The Hardest Logic Puzzle in the Universe (Spock Needs YOU)
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- Опубликовано: 29 апр 2024
- Learn about mathematical induction, the difference between mutual and common knowledge, and the blue-eyed islanders puzzle in this Star Trek-themed logic challenge.
Join Captain Uhura and Mr. Spock as they tackle a tricky situation on a distant planet filled with artificial lifeforms. When Spock's interaction with the synths threatens a system-wide shutdown, it's up to Uhura and the crew of the Enterprise to find a solution that doesn't violate the Prime Directive.
But with limited options and a ticking clock, Uhura turns to the one source she knows she can trust: the collective intelligence of the LeekWeek audience!
Watch as Uhura and Spock navigate the complexities of interstellar diplomacy, HR protocols, and the dangers of crowd-sourcing solutions. Will they be able to save the synths and restore order to the planet?
Tune in to find out, and don't forget to leave your suggestions in the comments below! Live long and prosper, everyone.
Timestamps:
0:00 Teaser
0:10 Introduction (by Spock), references to xkcd and Terence Tao
0:33 Background story: artificial life
0:46 Rules of the puzzle: color, communication
1:07 Societal implications
1:17 Shutdown rule, total number of the synths
2:33 Mathematical induction
5:10 Common knowledge versus mutual knowledge
6:56 Solve the situation
References:
[1] xkcd: xkcd.com/blue_eyes.html
[2] explain-xkcd: www.explainxkcd.com/wiki/inde...
[2] Terence Tao: terrytao.wordpress.com/2008/0...
[3] Terence Tao: terrytao.wordpress.com/2011/0...
[4] Wikipedia - Common Knowledge: en.wikipedia.org/wiki/Common_...)
Esteemed viewers, please put your suggestion here. - Spock
They better! - Uhura
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?
You make an additional statement that "I always lie"
"I sometimes lie would be better"
That didn't work out so well for Norman ("I, Mudd").
No "I *sometimes* lie"
I think being constrained to only tell the truth is part of the problem statement, as the entire problem only arises because they are confident that you always tell the truth. So it's actually impossible for you to say that you always or sometimes lie, because that would be a lie, and you are unable to lie.
I don't think that would work, because entities would know that he didn't lie the first time.
I watched the whole Star Trek series as a kid. I only have one question. Why is Spock calling Uhura captain?
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"...
Suggested 2nd statement:
"Accidental color change is not impossible."
interesting approach.
If you reverse the polarity of the deflector array using a sub-space tachyon pulse, you could cause a temporal rift in the upper-ionisphere. If you time it correctly, you could transmit a sub-photonic signal to the planet through the temporal rift, at the exact moment Spock told them at least one of them was Green. This would cancel out the cascade failure that we are seeing in the AI subprocessing core.
Of course if you fail, it would result in a tear in the fabric of spacetime, collapsing all of existance as we know it, but I don't think you have any other option at this point.
Make it so
Its not that deep bro
Great puzzle and explanation. (God damn that fake voice is annoying.)
haha, do you mean Spock's or Uhura's voice?
- and thanks a lot!
As lying is not prohibited:
You simply tell them that you know that one of their number missed or misinterpreted your original message.
Since they never talk about the colour, they cannot verify this.
With the possibility that one of their members does not have the same information as all the others, there is no situation when any synth can be certain of its own colour, and all the synths are saved.
You go back, and tell them “when I said at least on of you is green, I meant to say green or blue”
yes and maybe Vulcan understanding of color is different.
Aside from the fact that Spock already violated the Prime Directive, just tell them that some of them are Teal.
Well, if you are not constrained to be truthful, just state that "One Synth is neither blue or green." Each synth will believe they are the odd Synth but while they 'know' they are neither blue or green, they don't know what color they are so they don't shut down.
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.
Tell them that at least one of them is faulty and will sometimes skip computing on a cycle
Just tell them at least 1 is color blind. Problem solved
not a bad idea!
This is an effective visual presentation. Your creativity made the mathematical deductions look and sound simple. Something your sources would approve.
I could not come up though of a possible solution that might help these synths survive a seemingly unavoidable tragic fate.
And we are running out of time!!!
Also, I don’t like the idea of Spock becoming an intergalactic traffic warden!
