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😂

hahaha, that is awesome!

The puzzle with the logicians in a line guessing their own hat color has major differences though. They're allowed to devise a strategy before lining up to guess their hat colors and can communicate to some extent by adding extra meaning to their hat color guesses. With this particular one (often called the blue eyes puzzle), there's no communication, ever. They can only gain new information after so many days of inaction from everyone else because no one ever takes an action. You say "And when you look at what A knows about B's perspective (of C's perspective, and so on), this is true.", but dismiss this without explaining why, then go on to explain what B's perspective is. But A doesn't know what B's exact perspective is, because A doesn't know their own color. Put another way, you make the claim that everyone knows that their can't be at least two blue synths (i.e. 'From the full perspective of A, B, and C, there are only two possibilities. Either all are green, or all are green except the synth whose perspective we are considering.'). For that to be the case though, you're saying that A must know that B and C know their own colors, which they can't because we've already established that no one knows their own color.

@@rioc2802 My point is exactly what you are saying. In the video, when we look at what A's perspective of B's perspective is, then of course B doesn't know his own colour. And since A also does not know his own colour, then by A's perspective of B's perspective, both A and B's colours are unknown. This is of course true. From A's perspective. And that is the reason why, when we get to the end, A's perspective of B's perspective of C's perspective, there is the possibility that all of them might be blue. I didn't feel I needed to say that, since the video already says it. However, while this is true of A's perspective of B's perspective, it is NOT true of B's complete perspective. As I clearly said in my comment. Since B already know that A is green. And that is why this string of reasoning doesn't work. Since it only looks at PART of what the synths know. Yes. I do make the claim that everyone knows that there can't be at least two blue synths. And my basis for making that claim, is because everyone knows everyone else's colour. And everyone else is green. Thus everyone knows that there are only two possibilities. Either everyone is green, or everyone except themselves are green. But here is where you are wrong in your interpretation of what I am saying. I am NOT saying that B and C knows their own colour. What I AM saying, is that B and C knows A's colour. So: From A's perspective, there are only two possibilities. Either everyone is green, or B and C are green, and A is blue. From B's perspective, there are only two possibilities. Either everyone is green, or A and C are green, and B is blue. From C's perspective, there are only two possibilities. Either everyone is green, or A and B are green, and C is blue.

@@Tjalve70 "However, while this is true of A's perspective of B's perspective, it is NOT true of B's complete perspective. As I clearly said in my comment. Since B already know that A is green. " But A doesn't know what B's perspective is. A only knows one possible perspective that B may have. It's true that A knows the world can only be either all green synths (including A), or all green synths (except A). But A doesn't know if B or C know that. So it's possible and logically viable for A to consider B's perspective of C's perspective of D's perspective, and so on. You can't confirm one way or another what B and C know with 100% certainty because again, A doesn't know their own color. And that's what these perfectly logical creatures are ultimately theorizing; there's a non-zero chance that someone thinks someone else thinks someone else thinks someone looks out onto the world and sees all blue synths. Non-zero is greater than zero, which is all is needed for the puzzle to be stable at the start and slowly crumble after the visitor makes their announcement of their being at least one green synth.

@@rioc2802 Yes. A knows what B's perspective is, EXCEPT for A's colour. This is clearly stated in the video. And if you disagree with that, take it up with the guy who made the video.

@@Tjalve70 It seems we're just talking past each other. We both seem to agree on what the synths know and what they're capable of thinking of, yet you're disagreeing with the video, but I'm (mostly) agreeing with the video. So where's the disconnect? A can look at B and consider two possibilities for the world (2 green synths, or 1 green synth and 1 blue synth). A doesn't know which one is correct one way or the other and can't know what B knows, so these are just two realities that A needs to consider. Since A knows it's possible for B to consider a reality where there is 1 blue synth and 1 green synth, why couldn't A think B think that C (the green synth) look out onto the world and consider the possibility of 2 blue synths? You haven't explained why that's wrong and "I disagree" isn't a valid rebuttal.

