Problem in logic

[Jadzia]

Well, I agree with all of that. The issue is what sort of new information is the Guru providing?

I understand now.

Here is how it works. Be warned that this is not simple.

Let N = number of people with blue eyes, and suppose I (you) are one of the islanders.

Suppose I can see one person with blue eyes... but I wonder "can everyone else see that?"

Not necessarily. Only if N=2 or higher can *I* be sure that everyone knows that there is someone with blue eyes.

But can *everyone* say that? Not necessarily. Only if N=3 or higher can *I* be sure that everyone knows that everyone knows that there is someone with blue eyes.

But can *everyone* say that? Not necessarily. Only if N=4 or higher can *I* be sure that everyone knows that everyone knows that everyone knows that there is someone with blue eyes.

But can *everyone* say that? Not necessarily. Only if N=5 or higher can *I* be sure that everyone knows that everyone knows that everyone knows that everyone knows that there is someone with blue eyes.

...

and so on. Since N is only 100, no logician cannot be sure 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 everyone knows that there is someone with blue eyes.

What the guru does is provide this ^^ knowledge.

 

[Lindley]

I have no problem with the recursion stuff, I understand that just fine.

That case is a contradiction due to the guru's statement.

Here is the point where I need some clarification. It's a contradiction due to the content of the guru's statement. That content being that the guru sees someone with blue eyes. Some people are saying that the fact that guru is saying it gives it some extra meaning, which I don't get at all.

neozeks got it.

Are you familiar with the mathematical process of proof-by-contradiction? You make an initial assumption which you desire to prove false, and follow it logically until it results in a contradiction. Therefore, the initial assumption must have been false.

Under all of the cascading I'm-not-blue assumptions, the guru's statement is relevant and provides the eventual contradiction which renders all of those assumptions false.

Blue A assumes they are not blue. Within this assumption, Blue B assumes they are not blue. Within this assumption, Blue C assumes they are not blue, and so on. It doesn't matter that in reality blues B and C would also be seeing 99 other blues----under the set of active assumptions, they would be seeing 98 and 97 respectively. Once you activate enough assumptions, you get to the last blue who sees no others. This is not a real situation, it's a hypothetical one all within Blue A's mind (and symmetrically that of every other blue). The key is that every day that passes brings the contradiction up another level. IF all the remaining not-blue assumptions were true, all blues would leave on that day. Since they don't, the contradiction propagates up another level.

 

[Scout101]

And it would be nice if someone actually worked out in full the though processes in the 4 blues case. All the examples always stop at 3 blues and I have a nagging feeling the logic breaks somewhere at that point but I don't really know why.

Really is the same thing, people just stop after 2 or 3 because it's boring to write out 100 of them

1- if you don't see anyone with blue eyes, it's you. Free to go.

2 - if you only see 1 person with blue eyes, and they don't leave the first night, means that they can see 1 person also. Must be you, both leave night 2.

3 - if you see 2 people with blue eyes, they would have left on the 2nd night if they only saw 1 (see above), but since they are still there, they must ALSO see 2 people, meaning you're the 3rd, and you can leave.

4- you see 3 people with blue eyes. If your eyes weren't blue, they would have only seen 2, and left the night before. That means they see 3 also, and there are 4 of you. leave on night 4.

Logic just keeps repeating, but keeps taking an extra night because you don't know if there are X or X+1 until after you see whether they stick around on night X. If no one leaves on night 50, means they all see 50 people with blue eyes ALSO, meaning there are 51, leave on night 51.

It's tough to picture, but with all complicated systems, if you break them down to simpler terms and then build back up to the tough part, it's easier to understand.

 

[Gryffindorian]

It took me awhile to fully grasp the logic of this riddle. I can see why it's necessary for the guru to make the statement, "I see someone with blue eyes." What he's really saying is:

(1) "I'm setting the conditions for which individuals can determine his/her own eye color" AND

(2) "Only the blue-eyed people can leave the island."

 

[Yoda]

Still not agreeing.

Ok, let's try to unravel this thing from 3 blue, 3 brown, 1 green instead of 100/100/1.

These statements follow logically for everyone. Because everyone can see one common blue eyed or brown eyed person:
The Guru can see a blue eyed person.
The Guru can see a brown eyed person.

Along with lots of other possibilities, these perfect logicians would consider the assumptions "I do not have blue eyes" in the context for the first statement, and "I do not have brown eyes" in the context of the second statement.

So, for one of the random blue eyed people:

I see two blue eyed people, and I assume I do not have blue eyes. When each of those two works with that same assumption, they will see just one blue eyed person. When that one person considers the assumption that they're not blue, they would see a contradiction because we know that the Guru can see a blue person. So that assumption cannot be correct.