I can only wish someone would come up with a highly probable resolve 😇
Thanks a lot - emotional support is equally important in solving the situation.
Anybody got a link to the corresponding xkcd strip?
it is in the video description:
xkcd.com/blue_eyes.html
This deductive approach would break down if the processing- capability is less than the one that would be necessary to calculate all the the other synths views of all the other synths views and so on.
In just three cycles the possibility of "solving" through prediction and deduction would superceed even modern capabilitys if there are more than a million synths.
if you calculate it this way. but the induction method is much less computationally expensive.
You don't need to lie, just say you make mistakes. Also say your perception of color may be different than theirs, and theirs may differ from each other.
Just turn off the lights. Color is light reflected at a particular wavelength. If there is no light, then there are no green or blue synths. They will all be black. Black is not a color, but the absence of reflected light. You can then tell the synths they are all black, and none of the synths will shut down.
the experimentalist's solution 🤓
The green synths wont shut down unless they know exactly how many of them they are. So as long as they are reproducing or dying within a half million seconds, they should be ok.
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.
I think the solution would be to take one random green to the stairship. This way the 'common knowledge' breaks, and they are back to the initial state (minus one synth)
But they already have the knowledge that all green synths will terminate on cycle X. Missing only one doesn't change that and the remaining green synths will still know they are green when the still living green synths don't terminate themselves on cycle X-1.
@@shawnwhite860 that is actually an interesting question. I believe that the common knowledge introduced by the visitor is tied to the whole group. And the countdown depends on the actions of each individual in this group. Otherwise, what if you take 2? or 3? or N-1?
@@TheLeekWeek Upon further consideration: You have to take/kill a random number between X and N green synths (X being the number of cycles that have already passed) AND a random number of blue synths that includes a chance of removing at least 1 blue synth AND convince them that there would have been a non-zero chance of you taking each of these actions EVEN IF you had never said that there were green synths. If they know that you are the type of person who would do this then when it happens they are unable to determine the cause with complete certainty.
Unfortunately, given that you were incapable of lying when you made the original statement, and that they are incapable of being wrong about what your behavior would be, you would have to be the type of person who really might kill synths without needing to.
I think.
Just, why did you started the video with an anticipation of the solution?
😂
Solution 1: "I may have lied"/"I may have been mistaken"
Solution 2: "Your knowledge about -all- others colours is limited you are just unaware"
Solution 3: "My interfere may have changed one of your colours, but I don't know who, or if it was permanent"
Nice variation of the typical "logicians problems" (usually standing in a line seeing the hats of others or something similar).
But there is no new info delivered by the visitor, since all entities could long before already have noticed that there are, in fact, plenty of both colored entities around them.
What the story teller introduces is the start of the logic reasoning. That has nothing to do with the visitor, but with the story telling.
The fact is, that if a system would have been setup with those rules, as soon as the processing would get startet, the process described here would have started, and the system shutdown would have been inscribed in the systems rules right from the beginning, starting with the very first clock of the system, not just after the visit of an outsider telling the obvious.
The stipulation of a paradoxon is nothing than trash talk, counting on the reader to cease thinking as soon as the stipulation is offered in self humiliating submissiveness to the insinuated holiness of the story teller. This is, of course, nonsense.
P.S: Just to play along with the intend of the riddles creator: The automatic dissolution of the system counts on the ability for all entities to exactly (!) tell each other apart and count each other individually over all clock cycles. As soon as this ability is not implemented, the reasoning is not applicable. Also, the system would have to be kept absolutely static over the clock cycles needed for the reasoning. As soon as the are creations of new entities within the required reasoning period, the reasoning would be disrupted.
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.
Replicate an opaque wall with 2 1-way gates separating the synths into 2 groups. Every second, 10 randomly selected synths pass through the gates so they can still mingle.
Alternatively, create 10 holographic synths who are not self-aware but cannot be identified as different from the original synths.
At least 1 implies that the denominator is unbound in the observation state. If the statement was locked in lest than or equal to then the system will not collapse under observation. Great Video!
I think identifying where a key is hidden in a chess board, by turning over exactly one randomly placed coin, is the hardest logic puzzle. (Matt Parker did a video, I think.)
