on the island, and makes the following public announcement: "At least one of you has blue eyes". The outsider, furthermore, is known by all to be truthful, and all know that all know this, and so on: it is common knowledge that he is truthful, and thus it becomes common knowledge that there is at least one islander who has blue eyes (). The problem: finding the eventual outcome, assuming all persons on the island are completely logical (every participant's knowledge obeys the axiom schemata for epistemic logic) and that this too is common knowledge. Solution The answer is that, on the kth dawn after the announcement, all the blue-eyed people will leave the island. Proof The solution can be seen with an inductive argument. If k = 1 (that is, there is exactly one blue-eyed person), the person will recognize that they alone have blue eyes (by seeing only green eyes in the others) and leave at the first dawn. If k = 2, no one will leave at the first dawn, and the inaction (and the implied lack of knowledge for every agent) is observed by everyone, which then becomes common knowledge as well (). The two blue-eyed people, seeing only one person with blue eyes, and that no one left on the first dawn (and thus that k > 1; and also that the other blue-eyed person does not think that everyone except themself are not blue-eyed , so another blue-eyed person ), will leave on the second dawn. Inductively, it can be reasoned that no one will leave at the first k − 1 dawns if and only if there are at least k blue-eyed people. Those with blue eyes, seeing k − 1 blue-eyed people among the others and knowing there must be at least k, will reason that they must have blue eyes and leave. For k > 1, the outsider is only telling the island citizens what they already know: that there are blue-eyed people among them. However, before this fact is announced, the fact is not common knowledge, but instead mutual knowledge. For k = 2, it is merely "first-order" knowledge (). Each blue-eyed person knows that there is someone with blue eyes, but each blue eyed person does not know that the other blue-eyed person has this same knowledge. For k = 3, it is "second order" knowledge (). Each blue-eyed