Comments

  • An unintuitive logic puzzle
    No they don't because they could have a unique colour and being, unlike you, perfect logicians they know that, and therefore do not make the fallible guess that they do not have a unique eye colour, and so none of your predicted leavings happen and you will conclude that you must have eyes of every colou — unenlightened

    They don’t assume that they don’t have a unique eye colour. Rather, they infer it based on what the others don’t do. Notice that each step is a conditional. The implicit final step is “if everyone else has left then I have a unique eye colour”.

    Just as in your example; the blues and browns don’t assume that their eyes aren’t green or red. They figure it out.

    The point I am making is they that don’t need to wait for green to say anything. They already know that she sees blue and brown. If it helps they could just imagine her saying “I see blue” and “I see brown” and apply the same reasoning.

    It is bizzare to suggest that our perfect logicians all know this but must wait with bated breath for her to verbally express what they already know before they can start.

    They can start the moment they arrive at the island.
  • An unintuitive logic puzzle
    ↪unenlightened


    They don't need to know that they don't have a unique eye colour. If they don't have a unique eye colour then the reasoning will work, as demonstrated in the post here.

    Everyone does in fact correctly deduce their eye colour.

    Although I think the particular reasoning in that post only works if nobody has a unique eye colour. If somebody does have a unique eye colour then they can apply the reasoning in my original post.
  • An unintuitive logic puzzle
    Except that you can't because you might have Z coloured eyes and although you can see that the others don't have Z coloured eyes, they don't know that, and so they cannot make the deduction that you rely on them making, to make your deduction — unenlightened

    Which is why I also said "unless they have a unique eye colour", and is the Guru in the original example. She cannot determine the colour of her own eyes but the 100 blue and 100 brown can all determine their own eye colour by the 100th day, even without the Guru saying anything.
  • An unintuitive logic puzzle
    But since they do not know their eye colour they might all have unique eye colours and none of them can deduce their eye colour at all. — unenlightened

    That's why I said: for all if I see people with X-coloured eyes...

    So if there are at least 3 people with X-coloured eyes and at least 3 people with Y-coloured eyes (and at least 3 people with Z-coloured eyes, etc.) then everyone can deduce their eye colour without anyone saying anything.
  • An unintuitive logic puzzle
    Why would 1 brown leave on day 1 anyway, if guru says nothing? — flannel jesus

    He wouldn't, but that's irrelevant. It can be demonstrated that if everyone just follows the rule: for all , if I see people with X-coloured eyes and if they don't leave on day then I have X-coloured eyes, then they will correctly deduce their eye colour (unless they have a unique eye colour).

    Knowing this fact is all it takes for everyone on the island to deduce their eye colour (except those with a unique eye colour). And perfect logicians would know this fact.

    So say I see 4 blue, 5 brown, and 6 green

    I reason:

    1. If the 4 blues don't leave on day 4 then I am blue
    2. If the 5 browns don't leave on day 5 then I am brown
    3. If the 6 greens don't leave on day 6 then I am green

    If I did have blue eyes then the others with blue eyes would reason:

    1. If the 4 blues don't leave on day 4 then I am blue
    2. If the 5 browns don't leave on day 5 then I am brown
    3. If the 6 greens don't leave on day 6 then I am green

    And the browns would reason:

    4. If the 5 blues don't leave on day 5 then I am blue
    2. If the 4 browns don't leave on day 4 then I am brown
    3. If the 6 greens don't leave on day 6 then I am green

    And the greens would reason:

    4. If the 5 blues don't leave on day 5 then I am blue
    2. If the 5 browns don't leave on day 5 then I am brown
    3. If the 5 green don't leave on day 5 then I am green

    All 5 of us with blue eyes would deduce after day 4 that we have blue eyes and leave on day 5.
    And at the same time the 5 with brown eyes would deduce after day 4 that they have brown eyes and leave on day 5.
    And then finally the 6 greens would deduce that they have green eyes after day 5 and leave on day 6.

