This is one of those mathematical problems which is easier to solve in general than in the specific cases presented.  So suppose there are n people throwing a Frisbee and let's determine the longest sequence of throws without a repetition.  Several people noted that the length of the longest sequence of throws is bounded above by n(n-1), but that's a far cry from showing that this longest play can be achieved.  So let's do that.  Label the players 1, 2, ... , n.  With two players (n = 2) the solution is obvious, namely 121. By induction, we'll suppose that for a game with n-1 we have achieved a longest play of length (n-1)(n-2).  Player n now comes and wants to join the old n-1 players.  The old players do the same thing they did before, except that the first time any one of them receives the Frisbee, he/she throws it to player n who then throws it back again.  This creates a chain of length (n-1)(n-2) + 2(n-1) = n(n-1) having no repetition, which is the required maximum.  As an example, using this algorithm we would have the sequences 1312321, for n = 3, 1413431242321 for n=4 and finally 151454135343125242321 for n=5.

You are visitor number 2951 to this page.
ã2004 Alberto L. Delgado