I searched the forum for swiss discussions, but couldn't find this information in particular, and it's not in the Wiki either: What is the method for seeding the Round 1 Games in a Swiss System Tournament?
Based on a tournament we're in now, it looks like it's based on something like board ranking, but just looking for a confirmation.
Would also request a Wiki edit for whatever the confirmed answer is (just because it is not specified there either).
I vaguely remember us discussing an initial seeding, but I forget what was implemented and can't find the discussion thread.
Any OG players here know for sure. Because I'm hearing mixed reviews about whether R1 seeding actually happens.
Swiss tournaments have R1 seedings. I've looked back at several now and see they all have. I thought I'd played without but that was just due to the boards having little play on them. it's most visible in WGWF tourneys where almost all participants have public ranking and some very high ones.
If there are x R1 games then the x highest ranked players will be in different games. Using the example of 100 players and 10-player games, 1-10 will not see each other. But #10 will be lumped together with 11-19 and so not get an easy game. That's right, it's not random for the unseeded. Unseeded (11-100) get placed with people close to them in the rankings, as well as one seeded player. No idea why. The 10-19 game ahould probably, in fairness, have been something like, say, 10, 15, 18, 22, 34, 51, 58, 77, 85, 96. But that doesn't happen. Instead 10, 11, 12, 13, 14, 15, 16, 17, 18, 19.
hi - I've taken a look at the code and this is what happens with an example for 8 players
1. Sort players in order of their ranking score
12345678
2. Split the list of players into two groups, top half and bottom half
1234 and 5678
3. Reverse the order of the bottom half group of players
1234 and 8765
4. Merge the two groups into one to form the seed list
12348765
5. For each game, take a player off the top of the list and match them with one on the bottom of the list and continue until all the players are matched
Here are the resulting matchups:
1v5
2v6
3v7
4v8
The idea is that the top player from the top half plays the top player from the bottom half, then the 2nd best player plays the next 2nd best player from the bottom half.
For team games, we use the average (mean) ranking score for the team members to seed the teams.
For players unranked on the board (have never played in a public non-teamplay game that has finished) they get a ranking score of 1000 for seeding purposes.
As to why the seeding was done this way, it was taken from guidance on how to seed a Swiss system chess tournament. I don't have the original source URL unfortunately but this is the original text from my notes that was used to design the seeding system:
"The current system "seeds" players according to their rating. Players are listed from highest rating to lowest, and unrated players are listed at random at the bottom, then assigned a pairing number for the tournament. The top half of the list then plays the bottom half of the list (if there are 32 players in the section, #1 plays #17, #2 plays #18, etc.)"
Tom, that is a good description of a Swiss 1-on-1. It doesn't explain what we see in multiplayer.
Litotes wrote:Tom, that is a good description of a Swiss 1-on-1. It doesn't explain what we see in multiplayer.
I thought I covered that under teamplay:
For team games, we use the average (mean) ranking score for the team members to seed the teams.
Is that what you mean or something else?
I'm guessing that #5:
5. For each game, take a player off the top of the list and match them with one on the bottom of the list
is just tweaked for multiplayer with N people in a game:
5. For each game, take a player off the top of the list and match them with N-1 on the bottom of the list.
Ozyman wrote:I'm guessing that #5:
5. For each game, take a player off the top of the list and match them with one on the bottom of the list
is just tweaked for multiplayer with N people in a game:
5. For each game, take a player off the top of the list and match them with N-1 on the bottom of the list.
Thanks Ozy, that looks right.