    Everyone deduced the correct answer without anyone having to say anything.
  • An unintuitive logic puzzle
    And why n >= 3, rather than n >= 2? — flannel jesus

    Maybe also when .

    There are 2 brown, 2 blue, and 2 green.

    Each brown reasons that if the 1 brown doesn't leave on day 1 then he is brown, that if the 2 blues don't leave on day 2 then he is blue, and that if the 2 greens don't leave on day 2 then he is green.

    So when the other brown doesn't leave on day 1 he correctly deduces that he is brown.

    And then the same each for blue and green, all deducing the correct answer.
  • An unintuitive logic puzzle
    We may as well just imagine the guru making the statement, which means we may as well just imagine the guru, and this imaginary guru can make the statement about blue or brown — hypericin

    I don't even think we need to do that.

    It seems to be a simple mathematical fact that for all , if I see people with X-coloured eyes and if they don't leave on day then I have X-coloured eyes.

    So not only is the green person saying "I see blue" a red herring, but the green person being there at all is a red herring.

    Anyone who applies the above reasoning will correctly deduce their eye colour without anyone having to say anything, or even imagined to have said anything.

    I see 24 green, 36 blue, and 4 red.

    Therefore if the 4 red leave on day 4 but the 24 green don't leave on day 24 then I have green eyes.
  • An unintuitive logic puzzle
    These pair of premises don't make sense together. — flannel jesus

    Yes they do. Given that I know that green sees blue, I can just assume that she says so even if she doesn't, and so if helpful I can stipulate that in some hypothetical world in which I don't see blue (even though in reality I do see blue) that she says "I see blue" (even though in reality she doesn't say "I see blue").

    And again, the proof is in the pudding; as the above shows, all blues and all browns correctly deduce their eye colour on the third day.

    I can't explain this any clearer than I already have. So if you disagree then we're just going to have to agree to disagree.
  • An unintuitive logic puzzle
    What you're not understanding is that they could just add easily incorrectly deduce their eye colour. — flannel jesus

    No they won't. Let's take the example with 3 blue, 3 brown, and 1 green.

    Each blue's reasoning is:

    A1. Green sees blue
    A2. Therefore, if I don't see blue then I must be blue
    A3. Therefore, if I see one blue and he leaves on the first day then I must not be blue
    A4. Therefore, if I see one blue and he doesn't leave on the first day then I must be blue
    A3. Therefore, if I see two blue and they leave on the second day then I must not be blue
    A4. Therefore, if I see two blue and they don't leave on the second day then I must be blue
    A3. Therefore, if I see three blue and they leave on the third day then I must not be blue
    A4. Therefore, if I see three blue and they don't leave on the third day then I must be blue

    B1. Green sees brown
    B2. Therefore, if I don't see brown then I must be brown
    B3. Therefore, if I see one brown and he leaves on the first day then I must not be brown
    B4. Therefore, if I see one brown and he doesn't leave on the first day then I must be brown
    B3. Therefore, if I see two brown and they leave on the second day then I must not be brown
    B4. Therefore, if I see two brown and they don't leave on the second day then I must be brown
    B5. Therefore, if I see three brown and they leave on the third day then I must not be brown
    B6. Therefore, if I see three brown and they don't leave on the third day then I must be brown

    Given that each blue sees 2 blue and 3 brown, they can rule out some of these premises:

    A1. Green sees blue
    A2. Therefore, if I don't see blue then I must be blue
    A3. Therefore, if I see one blue and he leaves on the first day then I must not be blue
    A4. Therefore, if I see one blue and he doesn't leave on the first day then I must be blue
    A3. Therefore, if I see two blue and they leave on the second day then I must not be blue
    A4. Therefore, if I see two blue and they don't leave on the second day then I must be blue
    A3. Therefore, if I see three blue and they leave on the third day then I must not be blue
    A4. Therefore, if I see three blue and they don't leave on the third day then I must be blue

