Problem 5.31 Ryan H Kian L Jun Oh Y

Starting Out:  Define a Round Robin Tournament:  A tournament in which each player plays every other player. There can be no ties.  Suppose we have 3 players: Rock, Paper, and Scissors  If Rock beats Scissors, Scissors beats paper, and Paper beats Rock, who wins ?

Uh-Oh! Rock Paper Scissors

Is there a winner?  But a player x is a top player if:  For every player y, x beats y or beats some other player who beats y.  Who is the top player in our example?  There is no clear winner.

Every tournament with 2 players has a top player!  Let’s call our two players x and y.  One player must win.  Without loss of generality, call this player x.  Calling the winner x is totally arbitrary! xy

But what about 3 players?  We know that a two-player game between x and y must have a winner, x.  Let’s call our new player z.  Is there an easy way to figure out who is a top player?  How about cases?  Think: either x beats z, or loses to z.

YES! A 2 case model!  Case 1: x beats z  If x beats z, then x is still the top player.  Does it matter what happens between y and z ? z y x

Case 2: z beats x  If z beats x, then z is a top player.  Since x beat y, and z beat x, it does not matter what happens between y and z. xy z

The 3-Player Model  Generalized view Set A Set B x

Now it gets tricky...  What about four players?  Well, in a 3 player tournament we have:  A top player x  The set of players that x beat, called A  The (possibly empty) set of players that x lost to. We shall call this set B.  Since x is a top player, every player in B must have lost to at least one player in A.

Moving to 4 Players  Now a new player w joins.  Is x a top player here?  Is a two case model still appropriate?  Basically. Either x beats w, or w beats x. x y z w

Case 1: x beats w  Then w belongs in Set A. x is still a top player Set A Set B x W

Case 2a  If w beats x but loses to a member of A, then x is still a top player Set A Set B x w

Case 2b  If w beats x but does not lose to any member of A, then w is the new top player! Set A Set B w x

The relationship between A and B  Remember, for top player x we have  The set of players that x beat, called A  The set of players that x lost to, called B.  Since x is a top player, every player in B must have lost to at least one player in A.

So what does this mean?  If a 1 is a top player with k ≥ 2 players, we can symbolize sets A and B like this:  A = { a i : i = 2,..., j, where a 1 beats a i }  B = { a i: i = j + 1,... K, where a i beats a 1 }  Can B be an empty set? How about A?  Does a relationship exist between A and B?  Does every member of A need to beat someone in B?  How can we symbolize this with quantifiers?

Quantifier Time!  The relationship is this:  Can we formulate a relationship between members of A and B that starts off by quantifying elements of A?

Induction step  Assume: for some number of n players, every tournament has a top player.  Choose a tournament with n + 1 players.  Show that for every tournament with n+1 players, there exists a top player.  For now, ignore player n + 1’s results.  Within the n players we are observing, by assumption we have a top player, a 1.

Tournament with where k≥2  Case 1  a 1 beats a n+1  a 1 is a top player Set A Set B a1a1 a n+1

Tournament with where k≥2  Case 2. a n+1 beats a 1  Subcase 1: there exists an a i within set A that beats a n+1.  a 1 is a top player Set A Set B a n+1 a1a1 aiai

Tournament with where k≥2  Case 2. a n+1 beats a 1  Subcase 2: a n+1 beats everyone in set A  Do the games between a n+1 and Set B matter?  No, a n+1 is a top player! Set A Set B a1a1 a n+1

We’re Done!  We have shown the induction step  We have shown the basis step (for k = 2)  So as long as we have 2 or more players,  We always have a top player!

Q.E.D.!  A Ryan, Jun Oh, and Kian Production Copyright 2006 Honor code honored. (We Certify) Set A Set B a1a1