Death by infinity puzzles and the Axiom of Choice
ฝัง
- เผยแพร่เมื่อ 20 ม.ค. 2017
- In this video the Mathologer sets out to commit the perfect murder using infinitely many assassins and, subsequently, to get them off the hook in court. The story is broken up into three very tricky puzzles. Challenge yourself to figure them out before the Mathologer reveals his own solutions. Featuring Batman, the controversial Axiom of Choice and a guest appearance by the Banach-Tarski paradox.
The pictures that I used for the Banach-Tarski ball splitting action were grabbed off the brilliant VSauce's video on the Banach-Tarski paradox ( • The Banach-Tarski Paradox ) I mainly did this for easy reference since most people here will have seen this video and in this way would be able to connect easily with what I am talking about here.
I also mention that those sets that get pushed around in the Banach-Tarski paradox are constructed using the Axiom of Choice. Vsauce actually does not mention this although this is really a big deal as far as mathematics is concerned (understandable though since his video was already very long). Here is a link to the spot in the Vsauce video where the Axiom of Choice is envoked (although you have to have a really close look to see how :) • The Banach-Tarski Paradox
There is a very nice TEDed video about the finite version of the last of our puzzles: • Can you solve the pris... The solutions to both the infinite and the finite version are closely related.
Oh and today's t-shirt is from here shirt.woot.com/offers/infinite...
Thank you very much to Danil Dmitriev the official Mathologer translator for Russian for his subtitles.
Enjoy!
Burkard
![[UNCUT] รัฐสั่งหยุดปล่อยคลื่น! แต่หยุดพลังปะทุ “น้องหญิง” ไม่ได้ I คนดังนั่งเคลียร์](http://i.ytimg.com/vi/RmQh47eIiV0/mqdefault.jpg)
Very important: If there is anything about what I say that you are not sure about please ask. There are a lot of very knowledgable people roaming the comments section that you help out :)
There is a very nice TEDed video about the finite version of the last of our puzzles: th-cam.com/video/N5vJSNXPEwA/w-d-xo.html The solutions to both the infinite and the finite version are closely related.
I also mention that those sets that get pushed around in the Banach-Tarski paradox are constructed using the Axiom of Choice. Vsauce in his otherwise brilliant video on the Banach-Tarski paradox actually does not mention this although this is really a big deal as far as mathematics is concerned (understandable though since his video was already very long). Here is a link to the spot in the Vsauce video where the Axiom of Choice is envoked (although you have to have a really close look to see how :) th-cam.com/video/s86-Z-CbaHA/w-d-xo.htmlm2s
You just opened a can of worms, the cranks are going to flod here for the axiom of choice.
+EmperorZelos Hard to tell, but you could very well be right there. Have fun :)
Mathologer oh i will, you know it :)
I'd like to say i love that you are doing these topics to spread the joy of mathematics.
I love your videos, but I have to say, nothing was very well explained in this video. I found myself having to say to myself "Wait what?", and then watch to the end of the problem, and then go back and try to backfill what you meant to say when you set up the problem. If you want, I can help you with the scripts so that they make more sense and are easier to follow. Not all of your videos suffer from this problem, but lately, a few have been hard to follow for sure.
yes, this video wasn't well explained at all. I watched the TEDed video and it was instantaneously clear on how the strategy of survival worked. The end of the video started to get interesting, but you do not elaborate on the subject and how it ties together with this video's main subject. Much less satisfying than most of your previous work.
Want to know an anagram of banach tarski? Banach tarski banach tarski
Clever!
Minecraftster148790 it took me a while to figure this out.
nice.
:/ I don't get it
look up "the banach-tarski paradox"
This explanation is bogus because the assumption that Batman can be killed is incorrect.
Infinitely many Batverses.
There could be infinitely many batmans and infinitely many assassins, and even if you arranged for every single assassin to kill one batman, it's likely that you'll end up with (a finite set of) Batmen that weren't killed. So Batman wins here right?
Energy Core
Depending on the ratio between the set of Batmen and the set of mathsassins, either all Batmen wind up dead, or an infinite number survives.
David Wührer
Here's a situation where a finite number of batmen survive. Arrange the assassins one-to-one to the batmen. Then order them well. Then take each assassin and rearrange them so that the first assassin kills the second batman, and as such each assassin kills the assassin next. You get an arrangement such that every assassin is killing one batman, but only the very first batman does not get killed.
Energy Core
True, but there are still infinitely many assassins left for each of the finitely many Batmen, no matter how many there are. They don't stand a chance.
Do you really think infinite Assassins could kill Batman?
You would need at least twice as many!
:)
@@Mathologer reductio ad absurdum via ad infinidum lol
Infinity times two is still infinity
That's easy enough to accomplish. We simply divide them into finitely many groups (morning tea seems an opportune time for this), and then recombine into two equally numerous groups. After that, a simple lane merge on the way to the scene of the crime... Wait, doesn't that halve the number again?
Curses, we must have needed uncountably many assassins to employ Banach-Tarski Replication.
Batman isn't so tough. Even two fewer could get the job done.
Today - 21-012-2017 - Mathologer referenced VSauce, and VSauce referenced 3Blue1Brown. For the love of order and purity in the universe, I sincerely hope 3B1B releases a video today referencing Mathologer.
But then we'd have a circular reference and Excel would complain
He referenced minutephysics. I know its not mathematics, but hey i think we got the circular reference
You have to watch the bitcoin video.
I miss the olden days when VSauce uploaded videos
@@Danilego now those days are back
"it's simple. We Achilles the Batman."
It's the perfect murder because of biggest army politics. You have the biggest army with infinite assassins therefore you make the laws
That sounds great in theory, right up until time comes for you to serve up their first lunch... ;)
But you only have countably many assassins. What if the police force has uncountably many officers?
vytah *epic voice* In a world with infinite assassins batteling it out with infinite police officers one man needs to serve them lunch
+vytah Well, if need be all of this, except for the last puzzle, also works with uncountably many assassins :)
An interesting question would be: how many police officers should each highly-trained assassin be able to kill in order for countably many assassins to beat uncountably many police officers :)
A problem may arise when each assassin realises that there is no possibility he's going to kill Batman, and can as such go home so as to not risk being caught later.
SVVN
And Batman dies
Every conspirator is guilty of murder even if they back out.
@@pullt except when they all back out at once, I meant.
Ah, but you miss the part where every assassin who tries (and fails) to kill Batman becomes a Batman villain, and therefor will be let free/escape from jail after a couple weeks.