    B1. Green sees brown
    B2. Therefore, if I don't see brown then I must be brown
    B3. Therefore, if I see one brown and he leaves on the first day then I must not be brown
    B4. Therefore, if I see one brown and he doesn't leave on the first day then I must be brown
    B3. Therefore, if I see two brown and they leave on the second day then I must not be brown
    B4. Therefore, if I see two brown and they don't leave on the second day then I must be brown
    B5. Therefore, if I see three brown and they leave on the third day then I must not be brown
    B6. Therefore, if I see three brown and they don't leave on the third day then I must be brown

    Then, come the third day, they can rule out one more:

    A1. Green sees blue
    A2. Therefore, if I don't see blue then I must be blue
    A3. Therefore, if I see one blue and he leaves on the first day then I must not be blue
    A4. Therefore, if I see one blue and he doesn't leave on the first day then I must be blue
    A3. Therefore, if I see two blue and they leave on the second day then I must not be blue
    A4. Therefore, if I see two blue and they don't leave on the second day then I must be blue
    A3. Therefore, if I see three blue and they leave on the third day then I must not be blue
    A4. Therefore, if I see three blue and they don't leave on the third day then I must be blue

    B1. Green sees brown
    B2. Therefore, if I don't see brown then I must be brown
    B3. Therefore, if I see one brown and he leaves on the first day then I must not be brown
    B4. Therefore, if I see one brown and he doesn't leave on the first day then I must be brown
    B3. Therefore, if I see two brown and they leave on the second day then I must not be brown
    B4. Therefore, if I see two brown and they don't leave on the second day then I must be brown
    B5. Therefore, if I see three brown and they leave on the third day then I must not be brown
    B6. Therefore, if I see three brown and they don't leave on the third day then I must be brown

    Given that one of the As and one of the Bs must obtain, and given that only A4 is left of the As, blue knows on the third day that A4 must obtain and so that they are blue.

    The complimentary set of arguments will have all browns deducing that they are brown on the third day.

    All without green having to say anything.
  • An unintuitive logic puzzle
    Premise 2 is incorrect. — flannel jesus

    So you say, and yet if blues were to follow this reasoning and browns were to follow comparable reasoning then they would all correctly deduce their eye colour and leave on the 100th day — without green saying anything. I think the results speak for themselves.

    I've explained it as clearly as I can, so there's nothing else to add.
  • An unintuitive logic puzzle
    This doesn't work — flannel jesus

    It does work given that it allows me to correctly deduce my eye colour. What more proof do you need other than the results?

    Or is it just a coincidence?
  • An unintuitive logic puzzle
    If you want it as a step-by-step argument:

    P1. Green sees blue
    P2. Therefore, if I don't see blue then I must be blue
    P3. Therefore, if I see one blue and he leaves on the first day then I must not be blue
    P4. Therefore, if I see one blue and he doesn't leave on the first day then I must be blue
    etc.

    I know that P1 is true because I see 100 blue and I know that green can see them too, I know that the antecedent of P2 is false because I see 100 blue, I know that the antecedent of P3 is false because I see 100 blue, etc.

    This reasoning doesn't require green to actually say "I see blue" and will allow me to correctly deduce my eye colour.

    Therefore either the reasoning is sound or the correct "deduction" is a coincidence.
  • An unintuitive logic puzzle
    If green eyed person says nothing, what reason would the two blue have to leave on the second day? Without resting on the coattails of unenlightened — flannel jesus

    It's the same reasoning.

    Just as we can stipulate some hypothetical in which I don't see anyone with blue eyes, even though "in reality" I do, we can stipulate some hypothetical in which green says "I see blue" (and so I can know that she sees blue), even though "in reality" she doesn't.