'There are at least 3 green synths' isn't common knowledge at the start of the puzzle though. When you look at one of the other green synths, you don't know if they see 3 or 4 green synths (because you don't know your own color) and if they look at another green synth, you don't know if they think that other green synth sees 2, 3, or 4 green synths. The chain doesn't break down just because you don't agree with the hypothetical.

​@@rioc2802 I didn't say that "there are at least 3 green synths" is common knowledge. The point is that it will be common knowledge that there is more than one. And with that, the whole induction chain (which does work as advertised if there are fewer than five actual green synths) breaks down, no knowledge will be gained by anyone when there are no departures at T=1, and thus nothing will happen at T=2 through T=infinity either. As I said, feel free to map out the branching "what ifs" for the case of Greens=5 (the way the video author does at about 5:30 onward) and you'll see that it doesn't self-eliminate with five or more greens. To see the key point, note that at 6:38 he says, "after one second, all the scenarios with two blue synths [out of three total] are eliminated." But that only happens because everyone knows that if there is actually only one green synth, the lone green would have realized it and self-destructed. Since it didn't (in his sample scenario with three greens), everyone now has proof that the number of greens is two or more. Again, this can never happen in the greens=5 case, and everyone knows it from the start. Thus all of the subsequence potential eliminations (the inductive chain) DO NOT FOLLOW. Map out "what ifs" tree for Greens=5 and you'll see that at no point can the possible range of the number of greens be narrowed down for anyone in the population.

The fact that they know the approximate population distribution in itself doesn't actually matter. The problem is that all of the synths appear to be aware that their entire population only consists of blue and green synths and everyone already seems to know before hand that they're either blue or green, so 'At least one green synth' is already common knowledge before an outsider announces that. Other iterations on this sort of blue-eyes puzzle don't have this problem.

yes and maybe Vulcan understanding of color is different.

these are good points. And of course the story is a bit constructed. But yes, in principle we can think of the 'color' as some sort of parameter. The idea that they do not communicate about it, is to keep them alive. And the idea that they cant read their own color-parameter and the shutdown when they find out their own color is to keep the society discrimination-free. Think of it as a simulation of a new experimental society.

@@TheLeekWeek "keep the society discrimination-free" I do not think that is what we have here. In a discrimination free society any single individual would not care about everyone else's color or their own, rendering the color parameter meaningless. Here we have the opposite. Not only do they care about everyone else's color to the point of tracking it, they also act upon it, even if the act is upon oneself (self discrimination). Also, superficially they seem to not be narcissistic because they do not care enough about their own color to attempt to find it but then commit digital suicide when they do find out they are of the wrong color (compared to what they wished/expected)?! I find this experimental society weird :|

@@TheLeekWeek I am sorry if I am being annoying but this puzzle has been running on the back of my head and there is something else that is bothering me in this video. In the "mathematical induction" part of the video, I agree with the deduction for the case of one green and nine blue (1g9b) but I do not think that approach works for more than or equal to two green (@ 3:23 onward). In video you take a green one, say that he sees 1g8b, he sees the green does not shutdown so he concludes that he must be green. I think that he can only reach that conclusion if and only if the total number of green and blue is known by each of them (which is not indicated in the problem description). But, if the total amount of green and blue was known to them, then the statement from Spock was not needed since they could have reached that conclusion independently by simply counting how many green and blue they can see. As a counter example of the example above lets say that a blue one was taken instead. Following the same logic, he would see 2g7b, would see that the 2g had not shutdown and conclude he must be green! Which is wrong. This difference in evaluation, from my perspective, comes from the fact that their count is an approximation of the quantity of green and blue (~50% each according to the problem description). So, from my perspective, either 1) they know the exact amount of green and blue in the population and were already shutting down when Spock arrived or 2) they do not know the exact amount of green and blue in the population and can not reach a conclusion since only Spock, the visitor, knows that their approximate amount of green is 50% of a million I am not a mathematician so, if you have the time to reply, I would appreciate if you could comment on my logic. Maybe what I wrote makes no sense. Thank you. Either way, I forgot to mention before, great compelling video :)