It's not impossible for an assassin to get caught yet the chance is still 0%... Though that changes if instead of considering them deciding to leave in advance we consider an assassin seeing Batman and deciding not to try to kill him... But then again we know that no assassin will ever be in this situation, and yet Batman will still die with 100% chance... Dang it
Famous quote: “The Axiom of Choice is obviously true, the Well-ordering theorem is obviously false; and who can tell about Zorn’s Lemma?" Jerry Bona
I like Zorn's Lemma - Algebra is so much more beautiful with it than without.
:)
To me the axiom of choice is pretty much not questionable, if you want to work with infinite objects.
If you think about it, an equivalent definition of AC is that a product of empty sets is nonempty (since each element of the product constitutes a choice function).
so, imo, you either have to do without infinite objects (or at least without uncountably infinite objects) or you have to accept AC...
Terry, you could always say things like "Let R be a ring with a maximal ideal" or "Let K be a field with an algebraic closure" :)
The problem is that acording to the Law of Conservation of Ninjutsu, there is always a finite amount of fighting prowess that is evenly spread out across a group of Ninjas (or assassins if you will). So your army of infinitely many assassins are each infinitesimally effective. So Batman will just defeat them all.
:)
Even though the Assassins are infinitesimally effective it still takes some infinitesimal amount of energy to beat them, and Batman only has so much energy.
According to the same law, since Batman also uses a finite amount of ninjutsu, and there are an infinite amount of infinitesimal energies.. Banarch-Tarski strikes again and the universe is split into a multiverse where Batman is killed and is alive!
@@nawdawg4300 I'm pretty sure that's false, but I don't know if the proof involves uncountable infinities or a Silver Age comic.
@@nawdawg4300 this is why reqular folks need to wake up, understand who is the batman (kazarian mafia, masons, religions, incl science) and woosh, freed from eternal slavery. escalated quickly, huh?
If one of the assassins makes a wrong choice, those after him will still know their number because they will hear him getting executed for being wrong
I'm pretty sure he would see a pile of assassin's larger than the universe.
I think the axiom of choice, like almost anything else in mathematics, is a tool. It behaves in certain ways, it implies certain things about what it governs, and when you're doing math it pays to be conscious of when you are using it and to ponder what would change if you didn't.
You'll just end up with an infinite amount of knocked up and tied up assasins... because BATMAN!!!
The second assassin whose number differs from the series knows that the last guy's number is different (too), and since the first one said there was an even number of differences, the second assassin with the wrong number knows he has to have his number wrong too, cause he doesn't see any difference with all the other assassins and the series.
I know I didn't explain it very well, sorry, but it's just how it came out of my mouth... well, my fingers.
Close enough :)
It still works, and only the 1st guy has a 50-50 chance.
Guy 1 sees even number of differences, says 0, 50-50 chance.
Guys 2-59 see two differences, so they know to follow the sequence.
Guy 60 sees one difference and knows he is another difference.
Guys 61-79 now know there's an odd number of differences remaining, and since they see one, they are not a difference.
Guy 80 sees no difference so he knows the difference is in himself.
Guys 81 and forward know all differences are accounted for and follow the sequence without worries.
The first guy uses the code for "even number of differences", the second guy sees an even number of differences, so he knows his spot in the memorized sequence is correct. This repeats until you get to the guy who sees an odd number.
The only information given is whether there is an even number of differences, or an odd number of differences, and each person knows this fact, and the number of differences in front of them, as well as the answers from the assassins behind them, so when a persons answer from behind them is different to the memorized sequence, the known information changes.
to express msolec2000s example differently
p1 sees even number of differences (diff mod 2) = 0
p2-59 see even number of differences; (diff mod 2) = 0
p60 sees an odd number of differences (diff mod 2) = 1; and it differs from the know information
so after this there is an odd number of differences, it doesn't matter what the number is just that it is odd
p61-79 see an odd number of differences; so there is no difference from known information
p80 sees an even number of differences, this differs from known information so he knows there is a difference at his place, and this repeats for all the prisoners
Now your seeing the reason that the axiom of choice is so illogical
the Mathassin's Creed:
{thing} ≠ true; permit {ξ}
May the Mathologer of Understanding guide us.
:)
HAHAHAHAHAHAHA
Or 0/0
Omg. Hahaha!
The question is, of course, if Batman with infinite prep time could defeat an infinite number of infalible assassins.
That is a given by the Axiom of Batman.
In this case he would simply avoid stepping into the street at noon.
And that is probably also why you only see him at night.
All Batman needs to do is bypass all the rational points between 0 and 1 and visit only real ones.
Richard Kelly ...Aren't rational numbers a _subset_ of real numbers?
the trick is to send in an ice cream truck first. Since all the assassins know they won't be the one doing the assassination, they follow the truck, and the street is empty.
@@timothymclean they are, and moreover, q is dense in r
It's not the perfect murder because an infinite number of assassins has an infinite amount of mass. So the assassins, the planet they're standing on, and Batman collapses into a black hole. From the point of view of an outside observer we know Batman can't escape the event horizon, so he's dead. But neither can the assassins, so they all got the death penalty as well!
:)
Mathologer You also can't do it because the infinite assasins serve infinite space so that the assasins cant k45523ill Batman
Also police have to to through each layer to locate the 0 assassin.
Martin Heermance and due to time dilation batsman still lives onto infinity.
Nah, you can assume that every assasin is half the mass and size of the previous one. :)
+1 for the Michael Stevens's beard graphic at 12:00! Hilarious!
Could not resist :)
What did Batman ever do to you?
He's Batman.
He took it upon himself to try and save Gotham. Gotham cannot be saved, it must be destroyed.
A very interesting video. When you mentioned cheating death, it made me think of the prisoner execution problem, which has recently been puzzling me. (where the prisoner is promised to be executed without knowing when). I think that could make a good video too.
I was just looking for videos about the axiom of choice earlier today; what a coincidence! :-D
mathologer, your videos are very fun and educational. i am a fan for very long time, please continue making such good videos in the future, cheers
The Banach-Tarski paradox sounded SUPER COUNTER-INTUITIVE when I first heard about it, but after calmly comparing with the fact that it is possible to rearrange all points in [0,1] to make [0,2] (which is doubling the length and is SUPER TRIVIAL), I now think the only "weirdness" in this paradox is the "possible just by dividing the cube to finite sets and rearranging" part and not the "possible to double the volume" part.
The latter part, I think actually is the origin of the (false) counter-intuitiveness for many non-experts, while it's actually not really the main point. After realizing this, I'm more "pro-axiom of choice" now.