I see three brown eyed people, and I assume I do not have brown eyes. Each of three brown eyed people when they assume they are not brown eyed people will see two brown eyed people. When each of those two people works with that same assumption, that they aren't brown, they will see one person with brown eyes. When that one person considers the assumption that they're not brown, they would see a contradiction because we know the Guru can see a brown person. So that assumption cannot be correct.

Day 1 ferry comes and goes. Nobody gets on.

Still working with the assumption that I'm not blue, then that leaves the two blue eyed people. Each of them would now have a contradiction in their assumption that they're not blue, because the one blue person they saw did not leave on day 1. They would conclude that they must be blue, because otherwise the 1 blue person they saw would've left on day 1.

Still working with the assumption that I'm not brown, then that leave the three brown eyed people. Each of them working with the assumption that they're not brown would see two brown people, each of those working with that assumption would see 1 brown eyed person. They would conclude that they must be brown, because otherwise the 1 brown person they saw would've left on day 1.

Day 2 ferry comes. Nobody leaves.

Still working with the assumption that I'm not blue, then that leaves the other two blue eyed people. They didn't leave. They would have left if I wasn't blue, that's a contradiction. Therefore I am blue. I will leave and declare myself blue on the next ferry.

Even though we've determined that I'm blue, let's see continue the brown logic treating it as an assumption. Assuming I'm not brown, I see three brown people, each of them has seen the other two not leave on day 2. That's what they would have done if the third brown person who is assuming that he's not brown was in fact not brown. Therefore that person knows he is brown if I'm not brown.

Now, from each of the browns, the exact same logic would unfold, except with the roles of blue and brown reversed. They will reach the conclusion I am brown, I will leave and declare brown on the ferry 3 after the browns they see don't get on ferry 2.

Now for the perspective of the Guru. The Guru would treat blue AND brown as the longer of the two cases, since the Guru sees 3 of each. So, for the statement I am blue to be contradicted for the guru, the blues wouldn't get on the ferry on day 3. Likewise, for the statement I am brown to be contradicted, the browns would not get on the ferry on day 3. Since blues and browns do leave those statements can't be contradicted.

The guru is left alone on the island.

 

[boobatuba]

Yoda, that sounds pretty good. The problem is, the blues and browns don't know their own eye color so that cannot think differently when trying to solve the problem as you suggest.

When you consider what blue-eyed people are thinking, you only suggest that they consider other blue-eyed people. When you consider what brown-eyed people are thinking, you only suggest that they consider other brown-eyed people.

They can't do this. They have to consider all the colors they see, not just the one that corresponds to their own eye color (which they don't know). In your scenario, on the third day, the still cannot determine what eye color they have.

Still working with the assumption that I'm not blue, then that leaves the two blue eyed people. Each of them would now have a contradiction in their assumption that they're not blue, because the one blue person they saw did not leave on day 1. They would conclude that they must be blue, because otherwise the 1 blue person they saw would've left on day 1.

This is the crux of the matter. The person you describe cannot conclude that they must be blue simply because the one blue-eye person they saw would have left.

 

[Yoda]

Which leads me to the crux of the matter from my perspective. The Guru only adds something new if the number of blue eyed people is one. Then that person would know that the Guru was talking about them. If the number of blues is 2 or more then the statement "The guru sees someone with blue eyes" can be derived from being able to see a blue eyed person in common, even if you are the blue person.

Since the situation is blues = 100, the Guru is NOT saying it for the situation where blues = 1. The guru only says it once, the rest is a thought experiment of everyone involved. So, if you can logically deduce if you're blue with the statement from the guru, you should be able to logically deduce it with the exact same statement by deriving from looking at common blues rather than listening to the Guru.

So if the statement allows you to solve, when blues = 100, the same derived statement for browns allows you to solve when browns = 100. It only doesn't work if the initial conditions are blues = 1 or browns = 1.

 

[Lindley]

When that one person considers the assumption that they're not brown, they would see a contradiction because we know the Guru can see a brown person. So that assumption cannot be correct.

Here's the fault in your reasoning----the only reason we "know" the guru can see someone with brown eyes is because we can. Once you get down deep enough in the assume-not-brown branch of the decision tree, your assumptions have ruled out all browns but one, so under that set of assumptions that brown *cannot* know the guru can see a brown unless the guru says so. Which she doesn't. No contradiction exists until the guru speaks.

 

[Yoda]

How about this. Let's consider why the Guru is saying "I see a blue eyed person". What motivates the Guru? Maybe the Guru's rule is that if the Guru sees at least 2 blue people then she'll say "I see a blue eyed person". Since the initial conditions satisfy that rule, you can't rule it out.

 

[sojourner]

^not part of the conditions setup in the problem.

 

[Yoda]

I'm not saying it's part of the initial conditions, I'm saying it's not ruled out by the initial conditions.

