alt Hacker NewsA liar who always lies says "All my hats are green."
Comments
One complication is that in typical English, if you say "All my hats....", you are simultaneously making an existence statement that you have at least one hat... but the usual formal logic "forall" quantifier does NOT presume existence. Here's a formal proof that "forall" has a "surprise" meaning for those not well-versed in formal logic: https://us.metamath.org/mpeuni/alimp-surprise.html
I propose that when translating such statements to a formal logic, if that's what you really mean, use an "allsome" quantifier as I've described here: https://dwheeler.com/essays/allsome.html
It's really easy to forget to include an existence quantifier. Having notation specifically designed to automatically include it can avoid some problems.
I never liked this type of puzzle. It is not formal logic but more about the idiosyncrasies and conventions of the English language. I put this puzzle on par with Agatha Christie’s murder mysteries. It requires a suspension of disbelief and logic to be believable.
Someone who always lies means in the purest sense means you cannot trust anything they say. Even the word “hat” could mean they are talking about their pet cat that they like to carry on their head.
What the author would probably say is “All my hats are green” means the liar is either lying about All or Green. Either all their hats are some other color or only one hat is green. This means you have to assume the liar has a hat. How do we know that?
We only know that because of similar puzzles that came before. In other words this is not logic but more pattern recognition.
SPOILER
The statement translates to:
∀x ( IsAHatOfMine(x) => Green(x))
∀x (~IsAHatOfMine(x) ∨ Green(x))
The negation of that is (by repeated application of De Morgan's):
~∀x (~IsAHatOfMine(x) ∨ Green(x))
∃x ~(~IsAHatOfMine(x) ∨ Green(x))
∃x IsAHatOfMine(x) ∧ ~Green(x))
In ordinary English, the meaning of the original phrase, thus the answer to the puzzle, is different.
My take (possible spoiler):
If he had no hats, then his statement would technically be true. Therefore he has at least one hat.
He may have some green hats and some non-green hats, but must have at least one non-green hat. He could have any number of green hats, including zero, as long as he has at least one non-green hat.
So the only derived statement that we can conclude to be true is A.
1. it is not true that (for all hats h in {liar's hats}, h is green)
2. by the negation rule, there exists a hat h in {liar's hats}, such that it is not true that h is green.
3. there exists a hat in {liar's hats}
This is really basic first order logic guys.
"For all x in {}, P(x)" is always trivially true without regard for P.
Similarly, "there exists an x in {}, P(x)" is always trivially false.
This puzzle has appeared at least once on Youtube. This was two years ago, and comments are still being added - there are over 15,500 now. They are everything that you would expect on that platform, and such arguments as are being made now are pretty much the same as in the beginning. I hope HN does better.
Three gods A, B, and C are called, in no particular order, True, False, and Random. True always speaks truly, False always speaks falsely, but whether Random speaks truly or falsely is a completely random matter. Your task is to determine the identities of A, B, and C by asking three yes–no questions; each question must be put to exactly one god. The gods understand English, but will answer all questions in their own language, in which the words for yes and no are da and ja, in some order. You do not know which word means which.
Some clarifications. You can ask the same god repeatedly and you can make your questions and to whom you address them depend on previous answers. Think of Random as just flipping a fair coin to decide whether to tell the truth or lie. The puzzle is by George Boolos and titled The Hardest Logic Puzzle Ever for those looking for hints or whatever. Have fun.
I tried to figure it out but got stuck on the linguistic dilemma if he's lying about the concept of himself existing ("All of MY hats"). Then I decided I have better things to do.
Actually I would say by the rules of English we cannot conclude any of those multiple choice questions is the absolute case -
A) "The liar has at least one hat." cannot conclude because may have no hats, thus the lie is in the "all my hats"
B) "The liar has only one green hat." cannot conclude because may have 2+ green hats out of a 3+ set.
C) "The liar has no hats." cannot conclude that because he may have hats that are not green.
D) "The liar has at least one green hat." cannot conclude that because he may have no hats, or no green hats.
E) "The liar has no green hats." cannot conclude that because of the "all my" modifier means that he can have some green hats.
This is however different than what is true or not. Concluding from a set of multiple choice questions is not choosing ones that are potentially true, concluding is choosing something that is definitely true. There is not a single statement in that list of questions that is definitely true given the requirements, but all of the questions are potentially true.
on edit: all questions are potentially true, but not all potentially true at the same time of course - some of them lock out the others.
As a native English speaker, I'd have said "all my hats are green" is untrue when you don't own any hats, meaning (A) isn't actually implied to be true. The fact the guy is a liar means he's allowed to imply he owns hats when he doesn't, right??
The proposed solution misses the fact that you can't even derive that the liar has a hat because the subject of the predicate could be the lie. What if the liar had two green cars but any number or no hats? The lie is a lie then even if the statement is vacuous as there is too much ambiguity in the english language overall. Their hats could be any combination of colors or they could even be hatless if the lie was over the subject of the constraint, not the qualification of the subjects, as they were effectively miscommunicating what is green.
If everyone knows that the liar always lies, then can he effectively even lie? It’s like he speaks a language where everything means the opposite of what it means in English, and everyone understands this semantically inverted language. So then he might lie by just telling the truth.