Precisely, if you are allowed to move infinitely many points changing the volume can be trivial. The main point of the Banach-Tarsky paradox is changing the volume of a sphere by splitting it only in finetely many pieces, and moving them around with transformations that preserve the volume, such as rotations. The reason why volume isn't conserved is that these pieces are so weird that you can't associate a volume to them anymore (nor 0 nor positive, they are "non Lebesgue-measurable"), hence there is nothing for rotations and translations to conserve! You can move this pieces around in a way that they form now a set that has a volume, but there is no reason to believe that the original volume has been preserved.
If you split the sphere in 5 pieces each of which has volume, there's no way of rearranging them with rotations and translation to grow the volume.
@@99Megaluca99 And here's the thing: Banach-Tarski paradox is not true in the same form in two dimensions. It's not possible (not even assuming axiom of choice) to split a two-dimensional shape into finitely many subsets, move them around by translation and rotation, and get a shape with a different area. (The construction depends on the ability to rotate the set in two independent directions.) But it becomes possible when we don't restrict ourselves to length-preserving transformations - translations and rotations - but also allow area-preserving skew transformations. It's also possible - assuming axiom of choice - to split a one-dimensional interval into a countably infinite collection of subsets, move these around by translation, and get an interval of a different length.
Good news: this is math. We can take anything we wish as truth, so long as it does not contradict anything else we take to be true. Arguing about whether or not it makes physical sense is completely inconsequential until you construct a physical system where it might be applicable, which seems unlikely to happen with the types of things that make someone question the axiom of choice
Good news indeed :)
Good news everyone!
Saw the Assassin's Creed-style assassins in the thumbnail and was sold. XD
Wow, I just watched the Banach-Tarski Paradox by Vsauce and then I found this.
A fact that becomes quite mundane if you think about it. You mentioned and linked the Vsauce video in the description, so no wonder this got recommended.
I love this channel. Possibly my favorite maths channel. I hope there will be plenty of uploads in 2017 :D THANKS MATHOLOGER!
Glad you like what I do and thank you for saying so :)
Love your videos~
The most conterintuitive thing for me was that at 6:42 individual assasin's chances of dying are still %50, even if a finite number of an infinite number of assassins are dead no matter what happens. Though I get how this could happen, it's a shame you gloss over the fact so quickly.
I'm not sure if that probability can be rigorously defined in the first place. How does one compute the probability that the agreed-upon representative sequence from the equivalence class (the class of all infinite sequences differing from the actual sequence in a finite number of positions) matches the actual sequence in the nth position? It seems to me one would need to know explicitly the procedure through which the representative sequence is chosen (the choice function), but the axiom of choice has to be used here precisely because no choice function is known that can distinguish exactly one sequence from all the others in each equivalence class. There might also be issues with defining an appropriate measure on the set of all infinite sequences, but maybe that's possible.
Pick a sequence of 0's and 1's, that's the hat sequence. Now pick the sequence memorized. Each assassin knows the sequence beyond them, but not they own hat. You can never know if you are already in the "tail".
But that's not all, that's just one of the factors. The tail of the sequence do not have an expected position, which means that you cannot expect the tail to be in any position you can imagine. Pick a number (a position), for every number you can imagine, there are an (uncountably) infinite many other sequences in the box (that you picked the sequence) which the tail start later. The probability that you are the start of the tail is ("_almost_") 0%.
However, i don't know if this can justify the 50% statement.
I very much enjoy your videos, keep up the good work
Equivalence portioning can be used in software testing to reduce the number of test cases needing to be done when different values in a set are functionally equivalent when related to the logic being applied to each value.
Why is it 50/50 after using the close sequences strategy. Shouldn't it be more like 0 because only a finite number of them are being killed out of infinitely many?
Same question.
Where is the expected "start of the tail"? If you choose any finite number for it, you can see that there are infinitely many more possibilities that the tail starts after it. So it must be infinity. So the expected number of dead assassins is still infinity.
@@agsantana nope, there is no sequences in that but without some finite N after which all entries match. It could be 964326784333678853378984323789, but not infinitly many.
3:42 Love the Bill Wurtz feel
This was amazing, as always. Thank you :)
1:45 I called it! None of our assassins can get caught for killing Batman; since, no matter, which assassin the cops choose to suspect & investigate, there’s always infinitely many infallible assassins between him and Batman, who would have done the job before the suspect. 👌🏻
In the second problem, the probability that an individual dies cannot be 50%; in fact, this probability cannot be defined at all.
Indeed, assume for the sake of contradiction that the probability that an individual dies can be defined. By the fact that each individual is given a 0 or 1 hat independently and with 50% chance each, it is clear that the probability that any particular individual lives must be 50%. Thus, if I specify some set of K different individuals then the probability that all K of them live is (.5)^K. Therefore if I specify any INFINITE set of different individuals, then the probability P that all of them live must be 0: indeed, for each K, P must be less than or equal to the probability that the first K individuals in my infinite collection live, which is (.5)^K. Thus, P
The limit approches zero it's not actually zero, therefore the sum of the cardinality of these infinite sets cannot be assumed to be zero. That's a similar problem to Hempel's Raven Paradox.
Would you care to elaborate? I do not see the fault in bjo885:s reasoning nor the connection to the Raven Paradox. If I'm mistaken I would like to know it.
>it is clear that the probability that any particular individual lives must be 50%
Why? Not clear to me. Also intuitively I think that the farther an assassin is from the beginning, the lower his probability to die is.
An attempt at a different proof:
1) There are uncountably many binary sequences, let's assume each equally probable (i.e. uniformly distributed).
2) Thus, after binning into boxes, sequences in each box are uniformly distributed.
3) There are countably many sequences in each box.
4) Thus, sequences in a box form a uniform distribution on naturals, which does not exist
A more detailed explanation for the interested:
Regarding #3: Sequences in the box differ from the label on the box at finitely many places. Thus this set is actually equal to the set of binary sequences with a finite number of 1's. And a binary sequence with a finite number of 1's is nothing else than a natural number, written down in binary, and there are countably many natural numbers by definition.
Regarding #4: See e.g. math.stackexchange.com/questions/14167/probability-of-picking-a-random-natural-number
I'd say the Axiom of Choice is true as long as you don't have an Electoral College :(
"popular vote" is simply irrelevant.
the voting members of USA are THE STATES.
...when will the democrats learn?
