Conventionally when it comes to these kinds of puzzles, only solutions arrived at by thinking inside the box are considered satisfactory.
Did you try writing it out like I advised? Write all the patterns that three thinks are possible. Then write all the patterns that two thinks three could be thinking. Then continue that through however many people you want.
Not for a perfect logician. From the original statement of the puzzle:
Covered by:
Very nice. A picture is worth a thousand words, eh?
Yes, but as you note they are perfect logicians working expeditiously on the information. They all should collectively assume things that can be assumed, and try and leave as quickly as possible. From where I stand, they should always leave on day two.
So it shouldn’t be hard for you to take the case of three blue-eyed islanders and show the line of reasoning that on day 2 makes each of them certain of their eye color.
I like diagram and know if I get it it’ll be because of that but after going one deep he is assuming that he himself brown eyed but i see at that point he must also assume he’s not and where the thoughts would branch out and stop showing nothing on one branch or continue as presented in the diagram, I see the missing branch as a possibility and that’s what ruins the certainty of the patch chosen by all who accept the answer.
Everyone sees people with two blue eyes. They all know that the number of blue eyes is either 2 or 3. Therefore, they collectively think that if I think there are people with two blue eyes I leave day 1, and if I think three, we leave day 2. Since my compatriots didn’t leave day 1, I now know I have blue eyes. If I had brown eyes, they would have left day one.
Basically, if I see two people with blue eyes, I know there is absolutely no chance anyone is leaving day one. Therefore, day one does not provide information, and can simply be assumed.
No–remember, thought balloons don’t mean “I think the world is like this,” but rather, “For all I know, the world could be like this.” He’s not assuming he’s brown-eyed. He’s assuming he might be brown eyed.
If I see two blue-eyed people, I know no one will leave on day one. But I don’t know that everyone knows this–because for all I know, the other guys just see one blue-eyed islander. (And if they did see just one blue-eyed islander, they would think it possible for someone to leave on day one.) So I have to wait a day, not to see if anyone leaves, but rather to make sure the other guys realize no one’s leaving on the first day.
Check out my blue-eye/brown-eye thought-balloon diagram. It’s awesome.
Sure you do. We are all perfect logicians and have all made the same assumptions, and can assume that our compatriots have as well.
Tell me this. I see X number of blue eyed people, where X is 2 or more. What do I learn after day 1 when nobody leaves that I didn’t already know?
You need to wait each day for the same reason the guru needs to speak: to remove the uncertainty from the earlier steps in the induction.
After day one, you know for sure that everyone knows there is more than one blue eyed person. If you see two, before the first night you could imagine each acting like there was only one. That is no longer the case. On the second night, you watch to see if they behave as if there were two blue eyed people. If they do not (if they stay) you’ve learned that everyone knows there are more than two blue eyed people. You’ve also learned you’re one of them.
I already knew that. If I see 3 blue eyed people, I am absolutely certain that everyone knows there is more than one blue eyed person.
No we can’t–because we don’t know what our compatriots are seeing. So, for example, I assume B has blue eyes. But I can’t assume B assumes that–because B doesn’t know what colr B’s eyes are. That’s a trivial case, but there are more complex cases of non-shared assumptions. These more complex cases have been described in this thread, and turn out to lead to surprising conclusions in the puzzle scneario.
Here’s another explanation of one relevant complex non-shared assumption in the three-person case:
I do not know what color eyes I have.
So I do not know what color B sees when he looks at my eyes.
B does not know what color eyes he has.
So B does not know what color C sees when C looks at B’s eyes.
Since I don’t know what color B sees in my eyes, and B doesn’t know what color C sees in B’s eyes, it follows that when I am imagining B imagining what C sees, I can imagine that “C” (who is not really C but is C-as-imagined-by-B. And that “B” I just referred to in that last sentence isn’t really B but is B-as-imagined-by-me) thinks it possible that there are no blue eyes on the island.
Actual-C and Actual-B both know that this assumption on my part is false. They do not share this assumption with me.
Like I said in my previous post:
In the case of three people, A learns that B knows that C can see a blue-eyed person.
But not everyone has the same information. Blue eyed people always see one less blue eyed person than brown eyed ones do. And no one knows which he or she is.
For all I know, B thinks C might be the only blue-eyed islander. So for all I know, B thinks C might leave on the first day. I need to wait a day to make sure B doesn’t actually (or at least no longer actually) think(s) this–because after a day, I will then know that B saw that C didn’t leave.
Yes, this exactly.
But you don’t know that they each know that there are more than two. And if you suppose each of those three each see only two, you don’t know that the three know that the two know that there are more than one. Until after the first night.
And if they’re still around after the third night, you have blue eyes.
But you can always know what they think the possibilities are. If I see N blue eyed people, the possibilities are:
If they have blue eyes and I have blue eyes - They think N or N+1
If they have brown eyes and I have blue eyes- They think N+1 or N+2
If they have blue eyes and I have brown eyes- They think N-1 or N
If they have brown eyes and I have brown eyes- They think N or N+1
Since I don’t know my eye color, I can only go by their eye color. Thus, if I see blue eyes I know they think the possibilities are N-1, N, or N+1. If I see brown eyes, they think N, N+1, or N+2.
But agian, if you know something, then you know everyone else knows it too. So If I see three sets of blue eyes, I know everyone sees at least two, and therefore I know everyone knows that no one is leaving day 1.