​@@boam2943 For any sort of variation on the blue eyes problem (like this one), the general assumptions/rules include: - Everyone is a perfect logician. - No one is allowed to communicate with each other. - No one knows the distribution of colors among individuals in any capacity. You can look at 99 green synths and 100 blue synths, and be perfectly fine with believing you might b a red synth because for all you know, nothing says you aren't. For this particular variation, that means no one knows that there are only green and blue synths. - If an individual ever learns their own color, they must remove themselves from the group (killing themselves, leaving, etc.) - Everyone can see everyone else at all times, or at least, everyone sees everyone else at least once between cycles (seconds, days, etc.) - Everyone knows all of the above information. You don't need to fret about whether they understand the concept of what a color is because that's necessarily implied. The actual approximate distribution (i.e. about 50% green and blue) also doesn't matter, and shouldn't factor into their thinking because they're perfect logicians. They don't deal in approximations. They're primarily concerned with seeing if they can ever figure out their color through pure logician deduction alone. With that said, I don't agree with the full premise of the video, saying that all green synths eliminate themselves after X seconds where X is the number of green synths, followed one second later by all blue synths eliminating themselves. This implies that everyone already knew that the distribution only included green and blue synths at the start of the puzzle, so they already know that 'there must be at least one green and one blue synth' is common knowledge, which is what the visitor is meant to be providing (common knowledge of there being at least one green synth). They should really all be fully eliminated, or in the process of eliminating themselves prior to the visitor making their statement.

@@rioc2802 Thank you for detailing the rules. I never knew of this problem type before this video so my approach, on my first comment, was to treat it as an actual alien encounter, which is the premise of the video (Spock encounters an alien society which is blablabla). Your conclusion is similar to the one I give in my latest comment. If they know their exact distribution (exact number of green/blue), they should already be shutting down when Spock arrives. I believe that happens immediately to all in one cycle but I do not want to make that claim. Math is not my strongest subject. If they do not know their exact numbers, then they could not logically reach a conclusion and would not be shutting down (to prove I give a counter example to the logic used in the video).

maybe we can wait until they send Picard and Data to fix all this.

that saves a second, but still keeps the detrimental outcome.

@@TheLeekWeek What if you tell them there are many green synths?

you have to give all you got!

"The needs of the many outweigh the needs of the few". But we are still trying to save all of them.

@@TheLeekWeekWe can separate them, such that they don't see each other and then introduce them back to their "society" randomly, such that they don't know for how many days each one of them has been there.

😸

Ok, changing their colors can work 🖌

It's an interesting question of whether the statement "There are no green synths" is enough to trigger this cascade. Starting again with 1 green and some blue, the one green synth would see only blue, and use the statement to conclude it is also blue, and thus shutdown. On tick two, having seen one green synth shut down, the rest will realize it could see only blue, so they will also shut down. If it started with 2 green synths, then the green synths will realize they must be green on the second tick when the other green they could see didn't destruct. So I believe the statement "there are no green synths" is still enough to trigger the countdown, even though it is false.

The idea behind common knowledge is that yes, individually, you can look around and see at least 2 green creatures who see each other, therefore you know that everyone sees at least 1 green creature. Even in the case of just a total population of 2 green creatures, 'there is at least 1 green creature' is mutual knowledge. i.e. for knowledge p, all individuals are aware of p. Common knowledge goes a step further, stating that for knowledge p to be common knowledge, all entities in group E must know p, and all entities in E must know that everyone in E knows p, and know that everyone knows that everyone in E knows p, and so on. In the case of 2 green creatures (no other individuals) where you're one of the creatures, you know there is at least 1 green creature. You don't know that the other creature knows that there is least 1 green creature, because you don't know your own color. And you can never know your own color. In the case of 3 green creatures (no other individuals), you know there are at least 2 green creatures. You know each of the other 2 know there is at least 1 green creature. However, you don't know that they both know that you know that, because you don't know your own color. Importantly, this particular video isn't necessarily clear on its rules, as it can be interpreted that everyone already knows there are only green and blue creatures, meaning they know 'there is at least 1 green creature' is already common knowledge. In most iterations (often called the blue eyes puzzles), the only rule surrounding this is that everyone doesn't know their own eye color; you can see 99 green eyed people, 100 blue eyed people, and be perfectly content with believing you might be the only brown eyed person, because again, you can't look at other people and assume that they give you enough information on yourself if you can't communicate about whatever forbidden trait you can't discuss.

