"The famously difficult green-eyed logic puzzle" - I disagree with the proposed solution

In My Humble Opinion
61

Right, but why doesn’t my “two person objective perspective” act as an outside 3rd party telling everyone that “there is at least one Green eyed person” when my method (I think?) does exactly that for larger groups?

We are always assuming that these perfect logicians come to the same conclusions as any other person in the prison/group at least in time to act by the end of the same day, so they can always either leave on time, count up the relevant people on time, or save their lives in time.

For two people, they can be GG, BG, BB.

In GG, each person sees 1 G and they leave on day 2.

In GB, B sees G and leaves of day 2. G sees B(0) and would normally leave on day 1. Saving B from dieing by the guards as he sees G no longer there the next day, and can then know that he is a B.

In BB, both see B and both would normally leave on day 1 to die in the process.

Leaving on day 1 is too risky to try unless you both knew that both you and the other (or everyone in an arbitrarily large group) knew that at least 1 Green is in the group. And that everyone was on board with using the arithmetic of (#G’s seen +1=days waiting) starting on the same day.

I agree with you that they would sit there forever, and they would be right to. I agree that the 3rd person perspective of a truth teller speaking to everyone as a unified group, and telling them that “there is at least one Green eyed person” will work for all sizes of groups from 1-infinity no matter its make up of Green eyes.

That solution is true in all cases. My case is much more restricted, but only thanks to just how many in-group Greens there are vs the stated problem’s (and XKCD’s) out-group B’s, am I allowed to have my ugly crippled way of thinking work. I agree that my way does not work for groups smaller than 3, and needs at least 3 G’s for it to work.

I am assuming that these perfect logicians will reach every logical point on the same day at the same time as everyone else in the prison. And that they think perfectly fast such that they will reach my conclusion while still on the first day of incarceration. Otherwise, they will never have another truly guaranteed unified time/event of understanding and action until the 3rd party speaker tells everyone over the loudspeaker.

I also assume that these logicians are more like engineers, in that upon seeing a quick and dirty solution that is “eeh, close enough” rather than a thing of eternal truth and beauty like the already stated solution; they will pounce upon it to shorten their sentence. Rather than wait an indeterminate amount of time for the off chance that a Dictator will change his mind and allow someone to speak to them. (another question, are “perfect logicians” people who only accept the eternally true answer and forgo the imperfect technique that would save time/lives? Or are they also perfectly practical and willing to take some things on faith and trust their senses in order to save time/lives?)

That said, as we saw in the 2 person group above, the only things that actually kill a group are: 1. having no actual members of the G in-group while counting down, and 2. people in the group not being synchronized with the others while counting down. Thus, my logicians will use the first day of incarceration as a unified starting point. As they all realize this and are equally prone to action; just as the logicians in the true answer are all galvanized to action once they know everyone knows they know.

My “objective 2 point perspective” from my previous post allows everyone (to separately see through another’s subjective sight as well as their own) in a group to “objectively see” that there is at least 1 G in the group for every 2-person frame of reference. That sounds good enough for consensus right? I do not know my eye color, the guy across from me does not know his eye color, but we both can agree that the neighbor to my right is Green eyed. And every individual goes all around the circle of the group like this in their own head. And they only start their counting when they know that everyone can see at least 1 G. That universal knowledge of G is necessary to save the group from dieing and to break the natural deadlock of uncertainly that fewer G’s or super small >3 groups can’t get out of without outside help.

Thus I equate the 3rd party person’s public announcement, to a bunch of logicians coming to a similar “3rd party objectivity” concurrently in their own minds on the first day. Just as the real answer had everyone think at the same rate and start their logic chain on the same day after the announcement.

62

I think that long before this soothsaying stranger showed up I’d have examined my situation, weighed the idea of a lifetime stuck on this damned island vs the likelihood that I’m the only one without green eyes and, if the odds were greatly in my favor, taken my chances.

So an interesting question is, if you were to find yourself in this situation, how do would you calculate the odds that you were in fact different from everyone else? At first thought I’d say there’s a 1% chance, but that can’t really be justified can it?

Of course what I’d really do is look at my reflection in a cup of water or something, but that’s fighting the hypothetical so let’s ignore that.

63

The example I gave earlier was, “I’m sorry, I mistook you for someone else.”

If the color of eyes is the only consequential information on the island, and someone says this to me, I can take it as an indication that I have the same eye color as everyone else. Moreover, everyone can take it as an indication that everyone has the same eye color, from the vantage of the speaker. With the importance of eye color being so great in the prison, we can assume that this statement gives all of the information that the teller needs to impart. He would not say it if it was ambiguous. He would only say it if I could be confused with every other person he has seen.

The guy he tells this to then replies, “No problem, brother.”

And everyone leaves that night.