"The states" can't vote. People vote. And people don't vote for their states, they vote for their president. Up to 80% of the votes were meaningless (depending on which news or fake news you trust), this shouldn't be the case in the leading nation of the "free world". Trump is the rightful president and should be honoured from everybody, but also be controlled from the people as every president before. But the next vote has to be equal, just as it is written in the constituion. Even Trump said before Obama's second term, that the EC is nonsense. No country all around the world has such an artificial inequalizer in their voting system. I think the analogy with the Axiom of Choice comes pretty close to reality.
Well, I get your point, and historically, this "every state for themself" made sense. But among other, mostly weird things, the new president talks about unifieing the country (after splitting it...), so there is one man for everybody. People go and vote directly their president, why not count their votes directly? Well, it is, what it is now, but what is the deeper sense in today's world?
For your second part, I just can speak for my German country. We have indeed a different system, as I can see it in some form or another all around the globe. You are right, that our leader's party gets on average maybe 30 to 35% of the votes, but it's still a majority... In the US, you have your two big parties, of course the winner of the popular vote gets over 50% (that's still a math video, so yes, this has to be right). In Germany, we have lots of parties, but at the end, only 4 or 5 of them find their way to the parliament, where they vote the chancelor, who is typically the leader of the party with the most votes. The big difference is, that the parliament itself is seated with delegates based on their relative percentage, not with a winner takes it all mentality. So, if 35% vote the CDU, and therefor Angela Merkel, there will be about 35% delegates representing the CDU. (It's a really interesting way to do this the fairest way possible by using maths, even written in our election laws.) At the end of the day, this means, that every single vote, be it in Bavaria, Hesse or Berlin, has the same value. And we still have a federal system with lots of specific powers for the states. In the US in our days, republicans from California are as screwed as democrats in Texas, because their votes are simply thrown away. Or not even only that, in fact, they have been voted for their opponents maybe their whole life. I don't get it, sorry...
Mahissimo the smallest states entered the union on the condition they wouldn't just be outvoted by the big populous states. The electoral college is a compromise. If you get rid of it, most of the states would be better off leaving the union.
if the sequence doesn't fit, you must acquit
The hat problem was equivalent to a problem I had on one of my problem sheets during a countability course! Took me a while to get that one.
Re: Axiom of Choice - It is something that is intuitively obvious - and infinite sets are anything but intuitive. Personally, I am a constructivist when it comes to proofs, but then again I am also a computer scientist: if you can't construct a thing it is generally pretty useless (with the possible exception of Turing's Oracle)
kumoyuki As a computer scientist myself I share your bent toward construction. However, non-constructable things are useful in the construction of reductio ad absurdity. and these boxes are absurd.
Agree. Because most of my proofs were constructivist in nature, my professor tutor in algebra exclaimed in exasperation to one of my proofs: "Why do you have to do everything the hard way?". I should have replied: "Because I want to understand why it is true, not just know that it is true".
It is intuitively obvious that the Kuratowski-Zorn Lemma cannot be true. Yet it is provably equivalent to the Axiom of Choice. It seems to mean that intuition is not generally a good measure of judging what is true or false. It would be interesting if it is possible to create true Artificial Intelligence whether it would deem AC as true or false.
Intuitively, arriving at any mathmatical paradox implies that the problem has not been addressed in the correct manner. There is a fundamental disconnect between our perception of how things should be and reality. While we entertain notions of zero and infinity, neither absolute zero, nor infinity can exist as physical constructs. There are islands of stability in current maths but as soon as we try to pass a certain point it all decends into chaos and uncertainty in the same manner as a Mandelbrot set diagram. In order to get meaningful answers, we first have to frame the question correctly and to do that we need complete understanding of the problem. Still a long way to go.
Funny how you compare it to the Mandelbrot which as you zoom in the structure remains the same.
NotaWalrus The analogy was deliberately selected from a range of options. :-)
You guys need Gödel.
Can something that's both right and wrong exist?
Can something that's neither right or wrong exist?
Can something parially right and/or partially wrong exist?
The universe itself might be even wackier than Banach Tarski's paradox appears to some of us.
< The Full, Infinite, Whole, Complete >
"That is the whole, this is the whole;
from the whole, the whole becomes manifest;
taking away the whole from the whole,
the whole remains."
- fragment from the Isha Upanishad
----------------------------------------------------------------------------------------------------------------
9:30 Well; the 3rd guy hears: ”1”, corresponding to an odd number of differences. He sees 0 differences, which is an even number; and thus, he knows that the only difference must be on his spot. He refers to the 3rd number of the memorized sequence (1), and, knowing his hat number must be different; so, he calls out: ”0”; and, for the rest, the sequences coincide. 👌🏻
The axiom of choice was accepted, not because it seemed reasonable or necessary, but only because they'd have to throw out a bunch of theorems if they rejected it
But also if it didn't seem reasonable, then there wouldn't have been "a bunch of theorems" that would have to have been thrown out. The reason there were so many such theorems is because people would intuitively use the Axiom of Choice without realizing it.
N.J.Wildberger definitely doesn't ! In fact I thought it one of his most compelling arguments against 'real' numbers. Famous Problems 19d - modelling the continuum ?
Neither do I. Once you understand the relationship between Goedel's Incompleteness Theorem and the real numbers that have been constructed, it is hard to believe in the set of reals as a fully constructed set. A good tell that someone is hand-waving meta-logic is when they say things like "imagine a perfect logician", or "imagine that person A knows everything that person B knows", or uses a form of argument by induction on logical constructs that have not yet been fully constructed. Tao ties himself up in knots on the blue eyed islander problem like this, which goes to show the lengths and energy someone will go to with the mathematical version of cognitive dissonance.
Agreed
The problem with this is that batman is so popular they would have to bring him back
You can't kill the Batman, you can only temporarily exile him from the DC Universe.
Funny, I was reading the Wikipedia article on the axiom of choice yesterday (having no prior knowledge of the concept), thinking how interesting it was and almost wishing for such a video...
Great content, once again :D
Glad you liked it :) Personally this infinite hats puzzle is one of most fun and accessible setups in which the Axiom of Choice comes up.
Perfect murder with infinitely many infallible assassins concentrated to a point of infinite density where Batman will be at 12:00.
They can also memorize infinitely many numbers and do infinitely complex calculations. They must collectively form the entity referred to as "God"
Very efficient : )
does it matter if a finite or infinite/2 number of assassins die, since there's always infinitely more to replace them?
Well, having to deal with infinity corpses is a lot harder than a finite number of corpses XD
Well it does definitely matter to the assasins who were killed additionally when you kill infinitely many of them instead of just finitely many (also notice that infinite/2 is not an amount)
I'm glad I do engineering, we just let you guys do the tricky stuff XD
We do design things with your maths though :)
We aren't lazy, I promise.