the experimentalist's solution 🤓

not a bad idea!

this should work, even though just reproducing might not be enough - as the new synths could be seen as part of a new group (they would survive though) - but the knowledge within the old group is not affected.

true, but in this case I assumed they know that there are only two possible colors: green or blue. Maybe I should have stated it more clearly.

@@TheLeekWeek If everyone is already aware that there can only be green or blue, and everyone knows this fact to be common knowledge, is the visitor even necessary though? They'd be able to immediately reason that a single green synth would only see blue synths and immediately shut down at the start of the scenario, and vice versa because they'd have to already be aware that 'there is at least 1 green synth and 1 blue synth' is common knowledge. It's not like other retellings of the blue eyes puzzle where there may be 100 blue eyed and 100 green eyed people, but 'there are only blue eyed and green eyed people' is explicitly not known among the the group.

Of course the Synths knew about this danger in advance, and made some preparations. "There is at least one..." were the last words of the visitor, as he was struck down by the planetary logic-defence system.... Sadly, half a million *and one* seconds later all Synths shut down. 😢

@@rioc2802 I believe it is the same in the blue-eyed-puzzle. The visitor common knowledge, which was not there before: everyone knows that everyone knows that everyone knows... (N times).

@@TheLeekWeek The original iteration of the puzzle (at least, the one I always refer back to) is the one from xkcd: A group of people with assorted eye colors live on an island. They are all perfect logicians -- if a conclusion can be logically deduced, they will do it instantly. No one knows the color of their eyes. Every night at midnight, a ferry stops at the island. Any islanders who have figured out the color of their own eyes then leave the island, and the rest stay. Everyone can see everyone else at all times and keeps a count of the number of people they see with each eye color (excluding themselves), but they cannot otherwise communicate. Everyone on the island knows all the rules in this paragraph. On this island there are 100 blue-eyed people, 100 brown-eyed people, and the Guru (she happens to have green eyes). So any given blue-eyed person can see 100 people with brown eyes and 99 people with blue eyes (and one with green), but that does not tell him his own eye color; as far as he knows the totals could be 101 brown and 99 blue. Or 100 brown, 99 blue, and he could have red eyes. ================== The description makes it pretty clear that even though everyone only sees brown, blue, and green-eyed people that their own eye color doesn't have to be limited by what they see, therefore, there isn't common knowledge about there being at least one blue-eyed person. Your puzzle however does seem to make everyone aware that there are explicitly only green and blue synths, so 'there is at least one green' synth is already common knowledge; they would've all self destructed before the visitor appeared.

to keep the society harmonious and discrimination-free.

interesting approach.

Doesn't work. Each timestep the lowest number common knowledge ticks up by 1.

​@@donaldhobson8873 Correct. I guess my proposal was a method to convince them that you could lie without actually telling them a lie, but that won't work if everyone already knows that you are completely unable to lie. If you are actually able to lie then you just go with the much simpler solution of providing a proof of that fact (eg. saying that you were lying earlier, and that there are no green synths).

In a 5 green 5 blue scenario, and each green synth sees 4 green synths, but they’re still there after 4 clock cycles (ticks) all five will shut down by the fifth tick The short of it is, they all know the others color, so they all go through the motions of knowing how many ticks passed since the mutual knowledge is shared and how many synths exist currently, until a quantity observed- tick = 0 is met. Saying it will take 6 days we can assume the total population of the green synths if one tick is equal to an Earth second. 518400 ticks is total annihilation. One second before that, all the greens shut down. One second before that, Quantity observed - ticks equaled 0. 5184399 green synths x 2 for roughly 50/50 green to blue.

