I would like help understanding my flawed logic in my following reasoning:
The Blue-eyes Riddle is commonly expressed as
A group of people with assorted eye colors live on an island. They are all perfect logicians -- if a conclusion can be logically deduced, they will do it instantly. No one knows the color of their eyes. Every night at midnight, a ferry stops at the island. Any islanders who have figured out the color of their own eyes then leave the island, and the rest stay. Everyone can see everyone else at all times and keeps a count of the number of people they see with each eye color (excluding themselves), but they cannot otherwise communicate. Everyone on the island knows all the rules in this paragraph.
On this island there are 100 blue-eyed people, 100 brown-eyed people, and the Guru (she happens to have green eyes). So any given blue-eyed person can see 100 people with brown eyes and 99 people with blue eyes (and one with green), but that does not tell him his own eye color; as far as he knows the totals could be 101 brown and 99 blue. Or 100 brown, 99 blue, and he could have red eyes.
The Guru is allowed to speak once (let's say at noon), on one day in all their endless years on the island. Standing before the islanders, she says the following:
"I can see someone who has blue eyes."
Who leaves the island, and on what night?
There are no mirrors or reflecting surfaces, nothing dumb. It is not a trick question, and the answer is logical. It doesn't depend on tricky wording or anyone lying or guessing, and it doesn't involve people doing something silly like creating a sign language or doing genetics. The Guru is not making eye contact with anyone in particular; she's simply saying "I count at least one blue-eyed person on this island who isn't me."
And lastly, the answer is not "no one leaves."
The generally accepted solution states that all the blue-eyed people leave on the 100th day by the inductive reasoning of
A | B | C | D | |
---|---|---|---|---|
What A sees | ? | Blue | Blue | Blue |
What A knows B sees | ? | ? | Blue | Blue |
What A knows B knows C sees | ? | ? | ? | Blue |
What A knows B knows C knows D sees | ? | ? | ? | ? |
and hinges on the idea that someone could hypothetically think someone else does not know there are blue-eyed people on the island.
However, this does not account for parallel induction stated as
A | B | C | D | |
---|---|---|---|---|
What A knows B sees | ? | ? | Blue | Blue |
What A knows C sees | ? | Blue | ? | Blue |
What A knows D sees | ? | Blue | Blue | ? |
which when combined with the sequential induction collapses to
A knows ABCD knows ABCD knows >0 blue eyes
This combined induction covers all purely sequential inductions past k=4 blue eyed peoples. Therefore, I propose the following proof:
Theorem 0. If it can be proven that everybody knows that no one thinks there are no blue-eyed people then the oracle does not provide extra information and no one leaves the island
Theorem 1. If you see 1 blue-eyed person, you can prove that there are >0 blue-eyed people
Theorem 2. If you see 2 blue-eyed people, you can prove that everyone can prove there are >0 blue-eyed people
Theorem 3. If you see 3 blue-eyed people, you can prove that everyone can prove that everyone can prove there are >0 blue-eyed people
In summary, the state of "What A knows B knows C knows D sees there are 0 blue-eyed people" is not possible given that "ABCD knows ABCD knows there are >0 blue-eyed people" and only possible states should be taken into consideration.
***EDIT:
Great answers by everyone who have commented or answered. They are a very good explanation of the logic to arrive to the general solution. I, however, am failing to grasp why the fact that everyone knows that blue-eyes are directly observable does not provide any information to anybody. There seems to me to be a contradiction that everyone can observe something (and everyone know it can be observed) and still believing that somehow someone may not be able to observe it.