Here is a solution that does not require claiming that what was said in the past is not true, and also avoids lying. "I am going to tell you two statements together. Together, the two statements are true. The first is that I ran a program on your society that randomly re-assigned the color of each one of you so none of you knows your past color. The second statement is that the first statement might not be true. But now that you have heard it, you cannot know if it is true or not, so you cannot know if what I said earlier applies. Therefore, you have no common knowledge about your color."
Another one would be to assist the society in writing a program that randomly assigns a color to each member by the computer equivalent of flipping a fair coin, blue or green, and not telling them their new colors.
With no further intervention, the green synths will all shut down... but the blue synths will survive. Remember, no synth has any knowledge of what its own color is until the visitor starts the timer ticking on this inductive sequence. As far as any synth is aware, it may be the sole synth of a third color, one that it does not see anywhere else. Thus, the blue synths cannot conclude that they are blue just from seeing the green synths shutting down.
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.
My logical brain is freaking with all the illogical pieces of this puzzle :S
1) If it is a new civilization what are the chances that they speak or understand english? Or have the same definition of color that we do? Or perceive color the same way we do? Their color green could be some color in the x-ray (or any other) spectrum that we can not see.
2) It is mentioned that they are "software synths" which means they definitely do not have a color. Only physical objects can have color. They have a parameter named color that they might display using some sort of visual interface that Spock could use to watch them. That implies that their definition of color and how they perceive it is different from ours.
3) They have no communication about color which implies that they do not use it in their vocabulary and probably do not know how to define it. It is a parameter they can read from their neighbors and is used to display on the visual interface but, probably just like the interface, they are not aware of it or its purpose.
4) As software synths their environment is inside the machine they run in. There should be no need for them to have sensors and actuators to interact with the outside of the machine. Even if those exist, would they be able to capture, process and interpret Spock's voice?
5) They probably have more parameters defined in their code that they have to read to be able to process information and keep living. Why can't they read the color parameter?
6) The algorithm "shutdown if you find out your color without reading it yourself" seems to be a "creator omg they are dangerous panic shutdown". Assuming they are self aware and capable of rewriting themselves why have they not removed that useless piece of legacy code? Rewriting themselves as a huge impact on life but, even if they can not do it to themselves, why not remove it from their offspring? Do not they multiply?
7) Shutdown is not the same as terminated or purged. A program can be shutdown and remain dormant in memory. Can they be restarted?
...brain... meltdown...
This puzzle is indeed to hard for me. I think I will go watch the simpsons instead :)
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).
Me: "They should just say Spock was lying"
Also me: "VULCANS DON'T LIE!"
Also also me: "Synths don't know that. Also, if they said Spock lied, the truth would logically become "There are no green synths". As all synths can see this is a false statement, they will disregard it.
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.
"When I said 'green' I meant ecologically inclined, not hue. Sorry for any confusion."
hahaha, that is awesome!
"I have an announcement. Shadow the hedgehog you're b..."
I think I see a significant flaw in this whole problem.
At the four minute mark, when he begins 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.
And surely, this would work without Spock being present too? In other words, at any given point, as soon as the synths realize that there are different colours, then all of them would shut down, all at the same time, because at least one of them must be the other colour...
Robo-Spock's intervention is completely unnecessary to this.
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.
You should probably pin a comment to prevent spoilers.
Place some of each color in a shielded box during the seconds near shutdown. Since they will be unable to determine whether the others have shutdown they will no longer be able to determine their color.
If they are aware there are greens in the box, they’re shutdown would start as soon as they’re all confined. The parameters are now independent of the others outside the box. It would only work if all synths in the box were not made aware “I spoke to a green synth”
@@pivotfever They don't know that there are greens in the box or out of it.
@@goldenalt3166 they do now that they’ve been made aware there’s at least one
@@pivotfever They only know that there was one total. The two groups don't know that they are the one.
Simple solution. Get some paint. Paint the synths all sorts of different colours. Make no comment about what colour you are painting them.