@@pivotfever Yes, they all know every other synth's color. How does that impact knowing their own color if the probability of knowing an exact number of synths gives a 2^number probability of knowing the exact combination to know your own color? As soon as a synth can see more than one other green synth, knowing the phrase, "there's at least one green synth" doesn't necessarily mean each synth knows they are also a green synth.

⁠@@pace1195​​⁠​⁠You’re thinking about it the wrong way, and making it harder than it actually is. You don’t need to take 2 to any power. It has nothing to do with exponents. The only math involved is simple counting. Let’s say there are 69 greens. (The number of blues is irrelevant. There could be 69 of them, too, or there could be 72 billion of them, 2 of them, or none of them, at all). All the blues (if there are any) know that there are either 69 or 70 greens, depending on what they are, themselves. All 69 of the greens know that there are either 68 or 69 greens, depending on themselves. After the 68th second, each of the greens realizes they are green, too, because, if they were blue, and there were only 68 greens, total, all the others would have just disappeared. Therefore, during the 69th second, all 69 of them disappear. Then, all the blues see this happen, realize they are blue (because they can’t be green), and disappear the next second. All the blues would’ve known, all along, that they’d have to wait until the 69th second to see what happened, and find out what color they were, because they would’ve already known that there were 69 greens, at least. If no one had disappeared, during the 69th second, each of the blues would have disappeared the next second, regardless, thinking they were green, but that would never happen, though, because all the greens would’ve disappeared in the 69th second, and all the blues would’ve known that they were blue, just before they disappeared, one second later. As a matter of fact, every one of them would’ve already known, immediately (after Spock ran his mouth, and started the clock), which second they would be disappearing, assuming that they all knew exactly how many greens there were, other than themselves, and that they all knew that they all knew that. Every green would know that they would be disappearing during the 69th second, and every blue would know that they’d be gone in the 70th second. None of them would yet know what color they were, themselves, though, until right before they disappeared. The fact that nobody disappeared, during the 68th second, would let each of the greens know that there were 69 greens (rather than 68), meaning that they were green, themselves. The fact that the 69 greens disappeared, in the 69th second, would tell each of the others that there were only 69 greens (rather than 70), and, therefore, they were blue, themselves. I hope this helps. It definitely isn’t the easiest thing to understand. I had to think about it for a good while, myself, before I finally figured it out, but it makes sense, once you get it. Good luck. Godspeed. Live long and prosper 🖖🏻 and may the force be with you, always 👍🏿

@@pace1195 correct on the phrase part… makes me wonder how they accept information. If “there’s at least one green synth” starts a countdown, it could be for a few reasons. The first one could be... At least one, but not one, since no green synth shutdown at the 2nd tick. Not even the synth Spock spoke to. The second one might be …at least x=total greens accounted by a green synths, x=total greens accounted by a blue synth. This difference will await an answer until all greens visible by a green don’t disappear. There are more, but given that information I wonder if it’s enough to configure a phrase to stop the countdown

@@chriswebster24 I see what you are saying. All that matters are the initial conditions. 1. How many green synths there are. 2. How many cycles (seconds) until all the green synths disappear. 3. The starting time thanks to the statement by Spock. How could we stop everyone self destructing? Chaos. This is similar to the 3-body problem. In a 3-body problem, we can mathematically track a system, but we can't tell where it will go based on initial conditions due to mathematical chaos. Therefore, what is the one variable we might be able to change here? It must be time. If everyone saw a blue synth take out a green synth before the time was up, that number could be easily counted and tracked up to the appropriate ending time. Initially there were 69 green synths. Therefore we wait 69 cycles. Even if there are 68 + 1 premature green synth gone, the group can still count up to 69 cycles. So, we have to change the time. I don't know how synths tell time. Whatever method it is, turn it off for a random amount of time. Unless every synth is also a perfect time keeper, their personal time counts will deviate enough to cause chaos within the system enough such that the initial conditions beginning from the statement are no longer trackable. No initial conditions; no way to determine exactly how many green or blue synths there really are in society.

oh behave!

