下面是Leah找到的一个题目，转自墙外的某篇博客。讲到这个故事的起因，还是GEB中的递归概念。
There is an island upon which a tribe resides. The tribe consists of 1000 people, with various eye colours. Yet, their religion forbids them to know their own eye color, or even to discuss the topic; thus, each resident can (and does) see the eye colors of all other residents, but has no way of discovering his or her own (there are no reflective surfaces). If a tribesperson does discover his or her own eye color, then their religion compels them to commit ritual suicide at noon the following day in the village square for all to witness. All the tribespeople are highly logical and devout, and they all know that each other is also highly logical and devout (and they all know that they all know that each other is highly logical and devout, and so forth).
[for the purposes of this logic puzzle, “highly logical” means that any conclusion that can logically deduced from the information and observations available to an islander, will automatically be known to that islander.]
Of the 1000 islanders, it turns out that 100 of them have blue eyes and 900 of them have brown eyes, although the islanders are not initially aware of these statistics (each of them can of course only see 999 of the 1000 tribespeople).
One day, a blue-eyed foreigner visits to the island and wins the complete trust of the tribe.
One evening, he addresses the entire tribe to thank them for their hospitality.
However, not knowing the customs, the foreigner makes the mistake of mentioning eye color in his address, remarking “how unusual it is to see another blue-eyed person like myself in this region of the world”.
What effect, if anything, does this faux pas have on the tribe?
The interesting thing about this puzzle is that there are two quite plausible arguments here, which give opposing conclusions:
[Note: if you have not seen the puzzle before, I recommend thinking about it first before clicking ahead.]
Argument 1. The foreigner has no effect, because his comments do not tell the tribe anything that they do not already know (everyone in the tribe can already see that there are several blue-eyed people in their tribe).
Argument 2. 100 days after the address, all the blue eyed people commit suicide. This is proven as a special case of proposition. Suppose that the tribe had n blue-eyed people for some positive integer n. Then n days after the traveller’s address, all n blue-eyed people commit suicide.
Proof: We induct on n. When n=1, the single blue-eyed person realizes that the traveler is referring to him or her, and thus commits suicide on the next day. Now suppose inductively that n is larger than 1. Each blue-eyed person will reason as follows: “If I am not blue-eyed, then there will only be n-1 blue-eyed people on this island, and so they will all commit suicide n-1 days after the traveler’s address”. But when n-1 days pass, none of the blue-eyed people do so (because at that stage they have no evidence that they themselves are blue-eyed). After nobody commits suicide on the (n-1)^{st} day, each of the blue eyed people then realizes that they themselves must have blue eyes, and will then commit suicide on the n^{th} day.
Which argument is valid?
我也不确定自己的逻辑是没有破绽的，反正我就说说我的分析（欢迎讨论）：我觉得Argument 1是对的，关于Argument 2的解释仅在n<3的时候成立。
我现把我觉得比较重要的假设理一下。首先这个部落的人在外国人的到来之前都是好好活着的，并且他们是拥有逻辑推理的。那在外国人的明示之后，或者是没有任何影响，或者是族群里的蓝眼人集体自杀，其中的差别仅仅在于外国人说的那句话。那么那句话到底传达了什么结果信息呢？我觉得有两个层面，一是“部落里至少有一个人是蓝眼睛的”，二是“部落里的每一个人都知道部落里至少有一个人是蓝眼睛的”。但是这两条信息在这个外国人说穿之前，部落里的人都已经知道了（因为部落里蓝眼睛的人数大于2）。
其实也就是说，只有当外国人传达的两层信息是部落人本来不知道的，才能产生自杀的影响。当n=1的时候，那个唯一的蓝眼睛的人看到的都不是蓝眼睛的人，也就是说他不知道外国人的第一层信息。一旦得知第一层信息，就很容易推断出自己的情况。这个唯一的蓝眼睛的人自杀的开关，就在于这第一层信息的知晓与否。
当n=2的时候，所有的人（包括那两个蓝眼睛的人）本来就都知道第一层信息，即他们都能看到至少一个蓝眼睛的别人。但是其中一个蓝眼睛的人并不知道“另一个蓝眼睛的人知道部落里至少有一个人是蓝眼睛的”。而外国人的第二层信息捅破了这层纸。当第一天没有人自杀的时候，这所传达的讯息就是“另一个蓝眼人知道除了他自己本身外还有蓝眼人”。这两个蓝眼人的自杀的开关，就在于这第二层信息的知晓与否。
但是一旦n大于等于3的时候，这两层信息都失效了。因为此时，部落里的每个人都知道“部落里至少有一个人是蓝眼睛的”，并且每个人也都知道“部落里的每一个人都知道部落里至少有一个人是蓝眼睛的”。如果说光凭这个外国人所说的这段话，所有蓝眼人都要自杀了，那么这些信息部落里的人早就拥有了，他们为什么没有更早地自杀呢？
另外，关于自杀论的n从1到后面的推导也有问题。当第一天没有人自杀的时候，它的意义仅仅在于n不等于1。当n=x的时候，在蓝眼人的心目中部落里的蓝眼人总数要么是x要么是x-1。所以对于n=2的时候，他可以推断出n=2。但是如果n>3的话，大家老早就知道N不等于1了，那第一天没人自杀说明不了任何问题。此时的第一天没人自杀无法等同于n不等于1，同时第二天没人自杀也无法等同于n不等于2。比如说这道题目里的蓝眼人有100人，也就是说大家的共识已经是蓝眼人至少有99人，并且每个人都知道别人知道这回事。那就完全不能从前99天没人自杀这件事上推导出任何结果。
后面Leah补充说，不仅100天后所有蓝眼睛的人都自杀，第101天所有棕色眼睛的人也会自杀。这就更匪夷所思了，这另外的一些人怎么知道自己的眼睛不可能是黑色或是绿色或是其他任何颜色的呢？不过如果自杀论推论的真成立了，倒真的能算是GEB中的递归了吧。
28/10/11半夜突然想到的一个思路：
我们从三个蓝眼人的情况来看，要确定有三个蓝眼人的办法是通过以下的线索：
A）在蓝眼人的心目中蓝眼人的总数要么是2要么是3。
B）第二天没有人自杀，说明蓝眼人的总数不是2。
那么B）是如何推导出来的呢？B）是下面C）的反例，
C）第二天有人自杀的话，说明蓝眼人的总数是2。
那么C）是如何推导出来的呢？必须是通过以下的线索：
D）在蓝眼人的心目中蓝眼人的总数要么是1要么是2 。
E）第一天没有人自杀，说明蓝眼人的总数不是1。
所以说，要推导出一共有3个蓝眼人，必须满足A）、B）、C）、D）、E）这五个所有的线索。但是显然A）和D）是互相矛盾的，不可能同时满 足A）和D）然后结果有3个蓝眼人。所以当n>2以后的推导是不正确的。推导错误的原因就在于在每次推导中，嵌套了不一样的前提假设，而这是不合理 的。因为不管有多少个蓝眼人，其蓝眼人的总数是不变的，其假设也应该仅仅停留在“x或x-1”，不能从“1或2”、“2或3”这样上去。
我觉得我分析得好有道理啊！
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