Casual sentences and mathematics/logic don't go very well together, and lead to ambiguities and interpretation if there are no clear rules defined beforehand. This reminds me of those silly problems that circulate on TikTok with a series of additions and multiplications. The "correct" result depends on how you assume the operations precedence. Here, does "being a liar" mean that we have to take as a true statement the negation of "all my hats are green"? If so, is that "NOT all my hats are green"? How does that translate back to the mathematics realm? Is it {total_hats>0, green_hats>=1}, or is it {total_hats>=0, green_hats>=0}?
The phrase “All my hats are green” implicitly carries the premise “I have hats” in conversational language. Thus, if this statement were to be expressed as a logical proposition, it should be: “I have hats, and all my hats are green.”
This means there are two potential falsehoods in the statement: 1. “I have hats.” 2. “All my hats are green.”
Therefore, what we can deduce is: • He might not have any hats. • If he does have hats, at least one of them is not green.
However, considering the options provided by the author, it is clear they did not take into account the implicitly stated proposition (1) in conversational language. Instead, the author assumes that if he has no hats, then “All my hats are green” is true.
This interpretation, however, is conversationally unreasonable; otherwise, one could claim something equally absurd, such as “All my houses are worth over 100 million dollars” but actually has no house.
I answered it correctly, because it seems obvious to me, even though probably to some people it is not obvious.
From I remember from what we did in maths, a liar is usually used to have a NOT so we have something like NOT (All my hats are green) => There is at least one hat that is not green.
Yes, it is a counterintuitive aspect of mathematics that "for all x in X ..." is always true if X is an empty set, just like "A implies B" is always true if A is false.
I once passed a midterm by abusing the latter. The question was to prove "there exists x such that if |a - b| < x, then something". The professor forgot to specify that x should be positive, so my response was that the implication is trivially true for x = -5 because a modulus cannot be negative.
The actual proof for positive values of x was much harder but the professor respected my math hacking skills and gave me full points for that question.
{{ Spoiler }}
If you didn’t select A, then the test is great for weeding out less intelligent people; which means it’s a good test if that is the objective.
My guess is we're supposed to read this sentence as:
Ɐh G(h)
(for all my hats, the hat is green)
or whatever similar formulation:
Ɐx (H(x) ^ M(x)) → G(x)
(for all x, if x is a hat and x is mine, then x is green)
Either way, the general idea will be that negating the statement (making it a lie) will make it a negative existential quantifier:
Ǝh ~G(h)
(there exists one of my hats such that it is not green)
Or in the case of the alternate formulation:
Ǝh (H(x) ^ M(x)) ^ ~G(x)
(there exists an x, such that x is a hat and x is mine, and x is not green)
So I think we answer (A) The liar has at least one hat.
All that said, I think other commenters are rightly pointing out that this relies on a very questionable distinction between semantics - which is what we've formalized above - and pragmatics. In conversational pragmatics, "All my hats are green" means that I have at least one hat (probably at least 3, even, since the sentence didn't say "My only hat" or "Both my hats"). One might explain this by way of an implicit pragmatic conversational principle that all statements should be relevant and informative in some way, which vacuously true statements (like, "all grass growing on the moon is purple") are not (see the "Gricean maxims").
If we don't make this implausible distinction between semantics and pragmatics (implausible to me because it assumes that sentences in general are usefully analyzed as having "propositional" meanings which can be evaluated outside of any conversational context), we might cash out the statement as:
Ɐh G(h) ^ Ǝh G(h)
so we can conclude, since this is a lie, that:
Ǝh ~G(h) ∨ Ɐh ~G(h)
Which is consistent with the liar owning no hats, as in:
> "All my hats are green"
> "Liar! You don't own any hats"
Yes, the liar have some hats. There are two ways how to conclude this:
1. If they don't have any hats, then any sentence about them would be truth.
2. The negation of the sentence, which must be truth, is "not all my hats are green". For this to be true, there must be some hats that are not green, so there must be some hats in general.
Related is the raven paradox (https://en.wikipedia.org/wiki/Raven_paradox), where for the statement "All of my hats are green", seeing _anything_ non-green that is not one of my hats is evidence in favour of the overall statement.
I thought it would be a trick question. If this person is a known liar, how do we even know he has any hats to begin with?
Person could really have a neon colored bowling ball and the statement “All my hats are green” is still valid.
I suppose the solution of A is valid assuming the physical object itself is not a lie.
things.every(thing => thing.type == 'hat' && thing.color == 'green')
things.some(things => thing.type != 'hat' || thing.color != 'green')
So only "E) The liar has no green hats" is true
If you were to code this statement in JS, it would be:
hats.every((hat) => hat === 'green') === false
(A) is the only one of those potential answers that must be true to guarantee the above statement is satisfied.
Well, let's see...
NOT(FORALL(x): my-hat(x) -> green(x))
EXISTS(x): NOT(my-hat(x) -> green(x)) ; https://en.wikipedia.org/wiki/Universal_quantification#Negat...
