Postby **marzis** » Thu Oct 30, 2014 1:55 pm UTC

Hello everyone. I just want to quickly get out of the way the fact that I completely understand the common-knowledge based solution, the inability to know for certain that anyone on the island other than one's self that anyone else on the island knows for certain that there is at least 1 person on the island with Blue eyes, and how the Guru's statement "I can see at least one person with Blue eyes" leads to the establishment of a common-knowledge basis of 'at least 1' which then leads to all of the Blue eye'd people leaving on the 100th night (1st night is the one of the Guru's speaking). So, please don't bother trying to explain the common-knowledge aspect of things. My solution has nothing to do with it at all, and I'm fairly certain that if you try to use the common-knowledge solution to disprove mine, well, I won't get any further in understanding why my solution isn't correct (if it isn't) than I have so far on my own.

Here goes.

Initially, lets split the people on the island into three generic groups. A, B, and C. These groups are defined by their eye color, and everyone within the groups has the same eye color. I won't bother using Blue or Brown or Green anymore, because those are just arbitrarily chosen colors and can have no fundamental impact on the problem or solution.

Now, instead of bothering to assume to know what anyone else on the island can possibly know, we are going to look purely at what any one, single, individual can know, based entirely on their own eyes. This has nothing to do with common-knowledge, perfect-knowledge, or anything of that sort. It is simply looking at any one individual, wholly separate from everyone else on the island, and working out what that one individual can logically surmise based on what said individual can see. That is all.

What they can see:

Person A: 99 A, 100 B, and 1 C.

Person B: 100 A, 99 B, and 1 C.

Person C: 100 A, 100 B.

What they know:

Person A: 99 A, 100 B, 1 C, and one unknown (themselves).

Person B: 100 A, 99 B, 1 C, and one unknown (themselves).

Person C: 100 A, 100 B, and one unknown (themselves).

To reiterate the above: The line for Person A does not in any way assume to bother attempting to know or see, what the line for Person B will read. They are wholly uninterested with what anyone else on the island may or may not know or see.

What they can surmise:

Person A1: 99 A, 100 B, 1 C, and they themselves have eyes of A, thus 100 A, 100 B, 1 C.

Person A2: 99 A, 100 B, 1 C, and they themselves have eyes of B, thus 99 A, 101 B, 1 C.

Person A3: 99 A, 100 B, 1 C, and they themselves have eyes of X, thus 99 A, 100 B, 1 C, 1 X. (X in this case can be C, the function here is that X is NOT A or B).

Person B1: 100 A, 99 B, 1 C, and they themselves have eyes of A, thus 101 A, 99 B, 1 C.

Person B2: 100 A, 99 B, 1 C, and they themselves have eyes of B, thus 100 A, 100 B, 1 C.

Person B3: 100 A, 99 B, 1 C, and they themselves have eyes of X, thus 100 A, 99 B, 1 C, 1 X. (again, X is simply NOT A or B).

Taking, again, only what person A and B can surmise above, we can now take one further step, which does not depend on knowing with certainty that anyone on the Island can see anything, but rather, based on the above hypothetical situations that either person A or B have created, what can person A or B surmise about the other various participants in this hypothetical world?

I.e.,:

Person A1: Knows that if there are 100 A, 100 B, and 1 C, then: Every A will see: 99 A, 100 B, 1 C, and every B will see: 100 A, 99 B, and 1 C.

Person A2: Knows that if there are 99 A, 101 B, and 1 C, then: Every A will see: 98 A, 101 B, 1 C, and every B will see: 99 A, 100 B, and 1 C.

Person A3: Knows that if there are 99 A, 100 B, and 1 C, and 1 X, then: Every A will see: 98 A, 100 B, 1 C, and 1 X, and every B will see: 99 A, 99 B, and 1 C and 1 X.

Person B1: Knows that if there are 101 A, 99 B, and 1 C, then: Every A will see: 100 A, 99 B, 1 C, and every B will see: 101 A, 98 B, and 1 C.

Person B2: Knows that if there are 100 A, 100 B, and 1 C, then: Every A will see: 99 A, 100 B, 1 C, and every B will see: 99 A, 99 B, and 1 C.

Person B3: Knows that if there are 100 A, 99 B, and 1 C, and 1 X, then: Every A will see: 99 A, 99 B, 1 C, and 1 X, and every B will see: 100 A, 98 B, 1 C and 1 X.

No one in the situation above knows if they have A, B, or X, individually, just how 201 people on an island can be distributed based on a breakdown of 100/100/1 or 101/99/1/1.

As far as I can tell, so far, nothing in the above relies upon common-knowledge, as no one is attempting to know, with any ounce of certainty, that anyone else on the island knows anything else, since we are only working within a series of hypothetical compositions of the members of the island.

It is very important, I feel, to say that at this stage, no one on the island is any closer to leaving the island, nor at knowing which of the A1, A2, A3, or B1, B2, B3 situations above they fall into. But what they DO know, is this:

Person A: IF they have eye color X OR B, then: Everyone in group A can see 98 A.

Person B: IF they have eye color X OR A, then: Everyone in group B can see 98 B.

They know that they themselves are not members of said hypothetical group A or B listed above who can only see 98 A or 98 B, as they can see at least 99 of either A or B, but this still does not put them any closer to knowing their eye color, only what would be the reality of the distribution of the island based on their possible eye colors.