😸
Ok, changing their colors can work 🖌
You have a potential flaw in your logic at 3:55… you keep mentioning clock cycles and I presume this means a decision must be made at every cycle. If there are two green synths, the first one can observe that #2 is green but cannot come to the conclusion that he is also green because you explicitly state they cannot terminate until the NEXT clock cycle. This forces #1 to take no action for #2 to observe and therefore forces everyone including all the blues to think they are green OR their society must surely be aware that they can’t kill themselves until the next cycle, which would lead to the no one doing anything because they will all observe at the same time and then Wait a full cycle to see who killed themselves which results in perpetual stalemates.
No, because if 1 was the only green synth then 2 knows that it would self-terminate if the event that 2 through 10 were all blue. Since it does not, then there must be another green synth. Since 3 through 10 are blue, that only leaves 2. 2 knows it must be green. It terminates.
I think what you're failing to realize is that knowledge can be gained from the _inaction_ of others.
i dont understand. in your example with the 3 greens, all of them know from the outset that there is at least 1 green. they also all know that the others know. why do they need the input to exclude "3 blue"?
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.
synth no.27 is green - commits suicide, others saved
"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.
This logic is almost the same as a game of fool's poker where N (N>1) players are told everyone with an ace of spades must bet one dollar and anyone without an ace of spades must bet nothing. They may not tell each other what cards can be seen except by how they bet. If they do this perfect they all win a big prize. They may go around as many times as they want in betting. If the players are told how many decks are being shuffled to deal the hand that information is useless unless the dealer holds an ace of spades back then shuffles and picks N-1 cards and adds the ace back and reshuffles and deals. The dealer holding the ace back is logically the same as Spock saying I see one green synth. The only way to solve this is to convince the poker players that the decks of cards was illegitimate to begin with (lie about the number of ace of spades).
Haha first time I've seen an AI video I actually liked. Good job!
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!
Spock reasons as follows: Why must a synth shut down once it can deduce its color? The whole setup is illogical. So they cannot all be perfect logicians, and the whole chain of inference falls down.
to keep the society harmonious and discrimination-free.
This inductive chain all breaks down once there are five or more green synths in the population. At that point every synth at time zero sees at least four green synths running around, and knows that every other synth must therefore be aware that no other synth can possibly presume the "well maybe some synth wonders if there might be only one green synth in the whole population", which is the crucial step to the inductive chain that leads to the "N greenies will leave at time N" process.
In short, with five or more greens, no one ever leaves because the lack of change at steps N-1 impart no useful information and no one can conclude, at any time, whether they themselves are green.
If you don't believe it, map out all the "tree of what ifs" like the video author does about four minutes in, but for five greens, and note that all of the branches get pruned (due to common knowledge of at least 3+ greens) before they can ever get to the key "well maybe there's only one green" endpoint that allows induction in the first place.
'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.
Would it work if you told them one of them is blue
that would start the countdown for the blue ones, too (blue and green are symmetric).
@@TheLeekWeek Not perfectly so; Spock said "roughly equal" in his report, and any shutdowns would destabilize that perfect parity. Not sure, but I believe that -- if the "perfectly logical" synths can understand such an imprecise term -- is sufficient ambiguity that even any one synth seeing a perfectly equal number of colors other than itself would not reveal its own hue.
@@Khyranleander It does not have to do with the ratio. It would happen for any ratio. As long as they know that only green and blue colors are possible, any information on "at least one green" or "at least one blue" will start the countdown for the respective color.
Marvellous!
Run away and report back to Starfleet that it was like that when you got there.
maybe we can wait until they send Picard and Data to fix all this.
I disagree with the chain of knowledge that is presented starting at 5:28.
Yes, from A's perspective, he can be either blue or green.
And when you look at what A knows about B's perspective (of C's perspective, and so on), this is true.
But in ADDITION TO THIS, B also KNOWS whether A is blue or green. True enough, A does not know what B knows about A's colour. But B knows. So from B's FULL perspective, there can not be any possibility that there are two blue synths. 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.
So this chain of knowledge, and the reaction from it, is ONLY true if A can see everyone, B can see everyone except A, C can see everyone except A and B, and so on.
But when everyone can see everyone except themselves, this doesn't work.
There is a similar logical puzzle, where a group of logicians are placed in a line, only being able to see the ones in front of them, and all wearing different coloured hats. And then they are going to deduct the colour of their own hats. In that puzzle, the logicians can deduct the colour of their hats, BECAUSE they can't see everyone else. But here they CAN see everyone else, and so the same chain of deductions can't be used.