 

[boobatuba]

I'm not saying it's part of the initial conditions, I'm saying it's not ruled out by the initial conditions.

Well, if you could just make stuff up and say it's "not ruled out" it wouldn't be much of a logic problem, would it?

 

[Yoda]

I would say what you can and can't rule out is pretty important in a logic puzzle.

 

[boobatuba]

I would say what you can and can't rule out is pretty important in a logic puzzle.

Well, sure...as a general rule. What I'm saying is you can only use the information given in the parameters of the riddle to solve it. You can't just assume something about why the guru says what she says and then think it's ok because "it's not ruled out."

Besides, it's irrelevant. There's 100 blues, 100 browns, and 1 green. There's no reason to consider what the guru would say if there was only one blue-eyed person because we know that not to be the case.

 

[Yoda]

I'm not assuming it. I'm treating it as a possibility.

 

[Scout101]

It's not a possibility, you are GIVEN that there are 100 of blue, 100 brown, so the 'possibility' of what they could say if it was a different number isn't valid information.

It also wasn't ruled out that eye color randomly changes, or people might have one eye of each color, but your mind automatically rejected those as outside the needed parameters to solve the puzzle, so why are you trying to introduce other random obstacles and parameters?

If you're having a hard time solving the math, more variables rarely helps

 

[Yoda]

Here's the fault in your reasoning----the only reason we "know" the guru can see someone with brown eyes is because we can. Once you get down deep enough in the assume-not-brown branch of the decision tree, your assumptions have ruled out all browns but one, so under that set of assumptions that brown *cannot* know the guru can see a brown unless the guru says so. Which she doesn't. No contradiction exists until the guru speaks.

Well, sure...as a general rule. What I'm saying is you can only use the information given in the parameters of the riddle to solve it. You can't just assume something about why the guru says what she says and then think it's ok because "it's not ruled out."

Besides, it's irrelevant. There's 100 blues, 100 browns, and 1 green. There's no reason to consider what the guru would say if there was only one blue-eyed person because we know that not to be the case.

Ok, here's what I was trying to say, and it's related to both of those thoughts:

We don't know why the Guru said what was said, only that it's true. The way that everyone else can deduce that the Guru sees a blue person (without the Guru actually saying it) is "I see at least 2 blues therefore everyone can see a common blue." This is true for everyone in the 100/100/1 scenario.

In Lindley's explanation he says the knowledge from that statement isn't sufficient.

But, there's no way you can assume that the Guru isn't using that same logic, anymore than you could assume that the Guru is.

So it's still unclear to me how the Guru's statement is any 'better' than the derived "I see that everyone sees a common blue therefore the Guru sees a blue" statement.

It seems to me for an airtight induction proof you would need to know that for all numbers of blues >= 1 the Guru would make the same "I see a blue person" statement.

 

[Scout101]

Not sure how to phrase this differently...

Everyone KNOWS the Guru can see a blue before he speaks, they can all see at least 99 of them themselves.

What the Guru adds (and is otherwise forbidden) is the communication aspect. They have to try and solve for blue (or brown) on their own, but unless everyone is doing it, and solving for the same thing, it's impossible. You also have at least 3 variables (many more, really, if you don't limit it to just the blue, brown, and green that they can see. And because they can see a single green, it's possible that they could also be green, or the only one of some OTHER color).

the Guru reduces things to 2 groups (blue and not blue), and then gets everyone moving on that equation.

It's not new information, as he says something that everyone already knows. It's the communication that gets them all working the puzzle from the same angle that's key.

You're WAY overcomplicating this...

 

[Yoda]

I don't think Lindley's explanations agree with that.

With your explanation I maintain that everyone could solve for blue and not blue, and brown and not brown, and green and not green for that matter simultaneously, and come up with correct answers.

That is, for green, or all colors that they don't see, they could never leave the island because they can't see a common green (or other color) with everyone.

The blues could come to the conclusion that if no blues leaves on day 99, then they leave on day 100 as a blue. They would also come to the conclusion that if browns don't leave on day 100 they will leave as browns.

The browns would come to the exact opposite conclusion.

The green guru would conclude she could leave on day 101 as a blue if the blues don't leave on day 100, and she could leave on day 101 as a brown if the browns don't leave on day 100.

I think I showed how that worked adequately in an earlier post. If you solve for every possibility in front of you, you end up with a set of rules that don't conflict and that give you the right answer. As perfect logicians I imagine they could work that out themselves.

 

[Chaos Descending]

But the problem is that until the Guru actually speaks, no one knows what condition everyone else is solving for. Once the guru speak, it puts everyone on the same page. "Solve for blue". This simply CANNOT BE DONE until the Guru speaks. The Guru must speak, or no one can leave.