If there are x green and y blue synths, then the blue synths see x green and y-1 blue. If the simulation lasts x+1 ticks, then they would know they are green. but it doesn't, the green synths end themselves on tick x, thus informing all the blue synths that they are blue.

@@CodeKujo But all the green synths would realize their own colour all at the same time, since they are equal in all their properties. But since they do not know their colour, they are also equal to the blue ones in that regard. In other words, if ANY synth would recognize himself as green, ALL synths should recognize themselves as green, regardless of their actual colour.

@@Alexander_Kale The green synths are NOT equal to the blue ones, because green synths see a different number of green and blue synths than the blue synths do. Consider again the case of 2 green and 8 blue, on tick 2. From the perspective of a green synth (named Gary), it sees one green (named Greg) and 8 blue synths. it can deduce that Greg can see at least one other green synth because Greg didn't end itself last tick. But since Gary can't see any green synths besides Greg, Gary can conclude that it must be the green synth that Greg can see. From the perspective of a blue synth, it sees two green synths and seven blue synths. Only one tick has elapsed, and it knows that the two green synths can see each other, and so won't have ended themselves yet. Another way to think of this whole problem is that the statement "there is at least one green synth" is not new information, but rather a synchronization point that didn't exist before. From that point on, if you see *n* green synths and they haven't ended themselves on tick *n*, then you are green and you end yourself on tick *n+1*. If they have ended themselves on tick *n*, then you are blue... and you end yourself on tick *n+1*.

@@CodeKujo My understanding is that the number of green synths is effectively irrelevant, because they will all shut down at the same time, not one after the other. I assume this, because, all of them being identical, there would not be a mechanism to determine a sequence. So either they all shut down at the same time or not at all. But if that happens regardless of the number of greens, then the number of greens a blue one can see should be irrelevant and all the blues should come to the same conclusion, because of one property they share with the greens: they don't know therir colour. If there are 3 green synths, then green gary can see two other green synths and come to the conclusion as laid out in the video. All three of them will shut down at the same time. But if there are only two green synths, then blue stue will see exactly the same thing and necessarily come to the same conclusion, that being that he must be green despite this not being the case. And if stue can do this, then all the other blues should do the same. Am I being stupid here?.

​@@Alexander_Kale *"My understanding is that the number of green synths is effectively irrelevant, because they will all shut down at the same time, not one after the other"* Yes and no. They will all shut down at the same time, but when they shut down is based on how many green there are. *"But if that happens regardless of the number of greens, then the number of greens a blue one can see should be irrelevant and all the blues should come to the same conclusion,"* Again, you are disregarding the time. tick count is another piece of knowledge the synths have. The reason the blue synths don't come to the conclusion that they are green is because the green synths shut themselves down the tick before the blue synths would have come to that conclusion. The green synths shutting themselves down then informs the blue synths that they are blue. *"If there are 3 green synths, then green gary can see two other green synths and come to the conclusion as laid out in the video. All three of them will shut down at the same time."* Not just at the same time, but on tick 3. That is crucial. *"But if there are only two green synths, then blue stue will see exactly the same thing and necessarily come to the same conclusion,"* In this scenario, the green synths would have shut down on tick 2, so on tick 3, the blue synths have new information and know they are blue.

But before the visitor none of the entities knew the start time of their own birth so they didn't have synchronized clock cycles. The knowledge only propagates if all individuals know that all individuals know what each individual's clock cycle is.

Consider the case of 2 green and 2 blue synths. On the first tick, each green synth sees 1 green and 2 blue, so assumes it is neither. Each green synth also expects that the other green synth will see two blue and 1 neither, conclude that it is green and shut down. On the second tick, the two green synths will still see each other, and realize that the second statement must be a lie as the other green synth did not shut down. They will then thus conclude they must be green, and shut down. This extends to any number of green synths. As for the blue synths, once all the greens shut down the blues will all conclude they are neither. It's not clear to me if that's enough to trigger a shutdown.

