2 ms·

Well, the problem is fundamentally that the people should've been playing the game before anyone said anything, so there's some confusion as to why the game onl

by aston 19y ago

Well, the problem is fundamentally that the people should've been playing the game before anyone said anything, so there's some confusion as to why the game only starts when the guru mentions a fact that's already self-evident. That's a problem in both the grandparent link and xkcd's story.

yters 19y ago
You don't think the guru provides more information? I don't follow.
zoltz 19y ago
That's exactly the interesting point here: the foreigner states something self-evident (there is someone who has blue eyes), and yet it will kill all blue-eyed people, who would otherwise have been fine. Without the foreigner, the inductive chain cannot start at n=1.
- aston 19y ago
  Sure it can. Take the 100 brown, 100 blue problem xkcd states. If they all kept their eyes closed and then starting today opened them, the end result would be what you claim the foreigner would've instigated: people remove themselves when they figure out their eye color, and after enough time passes, if a bunch of people are still there who you believe should be gone, you must be part of the cause of uncertainty.
  - philh 19y ago
    No. Suppose there are only three people with blue eyes. A knows that B can see at least one blue-eyed person. And B knows that C can see at least one blue-eyed person. But A doesn't know that B knows that. A reasons: if I have brown eyes, B can only see one blue-eyed person, ie C. And if B can only see one blue-eyed person, he will reason "if I have brown eyes, C cannot see anybody with blue eyes". Thus, if I have brown eyes, B does not know whether C can see anybody with blue eyes. Or when there are four blue-eyed people, A's reasoning will be: if I have brown eyes, then B will only be able to see two blue-eyed people. He will then reason according to the above case. So if I have brown eyes, B doesn't know that C knows that D can see at least one blue-eyed person. By saying "I see a blue-eyed person", the foreigner removes this uncertainty.
    - aston 19y ago
      Hmm. I'm getting too tired to think really clearly about this. I could be way off here, but "B does not know whether C can see anybody with blue eyes." - that's not true. C can see A. And for the four person case, the same is true. If there is more than one blue eyed person, everyone knows there's at least one blue-eyed, person, so no new information via the foreigner. .... edit for the three person all blue eyes walk through, via the eyes of person A. Persons B and C have the same exact thoughts.: Day 0) Everyone gains consciousness together. I look around and see two people with blue eyes. Either I have blue eyes or not blue eyes, and I'll know soon enough which based on how confused the others are. Day 1) So everyone's still here. Great. They're confused too. Well, that's to be expected. B and C saw each others' blue eyes, but couldn't know whether or not their own matched, even if I were not blue eyed. Maybe they saw my blue eyes and needed even more time? Doesn't matter, tomorrow's D-day. Day 2) Oh shoot. Everyone's still here. If I were not blue eyed, these guys would've noticed yesterday that something was up and figured out that they individually had blue eyes. I guess that means I confused them an extra day, so I have blue eyes, too. Day 3) [silence] Also note: the game not only progresses without the outsider, but it relies on the fact more than one person shares an eye color.
      - philh 19y ago