Now you might complain that I’m changing the scenario. The scenario was that one guy speaks and he says, “One of you has green eyes.”

But if he was a perfect logician, he wouldn’t say that. The phrase that was said directly contradicts the information we have been given. Certainly, that phrase can lead to everyone escaping, but it takes n-days to accomplish. Why would a perfect logician choose a method that has a linear action time, instead of an instant one?

Or let’s say that he does tell them, “One of you has green eyes.”

Well so they all know that the number of people with green eyes is either 99 or 100. Does it make sense to wait 100 days to differentiate between two possibilities? No, Logically, you would say that if you see someone with non-green eyes, then you should leave today. If everyone you see has green eyes, then you should leave tomorrow.

So, the next morning, no one has left. The day after, everyone is gone.

We’re done in two days instead of a hundred.

If everyone thinks exactly the same and is seeking the most efficient solution, then they’re going to select a solution that reduces down to the minimum number of states necessary to make a determination.

But, like I said, if they’re seeking the most efficient solution, making a statement that reduces the number of states down to only one possible option is the most efficient, and that’s certainly possible while still being vague.

64

This is the key to me. They need a trigger point. Why couldn’t the trigger point be the first day it rained instead? It contains exactly as much new information about eye color as in the puzzle.

65

But it doesn’t. Again, go back to the example of three people. With three people, everyone knows that at least one person has green eyes. And everyone knows that everyone knows that at least one person has green eyes. So, on the surface, no new information is being conveyed.

But do you agree that the solution works for three people?

The new information is not about eye color, it’s about who knows about eye color. It’s about the publicness of that information. As mentioned in other posts, once that statement has been made, then any string of “Amy knows that Bob knows that Charles knows that Dave knows…” must always end up with “…that there is at least one person being with green eyes”, which was not previously true.
When I first heard this puzzle, the condition being identified was not eye color, but being-cheated-on-by-a-spouse. Something that is not neutral, but rather is clearly bad, and clearly something that everyone will refuse to believe could be true about himself unless proven otherwise. Imagining everyone having that mindset makes the puzzle seem more natural, I think.

So with the two person case, I’m sure that I’m not being cheated on. I know the other guy is being cheated on. Someone says out loud “at least one person is being cheated on”. I think to myself “well, that poor bastard, he finally knows the truth… tomorrow morning I will see that he has committed suicide in despair, as one does when one learns that one is being cheated on”. Then the next morning, he has not committed suicide. How is that possible? That’s only possible if being told that at least one person was being cheated on did NOT tell him that he was being cheated on. But that means that someone else is being cheated on. But… that must be ME. OH NO!!!", and so forth.

66

But that’s not the puzzle. The puzzle is not “you are visiting this prison island, what can you say to get everyone to escape”. If so, the answer is clearly “hey, here is a list of the names of the 100 people with green eyes…”. Rather, the puzzle states that the interloper said “at least one person has green eyes”. That’s what was said. That’s a given. So what happens next?
I think there are two additional qualifications which make the puzzle feel more correct:
(1) As mentioned in my last post, the quality being identified is a negative one, one which everyone will start out being certain could not be true of themself
(2) Only a small subset of the population has this characteristic (say, only 100 of 1000 total prisoners)

Neither one is necessary for the logic to work, but they both cut down on the temptation to go for lateral-thinking type solutions (ie, just guessing that if all 99 other people have green eyes, I probably do too…)

67

You can’t use the first day it rains, because that doesn’t convey that everyone knows that everyone knows that everyone knows that everyone knows that everyone knows that everyone knows that everyone knows that everyone knows that everyone knows that everyone knows that everyone knows that everyone knows that everyone knows that everyone knows that everyone knows that everyone knows that everyone knows that everyone knows that everyone knows that everyone knows that everyone knows that everyone knows that everyone knows that everyone knows that everyone knows that everyone knows that everyone knows that everyone knows that everyone knows that everyone knows that everyone knows that everyone knows that everyone knows that everyone knows that everyone knows that everyone knows that everyone knows that everyone knows that everyone knows that everyone knows that everyone knows that everyone knows that everyone knows that everyone knows that everyone knows that everyone knows that everyone knows that everyone knows that everyone knows that everyone knows that everyone knows that everyone knows that everyone knows that everyone knows that everyone knows that everyone knows that everyone knows that everyone knows that everyone knows that everyone knows that everyone knows that everyone knows that everyone knows that everyone knows that everyone knows that everyone knows that everyone knows that everyone knows that everyone knows that everyone knows that everyone knows that everyone knows that everyone knows that everyone knows that everyone knows that everyone knows that everyone knows that everyone knows that everyone knows that everyone knows that everyone knows that everyone knows that everyone knows that everyone knows that everyone knows that everyone knows that everyone knows that everyone knows that everyone knows that everyone knows that everyone knows that everyone knows that everyone knows that everyone knows that everyone knows that everyone knows that everyone knows that everyone knows that everyone knows that everyone knows that there is a green-eyed person on the island.