Jamie G Write some proofs you lazy bum.
Jamie G: What are you complaining about? They leave all the fun to us!
Haha so true. We wait for you to figure out how it's supposed to work, then we make one that works like that :)
But you do the things they have to cheat to do, like create a device that can move through an infinite number of positions in a finite amount of time despite Zeno's Paradox.
Batman will not be killed. He can only be killed by the first assassin he meets, no subsequent assassin will get the chance. As there is no first assassin, he will not be killed.
I think that if you try to convict a gang of *infinitely many* infallible assassins, you will have other problems than proving which one killed Batman.
it's high noon
oldcowbb what does HIGH noon even mean
But the real question is how does one convince an infinite assassins to kill one person
I'm not paying them.
Cast Splinter Twin on a Deceiver Exarch, boom infinite copies of the same assassin.
Wow I went deep.
Use an infinite megaphone loop to tell them about the plan. Like this but woth more megaphones so no sound is lost and shout in one of the megaphones
^ >
< v
Well, there is countably many of them. If you could convince all of them agree to kill Batman PROVIDED the previous one was willing, then you just have to convince any one of them; then you have murder by the principle of mathematical induction.
See the first part's easy, see the Fidget Spinner popularity conjecture, people will do anything if they think someone else will think it's cool. And, though the base case may be more difficult, you have an infinite number of chances to try.
So there are two important lemmas here:
1. Closeness (equality modulo finely many positions) is an equivalence relation. That is, each sequence belongs to exactly one putative "equivalence class".
2. Each infinite sequence can be associated with its equivalence class even if we lack knowledge of one of those places. We already know the sequence differs in at most a finite number of places, and finite + 1 is still finite, so this is trivial.
One thing that's left implicit is that each assassin knows his position in the sequence. This follows from hearing N prior assassins say 0 or 1.
2 also implies that every assassin knows the correct equivalence class since they are lacking knowledge of no more than N+1 digits where N is their position in the sequence (starting at 0, of course). This means that they know the class even if the assassins before him in the sequence mess up and say the wrong number. However, each assassin after the 0th will say the wrong number unless either they also know whether each prior assassin lives or dies (in which case they know which number was really on that hat) OR that set of assassins makes an even number of mistakes.
I was thinking about how there are uncountably many boxes of sequences in this puzzle, and in light of your "Transcendental numbers powered by Cantor's infinities" video, I was wondering if I could prove this using Cantor's diagonal argument.
The natural approach would be assume there are countably many boxes, choose a representative from each box, list those in order, and then construct a new sequence by making the sequence differ in the nth place from the nth sequence. Of course, this technique does not work here, since it would only guarantee that the newly constructed sequence differs from any sequence in 1 place. It could still be in the same box. So I came up with the following modification.
Assume there are countably many boxes. For each box, choose a representative sequence. List these sequences. For each sequence in the list, we may associate it with a prime number p. We do this by associating the nth sequence in the list with the nth prime number. (By Euclid's argument that there are infinitely many primes, we are guaranteed that every sequence in the list has a prime number associated to it.)
Now, we construct a new sequence according to the following algorithm:
To determine whether a 0 or 1 goes in position n:
1. Consider the prime factoization of n. If n is _not_ a pure power of a prime, place a 0 in position n.
2. If n _is_ a pure power of a prime, i.e., n = p^k for some prime number p, then consider _which_ prime number p it is. In other words, it is the mth prime number for some positive integer m.
3. Consider the mth sequence on your list. Look at the number in position n. If it is a 0, make position n in your constructed sequence a 1; if is a 1, make position n in your constructed sequence a 0.
Now, our constructed sequence must differ from each sequence in the list in infinitely many places. This is because for each positive integer m, there are infinitely many powers of the mth prime. Hence, the constructed sequence differs from the mth sequence in the list at all (infinitely many) positions given by powers of the mth prime. Therefore, the constructed sequence is not in the same box as any of the sequences on the list. Hence, we have demonstrated that we missed a box. Therefore, there are uncountably many boxes.
but infinite number of asassins will fill the entire universe , even if the asassins is smaller than a quark....
That is only true if they are all equally large. They would for example only fill a finite part of space if the volume of assasin N is 1/2^N ( for that matter any geometric progression with ratio r for which 0
@@benjaminbedert1760 which works in theoretical maths, but out here in the real world an assassin can't be smaller than a planck length in any direction. Or else by observing themselves they'll create a black hole.
3:51 You probably shouldn't say "list"! The set of infinite sequences of 0's and 1's is not listable!
Well, it is, you just well-order it :)
Well, that doesn't suffice. listable essentially is the same as recursive enumerable (i.e. you are well below anything using the axiom of choice). And you cannot find a (finite) recursive algorithm that outputs exactly the elements of any given such equivalence class, because this would yield a well-ordering of the class which probably could yield a proof of countable choice in ZF.
Yeah... ummm... Set Theory was a long time ago for me, but I seem to recall finding out that the Reals could be well-ordered but were still UnCountable... and I never really DID understand that part. (At least not intuitively.) I believe it, I just don't understand it. :) [How can everything have an "immediate successor", but you STILL can't put them into 1-1 correspondence with the natural numbers?...it's odd. Or even. Or just strange. :) ]
The Scattered ANY real number has infinite successors (not just one single "immediate successor"). Actually, NO REAL number has one immediate successor or immediate predecessor, only a proximal to both a minimum and a maximum in a given interval, arbitrarily "close" to the number. This is the way the set is considered as well-ordered.
The same goes for Rational numbers (a sub-set of Reals). This means that both sets of numbers have a DENSITY strictly higher than 0, as opposed to the set of Integers, wich density is strictly equal to 0.
Even though, Rational numbers are countable, while Reals are not.
The surprising proof of this was established by Cantor in his transcendental work on sets and infinite collections. The very beginning of Set Theory.
@@venturarodriguezvallejo1567 however the reals are well orderable, assuming AoC, just that that order will respect the "natural" ordering that is usual to them
Assuming AoC any set is well orderable, the reals, the rationals, the set of functions from R to R,...
For me the axiom of choice is a no brainer. Just because we can't explicitly point to any choice function it doesn't mean it doesn't exist. And it seems to me very intuitive that there's at least one,so I'm ok with using it. It has lot of very beautiful consequences and I can't imagine myself doing algebra without assuming there are maximal ideals in any ring.