But what each islander individually knows, is relayed above. They should each, individually, be able to come to this conclusion on their own, without having to know with any certainty that anyone else on the island can come to that same conclusion. They don't care about what anyone else thinks, as they aren't of one mind or anything, they're just individuals observing and making conjectures about how the island might be distributed. So far I can't see how these assumptions above can depend on common-knowledge.

Smaller Island example: Three people on an island, 2 A (A' and A") and 1 B.

A' can see 1 A", 1 B. B can see 1 A' and 1 A".

If B has A, then A' and A" can see 2 A. If B has B, then A' can see 1 A" and 1 B. If A' has A, then A" can see 1 A' and 1 B. If A' has B, then A" can see 2 B. If A" has B, then A' can see 2 B. This may not be complete, but this is just to illustrate a point. They are not attempting to know anything, they are just laying out the various ways in which 3 people, of which they themselves are an unknown, can populate an island, and what each member of said island might see if any one of those combinations was the case.

Ok, onto the Guru.

If the various members of A and B can make the above conjectures about the distribution of the 201 members on the island, then it is logical to assume that, as they do so, so does the Guru (being a perfect logician as well).

And this is where I run into trouble with the common-knowledge solutions: They believe that the Guru's statement (as interpreted for the common-knowledge solution) is a logical one. [if the problem was built to be a common-knowledge exercise, then that's fine. but I don't know that]

If my above conjectures about the ways in which the respective A and B members of the island can surmise they are distributed is accurate, then our C would know those various ways as well, owing to the fact that she can see the 100 A and 100 B, and knows that any one of them doesn't know the color of their own eyes, and would subsequently surmise the various A1, A2, A3, and B1, B2, B3 possibilities that each respective A or B individual could. She would know also that she can speak, can speak only once, and that no one else on the island can speak, and of course, knows that she herself is a perfect logician, along with everyone else on the island.

So the Guru, knowing all this, decides to provide information only to one group of people (in the original: the Blue eyed people)? Decides, somehow, to choose A and not B? How is this logical? She has chosen to give information that, while being applicable to every Blue eyed person, is wholly irrelevant to every Brown eyed person. Every Brown eyed person can do nothing with the information she gives, AS INTERPRETED, in the common-knowledge solutions. And to me, this feels wrong, since I cannot find a logical way in which to say 'A' instead of 'B'.

And so we come to my interpretation of the Guru's statement, given all that she can see, and all that she has surmised along with the various members of A and B.

The Guru can provide, as referenced by either an A, or a B, person, 3 pieces of information.

Information 1: They have A color eyes.

Information 2: They have B color eyes.

Information 3: They have X color eyes. (This is equivalent to: 'They do not have A or B color eyes' as X is defined as not A or B)

These are the three possibilities that everyone on the island knows to be the case. Each person on the island can see everyone else, so, individually, they know that the only piece of information they need is their OWN eye color. They don't need to know that everyone else on the island knows what everyone else on the island knows. They just need the answer to their own, individual, selfish, isolated question, which is: What color are my eyes?

The Guru knows this, and says "I can see at least one person with Blue eyes."

The members of A and B instantly start to process this information relative to their three questions. Has the Guru provided Information 1? Information 2? Information 3? No. She has answered a question they were uninterested in, if they have Brown eyes, according to the common-knowledge solution.

My analysis of the Guru's statement is as follows:

The Guru is incapable of an illogical action, being a perfect logician, she just wouldn't ever conceive of it. Therefore she could never choose to provide information to A over B, since A is = to B in terms of number (100 vs 100) and we don't have any sort of history of the island or a valuation system against which to draw any sort of bias towards a particular group or individual. Therefore any statement she makes cannot be focused on only group A or group B, and similarly, cannot be provided to and kept isolated to only one member of group A or group B. Her statement must be equally applicable to every member on the island, since she is a perfectly logical machine.

Since the Guru must provide information to A and B equally, as any other option would be providing arbitrary preferential treatment, which is not a logical process, when she says "I can see at least one person with Blue eyes" her statement must be read as being equivalent to "I can see at least one person with Brown eyes" which must be read as being equivalent to "I cannot see anyone who does not have Blue or Brown eyes".

Once she says this, everyone has received the answer to Information 3, which is: "I do not have X color eyes" i.e., "I must have A or B color eyes".

[of special note: everyone on the island of either A or B has conceived the following possible results before the Guru speaks, and therefore, the Guru knows that everyone of A and B knows the possible results before she speaks, as they are all perfect logicians, the above hypothetical distributions upon the island are logical and do not depend on anyone other than themselves knowing anything, and rely entirely on what they can see, only]

Everyone of color A will say: If I have color B, then color A will see 98 A, 101 B, and now knowing they have either A or B, will know they have A, and will leave the island on the first night, as no one on the island can see fewer than 99 of either A or B. After the first night, if they see that the 99 A have left, they will know that each of them saw 98 A, and 101 B, and then on night 2 the 101 B will leave the island.

Everyone of color B will say: If I have color A, then color B will see 98 B, 101 A, and now knowing they have either A or B, will know they have A, and will leave the island on the first night, etc etc etc.

Everyone on the island will know also, that if NO ONE leaves the island on the first night, then NO ONE saw 98 of either A or B, and therefore the distribution is 100/100, (see above) and everyone will leave the island on night 2. (this lines up with the 'everyone of color A will say: If I have color A, then....' and 'everyone of color B will say: If I have color B, then....'

And so, that's it. The Guru's statement has to be perfectly logical, and I cannot see how a perfect logician can choose to provide information that is irrelevant to 1/2 of a group of recipients (i.e., common-knowledge solution)

Have at me.