11/27   Let's talk about ilyas' blue-eyed problem. Again. I still don't see
        how the base case works. Assume 1 blue eyed person only. How would
        that person know that he should commit suicide? Wouldn't the brown
        eyed person think the same way and commit suicide? Think induction.
        Tom's basis is flawed.
        \_ The big problem with this whole thing is that it assumes that
           all these people have taken math 55 and understood it.
        \_ Please report to the food vats immediately.
        \_ No.  The key is that there exists at least one blue eyed person.
           If only 1 blue eyed person exists, he will notice that as far
           as he can tell, no one has blue eyes.  Since at least one person
           must have blue eyes, it must be him/her.  So (s)he kills himself.
           \_ Added to that, a brown eyed person would see that there is a
              person with blue eyes, and would expect him/her to commit
              suicide after the first day. And when (s)he does, the brown
              eyed person concludes that that person was the only blue
              eyed person in the town.
           \_ Can you explain the case of three blue-eyed people, both for
              blue and brown-eyed people. Won't everyone expect someone
              else to commit suicide?
              \_ Another way to explain it:
                 [someone else motd-mashed the first explanation]

                 For two blue-eyed ppl, a blue-eyed person would see one
                 blue-eyed person on the first day.  At the end of the
                 second day, both blue-eyed people would kill
                 themselves.

                 For three blue-eyed ppl, a blue-eyed person would see
                 two blue-eyed people on the first day.  He would see
                 them again on the second day.  Assuming that there are
                 only two blue-eyed people total, they would kill
                 themselves at the end of the second day.  But we are
                 saying that our given person is blue-eyed, and there
                 are three blue-eyed ppl.  On the third day, this
                 blue-eyed person would STILL see two blue-eyed people.
                 At the end of the third day, all three blue-eyed people
                 conclude that there must be three blue-eyed people, and
                 they must be one of them, and kill themselves.
                 \_ Ah, ok, I finally get it. Thanx! -stupid Math 55 flunkie

                 For two blues and one brown, from a blue's point of view
                 there is one other blue and one brown. Now at the second
                 day, each blue knows he's a blue, because if he was brown
                 the other blue would have known he was the only blue since
                 there is at least one blue and the other two are brown.
                 So both blues kill themselves.  From the brown's point of
                 view, he knows he's a brown by the third day, since the
                 blues wouldn't have killed themselves yet had he been blue.
                 --0x48
        \_ Take math1a or 55 or high school math. learn induction. 0xAFB
        \_ one of the many problems is with ilyas' question. He should add
           that each person can see every other person ONCE PER DAY. This
           question really reflects ilyas' intelligence.
           \_ 1.) The kingdom was small: any person in the country could meet
                  all others in one day.
              ?
              \_ it says 'could', but that doesn't mean it is necessarily so.
                 \_ Rule 1 suggests that they could if they wanted.
                    Rule 2 suggests that everyone really wants to.
                    The weakness perhaps is we have a lot of inferences.
                    \_ The weakness is that it's a stupid problem.  -tom
                       \_ I thought the puzzle was a good one and I learned
                          some things.  Of course, the problem did have
                          problems.