BQM 67

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BQM 67: Fair komi adjustment to board size?

How should fair (e.g. 6.5 points) komi for 19x19 games be adjusted to non-19x19 games?

[ext] http://www.dragongoserver.net/phorum/read.php?f=4&i=1034&t=1034 states, that komi is independent of board size (if board size does not drop below 9x9). Is that correct? Proved by statistics? Or how should komi be adjusted to non-default board sizes? -- Frs, Feb 2003

Bill: Komi depends on board size.
I believe that we have discussed that somewhere here on SL. :-)

Nico: I have not found anything satisfactory regarding this issue on SL. Please can someone come up with a more definite answer to this question, especially for 9x9 and 13x13 boards ?

Charles Komi obviously does depend on the size of the board, for very small boards. I don't feel that it can differ much between 13x13 and 19x19 (some discussion at handicap for smaller board sizes which explains why I think that); but if someone told me that there are 13x13 go hustlers who can take Black and give 10 komi then I wouldn't be really surprised. My idea of the graph of komi against board size N is that it decreases rapidly when N is small, is effectively flat from 13x13 to 19x19, and is probably a little less for N very large. But who knows?

Anthony? I did a quick calculation and komi of 5.5 is 2% of the board area if the board is 19x19. However, it is 7% of the board for a 9x9. That's a lot to give for going first.

Alex Weldon: Yes, but the advantage of going first is inversely proportional to board size. So Black in 9x9 has a HUGE advantage (if there's no komi, it's virtually impossible for W to win if both players are sufficiently strong and of equal rank), so White's komi should be equally huge. For a very large board, 6.5 komi would be almost negligible, but a large board would make for a very long game, and Black's first move advantage would be accordingly diluted. It makes sense to me that, except at the very low end of the scale, komi wouldn't change much. If we want to be more precise, you could probably fit something like Alog(Bn) + C to it pretty closely.


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