EXISTS(x): NOT(NOT(my-hat(x)) OR green(x)) ; https://en.wikipedia.org/wiki/Material_implication_(rule_of_...
EXISTS(x): NOT(NOT(my-hat(x)) AND NOT(green(x)) ; https://en.wikipedia.org/wiki/De_Morgan%27s_laws
EXISTS(x): my-hat(x) AND NOT(green(x)) ; https://en.wikipedia.org/wiki/Double_negation#Elimination_an... (note that this assumes non-intuitionistic logic)
EXISTS(x): my-hat(x) ; https://en.wikipedia.org/wiki/Conjunction_elimination
So yeah, the liar must have at least one (non-green) hat.
It is either that:
a) they don't have any hats, i.e. that they are lying about having any hats at all.
b) they have some number of hats (n >= 1) and at least one is not green, because they are lying about the color of all their hats.
I would be happy to learn how I am wrong.
Isn't the answer D? From predicate logic, the negation of forall() is existsone().
A liar who always lies lies always about everything.
Therefore he lied when he said "All". He isn't talking about all of his hats.
He lied about "my", meaning that the hats aren't his. They belong to someone else.
He lied about "hats", really, they aren't hats at all!
He also lied about "are", because he is not talking about the present, but the past and the future states of the "hats".
Finally, he also lied about "green", the "hats" are anything but green, which proves, conclusively, that when he said 'green' he was actually referring to the very specific shade of green found only in government-issue pickle jars from 1972.
!! Spoil:
Since we cannot refer to the liar, we can refer to the extended puzzle's author.
The author states that the goal of the extended puzzle is to determine if there is, 'IF ANY' (!!), a correct statement among [A, B, C, D, E]. Thus, there can be zero or at most one statement we can conclude as being true for sure.
The liar didn't said if he has hats. Maybe he has 0. Maybe 1. Maybe n. We just don't know.
'A: The liar has at least one hat.' > We cannot conclude this statement as sure, since maybe the liar has in fact 0 hats.
'B: The liar has only one green hat.' > He has maybe 0 hats. Or maybe n | n>1.
'C: The liar has no hats.' > He has maybe 1 hat. Or maybe n | n>1.
'D: The liar has at least one green hat.' > He might not have any hats at all.
'E: The liar has no green hats.' > Since the liar may have 2 hats, one could be green and the proposition could still be false, as it is a lie.
Since we cannot conclude any of the statements as being definitively true, the extended puzzle's answer is none of them are true for certain. It depends on how many hats the liar has.
I thought A. ChatGPT said E. ChatGPT o1 said A.
Also, the novel referenced (Intermezzo) by Sally Rooney is quite good. It makes a few references to analytic philosophy, mostly Wittgenstein. It concerns two brothers whose relationship has broken down. One of them is a "wordcel" and the other is a "shape rotator".
Mr liar.
"All my hats are green"
Is lie.
They are sombreros. Some you borrowed from me. You have no head!
If we're doing "provoke arguments on the internet using confusingly-stated constructs", they should have some self respect and use the venerable 6÷2(1+2).
And then read https://xkcd.com/169/
Anyway, I'm sure there'll be a YouTube video about this with an AI voiceover soon.
This is why we need while-else. Not the pythonic version, mind you.
I kind of disagree
1) “All my hats are green”
2) We assume that “always lies” means that (1) is false.
3) If the liar has at least one hat that is not green satisfies (2)
4) If the liar doesn’t have any hats also satisfies (2)
If liar is lying so the not false statement can also be: "Not all not my not hats are not not green".
A way to think about it is, if you went into the liars house to prove him wrong, you could find one non-green hat to prove him wrong. If he has no hats you wouldn't be able to prove him wrong.
Which reminds me of a quote from the British TV series Yes Minister: "A good speech isn't one where we can prove he's telling the truth. It's one in which nobody else can prove he's lying!"
Perhaps they aren’t even his hats at all and the “my” part is the lie.
That entirely depends whether an empty set trivially fulfills every possible predicate.
And if I remember correctly it's not something that's obvious although for convenience it's often assumed.
If all statements made by the liar are false, then the statement 'my hats' which is a statement made by the liar about his possession of hats, must be false, right? If the liar claims possession of hats, the truth must be that the liar does not possess any hats.
What if the liar said "I own no hats, but if I did own hats, not a single one of them would be green"? Can we then conclude the liar owns a green hat?
Formal logic generates contradictions easily and confuses people when expressed in word problem format, because humans aren't that logical - liars may only lie 10% of the time, when they see a benefit. It's like rational actor theory in economics - people may only act rationally 50% of the time.
taking this seriously is a midwit trap
Also please solve it without using ChatGPT o1 or any other LLMs please?
The puzzle is referring to the concept of a vacuous truth (https://en.wikipedia.org/wiki/Vacuous_truth).
In most logic frameworks, the All function (upside down A in standard logic notation) is true if and only if no statement within the set is false (i.e. All his hats are green if he has no hats). This is for several reasons:
- it allows for more coherent empty set functions. For example if we take the power set of a set, that power set has the same All value as the standard set (since the power set includes the empty set)
- it allows for early stopping on false statements. So you can define the statement as a lazy executor of all child conditions