Only saying aloud, to everyone at once, “There’s a green-eyed person” does that.

Do you understand that in a prison of four greens (Alice, Bob, Charlie, Dave) and 100 browns, that Alice considers it a possibility that Bob thinks it’s *possible *that Charlie thinks perhaps Dave might think there are no green eyed people at all?

Do you understand that Dave thinks it possible that only 3 of the 104 people have green eyes?

Do you understand that Charlie, when asked about Dave’s thoughts, thinks Dave might assess there’s only 2, because Charlie isn’t counting himself or Dave?

Do you understand that Bob realizes all of this, knows that Charlie could think that Dave thinks a minimum of 2, but without knowing his own color, thinks it possible that Charlie actually assessed 1 as the lowest number?

Do you understand that Alice, repeating this logic, thinks perhaps Bob said that Charlie’s lowest guess of what Dave said was 0?

68

By the way, this puzzle was discussed, along with a variety of other puzzles, in a thread from September 2015.

Here’s one of my posts on the topic:

69

Of course it does. No one is disputing this, or claiming that this is ever a realistic model of the way groups of people would ever possibly act in real life. Heck, the puzzle would probably be better off being told about super-intelligent AIs or something, such that all of our instincts about human behavior could be set aside.

But again, I feel that you’re fighting the hypothetical. The puzzle isn’t uninteresting or meaningless just because it operates in the land of purely make-believe.

70

This was a great help in understanding (though for my brain it just helped to write out the ranges of possibilities, ie:


Amy sees 4 people with green eyes.
Amy knows that Beth can see 3-4 people with green eyes.
Amy knows that Beth knows that Clarice can see 2-4 people with green eyes.
Amy knows that Beth knows that Clarice knows that Daphne can see 1-4 people with green eyes.
Amy knows . . . that Daphne knows that Erica can see 0-4 people with green eyes.

The statement ‘everyone can see someone with green eyes’ cannot coexist with the preceding train of logic.

I’m not quite making the next leap though. How do I turn that knowledge into knowing that I have green eyes?

71

Are there some rules that I’m not aware of, like “the green-eyed prisoners are all blind”? Why can’t they look in each others’ eyes and confirm that everyone has green eyes?

I know it sidesteps the whole logic problem, but to me it’s a perfectly logical solution.

72

Everyone on the island can see everyone else. No one on the island can speak to each other or convey information ever. There are no mirrors.

73

Asked and answered in the linked-to thread. Begin with my linked post and review the rejoinders and surrejoinders.

74

Well then, they leave in two days, as I explained. (Or faster, probably, as I doubt that I came up with the optimal solution.)

75

Sorry, I do not really agree that everyone has to take into account everyone else’s viewpoint simultaneously, or even progress down the list of people in the group adding those viewpoints in succession until “0 known” is reached.

I assume that everyone has working eyes and common sense. In your example of "four greens (Alice, Bob, Charlie, Dave) and 100 browns,” Alice sees 3 Greens. She knows that all 104 people other than Bob, Charlie, and Dave can see at least the same 3 Greens that she does. She also knows that those 3 Greens see, from their individual viewpoint, at least 2 Greens.

That information (that there is at least 1 Green to make the algorithm a safe action to take) is universally known to the group, and everyone knows that everyone knows that everyone knows that (as long as they are all equal logicians). Thus, it is enough information to start the sifting algorithm immediately. Because, the algorithm works on what is actually seen by the individual, Alice Bob Charlie Dave will wait 3+1=4 days and the other 100 will wait 4+1=5 days and be safe as they see the Greens have left a day before them.

Alice is unknown to herself (Brown), sees Bob G, Charlie G, Dave G

Alice knows that at worst Bob (unknown to his self) sees Charlie G, David G, and Alice Brown

The same applies to the other Greens, their viewpoints always giving at worst 2 G’s to see.

This has to work from every viewpoint, even from Alice’s when she is using Bob and Charlie as the pair. Thus Alice (assumed B) thinks Bob (assumed B) and Charlie (assumed B) can only see David (Objectively G). But this is still enough to succeed with. As when Alice does this from Bob and David’s viewpoints, Charlie becomes the objective G.

Thank you for making me think about this again, now I see that for my idea to work it needs at least 4 Greens in a group, not 3. My above posts about working in sub 4 groups were wrong.

No one ever has to know their own eye color, nor even trust that others know their own eye color. All they have to do is trust the 3rd party objectivity of the group’s sight and agree that there is someone Green outside of the group that will nevertheless, count as inside the group and know that himself when the time comes to synchronize their algorithm.