    And we can do this because we know "in reality" that green sees blue even if she doesn't say so.
  • An unintuitive logic puzzle
    There's no reason in your scenario that anybody could figure out their own eye colour. — flannel jesus

    It’s explained in the post.
  • An unintuitive logic puzzle
    ↪flannel jesus
    I already answered. I don’t know. But the reasoning nonetheless allows all blues and browns to correctly deduce their eye colour and leave on the 100th day, answering the question in the OP.
  • An unintuitive logic puzzle
    your reasoning is still based on nothing other than unenlighteneds reasoning, and he's already told you his reasoning is based on the guru saying something. — flannel jesus

    And as I have repeatedly explained, it doesn’t actually require the Guru to say anything. It’s a red herring. It might appear to be necessary, but counterintuitively it isn’t.

    As evidenced by the fact that my reasoning allows all blues and browns to correctly deduce their eye colour and leave on the 100th day, answering the question in the OP.
  • An unintuitive logic puzzle
    ↪flannel jesus


    As I said, I’m not sure. But it appears to be a fact that if the blue-eyed people reason in such a way then they correctly deduce that they have blue eyes and that if the brown-eyed people reason in such a way then they correctly deduce that they have brown eyes.

    Therefore either the reasoning is sound or it’s a coincidence.

    Just as you don’t appear to be sure why green must verbally express what everyone already knows to be true.
  • An unintuitive logic puzzle
    you haven't justified that it does, is why — flannel jesus

    I explained the reasoning that each person performs and the conclusion they draw from it; a conclusion that is correct.

    I don’t understand what else you’re looking for.
  • An unintuitive logic puzzle
    the reasoning is incomplete — flannel jesus

    How can it be incomplete it it allows all brown and blue to leave knowing their eye colour?
  • An unintuitive logic puzzle
    ↪flannel jesus


    The one with 3 brown, 3 blue, and 1 green
  • An unintuitive logic puzzle
    You're saying "and yet" as if you've demonstrated that. You haven't — flannel jesus

    I demonstrated it in the example above.
  • An unintuitive logic puzzle
    I don't think it's sound or a coincidence. I don't think it's correct. — flannel jesus

    And yet every blue-eyed person leaves knowing they have blue eyes and every brown-eyed person leaves knowing they have brown eyes. So what do you mean by it “not being correct”?

    You have to follow the logic carefully one step at a time to find that out. It's very subtle and honestly strange - that what makes this such a good logic puzzle. It's completely counterintuitive, but also, once you fully grok it, undeniably true. That gives it this really unique flavour as a puzzle. — flannel jesus

    And I think that it goes even further: it may be counterintuitive, but one can get to the correct answer without green saying anything.
  • An unintuitive logic puzzle
    If it doesn't matter what the green eyed person says, why is his presence required at all? — flannel jesus

    I’m not sure, but my reasoning does allow all brown and all blue to leave knowing their eye colour, so either it’s sound or it’s a very lucky coincidence.

    But as a question to you, why would it require green verbally expressing what everyone already knows?
  • An unintuitive logic puzzle
    ↪flannel jesus


    I don’t think that’s a comparable scenario. I think a minimal example requires 3 blue, 3 brown, and 1 green.

    Each blue reasons: green sees blue, and so if the two blue I see don’t leave on the second day then I must be blue and green sees brown, and so if three brown I see don’t leave on the third day then I must be brown.

    Therefore on the third day each blue knows they are blue and leaves

    Each brown reasons: green sees brown, and so if the two brown I see don’t leave on the second day then I must be brown and green sees blue and so if the three blue I see don’t leave on the third day then I must be blue.

    Therefore on the third day each brown knows they are brown and leaves.

    Each person (other than green) leaves knowing their eye colour, all without anyone saying anything. The proof is in the pudding, as it were.

    Or is it just a coincidence?
  • An unintuitive logic puzzle
    No. It does depend on the guru saying so unless everyone already knows that everyone already knows at the same time, as I suggested above and you ignored. This is the extra information that the guru imparts: she doesn't inform them about what she sees, but she puts everyone in a synchronised state of knowing each other's knowing. That is what is required for the nested hypotheticals to begin. — unenlightened

    She doesn’t need to say anything for perfect logicians to be in a synchronised state. At every moment they are in a synchronised state and will apply the same reasoning.