        "B does not know whether C can see anybody with blue eyes." - that's not true. C can see A. Yes. But what you quoted was part of A's reasoning. A doesn't know that he has blue eyes. If he doesn't, then C can only see B with blue eyes. And if C can only see B, then B doesn't know if C can see anyone. So A doesn't know whether B knows C can see a blue-eyed person (implying A has blue eyes) or not (implying he doesn't). It's not about what people know directly. It's about what people know about other people's knowledge. In the three-person case, I know that there is at least one blue person. I know that everyone knows it. But I don't know that everyone knows that everyone knows it. If K(0) is "I know there is at least one blue-eyed person", and K(n) is "I know that everyone knows K(n-1)", then (assuming I have blue eyes), K(0) is false only if I am the only blue-eyed person. K(1) is false if there are one or two of us. In general, K(n) is false if there are n+1 or fewer people with blue eyes. The foreigner makes K(n) true for all n.
      - 19y ago
- aston 19y ago
  Also, this problem isn't solved by induction. It's solved by recognizing the pattern. You don't need to establish the base case via the outside observer. All you need to know is that the day-based formula works for any number of islanders. It'd be like proving x^2 - 1 = (x+1)(x-1) for all x by starting with 1 and walking up the integers... Which is to say, not a bad way to prove it, but not necessary when you can logic it out otherwise.
  - deleted 19y ago
    [deleted]
mechanical_fish 19y ago
Well, the problem is fundamentally that the people should've been playing the game before anyone said anything, No, that's not true. The proof-by-induction doesn't work without the guru, because the initial case, N=1, is a failure. If there is only one blue-eyed person (call him "Adam"), he looks around and see a lot of brown-eyed people but has no way to know whether his own eyes are blue, or not. Meanwhile, the other 999 people see a single blue-eyed person, Adam, who isn't leaving the island, but they think it's just because Adam doesn't realize that there are any blue-eyed people at all. Once the guru makes the observation then the chain of induction is set in motion. So, yes, that observation most certainly does convey information. The real question is: why do our minds persist in believing that the guru's statement conveys no information? IMHO, it's a matter of psychology and anthropology. Proof-by-induction is, literally speaking, an inhuman trick. No creature on Earth can do it except humans, and most humans can't do it without special training. (We call those trained humans "math students".) Because there weren't any math students around during most of evolutionary history, our mental models of other people's behavior do not include proof-by-induction. When we see a problem like this -- one which involves an island containing 1,000 math students who apply inductive reasoning flawlessly and continuously -- our mental model of behavior just goes haywire. This scenario is literally outside our mind's design parameters, so it's completely nonintuitive. Just solving the problem at all requires a heroic mindhack -- you have to plod along through a set recipe, starting with the N=1 case and proceeding by a tedious chain of cases. It's fun to imagine a creature who could just see the answer to this problem, in the same way that many, many humans can spot a fleeting facial expression on the face of a Survivor contestant and just know, immediately, who will be voted off the island. (That is a task that our brain was designed for.) Perhaps some savants can do it. Perhaps Ramanujan could have done it. Perhaps Garry Kasparov can do it. Perhaps you can do it (but, if you can, it may be hard to explain how...)

