You have a twentysided die (a regular icosahedron) with an arrow on each face. You play a little game with the die. You place the die flat on the table. You rotate the die in the direction of the arrow on the top face. This step is repeated each time by looking at the arrow on the top face of the die. The game is over when you see the same arrow pointing in the same direction twice.
If you can choose the directions of the arrows and the starting position of the die, what is the longest this game can last?
What is the farthest the die can go from the start to the end of the game?
Below is shown a 15move sequence of numbers showing on the top. As it's odd, the pointing of the top face will be reversed, and in fact the direction toward the next face will rotate 60 degrees from the previous time it was up. The cycle therefore has 6*15=90 moves. There can be three moves that lead into this at position 2, making a total of 93 moves.
/ \ / \ / \ / \ / \
/ \ / \ / \ / \ / \
/ \ / 1 \ / 2 \ / 3 \ / 4 \

\ 13 / \ 15 / \ / \ 7 / \ 5 / \
\ / \ / \ / \ / \ / \
\ / 14 \ / \ / 8 \ / 6 \ / 12 \

\ / \ / \ 9 / \ 10 / \ 11 /
\ / \ / \ / \ / \ /
\ / \ / \ / \ / \ /
If I understand the arrowheads correctly they point in the opposite direction to what the next face is to be. Each time through the cycle of 15 they will be rotated 60 degrees from the previous time.

Posted by Charlie
on 20060406 22:06:43 