Incompatibility between Swiss and Round-Robin


Round-robin tournaments and Swiss pairing are not compatible. If you want to play round-robin, you should make a complete pairing for all rounds in advance. You cannot just keep pairing Swiss and hope that it works, at some point you will run into the problem that there no longer exists a pairing where every player gets an opponent they have not met yet. This page demonstrates the problem by pairing a group of 6 players according to the Swiss method and showing that there is no possible pairing for round 4.

The setup

There are 6 players: Alice, Billy, Chris, David, Erica, Fritz. We want them all to play each other, but try to make that happen by Swiss pairing.

First round

Since we're playing Swiss, any pairing will do.

Pairing and results

Black White Result
Alice David 1-0
Billy Erica 1-0
Chris Fritz 0-1

Standings after round 1:

# Name Pt R1 SOS
1 Alice 1 4+ 0
2 Billy 1 5+ 0
3 Fritz 1 6+ 0
4 David 0 1- 1
5 Erica 0 2- 1
6 Chris 0 3- 1

Second round

In this round, pretty much any pairing within each group is possible, as there is no differentiation by SOS yet. The only criterion to declare a pairing undesirable is if both players held the same color in the first game. So Alice - Billy and David - Erica are considered undesirable, and are not in the pairing.

Pairing and results

Black White Result
Fritz Alice 0-1
David Billy 1-0
Erica Chris 1-0

Standings after round 2

# Name Pt R1 R2 SOS
1 Alice 2 2+ 4+ 2
2 David 1 1- 3+ 3
3 Billy 1 5+ 2- 2
4 Fritz 1 6+ 1- 2
5 Erica 1 3- 6+ 1
6 Chris 0 4- 5- 2

Third round

Now there is some differentiation by SOS, so we can use it for pairing. From the middle group, we want to pair Erica up (lowest SOS) and David down (highest SOS).

Pairing and results

Black White Result
Alice Erica 1-0
David Chris 1-0
Billy Fritz 1-0

Standings after round 3

# Name Pt R1 R2 R3 SOS
1 Alice 3 2+ 4+ 5+ 4
2 David 2 1- 3+ 6+ 5
3 Billy 2 5+ 2- 4+ 4
4 Fritz 1 6+ 1- 3- 5
5 Erica 1 3- 6+ 1- 5
6 Chris 0 4- 5- 2- 4

The problem

The next round is a problem. Alice, Billy and Chris have all played against David, Erica and Fritz in some order, and have not played each other. And the same is true vice versa. David, Erica and Fritz have all played against Alice, Billy and Chris in some order, and have not played each other.

So for the next round we must pair Alice, Billy and Chris amongst themselves, and David, Erica and Fritz amongst themselves. Since both these groups have an odd number of players, no such pairing is possible. (Note that the results of round 3 are irrelevant. Once the round 3 pairing happens, round 4 is impossible to pair regardless of the outcomes of round 3.)

So now we have demonstrated that it is not guaranteed to be possible to create a Round-Robin tournament with Swiss pairing. No special tricks were used, all pairings were optimal according to commonly used Swiss pairing rules.

Incompatibility between Swiss and Round-Robin last edited by HermanHiddema on March 14, 2014 - 16:09
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