by aston 19y ago

yters 19y ago

You don't think the guru provides more information? I don't follow.

zoltz 19y ago

That's exactly the interesting point here: the foreigner states something self-evident (there is someone who has blue eyes), and yet it will kill all blue-eyed people, who would otherwise have been fine. Without the foreigner, the inductive chain cannot start at n=1.

aston 19y ago
Sure it can. Take the 100 brown, 100 blue problem xkcd states. If they all kept their eyes closed and then starting today opened them, the end result would be what you claim the foreigner would've instigated: people remove themselves when they figure out their eye color, and after enough time passes, if a bunch of people are still there who you believe should be gone, you must be part of the cause of uncertainty.
- philh 19y ago
  No. Suppose there are only three people with blue eyes. A knows that B can see at least one blue-eyed person. And B knows that C can see at least one blue-eyed person. But A doesn't know that B knows that. A reasons: if I have brown eyes, B can only see one blue-eyed person, ie C. And if B can only see one blue-eyed person, he will reason "if I have brown eyes, C cannot see anybody with blue eyes". Thus, if I have brown eyes, B does not know whether C can see anybody with blue eyes. Or when there are four blue-eyed people, A's reasoning will be: if I have brown eyes, then B will only be able to see two blue-eyed people. He will then reason according to the above case. So if I have brown eyes, B doesn't know that C knows that D can see at least one blue-eyed person. By saying "I see a blue-eyed person", the foreigner removes this uncertainty.
  - aston 19y ago
    Hmm. I'm getting too tired to think really clearly about this. I could be way off here, but "B does not know whether C can see anybody with blue eyes." - that's not true. C can see A. And for the four person case, the same is true. If there is more than one blue eyed person, everyone knows there's at least one blue-eyed, person, so no new information via the foreigner. .... edit for the three person all blue eyes walk through, via the eyes of person A. Persons B and C have the same exact thoughts.: Day 0) Everyone gains consciousness together. I look around and see two people with blue eyes. Either I have blue eyes or not blue eyes, and I'll know soon enough which based on how confused the others are. Day 1) So everyone's still here. Great. They're confused too. Well, that's to be expected. B and C saw each others' blue eyes, but couldn't know whether or not their own matched, even if I were not blue eyed. Maybe they saw my blue eyes and needed even more time? Doesn't matter, tomorrow's D-day. Day 2) Oh shoot. Everyone's still here. If I were not blue eyed, these guys would've noticed yesterday that something was up and figured out that they individually had blue eyes. I guess that means I confused them an extra day, so I have blue eyes, too. Day 3) [silence] Also note: the game not only progresses without the outsider, but it relies on the fact more than one person shares an eye color.
    - philh 19y ago
      "B does not know whether C can see anybody with blue eyes." - that's not true. C can see A. Yes. But what you quoted was part of A's reasoning. A doesn't know that he has blue eyes. If he doesn't, then C can only see B with blue eyes. And if C can only see B, then B doesn't know if C can see anyone. So A doesn't know whether B knows C can see a blue-eyed person (implying A has blue eyes) or not (implying he doesn't). It's not about what people know directly. It's about what people know about other people's knowledge. In the three-person case, I know that there is at least one blue person. I know that everyone knows it. But I don't know that everyone knows that everyone knows it. If K(0) is "I know there is at least one blue-eyed person", and K(n) is "I know that everyone knows K(n-1)", then (assuming I have blue eyes), K(0) is false only if I am the only blue-eyed person. K(1) is false if there are one or two of us. In general, K(n) is false if there are n+1 or fewer people with blue eyes. The foreigner makes K(n) true for all n.
    - 19y ago
aston 19y ago
Also, this problem isn't solved by induction. It's solved by recognizing the pattern. You don't need to establish the base case via the outside observer. All you need to know is that the day-based formula works for any number of islanders. It'd be like proving x^2 - 1 = (x+1)(x-1) for all x by starting with 1 and walking up the integers... Which is to say, not a bad way to prove it, but not necessary when you can logic it out otherwise.
- deleted 19y ago
  [deleted]

mechanical_fish 19y ago

Well, the problem is fundamentally that the people should've been playing the game before anyone said anything, No, that's not true. The proof-by-induction doesn't work without the guru, because the initial case, N=1, is a failure. If there is only one blue-eyed person (call him "Adam"), he looks around and see a lot of brown-eyed people but has no way to know whether his own eyes are blue, or not. Meanwhile, the other 999 people see a single blue-eyed person, Adam, who isn't leaving the island, but they think it's just because Adam doesn't realize that there are any blue-eyed people at all. Once the guru makes the observation then the chain of induction is set in motion. So, yes, that observation most certainly does convey information. The real question is: why do our minds persist in believing that the guru's statement conveys no information? IMHO, it's a matter of psychology and anthropology. Proof-by-induction is, literally speaking, an inhuman trick. No creature on Earth can do it except humans, and most humans can't do it without special training. (We call those trained humans "math students".) Because there weren't any math students around during most of evolutionary history, our mental models of other people's behavior do not include proof-by-induction. When we see a problem like this -- one which involves an island containing 1,000 math students who apply inductive reasoning flawlessly and continuously -- our mental model of behavior just goes haywire. This scenario is literally outside our mind's design parameters, so it's completely nonintuitive. Just solving the problem at all requires a heroic mindhack -- you have to plod along through a set recipe, starting with the N=1 case and proceeding by a tedious chain of cases. It's fun to imagine a creature who could just see the answer to this problem, in the same way that many, many humans can spot a fleeting facial expression on the face of a Survivor contestant and just know, immediately, who will be voted off the island. (That is a task that our brain was designed for.) Perhaps some savants can do it. Perhaps Ramanujan could have done it. Perhaps Garry Kasparov can do it. Perhaps you can do it (but, if you can, it may be hard to explain how...)