In the tournament that started this thread, it’s interesting to speculate on the thought processes the organizers must have gone through at the various stages:
Thought #1: “Only 6 players? Then we need to watch out for the 6-player trap. Actually, though, we also have 6 rounds, so there are going to be repeat pairings anyway. So let’s not worry about the trap. If some of the repeat pairings happen already in round 4, it’s not the end of the world.”
Thought #2: “Oh, now we have 7 players? Perfect! Then we can make it a complete round robin, and everything will work itself out automatically!”
Thought #3: “Wait a minute, then we’ll need a 7th round, because everybody will be getting a bye at some point. Let’s just start off pretending it’s a round robin, and stop after 6 rounds.”
Of course, Thought #1 is highly questionable, and Thoughts #2 and #3 are downright wrong. There are just too many “they’ve-already-played-each-other” pairing traps. The Crenshaw-Berger round-robin tables (chapter 12 in the 6th edition rulebook) must be used.
In looking at the Crenshaw-Berger table for 7-8 players last night, I noticed a few things:
- If the rounds are played in order, then Player 8 alternates colors every round, and
- Players 1-7 each alternate colors every round, except for the round in which they face Player 8, at which time they each have the same color as either the previous or the following round, and
- Therefore, if there are only 7 players and Player 8 is the bye, then all players alternate colors in every round.
Therefore, when using the table for 7-8 players when there are 7 players, it is highly desirable that Player 8 be the bye.
It is not important that Players 1-7 be in rating order. In fact, rating order is highly undesirable for a couple of reasons. The players should be assigned numbers 1-7 randomly, without regard to ratings.
But Player 8 should still be the bye. That way, players 1-7 each get 3 whites and 3 blacks. If a different number (1-7) is designated as the bye, then at least 1 player (probably 2, maybe 3) will end up with 4 whites and 2 blacks, and a similar number (1, 2, or 3) will end up with 2 whites and 4 blacks. (Work it out for yourself from the table, if you don’t believe me.)
As Ken Sloan pointed out, when using the round robin table to pair what started out as a Swiss, you don’t even need to have assigned numbers 1-7 to the players before round 1. You can do it during or after round 1. Just assign numbers 5 and 3 (white and black respectively) to the players in one of the games, likewise numbers 6 and 2 to the players in a second game, and 7 and 1 to the players in the third game. Assign number 4 to the player who is sitting out this round, i.e. to the player who is “playing” number 8 (the bye).
That way, you align your already-made “Swiss” pairings with the pairings laid out in round 1 of the round robin table.
Then pair all future rounds following the table, round by round. You can even post all the pairings (for all the rounds) when you post round 2, or even while round 1 is still being played.
Even if rounds 1 and 2 have both been paired, before the decision to “go round robin” was made, it should still always be possible to fit the players perfectly (including colors) into two of the rounds in the table. These two rounds might not be 1 and 2, though. They might be 1 and 5, or 3 and 6, or whatever. Doesn’t matter. Then the remaining four (or five) rounds can be played in any order. You might even decide, when choosing which round number to play next, to choose the round where the pairings look most like the ones a Swiss algorithm would have made. Don’t change any colors. Use the colors from the table. Otherwise, you will probably paint yourself into a corner for future rounds.
Don’t wait any longer than round 2 to “go round robin”. If you also pair round 3 Swiss-style before looking at the table, you will likely be out of luck, i.e. you may have already fallen into the 6-player trap (or its 7-player counterpart), and you won’t notice it until you try to pair round 4.
Bill Smythe