It seems odd to have so much explanation on the technical strategies the assassins would use at every step of the execution, but hand-wave past the underlying math, at least considering that the video is only 12 minutes. It's a really interesting logical result that you can achieve these low finite numbers of deaths, but you just have to take the interesting math that leads to it(dividing up infinitely many numbers into these classes, the choosing, and problems with that choosing) for granted. Even the titular Axiom of Choice is only really mentioned at the very end of the video, not in particularly much detail, and the relation between Banach-Tarski and it isn't explained.
Isn't the first one just Zeno's paradox about turtle but formulated in terms of assassins and batman?
Антоша Пушкин I believe so
It's actually a variation on Benardete's paradox:
"A man walks a mile from a point α. But there is an infinity of gods each of whom, unknown to the others, intends to obstruct him. One of them will raise a barrier to stop his further advance if he reaches the half-mile point, a second if he reaches the quarter-mile point, a third if he goes one-eighth of a mile, and so on ad infinitum. So he cannot even get started, because however short a distance he travels he will already have been stopped by a barrier. But in that case no barrier will rise, so that there is nothing to stop him setting off. He has been forced to stay where he is by the mere unfulfilled intentions of the gods."
Also a nice one :)
If we don't accept the Axiom of Choice, then we are forced to conclude that not every vector space has a basis. This seems very, very bad.
I don't see what is "bad" about ℝ as a vector space over ℚ not having a basis.
That’s why it is an axiom, the basis is assumed
You mean "if we assume the negation of the axiom of choice", right?
When I saw this in my recommandation and I looked at the picture and the title I was like, is this some gametheory video? Then I see mathloger... So here I am
The 2nd assassin whose number is different heard that there's an even number of differences, but he also heard that the person behind him said that he was different. Because of this, he knows that there's an odd amount of differences in front of that person. Since he only sees an even amount of differences (in this case, 0), he must be one of those differences.
I am not sure how I feel about the new format, I think that the videos were more engaging when you are on screen. :)
Well, as I said, just an experiment and definitely not something I plan to do a lot from now on :)
Mathologer This time, it seems to me it was neccesary to visualize property the subject. Good choice. ;-)
Maybe, my only objection would be about the speed of your explanations. Too fast.
This is my guess for the assassins lined up facing to one side that can guess their number one at a time. Haven’t watched the answer yet. edit: this isn’t the answer in the video, so no spoilers :)
The first assassin can give some type of parity bit, like saying “one!” if the number of assassins with ones on their heads ahead of him are odd (thus making it an even parity). That way, when it comes to a certain assassins turn, he can just add up all the previous assassins one’s with all the one’s ahead of him and yell one or zero to maintain that even parity.
This should work flawlessly for a finite number of assassins, but i’m not so sure for infinite because I wonder if it could ever be even or odd?
In the English captions, the caption that comes straight after 2:34 is not visible for long enough, and the captions are locked so I cannot fix this.
Thx for pointing that Banach-Tarski video out, I remember doing the proof in a course. Even though I understood the math, the illustrations and thought experiments provided by that zsauce guy and of COURSE you, makes one think of it in a different light. Choice axiom is probably a bad idea. but I don't wanna live in a world without Zorns lemma, just sounds grey and boring :p Keep up the nice videos!
life lessons from mathologer: it is a bad idea to try to be intimitate with someone who has nothing in common with you because "being CLOSE is an equivalence relation."
I don't understand the idea behind the boxes at all. First of all, they still have 50% chance of dying. You seem to be saying that 50% of infinity is somehow suddenly finite. I don't see how you could get away with that.
Further; the details of how they could divide their sequences without ambigiouity don't make sense to me:
Let's assume that they all get together and agree that a certain (randomly chosen) sequence, A, will make up the backbone of the first box (which they write "A" on the lid): if they see any "close" series to this one, they all agree that this is the one they will shout.
Imagine then that they find another sequence, B, that is not close to A. They put this into another box marked "B".
Then I, being a fan of Batman, come along. I'm feeling a bit trollish, so I decide to "help" the assassins: I offer a hand in sorting out all sequences into their correct box.
(They first answer that they are infinitely many so my contribution will be too small to matter, but I argue that the task is so large that they need all the help they can get.)
They agree, and so I start with sequence A1: it is the same as sequence A, except at the first place that A has a difference from B, I swap that number for what is in B. This is still close to A, so it goes into the same box. I then take A1 and pick the first number where it is different from B and swap that entry for what is in B, and call the new sequence A2. Then I take A2 and pick the first number differing from B and swap it out for what is in that place in B and call it A3. All of these are of course close to A, so they go in the box labled "A". I continue with this exercise an infinite amount of times until eventually I end up with An = B.
But: as I create them, each is just one number removed from entries already existing in box "A". And because of the identiy that if X=Y and Y=Z then X=Z, all of the entries must be put into box "A". Eventually I get close to B, and by similar argument can lump B in with A.
I continue working through the night, building simliar bridges to all other of the asssassins' boxes. Now the assassins will have the choice that they either have to
1) shout the same sequence no matter what they see around themselves (essentially they have just all agreed beforehand what to shout), or
2) Since any member of A is also a member of B, they could not agree on which box a particular sequece would belong to and end up shoulting numbers from different sequences with equally depressing results.
But as I watch infinitely many assassins get executed, I wonder: how could they possibly have made any meaningful partition to begin with?
Well, I really enjoyed reading your comment.
However, in mathematics the default approach is to NOT do the sorting one sequence at a time. The sorting is implicit in terms of our definition of when two sequences are close.
The key point here is that if A is a sequence that is close to a sequence B and B is a sequence that is close to C then it follows that A is close to C.
Now take any sequence A and consider the set of all sequences close to it and call this set BOX(A). Now you can convince yourself that for different sequences A and B either BOX(A)=BOX(B) or BOX(A) disjoint to BOX(B). The first happens exactly if A and B are close and the second happens otherwise. And this then means that these sets split up the sequences as required.
Maybe the right picture to keep in mind is this: Imagine all 0,1 sequences hovering in mid-air. The head mathassin issues the BOX IT command. Immediately any two sequences that are close are attracted to each other and otherwise repel each other. Now as you watch the sequences assemble themselves into the disjoint set/boxes. True Mathemagic in action.
Now we chose one sequence from each box and everybody memorises it, etc.
Hmmm... thank you for your answer, but I'm still don't quite I understand the logic behind this. I'm essentially arguing that you can _always_ show that BOX(A) = BOX(B) for two boxes starting with _any_ sequence A or B through a chain of close sequences.
Sure, the chain of sequences would have to be infinite, but each box has infinite number of sequences anyway, so that's not a problem. (Or are you saying that each box has finite number of sequences in them?)
In other words: Even if B is _not_ close to A; still BOX(B) = BOX(A).
