Swiss Pairing Question

Running Chess Tournaments
21

Deanna, your first two examples have an incorrect initial setup. As noted in another thread, seeing the full history makes it much easier to answer the questions. Also, the natural pairing for the odd man is to have the bottom player in the upper score group play the top player in the lower score group, and your listed round two pairings did not do that.

As a side note, the pairing ranges are greater than the normal ranges the USCF rulebook was written for, so adjustments to the 200-point and 80-point limits may be in order.

In A, the round one pairings would have been (colors unknown)
4213 (1) - 2855 (0)
3171 (0) - 2065 (1)
2739 (0) - 1892 (1)

In any case the natural pairings for round two would be
4213 - 2065 (top two 1s)
1892 - 3171 (bottom 1 vs top 0)
2855 - 2739 (bottom two 0s)

If you opt for Harkness pairings (drop the middle) then the raw
4213 - 1892
2065 - 3171
2855 - 2739
would be changed (to avoid the repetition on board two) to
4213 - 1892
2065 - 2855
3171 - 2739

In your second example the first round would have been

4055 (0) - 2329 (1)
3147 (1) - 2088 (0)
2913 (0) - 1949 (1)

The natural round two pairings would are
3147 - 2329
1949 - 4055
2913 - 2088

Harkness would be modified to
3147 - 1949
2329 - 2913
4055 - 2088

Your third example is incomplete. If I make the assumption that you really meant round 2 (since only three points have been awarded so far in the tournament) then I run into a problem in that E and F should have played each other in round one and thus should not both have zero points. The only possibility I can see is that the round one pairings were:
3322 (1) - 2701 (0)
2956 (1) - 1981 (0)
1997 (1) - 1000 (0) (the 1000 was originally going to have a bye, and then after the round had started the 1997 player showed up to play)

The natural round 2 pairings would be
3322 - 2956
1997 - 2701
1981 - 1000

22

Jw, thank you for your response. Allow me to provide additional details including the Round 1 pairs; I have gone back to the pairings for Round 1 and we do not get the same pairings for Round 1.

My understanding is that in order to determine Round 1 pairings, we should begin by listing the players in order from highest rating to lowest rating.

For example A, this gives us:
4213
3171
2855
2739
2065
1892

We then subdivide this group into S1 and S2.

Accordingly, we get:

S1: 4213, 3171, 2855
S2: 2739, 2065, 1892

Since this group is even (6 players), the top of S1 plays top of S2 and so on which gives us the following Round 1 pairings:

4213 v 2739
3171 v 2065
2855 v 1892

If there had been 7 players with the rating of the 7th player being 1500, this player would be awarded the Bye (this being Round 1), and the other pairings would remain as given above.

Do we agree so far? If not, please kindly indicate what rules you are using to generate Round 1.

Now assuming we agree on Round 1 pairings, assume winners are:
4213, 2065, and 1892; these comprose the 1-0 score group.

The 0-1 score group includes: 3171, 2855, and 2739.

Both the 1-0 and the 0-1 score groups are odd.

Thus, the lowest rated in the 1-0 group should play the highest rated in the 0-1 group.

We would then get the following pairings for Round 2:

4213 v 2065
1892 v 3171
2855 v 2739

These are the same pairings you stated for Round 2.

I also agree that we opted for Harkness pairings (drop the middle) then the raw pairings would be:

4213 - 1892
2065 - 3171
2855 - 2739

While I do not recommend the Harkness system; I always drop the lowest; this is where my question initiated from since 2065 v 3171 is a repeated pairing.

Do we all agree that we would swap 3171 with 2855 to give us:

4213 v 1892
2065 v 2855
3171 v 2739

Or would someone argue that we should swap 1892 with 2065 (assuming that this would not create a repeated pairing) to keep the rating difference as low as possible?

Sorry for the confusion in my first post. Assuming we agree thus far, I will re-present my third example, in a separate post.

I am not sure why we seem to not agree on the correct pairings for Round 1; I thought that was pretty much agreed upon.

There is one problem that I am aware of with our system; that being that the rule is when a score group is odd, the lowest rated in the top score group is paired with the top rated of the next lowest score group … our system simply puts the lowest rated in the top score group on TOP of the next lowest score group, to make the next score group even and pairs them up as normal. I know this is incorrect … this is an issue that I am trying to resolve, but the programmers keep telling me that it is simpler this way; when I suggest WinTD or SwissSyst or even tell them why can’t your system do it when these other programs can, I am told that they spent lots of money developing and fine tuning the programs over numbers of years at great expense.

I believe this is a battle I may never win; although I will keep trying; I was lucky to get them to do Round 1 pairings with the lowest rated getting the Bye instead of, for 7 players, 1 vs 5, 2 vs 6, 3 vs 7, 4 vs Bye (where the numbers indicate the players order/seed from high rating to low rating). The correct pairings would be 1 vs 4, 2 vs 5, 3 vs 6, 7 vs Bye. This alone took alot to convince the owner and programmers as they insisted that giving 4 the Bye was more logical mathematically and would result in the same pairings if there were 8 players. I won that battle, but I do not think I will win the other battle … unless they are willing to redesign the system, and I think there would need to be some strong and compelling reason to make them do that (they seem to care only about the top players being paired correctly).

