The Hatter, the Hare, the Rabbit, the Mouse, Alice and I met for a Tea Party. The table had two sides, with three chairs on each side, numbered like so:

Side A: 1 2 3
Side B: 6 5 4

Chair one was across from chair 6, chair 2 across from chair 5, etc. We would sit down one at a time, in any chair we choose. The first guest to sit down was called the first guest. After the second guest sat down, however, everyone moved clockwise to the next chair. This behavior was continued after the fourth and sixth guest sat down.

My question is, after everyone had been seated, where was I seated?

(1) The first guest sat in chair 1.
(2) The mouse sat down last.
(3) After the first time everyone moved, no one had been on side B yet.
(4) The Hatter and Alice choose the same starting chair.
(5) When I was in chair 1, the Rabbit was in chair 6.
(6) I started on side B, but didn't end there.
(7) The hare moved all three times.
(8) Alice ended up between the mouse and the hare.
(9) I sat down after the rabbit but before the hatter.
(10) The mouse sat down in an odd numbered chair.

Also, a bonus question. Who am I?

I'm sure this has been solved below, but here it is again:

Guest #:     1          2          3          4          5          6

Name:        Hare     Alice     Rabbit   Me        Hatter   Mouse

Chair:         1          2

1st move:    2          3         5           6

2nd move:   3          4         6           1          2          5

3rd move:    4          5         1           2          3          6

I'll go a step farther and show that this is the unique solution under the given constraints.  By (9) and (2), there were at least two guests seated after me.  By (1) and (3), The first guest sat at 1, and the second at 2.  Thus, by (6), I can't have been either of these.  By (5), the Rabbit must be to my right [note that I must sit at 1 at some point by (6)].  By (9), the Rabbit was seated before me, and since neither of the first two guests ever reaches 6, the Rabbit must be third, and I must be fourth.  For (9) to hold, i must have sat in 5 or 6.  If i had sat in 5, the Rabbit would have sat in 4.  This means, after 2 moves, the only odd seat left is 1, so thats where the mouse sat by (10), which leaves spot 2 for the Hatter [who must be after me by (9)].  But this puts the hatter and the Rabbit on opposite sides of the table for all rotations, which violates (8).  Therefore, I must have been fourth and sat at 6, which forces the solution above.  So this is the only possible solution.

As for the bonus, am i the Cheshire Cat?

Posted by josh on 2005-06-28 19:41:36