Donna, you have run afoul of the infamous Dennis Keen trap.

Dennis Keen was a middle-level TD in the 1970’s, now deceased. One fateful weekend, the top section of his 5-round Swiss drew only six players. Blissfully, he paired the first 3 rounds as a Swiss, only to discover that there were NO possible pairings in round 4.

Many TDs, both more and less illustrious than Dennis Keen, have fallen into this same trap many times.

The trap lurks whenever there are six players in either a 4- or a 5-round Swiss. It also lurks whenever there are five players, because then the sixth player, Mr. Bye, is supposed to be paired according to the same rules as the other five – do not pair him against the same opponent twice.

In your tournament, you have created two “camps”, B-C-D and A-E-bye, who have played exclusively inter-camp pairings, no intra-camp pairings. So if now B plays C, whom does D play? Or if B plays D, whom does C play? Or if C plays D, whom does B play? And similarly for the other camp.

The following theorems are true of six-player tournaments of either 4 or 5 rounds. The proofs are left as exercises for the reader. By extension, the theorems are also true of five-player tournaments, with Mr. Bye playing the role of the sixth player.

Theorem 1. The first 2 rounds may be paired any way you wish, without causing problems in later rounds.

Theorem 2. After round 3, there is trouble for round 4 if, and only if, there now exist 2 three-player camps who have played exclusively inter-camp pairings, no intra-camp pairings.

Theorem 3. The easiest way to fall into the trap is to make all the colors alternate for all the players in both rounds 2 and 3. (In that case the two camps are those who started with white, and those who started with black.)

Theorem 4. Immediately after making tentative round 3 pairings, check to see whether round 4 pairings exist. (Don’t worry about score or color, just make sure nobody is facing the same opponent twice.) If round 4 pairings don’t exist, change the round 3 pairings.

Theorem 5. If round 4 pairings exist, then round 5 pairings will automatically exist. (Just pair each player against the one opponent he has not yet faced.)

Theorem 6. After 2 rounds, there are exactly four ways to pair round 3 (not including color reversals). Three of these are OK. Only one leads to trouble in round 4.

Theorem 7. After 2 rounds, draw a regular hexagon on paper. Write the names of the six players at the vertices, in such a way that two players are at adjacent vertices if and only if they have already played each other. Then, the round 3 pairing that leads to trouble is the one with three diagonal (diametrically opposite) pairings. The OK pairings are those with one diagonal and two non-diagonal pairings.

Try it, you’ll like it.

Bill Smythe