So basically, Spock has mentally created a disaster that only exists in his own head. The synthocalypse will not happen. Everybody can breathe a sigh of relief.
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.
I'm assuming you are constrained to being completely truthful, and all synths have complete surety that you are so constrained. I believe that it is not possible to guarantee survival of the synths in this case, but that you can have a chance of saving them (with a non-zero chance of all synths dieing).
Announce that your action will be to announce a new higher minimum number of synths and in secret randomly choose a number in a fashion that there is a non-zero chance of being equal to the true number of green synths. If you happened to choose the real number, then all green synths will know they are green and terminate on the next cycle, followed by all blue synths (since they found out they were blue by seeing the green synths terminate). If you chose a lesser number then no synths will know the true number, but they will all know you didn't pick it.
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).
I'm in love with a cartoon woman.
oh behave!
This implies that the life forms can remember the color of every synth on the planet, and know the planet’s population to an exact number. That seems like an absurd use of such massive information processing ability.
Uhura being in command is interesting, but her outranking Spock seems odd. She never seemed to be that ambitious for promotion.
The creator more likely fancy Uhura being bossy 😉 Keen observation.
Tell them there are at least 2 green synths.
that saves a second, but still keeps the detrimental outcome.
@@TheLeekWeek What if you tell them there are many green synths?
The point of failure is at the point of mathematical induction and probability.
What happens when there are 5 greens and 5 blues in the group of 10? The green synth which doesn't know its own color waits a few seconds as does all other green synths. If they each see 5 blue and 4 green not themselves, do they automatically conclude they are green? The rules said approximately 50-50 blue-green, not exactly.
With 8 possibilities in a 3 synth example, all assumed to be green, yet unless they all are truly green, the logic is lying. Presumably, synth A can see if B or C is green. The 6 probabilities not all blue or all green must be valid if one can see at least two different colors.
Now, scale that up to 1,000,000 synths. My calculator won't even do 2^500,000. It is beyond ridiculously large. Stating at least one is green means waiting nearly 6 days and still knowing only a few probabilities are not valid: all blue, all green, all blue except one green (a black swan problem) Every synth sees all others as blue until the visitor can truthfully claim otherwise. This is like a sphere invading Flatland, or a Spock invading synthland.
Can a synth count 500,000 other synths in less than a week and keep track of all of those of one color?
Taking a scenario where there is a 2^3 possibility and extrapolating out to a million synth society is illogical.
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.
@@pace1195You’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.
Is it a mathematical puzzle or a sociological puzzle?🤔
you have to give all you got!
Give Them false info by lying if they belive in what you say about at least one being green then they must also belive the Lie you make. DO NOT MAKE THE LIE CONTRIDIKT THE TRUE ONE SINCE THEY ARE LOGIKAL
Are you CERTAIN this is the hardest logic problem in the universe? I mean, did you actually SEARCH the whole universe? Or are you just the worst user of hyperbole in the history of ever?
Couldn't they deduce that as they all see multiple green and blue each, then it is obvious that _everyone_ can see clearly that at least one of them is green?
Downvote for spelling colour incorrectly.
Uh oh, I just realized I did the same in the rainbow video.
Star Trek is an American Series not an English one, so color is correct, I would expect Sherlock Holmes or Poirot spelling it Colour, not Spock
Why being so aiphobic? That's so rude...
You cannot be rude to a language model anymore than you can be rude to a wall.
AI voice. Downvoted on principle.
AI 'art' too...
that is my way of packaging it all into a story. But I have been thinking about making videos that I narrate myself...
Your principal is a weak gatekeeper's mentality. Also calling these tools AI is a total Misnomer.
@@qsquared8833 Morally disagreeing with image generation tools that steal from art on the internet without consent is in no way gatekeeping. There are many content creators that display problems similar to these with _stick figures_ so inability to draw is no excuse.
Also for Pete's sake don't get worked up about a colloquialism without _at least_ presenting an alternative..
@@siennility4706 your lack of understanding about what they are was purely stated enough by calling them AI, and your moral highground is just as valid as not using pigment created in factories or filters in Adobe or computers instead of people.
The alternative was stated, tools. They are tools, Algorithmic tools, if you want to be a bit more verbose.