I appreciate your help and suggestions. If you have strong arguments that would help me convince the programmers to make what appears to be a major change and one they think is more complicated to program, please share; I very much want to follow the correct pairing rules; there are actually many Swiss Systems out there; and different games use different systems. For example Scrabble uses Speed Pairings and Portland Swiss and Modified Swiss (in the Modified Swiss version, if the first score group is odd, they move the top player of the second score group to the first score group to get 1 v 4, 2 v 5, 3 v 6 (player 6 was moved to the first score group). In my view, this is worst than moving the lowest player from the first score group to the top of the second score group (as we do) as it is more likely to affect the outcome of the tournament.

For those of you that love to have debates over pairing rules, you may want to read pages 38 and 39 from the Scrabble Director’s Manual found at scrabble-assoc.com/director/ … 051201.pdf
(PS the other Swiss Systems are also interesting).

Regards.

DMDesiderata
Leagues Support Director
Swiss Training Director
Leaga.com

23

Deanna, you caught my error on the first pairing by correcting my switch of the 2739 and 2855. I should have written them down instead of bouncing up and down on the screen to do them.

For the pairing interchange issue, there is another recent thread that goes into that in depth. I would keep the board one pairings the same.

As far as pairing the dropped-down player AS the top of the next scoregroup instead of WITH the top of the next scoregroup, that is a cousin of the Harkness method of dropping the middle (which is effectively the same as pairing the top of the next scoregroup AS the bottom of the higher scoregroup instead of WITH the bottom of the higher scoregroup). As a person who in the ‘80s paired using the Harkness method before I even got around to reading the rulebook, I’m not the best person to say that your programmers’ concept is illogical.

24

One of the virtues of standard pairings is that they are more likely to have larger ratings mismatches against the dropped-down higher score group player, thus reducing perfect scores more rapidly. If there are no upsets and the dropped-down player is paired down, then the opponent is close in rating and the chance of a draw is higher, still reducing perfect scores more rapidly.

Harkness is more likely to have similar rating differences for all matches, thus not over-penalizing the bottom of a score group or over-rewarding the top of the next score group. This can sometimes have unusual effects in the bottom score groups.

You programmers concept is similar to Harkness, but can sometimes have unusual effects in the top score groups.

The following is a comparison of standard pairings, harkness and your programmers concept. It has 20 players and assumes no upsets. I have the pairings for four rounds. Colors are ignored. Even with no upsets, standard pairings reduce the perfect scores more rapidly.

Round 1
(0-0) 2000-1000, 1900-900, 1800-800, 1700-700. 1600-600, 1500-500, 1400-400, 1300-300, 1200-200, 1100-100.

Round 2
(1-0) 2000-1500, 1900-1400, 1800-1300, 1700-1200, 1600-1100
(0-1) 1000-500, 900-400. 800-300. 700-200, 600-100

Round 3 standard
(2-0) 2000-1800, 1900-1700
(x) 1600-1500
(1-1) 1400-1000, 1300-900, 1200-800, 1100-700
(x) 600-500
(0-1) 400-200, 300-100

Round 3 Harkness
(2-0) 2000-1700, 1900-1600
(x) 1800-1500
(1-1) 1400-900, 1300-800, 1200-700, 1100-600
(x) 1000-500
(0-2) 400-200, 300-100

Round 3 Programmer
(2-0) 2000-1800, 1900-1700
(x) 1600-1100
(1-1) 1500-1000, 1400-900, 1300-800, 1200-700
(x) 600-300
(0-1) 500-200, 400-100

Round 4 standard
(3-0) 2000-1900
(x) 1600-1800
(2-1) 1700-1200, 1400-1100, 1300-600
(1-2) 1500-800, 1000-700, 900-400
(x) 300-500
(0-3) 200-100
finishing with one each 4-0 (2000), 0-4 (100), six each 3-1 (1900-1600, 1400, 1300), 2-2 (1500, 1200-900, 600), 1-3 (800, 700, 500-200).

Round 4 Harkness
(3-0) 2000-1800
(x) 1900-1700
(2-1) 1600-1200, 1400-1100, 1300-1000
(1-2) 1500-600, 900-400, 800-300
(x) 700-500
(0-3) 200-100
finishing with two each 4-0 (2000, 1900), 0-4 (500, 100), four each 3-1 (1800, 1600, 1400, 1300), 1-3 (600, 400-200), eight 2-2 (1700, 1500, 1200-700)

Round 4 Programmer
(3-0) 2000-1900
(x) 1600-1400
(2-1) 1800-1300, 1700-1200, 1500-600
(1-2) 1100-800, 1000-700, 900-500
(x) 400-200
(0-3) 300-100
finishing with two each 4-0 (2000, 1600), 0-4 (200, 100), four each 3-1 (1900-1700, 1500), 1-3 (800, 700, 500, 300), eight 2-2 (1400-900, 600, 400)

25

Deanna,

I think your programmer is resisting “lowest in group A vs highest in group B” because it would, in effect, create an additional score group between A and B.

By contrast, with “upside-down Harkness” (lowest in A vs middle in B), the effect can be achieved by simply moving the lowest player in group A to the top of group B. That way he’d naturally be paired against the middle of B.

The same is true of “rightside-up Harkness” (middle in A vs highest in B). By moving the highest player in group B to the bottom of group A, he’d naturally be paired against the middle of A.

But doing it the “right” way (lowest A vs highest B) essentially creates an extra, intermediate, score group consisting only of the two players (one from A, one from B) who are being paired against each other.

Maybe you can appeal to your programmer by pointing out that, no matter how many score groups there are, the same programming code (subroutine) could be invoked each time. The program would simply have to loop more times.

Bill Smythe