan. If they don’t match you have found the heavy coin on the side that swings down; if they match, the unweighed coin from this weighing is the heavy one from the original group.
So: three weighings.
This might be one of those that comes up as “you all know how to do this” before complicating the picture; do you agree?
I don’t know what the one-weighing problem you saw was, but here’s one!
You have ten bags of coins, ten coins in each bag. You are told all the coins are supposed to weigh 100 grams, but due to a manufacturing error all 10 coins in one of the bags were made to weigh 110 grams each. You have a weighing scale (not a balance) with a big dial and numerical readout. Or it can be digital, fine! Now how to identify the bag with the heavy coins?
One weighing will suffice.
@Mitch: The Scientific American Book of Mathematical Puzzles and Diversions by Martin Gardner. Kilby mentioned it on the previous page of comments. It has the second coin weighing problem you give, that I was referring to, where as you note, the solution can be done in one weighing, though the problem as presented just innocuously asks how many weighings do you need to do, leading you to expect, with ten piles of coins, something on the order of 5 weighings, and if you’re clever you might be able to reduce that, but you don’t really expect the answer to be “one”.
As an example of the “introductory” puzzle, in the chapter entitled “Nine Problems”, the first is introduced with the following: A guy walks a mile due south, a mile due east, and then a mile due north, and arrives at the same spot he started, and then he shoots a bear — what color was the bear? It is assumed everyone can figure this out (it doesn’t sound like it was previously given puzzle — it sounds like it’s just trivial). Then he goes on from that answer to the actual puzzle…
(So walking the way he did to arrive where he started, he must be on the North Pole, so the bear must be white, being a polar bear.) The actual puzzle is: is there any other spot on earth from which you can walk due south for a mile, due east for a mile, due north for a mile, and arrive at the same spot you started? (There are actually an infinite number of them, but they are particularly confined, and you need to discover and describe where on Earth this is…)
The hat problem I had the answer “white” for the logic you gave (no, really, I did!).
Ah yes, “the man walks a mile south…” puzzle — one of the all time greats!
For the southern circumpolar set of solutions, the story can no longer involve a polar bear!
In the book, Gardner points out that the single weighing solution will still work even if there are eleven bags (or stacks) of ten coins each.
larK – The President is President if the Vice President dies.