In essense, when the mathassin big boss shouts "BOX IT!", I imagine that while the sequences will indeed float around repel each other and attract each other based on closeness, they will eventually all be forced to form an infinitely large "cirlce": Even as most sequences are infinitely far from each other, they are still _all_ topologically connected.
No two boxes could then possibly be disjoint unless you start excluding sequences from the excercise.
(Or I guess more like an infinite dimensional n-sphere (I suppose whoever they chose as leader would be that good....)? Anyway, I hope you understand my thinking)
+Snagabot " I continue with this exercise an infinite amount of times until eventually I end up with An = B."
No, An was the result of exactly n replacements, so An =/= B for any integer n.
"Eventually I get close to B, ..."
No. That is wrong for either meaning of the word "close" relevant here. There is no reason to believe that your sequence A1, A2, A3, ... converges to B. Each An is no closer to B than the one before.
The series An = .1 + .01 + .001 + ...+ .1^n converges to .111 ... = 1.0 as n approaches infinity because an integer k can always be found to make (1-Ak) as small as you like. That isn't true for your sequence because the distance from An to B is constant.
Snagabott The problem is very simple. "I continue with this an INFINITE amount os times" so you broke the rules, because the diference needs to be FINITE! :). You can walk from a sequence A to a sequence B from another box, while someone walk B to A. You will never meet in a FINITE number of steps, so no problem ^^ (Sorry for the bad english)
@Marcos No matter how you look at it each box will have to contain an infinite number of sequences anyway, which is why I'm not bothered by having to put an infinite number of sequences in the box to achieve what I want - so do the assassins. Or are you contending that each box will only contain a finite number of sequences?
Second, there is the property that if A is close to B, and B is close to C, then A is close to C.
This is a special case of the identity A1=A2, A2=A3, A3=A4, ..., An-1=An => A1=An
I have never heard of any axiom or theorem stopping this from being true just because we allow n to grow infinitely large. Is there one?
That is why I find this manner of box partition inherently self-contradicory. It just seems like a nonsensical task to me. The only two meaningful sets I can imagine ending up with after doing the BOXIT spell, are the empty set and the set of all sequences, neither of which satisfies the original requirement.
Proving continuum hypothesis , proving inconsistency in ZFC , constructing ZFC from naive set specification , resolving Russell's paradox , constructing infinite number system , construct and ensure overall consistent mathematical universe and developing arithmetic system - edition 8
May 2024
DOI: 10.13140/RG.2.2.21713.75361
LicenseCC BY-NC-ND 4.0
Practical application, of course I have to make some adjustments but this is actually kind of helpful.
Excellent video! You really are a fantastic teacher! I’m not sure why the directional assassin problem requires all the complexity of memorizing all infinite binary sequences and placing them in an infinite number of boxes. My solution is to have the first assassin simply call out the mod 2 sum of all of the assassins he can see. The next assassin also computes the mod 2 sum of all the assassins he sees and lastly performs a mod 2 addition with the previous assassin’s called out number. In this manner (and with no memorization or need for the Axiom of Choice) it seems to accomplish the same result - 50% survival for the first assassin and 100% survival of all others. As I thought more about this, I guessed that you might say that the sum of an infinite number of binary digits (mod 2 or otherwise) does not converge (I really loved your videos on infinite sums!). But I would argue that if you don’t think the infinite sum converges, why do you think an assassin could memorize (and bundle) an infinite number of infinite sequences?
The problem with your approach is that if there are an infinite number of assassins in front of you the mod 2 sum does not make any sense if there are infinitely many 1s. In fact, in some (measure theoretic) sense it is certain that you'll see an infinite number of 1s and an infinite number of 0s in front of you :)
Because the mail can't handle all the letters lawyers would write and no-one could afford to pay for them anyway! Ah, so this is just the old mathematician vs. engineer joke, with the naked lady/man/tasty treat/whatever - they can only move half way from their current position to the prize each second, the mathematician just storms off because he can't ever reach the prize, the engineer proceeds anyway claiming he'll be close enough for all 'practical purposes'.
By the time the final question got asked I believe I had the correct idea in my mind.
You would want to use some form of encoding where a change in bits and a continuation of bits is equal to the bit you are currently on. You can't just say the same that's infront of you, as you die. But you need to learn your own hat from somewhere behind and also give away the next hat while your own call. Not quite sure if just the first guy as to gamble this way tho.
It's not the full solution - but I will watch it now
I'd like to see one million assassins standing on a pin.
Nothing clever say; just thank you for a really entertaining video.
The third assassin sees even no of differences despite knowing the previous assassin saw an odd no of differences since differing from sequence therefore he must differ and be a 0
I don't get this on a such a level that I can't think of what questions to ask.
How about a video on the relationship between the Axiom of Choice and The Continuum Hypothesis - I remember that being interesting but can't remember what it is (I know, I know, easy to look up these days, but it's more fun to watch Mathologer videos...)
as to whether an assassin that can see down the line and hear the choices behind him being made screwing up makes a difference in whether the rest of the assassins can still choose correctly, my guess is that it depends upon whether they know if the assassin preceding them chose wrong or not. if they are alerted to the mistake before they have to choose, then the sequence does not change. if they aren't alerted to the mistake, then, for their choice, the sequence has changed, and they would be choosing based on a sequence which is difference from the cogent sequence
Exactly :)
At 6:33, how can it still be a 50% chance? If only finitely many are killed out of infinity doesn't that necessarily that only a negligible percentage will be killed. Even if it were a stupendously large finite number of people killed - like graham's number or rayo's number or even the one raised to the power of the other - wouldn't it necessarily still be much closer to zero than to infinity
great information for any system to consider
"Choice" is not a rigorously defined math activity.
It's not the fault of the Axiom of Choice, but ours for measuring weird sets.
But in set theory without axiom of choice, you can't prove that this weird set actually exists. The entirety of axiom choice is not necessary - for example, existence of non-measurable sets follows from existence of well-ordering relation on the reals. It's also consistent that real numbers can't be well-ordered, but non-measurable sets still exist (and the Banach-Tarski paradox is true).
@@MikeRosoftJH
You seem to contradict yourself in the end here. : /
Do you mean that "non-measurable" sets are consistent with not assuming or rejecting the Axiom of Choice?
@@MrCmon113 Well, yes. Axiom of choice implies (and is equivalent to) that every set can be well-ordered. This means: if axiom of choice is false, then there exists some set which can't be well-ordered. But that doesn't tell which set it is - it might be the set of all real numbers, it might be the set of all functions on reals, it might be a completely different set. (That real numbers can be well-ordered in no way implies axiom of choice.)