What is happening is that everyone is starting their own sub-group of two people. These two are in a natural deadlock, and so they look outside (into the greater 104-2 102 population) to see if they see a Green. They do, they see 4 (or at worst 2, or at worst worst when a G extends their view to take in a GG pair, 1). They know that these Greens are a part of their greater whole group, and as such are certain that they will include all 104 people in their algorithm when the time comes. This looking outside of the sub-group for a guarantee within the group as a whole that at least 1 Green exists, is like the loudspeaker guarantee that at least 1 G exists.

But knowing that in just that 2 person group means as much as if the 3rd party person just whispered to a 2 person group that a G exists. There needs to be a way for everyone in the full group to know that at least 1 G objectively exists, especially for the “outside” G they just used; if he has no other G to see, he will never activate his end of the algorithm. Thus, every person must pair up every possible pairing of two people in their own heads (eventually forcing 3 Greens to see themselves as Browns), and see through their worst case interpretation of events, in order to see if there is an “outside” G for every pair. Only then do the whispers become a loudspeaker. And only after everyone knows that everyone can see at least 1 G objectively, then can the algorithm begin on that day. (luckily our perfect logicians can do that)

76

And your explanation is wrong. Each one of them knows that the number of green eyes is either 100 or 99. But they don’t know that everyone else knows that. Alex knows that the number is 100 or 99. But from Alex’s perspective, Brian knows that the number is either 100-or-99 or 99-or-98. And what does Alex think that Brian thinks that Chris thinks? Alex thinks that Brian thinks that Chris thinks that the number is 100-or-99 or 99-or-98 or 98-or-97. And so forth.

If someone came out and told them all “every one of you knows that the number of people with green eyes is either 100 or 99”, then that’s tantamount to just telling them all they have green eyes, and they’ll all leave the next morning. But that is information that they don’t have.

Whatever your solution is, please explain it in detail for the case of either 3 or 4 people.

77

I don’t think that reducing the days waiting from X to 1 or 2 would work.

The purpose of the math is to allow any number of “saved” and “damned” individual peoples (who do not also know their own status) within the same group to sift themselves out safely, as long as 1 saved really does exist.

Take the 104 example: 4 Green, 100 Brown. The Brown individuals see 4 Green and wait 4+1=5 days. The Green see 3 other Greens and wait 3+1=4 days. Their leaving a day earlier warns the Browns that they are not Green and so they can live, rather than die to the guards.

If we instead say,
[QUOTE=Sage Rat]
logically, you would say that if you see someone with non-green eyes, then you should leave today. If everyone you see has green eyes, then you should leave tomorrow.
[/QUOTE]

Then everyone, who sees both 99-100 Browns and 3-4 Greens would all leave on the same first day, and 100 people would die. I agree that in the all Green eyed prisoner case, or even if 1 is Brown eyed, your solution would work. But what about if there are 2 Brown eye people and 98 green?

Now I do think you are on to something… as long as the ratios stay the same, then the algorithm holds true in all cases. Brown 5/2=2.5, Green 4/2=2, 2.5/2=1.25, 2/2=1. 1.25 round up =2 for Brown, 1 for Green.

97Green, 3 Brown… Brown see 97, Green see 96.
97/2=48.5/2=24.25/2=12.125/2=6.0625 Round up 7 day wait for Brown
96/2=48/2=24/2=12/2=6 day wait for Green

or even further 7/2=3.5/2=1.75 =2
6/2=3/2=1.5 =2…hmm no, maybe staying around 3 would be fine, but lets keep it at first single digit just to be safe.

If our rule becomes: whatever the number you count for the Greens, keep dividing it by 2 never toss remainders, always calculate with them until you get a single digit number. If there is a remainder, round up no matter what. And leave on that day. Then that would save everyone in much less time. Good idea Sage Rat. I wonder why this simple time saving solution was not thought up by our perfect logicians.

78

Ignore me, I am an idiot. If instead we had Brown sees 96 and Green sees 95, then 95 would reduce to 5.9375, which would have both Greens and Browns leave on the 6th day. Maybe if we add an additional addendum to whether or not the counted amount was even or odd, it might work…but, too much work to figure it all out. And too easy to screw up. I’ll just wait the extra days.

79

If no one on the island can speak to one another then no one can ask to leave.

80

Maybe, maybe not. Depends on the interpretation.

Note that the very fact that this is being debated so earnestly by people who are fairly competent at logic itself exposes the flaw in things. If it was really that clear, we wouldn’t be having this discussion!

(This problem, in the 50 cheating wives form, was brought up and discussed in my Computer Science research crowd many years ago. It was supposed to be an example of how not getting information was information. But, in addition to being sexist, it was a really bad example. Esp. when there were many simpler working examples known at the time. So, I’ve heard all the naive arguments already.)