    And if new people arrive or are killed by tigers during this process they’ll adapt their premises (e.g “the Guru sees either N or N+1 blue-eyed people”) as needed.

    But the factual knowledge that I can see multiple blue eyes and thus already know that the guru can see blue eyes cannot be imported into the counterfactual hypothetical wherein the blue eyed person would know no such thing because he would not himself see blue eyes, and thus could not know therefore that the guru saw blue eyes ... wait for it ... UNLESS SHE SAID SO. — unenlightened

    She doesn’t need to have actually said something in the “real” world in which I see 100 blue-eyed people for me to stipulate that she said something in this counterfactual hypothetical world in which I don’t see any blue-eyed people.

    In the “real” world I know that she sees at least one blue-eyed person even without her saying so, and so if it helps I can just assume that she says so even if she doesn’t.
  • An unintuitive logic puzzle
    If there's an island with 2 people and the guru and he doesn't say anything, and there's no telepathy, nobody knows anything — flannel jesus

    And if in this scenario I have brown eyes then the Guru wouldn’t say “I see someone with blue eyes”, and yet we are allowed for the sake of argument to assume that she does.

    Given that the Guru does in fact say it when there are 201 people we are allowed to assume that she still says it in a hypothetical scenario with 2 people.

    And given that we all know what the Guru sees without her saying it when there are 201 people we are allowed to assume that we still do in some hypothetical scenario with 2 people.

    The reasoning is sound in either case and gets us to the correct answer.
  • An unintuitive logic puzzle
    ↪flannel jesus


    I’m not stipulating random things, as shown by the fact that if the people on the island were to apply my reasoning then they would all correctly deduce the colour of their eyes. That’s not just some happy coincidence; it’s because the reasoning is sound.
  • An unintuitive logic puzzle
    If there's only one guy with blue eyes, he would only know that the guru sees blue eyes if the guru told him. — flannel jesus

    In practice, perhaps, but the logic doesn't require that the Guru say anything. The logic only requires that I know that the Guru sees at least one blue-eyed person and one brown-eyed person.

    Given that our reasoning stipulates (contrary to the facts) that I don't see any other blue-eyed person but that the Guru still says "I see at least one blue-eyed person", it can also stipulate (contrary to the facts) that I don't see any other blue-eyed person but that I know that the Guru still sees at least one blue-eyed person.

    And I can stipulate all of this even if I in fact have brown eyes.
  • An unintuitive logic puzzle
    Not in the scenario with one blue eyed person they don't — flannel jesus

    I see 100 people with blue eyes and (unknown to me) I have brown eyes. The Guru says "I see at least one blue-eyed person". Now I imagine a scenario where these 100 blue-eyed people don't exist.

    If I can still assume that the Guru says "I see at least one blue-eyed person" then I can still assume that the Guru sees at least one blue-eyed person, but doesn't say so.
  • An unintuitive logic puzzle
    but how would he know the guru knows that? The guru didn't say anything. He has no idea what the guru knows — flannel jesus

    Every person on the island already knows that the Guru sees at least one person with blue eyes and one person with brown eyes, whether or not she says so, as explained here. Given this fact:

    If I don't see anyone with blue eyes then I have blue eyes, else if I don't see anyone with brown eyes then I have brown eyes.

    This reasoning is valid even though I do in fact see others with blue eyes and brown eyes.
  • An unintuitive logic puzzle
    How does the blue eyed person know they have blue eyes in that scenario? What's the single blue eyed persons reasoning in that scenario? — flannel jesus

    The reasoning is: if the Guru sees at least one person with blue eyes and if I don't see anybody with blue eyes then I have blue eyes.
  • An unintuitive logic puzzle
    The first sentence of his reasoning clearly depends on the guru saying what she said. — flannel jesus

    No it doesn't. It only depends on "the Guru sees at least one blue-eyed person" being true. It doesn't depend on her saying so.
  • An unintuitive logic puzzle
    ok so your reasoning is different from unenlighteneds then. Can you tell us what it is? — flannel jesus

    It's the same. Here are unenlightened's exact and complete words:

    Reveal
    If there was only 1 person w. blue eyes, that person would see no blue eyes and therefore know they had blue eyes and leave that night.
    If there were 2, they would both see one person w blue eyes and when they did not leave the first night, they would both know the second night that they must have blue eyes and leave.