Okay. Now suppose that real numbers can be well-ordered. In that case it can be easily shown that the non-measurable set exists: pick a well-ordering of reals, split the interval into equivalence classes, and from each class pick the minimum according to the well-ordering. But what if continuum can't be well-ordered? Then it's still consistent that the non-measurable set exists.
What resources do you recommend if one would want to read further into the Axiom of Choice?
That aside, the only solution that my high school educated brain can come up with is that in a problem with an infinite amount of solutions, the predictability of those solutions (at least to the assassins) can only be as predictable as those willing to die for the cause. For example, if we reached assassin #2045, and he sees that there are an odd number of pattern changes as far as he can see, what isn't to say that from his position that the solution of the 1's and 0's has changed and assassin #2045 would get killed and those assassins who previously had their pattern change now had the correct hats?
3:04 What about the following strategy that avoids the axion of choice?
Assassin in position k looks at 4^k assassins in front of him and 4^k behind him and searches for the smallest least common string of length k bits (so 2^k options) that he could add to by picking 0 or 1 appropriately?
Since most numbers are normal, and since we can look at the problem as getting a random number in the range [0,1] in binary mostly right. The probability that infinitely many assassins are dead should be 0 given that the hats have no patterns to them.
First, this violates the rules. A guy can see all guys in front of him, but none behind him. Second, what information would it get him to decide on a finite subset? There are infinitely many sequences matching that finite subset where he has a 0, and infinitely many matching the same subset where he has 1.
@@MikeRosoftJH This doesn't violate the rules for the other game he defined. Also, I use the fact that most numbers are normal.
@l954 What "other game"? At 3:04 there isn't any "other game" defined. And again, 1) a guy at position k doesn't have 4^k guys behind him (but only k-1), and 2) any guy only sees the guys in front of him, but not the guys behind him. And sure, almost all sequences are normal (in the sense that the measure of the set of non-normal sequences is 0). What exactly follows from there (even assuming - contrary to the original setup - that any guy sees all guys except himself)? No finite segment of the sequence affects whether or not a sequence is normal; for example, given a normal sequence, a sequence which differs from it by setting the first googolplex of elements to 1 is still normal.
The best I come up for puzzle#3 after a minute of thinking: If the first guy answers ((hat#2 + hat#3) modulo 2), then both #2 and #3 walk free; in general if he answers ((hat#2+...+hat#N) modulo 2), all up to N walk free. So maybe the sum of the infinitely many binary digits modulo 2 would work? That number doesn't seem defined though (What would it be for the sequence 111.... ?).
Thank you Burkard. It's good to finally have an awesome Australian mathematician! Happy Australia day btw :)
:)
Hi Mathologer. Could you make a video explaining John Conway's surreal numbers? How can it "create new numbers"?
Marcos K Whether they are "numbers" is really a philosophical or linguistic question rather than a mathematical one.
I'm a simple mathematician, i read axiom of choice, i see the video
The most confusing thing about the Batman-assassin illustration is that first we see Batman on the left and are told he'll walk down the street, the natural interpretation being that he'll walk toward the right. So we first place an assassin *right* where Batman starts on the left. Ok, so he's dead. Then for some reason we place assassin's to the right fo that? Which Batman would never get to? ....Only then do we notice that the presenter flipped everything left-to-right, with "0" being on the right and "1" on the left (in defiance of nearly every presentation of the real numbers ever seen), and that --- apparently, though it's not been stated -- Batman is in fact supposed to start on the right and walk to the left?? ("Because to get to number 1, he has to pass number 2 alive". No, not if he got killed by 1 before he got to 2.)
I guess some people are just very easily confused ... :)
You can use an arbitrary formula to choose. Like use the series which starts with 1, if multiple do then the series starting with 11 and etc, if all the options have a 0 at a point then skip over that point and continue on.
Every box will contain sequences that start with arbitrary long strings of 1s :)
The proble is that hower finitely long your string of 1s is, there will always be infinitely many sequences in the box which will star with that exact string of 1s
Mathologer so why not go for the longest? Is there no longest?
Firaro Nope there isnt, exactly as there is no biggest natural number
Transcendental Batman lives :-). If Batman were imaginary the assassins would have a complex task. The strategy chosen by the assassins is far too rational. Batman is real but completely irrational so can easily bypass the assassins. My friend Al (Al F that is) assures me that even though Batman is only 1 the infinite number of assassins amount to 0. Believe it or not, he adds, the axiom of choice is about whether or not there is anything between Batman and the assassins.
OK OK OK. Mathassins take the finite-deaths challenge. Without loss of generality, the Templar chooses to kill an arbitrary finite number of Mathassins first, then let the rest go. He now begins his algorithm: approach the smallest-numbered living Mathassin, ask for the Mathassin's ordinal, repeat the phrase "My finite number is bigger than yours", kill the Mathassin, then repeat. At some point the Mathassins become concerned that the Templar might be cheating. How many are left when they prove it?
I would love to see an introduction to the proof of Fermat's Last Theorem by Andrew Wiles from Mathologer!
Could you do a video about "game numbers" (number-like objects that are actually games of infinite lengths between 2 players)? I find it interesting how we can define a whole complete lot of "winning strategies for player 1" and "winning strategies for player 2" - and then screw them both simultaneously by never subscribing to either of strategies, always making moves that break out of either strategy until we have an (infinte) collection of moves that are neither 1-winning or 2-winning strategy, since our both sets of winning strategies are complete. It is an interesting theme, but the math for this is a bit hard, even in Wikipedia, so basically it is "mind = blown" stuff for me. Could you explain this theme in layman terms?
Assassin #3 will know there is an even number of differences from the box sequence (from #1). And since #3 knows the box sequence, he knows #2 gave an answer which did not match the box sequence (meaning #3 knows there is 1 change to the box sequence already). Therefor, #3 will knows that: if he can see an odd number of changes to the box sequence, then his hat will match the sequence. However, he sees no changes to the box sequence and therefor knows that his hat does not match the box sequence.
But, the real answer is all the assassins die, because it takes an infinite amount of time for assassin #1 to look at the infinitely many hats. So, he doesn't answer the question before they all starve in a finite amount of time.
Facing certain assignation, Batman kills himself. Every infallible assassin fails to kill his target. The league of infallible assassins has failed. Faced with this untenable paradox all the assassins kill themselves.
The funny thing is, the reason they can't easily choose an element from each box is very similar to the reason why batman could die without a murderer. There isn't always smallest sequence just like there isn't a final mathassin in the alley.