recursive knockout
alex
I'm looking for any existing references to what I describe as a
"recursive knockout" system.
It is a simple, non-graphical way to determine one or more winners in
a knockout system, with the difference that 2nd, 3rd etc... places can
also be determined. The mathematical background most probaby already
exists, but the method does not seem to be described anywhere.
Nevertheless, I assume I'm not the first one to describe this system.
Basics are here: http://senseis.xmp.net/?Axd/RecursiveKnockout
-alex-
dmartin
alex
> Humm Comparaison with swiss-pairing?http://senseis.xmp.net/?SwissPairing
If I understand well, Swiss pairing also requires to track extra
information if you want to identify lower winners (2nd, 3rd,
4th, ...): suppose you have the winner with 4/4 wins, a bunch of
players with 3/4, a bunch with 2/4, ...
Assume the winner is the single one with 4/4 victories (in reality
this will depend on the number of rounds etc.).
To split out the group of 3/4 players, additional games are required,
resulting in groups 4/5 and 3/5. And then you let all players with 4
victories play each other, all those with 3 victories, etc... Winners
from one group (eg group of 4 victories) then proceed to the higher
group (group of 5 victories), but you must remain careful, because
other players with more victories may not claim a tournament prize
that is awarded to the winner with 4 winning games.
-alex-
dmartin
> On Aug 18, 2:53 pm, dmartin <dmar...@irisa.fr> wrote:
>
>> Humm Comparaison with swiss-pairing?http://senseis.xmp.net/?SwissPairing
>
> If I understand well, Swiss pairing also requires to track extra
> information if you want to identify lower winners (2nd, 3rd,
> 4th, ...): suppose you have the winner with 4/4 wins, a bunch of
> players with 3/4, a bunch with 2/4, ...
>
I don't know how it compares with "recursive knockout". Should be an
interesting things to know.
Let's use an exemple of a 8 persons tournament with awards for 4 people
On classic knock-out: Let's name the player's as the following:
/ 8
/ 4 -
/ \ 4
/- 2 -
/ \ / 6
/ \ 2 -
/ \ 2
1 -
\ / 7
\ / 3 -
\ / \ 3
\- 1 -
\ / 5
\ 1 -
\ 1
To be optimal:
1st should be player 1, 2nd place need a competition between 2, 3, and
5, it's getting intricate when trying to determine further spots
On recursive knock-out... We suppose the draws for equals score to give
the same as the knock-out before.
Format: opp:scores
R1 R2 R3
1 5:0 3:0 2:0
2 6:0 4:0 1:1
3 7:0 1:1 4:2
4 8:0 2:1 3:3
5 1:1 7:2 6:4
6 2:1 8:2 5:5
7 3:1 5:3 8:6
8 4:1 6:3 7:7
This give a complete ranking, however match between 3 and 5 was not made
and 5, having fought the stronger opponent might be the real second.
On swiss-paring:
Format: opp:score
R1 R2 R3
1 5:1 3:2 2:3
2 6:1 4:2 1:2
3 7:1 1:1 4:2
4 8:1 2:1 3:1
5 1:0 7:1 6:2
6 2:0 8:1 5:1
7 3:0 5:0 8:1
8 4:0 6:0 7:0
The swiss pairing give the same finals except that 3 can play 5 or 6 too
in round 3, not only 4. Also it shows that 5 should have a chance to
play 2 or 3 for second spot. The next round will probably needs SOS &
SOSOS for tiebreakers.
The swiss pairing here tiebreaks does not exclude 5 because he lost the
first round.
> Assume the winner is the single one with 4/4 victories (in reality
> this will depend on the number of rounds etc.).
>
> To split out the group of 3/4 players, additional games are required,
> resulting in groups 4/5 and 3/5. And then you let all players with 4
> victories play each other, all those with 3 victories, etc... Winners
> from one group (eg group of 4 victories) then proceed to the higher
> group (group of 5 victories), but you must remain careful, because
> other players with more victories may not claim a tournament prize
> that is awarded to the winner with 4 winning games.
>
> -alex-
Humm sorry, I have hard time understanding what you want to point out
here. Usually you decide n rounds for the swiss pairing (enough to find
a winner), then you use the results. Not the contrary.
An example with 8 players, tiebreakers for pairing was random:
(parenthesis give the score after the round)
R1 2(1) won vs 4(0)
3(1) won vs 5(0)
1(1) won vs 7(0)
6(1) won vs 8(0)
R2 2(2) won vs 6(1)
1(2) won vs 3(1)
5(1) won vs 7(0)
4(1) won vs 8(0)
R3 1(3) won vs 2(2)
3(2) won vs 6(1)
4(2) won vs 5(1)
7(1) won vs 8(0)
Rank Score SOS SOSOS
1 3 5
2 2 6
3 2 5
4 2 3
5 1 5
6 1 4 16
7 1 4 15
8 0 5
So after 3 rounds there was a total split.
If we use recursive knock out:
R1 R2 R3
1 7:0 3:0 2:0
2 4:0 6:0 1:1
3 5:0 1:1 6:2
4 2:1 8:2 5:4
5 3:1 7:2 4:5
6 8:0 2:1 3:3
7 1:1 5:3 8:6
8 6:1 4:3 7:7
1/ That gives the same draws
2/ That gives 1 2 3 6 4 5 7 8
6 beeing 4th while he only beat the weaker player.
So that's all I have to say for today,
Regards,
-- Damien