​@@CodeKujo How can they knkw the 2nd statement is a lie and not the first one instead (or both) ?

@@quentind1924 because they can see green synths. They already knew there was at least one green synth, which is why at first you would think Spock did not change the situation.

@@CodeKujo Once they see that at least one of the statements is a lie, they can no longer consider that the green one is true. And the fact that they see all see green synths doesn’t change anything just like they couldn’t know their color before we came up and say it

@@CodeKujo Once all of the green synths have shut down, as you said, the blue synths will conclude that they are the "other" colour. They will then expect all of the other synths to shut down, since all of the blue synths should see one "other" and the rest will be blue. Since that doesn't happen, they will ALSO conclude that the second statement is a lie, and that they are blue. And they will shut down. Congratulations. You delayed the apocalypse by 2 seconds.

I have a logical theory.

@@taotoo2Yeah, me too. Actually being qualified for a position doesn't mean much anymore as "other factors" are taken into consideration. Plus there are agendas and quotas to adhere to.

Because come on Kirk had enough screen time already.

an alternative timeline. Such as the more recent Star Trek movies "Into Darkness", "Beyond"...

They don't (initially) know that _others know that others know_.

@@vekyll why would you say that? everyone sees 2 green and know immediately that everyone sees at least 1 green.

You need to go one level deeper: Do all know that all know that all know? See the example with: A knows that B knows that C knows (three levels deep).

please be more concrete. in your example of 3 greens. everyone knows from the beginning that there is at least 1 green. and everyone knows that everyone else knows (if 1 see 2 green, they must see at least 1 green). so how can the information "at least 1 is green" possibly add anything?

@@p0gr you go two levels deep, but we need to go all three levels deep. "Everyone knows that everyone knows" is two levels deep. But "Everyone knows that everyone knows that everyone knows" is three levels deep. And in the third level you cant be sure that everyone knows anymore, i.e. A can't be sure that B knows that C knows that there is at least one green synth.

Simply state, there is one additional synth which can not be seen by any of the other synths. Or state that a third, unspecified color exists, in which case every synth will then believe it is the 3rd color, but since that color is unspecified it will not know if it is yellow, or purple, or blue, or orange or gold, or brown, or white, or black, etc, and therefore be unable to shut down.

I think I see a significant flaw in this whole problem. At the four minute mark, when you begin discussing n=3, and one of the three green ones concludes that it too must be green - surely all the blue ones would come to the same conclusion, except in their estimation, N would equal four? They would all conclude that they are green, regardless what their actual colour was, and all shut down, all at the same time.

@@Alexander_Kale They would, but one second later! Because they see three other green synths. A three-green-synth structure will shut down after 3 seconds. On the other hand, each green synth sees two other green synths. A two-green-synth-structure is expected to shut down after 2 seconds. It does not, so the green synth concludes it is green.

@@TheLeekWeek yeah all blues would turn themselves off one second later whether they realize they are blue or incorrectly think they are green, it's sort of a moot point which they think they are except that once they observe all the greens turn off they must realize they are blue.

@@qsquared8833 well it is a difference in reasoning. but where do you think there is a flaw now?

it is in the video description: xkcd.com/blue_eyes.html

The creator more likely fancy Uhura being bossy 😉 Keen observation.

thanks a lot! I am actually somewhat surprised about all the resistance to the use of AI in videos.

​@TheLeekWeek I think most of the pushback comes from channels just telling their bot the 'make videos on [insert topic here]' and dumping them on RUclips in mass with minimal if any human intervention. Yes, the visuals and voices (and maybe the script) were AI, but a person editing it. Well done!

@@matthewmiller6979 thanks - I have not even thought of this. It takes me a lot of work to create these videos. All the ideas and the script require a lot of thought. And even though AI generates the images, I usually create about 50 to 100 to get one that matches my expectations. And I still edit the selected images in photoshop. And then the whole video editing... Maybe I can find a way to make my vids look less AI-like. Thanks a lot for your explanations!