    Notice that he doesn't mention the Guru or what she says at all.
  • An unintuitive logic puzzle
    ↪flannel jesus


    The first step in the reasoning is "the Guru sees at least one person with brown eyes". She doesn't need to say "I see at least one person with brown eyes" for this first step to be true. Her saying so is a red herring.
  • An unintuitive logic puzzle
    Step 1 of his reasoning completely relies on the guru saying what he said. Can you see that? — flannel jesus

    That's the red herring; it doesn't. Everyone already knows that she sees at least one brown-eyed person, so her expressing this fact verbally provides no new information.
  • An unintuitive logic puzzle
    Also I think the brown eyed people would not know their eye-colour for another 99 days after the blue eyes left, but only that they themselves didn't have blue eyes. — unenlightened

    Why do you think that? Imagine the Guru were to have said "I see at least one blue-eyed person and at least one brown-eyed person".

    But as I said to flannel, the Guru doesn't even need to say it because everyone already knows that she sees at least one blue-eyed person and at least one brown-eyed person, and so her saying it is a red herring.
  • An unintuitive logic puzzle
    The reasoning for blue eyed people specifically works because the guru said he sees blue eyes. — flannel jesus

    The Guru doesn't need to say it. Him saying it is a red herring. As perfect logicians, every blue-eyed person already knows that the Guru sees at least one blue-eyed person and every brown-eyed person already knows that the Guru sees at least one brown-eyed person.
  • An unintuitive logic puzzle
    What's the reasoning for brown eyed people? Unenlightened gave reasoning for blue-eyed people — flannel jesus

    The same, just change "blue" for "brown".
  • An unintuitive logic puzzle
    ↪flannel jesus


    Reveal
    If they're perfect logicians then on the first day that they arrived on the island, even before the Guru speaks:

    1. The Guru knows that 100 blue-eyed people each see either 99 or 100 blue-eyed people and either 100 or 101 brown-eyed people and that 100 brown-eyed people each see either 99 or 100 brown-eyed people and either 100 or 101 blue-eyed people.

    2. Every blue-eyed person knows that 99 blue-eyed people each see either 98 or 99 blue-eyed people, either 100 or 101 brown-eyed people, and either 1 or 2 green-eyed people, that 100 brown-eyed people each see either 99 or 100 brown-eyed people, either 100 or 101 blue-eyed people, and either 1 or 2 green-eyed people, and that the Guru sees either 99 or 100 blue-eyed people, either 100 or 101 brown-eyed people, and either 0 or 1 green-eyed person.

    3. Every brown-eyed person knows that 99 brown-eyed people each see either 98 or 99 brown-eyed people, either 100 or 101 blue-eyed people, and either 1 or 2 green-eyed people, that 100 blue-eyed people each see either 99 or 100 blue-eyed people, either 100 or 101 brown-eyed people, and either 1 or 2 green-eyed people, and that the Guru sees either 99 or 100 brown-eyed people, either 100 or 101 blue-eyed people, and either 0 or 1 green-eyed person.

    Using the reasoning given by unenlightened, on day 100 every blue-eyed person would leave knowing that they have blue eyes, every brown-eyed person would leave knowing that they have brown eyes, and the Guru would stay knowing that they have neither blue nor brown eyes.

    The Guru doesn't need to say anything; her saying either "I see at least one blue-eyed person" or "I see at least one brown-eyed person" provides no new information and is a red herring.
  • An unintuitive logic puzzle
    I was wondering if this forum has SPOILER technology — flannel jesus

    [hide]this is hidden[/hide]

    Reveal
    this is hidden
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