Our small club in western NJ runs a long round robin tournament around this time every year. We usually get 10-12 players. This year 18 players showed up, causing a major dilemma.
We have just 13 Mondays to play our games. For reasons I cannot get into the director insists we play a round robin format. Is there any algorithm for constructing a “semi round robin” in which 18 players play only 13, 12, or 11 games? Obviously we would not play everyone in the field.
We have 13 Mondays remaining before the next event begins. We could live with 11-13 games and use the spare nights for makeup games.
You could divide the field into two halves. The simplest way would be:
1 2-3
4-5 6-7
8-9 10-11
…
(though the options for dividing equally has been discussed in another thread)
Then play the first 9 rounds in these two 9-player RRs. To finish, set up three 6-player RRs by taking three from each of these preliminaries - top 3 from each, middle 3 from each, bottom 3 from each. You already have six game results from the preliminaries that can carry-over to these finals, so you only need 3 more rounds against the players from the other preliminary to get all 15 games needed for this 6-player RR.
So each player plays 11 games (8 in the preliminaries with a bye and 3 in the finals with 2 carry-overs from the preliminaries) over 12 weeks.
You could run the tournament like a hybrid Swiss/Round-robin.
Pair the top board (don’t worry about colors), then go to the Round-robin tables in your Rulebook and look for the round with that pairing. Pair the remaining players accordingly. You should be able to repeat this process for your 12 remaining nights.
Not everyone will get to play against everyone, but most of your top players will still have the opportunity to play the other top players and you should have a legitimate first place finisher that way.
What are you planning on doing about players missing some of the Monday Nights? Plus I would imagine you will get some dropouts along the way.
But back to the original question. You could make it an elimination event. Say for example after 6 rounds any players with less than 2.5 points are out. Then have a seperate section with those players playing each other.
We used to run the Lincoln City Championship as a two stage event. We started with a 4 round Swiss, which was then used to pair round robin finals, usually 6 players per group.
Yes, this often meant that players in the latter rounds of the prelim faced each other again in the finals.
I was actually looking for an algorithm that would create 13 “fair” games in some logical manner, in one field.
Our club in fact does run a combination swiss/RR event, the Hackettstown Summer Holland. We begin with a five-round swiss qualifier, followed by two or three parallel RR events depending on the number of players. Unfortunately we usually don’t have 18 players for the summer tournament. We’ve had so few players at times that we get the crazy situation where a 1900 player plays 1200-1600 opponents for the entire summer, or a 1400 player plays only A-players.
By the way, has anyone here ever heard of the “holland” format?
I believe the director is going to create two groups, comprised of the top six players and the bottom twelve. The former will play a double RR, the latter a single RR. That leaves 1 and 2 (or 2 and 3) weeks for makeup games, respectively, for the two groups.
This situation arose because two players entered late. The director was happy with a 2 x 8 double RR, but the players insisted he rethink it so the other two guys could join.
With 18 players and 13 rounds, Swiss pairings are likely to become difficult, or even impossible (can’t avoid repeat pairings) in the later rounds.
Divide the players into two 9-player groups. To keep the two groups about equal in strength, put the odd-numbered (by rating) players in one group, the even in the other. Better yet, put 1-4-5-8-9-12-13-16-17 in one group, and 2-3-6-7-10-11-14-15-18 in the other. For the first 9 rounds, pair players in group A vs those in group B. Pair the remaining 4 rounds as a Swiss, making exclusively intra-group pairings.
Unless you have dropouts, I don’t see how that Swiss pairings could become “impossible.” You would have to pair people with widely varying scores, but so what? That’s exactly what happens in a round-robin.
For odd N, any time you have N+1 rounds and 2N players you run the risk of having to pair the same players twice. Look at the example he gave, the 10th round intra-group pairings are impossible - 9 players gives you 4 games within the group and one player who has to play the other group - and he’s already played all of them.
The case that I first saw before computer pairings was with 6 players over 4 rounds. And the computers tend to fall into the same trap, so check in the Nth round to make sure you haven’t divided the players into 2 groups before posting the pairings.
Try this: Pair a 6-player tournament as a Swiss. Make all the colors alternate in round 2, and again in round 3. Then tell me how you would pair round 4.
This is a guaranteed algorithm for impossible pairings.
That’s true, but in order to meet your second condition you have to tinker with the pairings in a way that violates Swiss rules. What you’re doing is making alternation of colors in round three more important than score. So yes, it’s possible to construct a case in which no legal pairings can be made in round 4. But I’m still waiting for a demonstration that this can happen if you follow the proper pairing rules. I’m not saying it’s impossible, but it’s up to you to provide an example.
Here’s where I discovered the problem personally: main.uschess.org/assets/msa_joom … 12062880.5
I was using a DOS version of SwissSys and with 5 sections going 4 rounds in one day I didn’t think to look ahead when a reasonable pairing came up in round 3.
At the time of the tournament, the ratings were:
Piper 909
Zhai 908
Gursoz 879
Kumar 849
Kerr unr
Jones unr
Rd 1
1 L 4
5 L 2
3 W 6
Rd 2
2 L 3
4 D 5
6 L 1
The natural pairings are:
Rd 3
3(2 points) vs 4(1.5 points)
1(1 point) vs 2(1 point)
5(.5 point) vs 6(0 points)
but now the fourth round requires a duplicate pairing, because the players have been divided into even and odd groups.
Interesting. The way to fix it is to transpose 1 and 4 in round three, but in order to see that you have to anticipate the problem in round 4. If the number of rounds is less than the number of players there must always be a way to do this, but it may not be obvious.
Having run a 12 player round robin over the course of a number of weeks, I can tell you it’s a major pain in the butt. People don’t show up, matches have to be rescheduled, and people get aggravated and drop out. We had 12 players initially and then another 4 showed up the following week. I wanted to split the original 12 players into a group of 8 and 4 and add the additional 4 players to have two 8 player round robins. The person who came up with the idea, would have none of that. It took almost 4 months to complete, and tied up a large group of players for that period of time. If the guy wants to do another round robin this summer, I’m insisisting that it be split up if more then 10 players.