GeneralisedKnockout/Bye Discussion

Sub-page of GeneralisedKnockout

The selection of byes is a weak point in knockout systems with other than 2^N participants.

Table of contents

Following is a path that is being explored. Remember that the "spirit" of knockout is the repeated elimination (rather: reclassification) across rounds. Also, one should not lose the chance to reach finals because of not being able to play a round.

Assume the situation where all players are equal. Who will get the bye? Introducing round robin to differentiate somehow the players is against the spirit of this system. Random selection does not feel right either.

1. Play for bye

Here is a proposed alternative. Take a situation with 7 identical strength players A-B-C-D-E-F-G. A bye is decided by a game between the two strongest players, or two random players if all players are equal.

  • winner receives the bye;
  • loser remains in the same score category;
  • remaining players of that category (i.e. including the loser) play normally.
GKO byes requires 3+2 rounds
ID score bye R1 R1 score bye R2 R2 score R3 score games played
A 0 B+ bye 1 . C+ 3 (B+) #1 7 2
B 0 A- E+ 1 . D+ 3 (A-) 6 3
C 0 . F+ 1 . A- 2 D+ 5 3
D 0 . G+ 1 . B- 2 C- 4 3
E 0 . B- 0 F+ bye 1 (F+)#1 3 2
F 0 . C- 0 E- G+ 1 (E-) 2 3
G 0 . D- 0 . F- 0 bye 1 2

Notes: #1: match AB has already been played in bye R1, same for EF in bye R2

In cases where a byed player meets that opponent again, the result is already determined. Also, maybe some form of telescoping is possible: players do something in each round (rather than have separate bye rounds).

Inconvenience is that intermediate rounds are introduced where byes are decided, while other players wait.

2. Random bye

willemien I think it is a bit a contradictionary having a game to decide who gets a bye. is almost giving everybody else a bye (they have to wait till the game is over. randomize the bye is i think the only solution. In the above example you have in practice two more rounds (with only 1 game).

axd But the idea of a bye is to give someone a free ticket to the next round; a random bye is less acceptable than a bye that is deserved.

Randomness can be reduced by using more criteria such as strength, earliest registration, last years results, or some combination of these.

3. Pre-selection to avoid bye issue

To avoid byes, a preselection round could filter participants away so that the actual tournament starts with 2^N players. Yet this is not ideal, as the idea remains to keep a maximum of participants busy in as many rounds as possible.

4. Recursive split

Another fuzzy option to explore is to extend the preselection idea into a split of the player field in a "2^n" ("high section") and a "remainder" ("low") section, so that bye issues occur only in the "low" section. E.g. for 17 players, the ideal split would not be 16+1, but 8+9 to avoid a single player ending up alone for the remainder of the event. Another example: 24=16+8 (rather than 12+12 giving bye issues occuring all over the grid).

This would be repeated in each section where a non-2^N field of players exists.

Problem is the byes: to take the example 8+9, 16 players will compete for the 8 places, one will not - and will be excluded from the finals. A bye should be a free ticket to the next round, rather than a definitive loss. So it looks like byes should be concentrated in the top half (9+8) (which contradicts the idea to let byes occur in the "low" section).

5. Byes are lucky

Teaching Go To Newcomers/Minimalism made me think that another option exist. Indeed, imagine a case with beginners learning the game; why not organise a gentle kind of tournament?

  • this will bring the similar strength players together, making the game even more exciting for them
  • this can trigger competition at the earliest stage

Byes should be avoided in such scenario.

Here it is: if we forget the fact that byes are meant to be used in "ruthless tournaments", we could give a win to the loser of a round.

Let's first assume that there is an even total number of participants. The idea is that the outcome of a round will split a category in an even or odd number of winners and losers.

  • In case a new category consists of an even number of particpants, proceed as usual.
  • If the number is odd, sort the losers from strongest to weakest, and let the strongest loser jump to the winner category (i.e., give him/her a point); as a result, the winner category (who also consisted of an odd number of players) now is also even.
GKO-lucky losers

(TODO: need a bigger number of participants for a better demonstration)

ID score R1 score R2 score R3 score
A 0 D+ 1 C+ 3 B+ 7
B 0 E+ 1 D+ 3 A- 6
C 0 F+ 1 A- 2 D+ 5
D 0 A- 1 B- 2 C- 4
E 0 B- 0 F+ 1 bye 3
F 0 C- 0 E- 0 bye 0

This can be repeated whenever it happens, and will reduce the number of byes. One must be careful not to reassign the same opponent to such "lucky losers".

In case of an odd number of participants, one "real" bye will have to be assigned in turn to different participants.

Note

  • this scheme will introduce an "imbalance" in number of rounds. E.g. in case of 6 participants, the first round will give 3 winners and 3 losers, of which one player will become a winner.

As a result, the winner "half" will consist of 4 players - requiring 2 rounds to complete, while the loser half will have only 2 players, which requires only 1 round to complete.

  • chosing the lucky loser might be tricky

goban pairing?

The previous option ("byes are lucky") leads to a new idea: (to put later in a separate page): currently, pairings are published in the form of lists of A-B lines (player A against player B). I have the impression that a dual form exists where it is sufficient to number the gobans and publish a table of anonymous participants.

  • avoids having to register participants and juggle with name cards (not really significant during initiation sessions).
  • a preprinted grid can be used - assuming the number of participants is known
  • the final order is defined by the goban order
table pairings, 8 players
R1 R2 R3
goban# Winner Loser Winner Loser (final round)
1 1 3 1 2 decides places 1,2
2 2 4 1 2 decides places 3,4
3 1 3 3 4 decides places 5,6
4 2 4 3 4 decides places 7,8

How to read this table: e.g. the player who won round 1 while sitting at goban 3 goes to goban 1; his opponent stays at goban 3.

TODO:

  • include example of real byes with larger player field
  • describe the general algorithm

The idea is that for each round, the pairing table shows which goban the winner and loser must go to for the next round.

table pairings, 26 players
R1 R2 R3 R4
goban# Winner Loser Winner Loser Winner Loser Winner Loser (final round)
1 1 7 1 4 1 3 1 2
2 2 8 2 5 2 4 1 2
3 3 9 3 6 1 3 3 4
4 4 10 4 7 2 4 3 4
5 5 11 1 5 5 6 5 6
6 6 12 2 6 6 7 5 6
7 7 13 3 7 5 7
8 1 8 8 11 8 9
9 2 9 9 12 9 10
10 3 10 10 13 8 10
11 4 11 8 11 11 12
12 5 12 9 12 12 13
13 6 13 10 13 11 13
  • As a result of the "cutting in half" and selection of lucky byes, some participants that end up in the lower regions of this table will not be able to reach the last rounds, but will remain with the same opponent. This should be improved.
  • At this moment, it seems that the best way to handle an odd number of participants is to use the current approach by adding a "ghost" participant - which makes the count even; the opponent of the "ghost player" actually receives the bye.
  • Note that Swiss might make things much more easy - but here we try to keep the recursive knockout principle; also, this system does not need to do any kind of "bookkeeping".

GeneralisedKnockout/Bye Discussion last edited by axd on November 23, 2012 - 20:18
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