Dividing players into "winners" and "losers": how to prove that greedy solution gives optimal result?

Question

I have a problem that states the following:

n players (where n is even) are to a play games against each other. Everyone will not necessarily play but a player can only play against someone else once. If two people do decide to play against each other, we have one loser and one winner. I then wish to partition my n players into two sets of size n/2: winners (W) and losers (L). I want all players in my winner set to have never lost against someone in my losers set.

This is impossible ex. for 4 players and games p1 won against p2, p2 won against p3, p3 won against p4 and p4 won against p1 then there is no way to partition the players into W and L. I do the next best thing, which is I wish to minimise my error: the number of pairs of players where a player in W has lost to a player in L (not playing against each other is not a loss).

I (think) I found a greedy solution to this problem. I simply sort the players by their number of losses and place the people with the least loses in my W set and fill in the rest to L. How do I go about proving that my greedy approach is in fact optimal? I have done several random tests and I can show that my approach will give a feasible solution but I don't know how to show that this does in fact minimise my error.

Answer 1

Consider the following outcomes:

Winner    Loser
Adam      John
Bob       John
John      Charles
John      David
John      Ernest
John      Frank
John      George

We tally up the losses and sort in ascending order:

Player    Losses
Adam      0
Bob       0
Charles   1
David     1
Ernest    1
Frank     1
George    1
John      2

Your algorithm divides the players as follows:

Winners    Losers
Adam       Ernest
Bob        Frank
Charles    George
David      John

The errors are (Charles, John) and (David, John); there are two errors. Consider instead the following division:

Winners    Losers
Adam       David
Bob        Ernest
Charles    Frank
John       George

There is no error in this division: there is no winner who lost to a loser. This is a better division, with less error; so your algorithm, as stated, is not optimal.

The fundamental problem with your algorithm is that it considers only the number of losses; prolific players can appear worse to this algorithm than they really are simply because they have more losses than others, despite possibly having many more wins.

Answer 2

Your greedy algorithm is not optimal. It fails for:

 W      L
===    ===
 A  vs  x
 B  vs  y
 C  vs  z
 B  vs  A
 C  vs  A
 x  vs  y

The optimal partition is W=(A,B,C), L=(x,y,z), but you will put A in the loser set, because he has 2 losses.

You say you did some randomized tests. How did you validate that your greedy algorithm produced the correct results for these tests?