# H vs K? [#1182]

axd: H vs K? (2007-11-02 15:29) [#4010]

I'm looking for an equivalence/conversion table involving komi, handicap and board size: eg on a 19x19, how much komi can replace 9H (I think it must be something like 9x10=90K)? On a 13x13 with 4H? etc.

The idea is to replace handicaps with a reverse komi; but how much?

Ideally we could create following table that lists the equivalent reverse komi for a given size and handicap:

```size  0H    1H    2H     3H
----  ----  ----  ----  ----
19     6.5?  10?   20?  30?
13     10?  ....  ....  ....
9     20?  ....  ....  ....
7    ....  ....  ....  ....
```

(I have no idea of the values, so I just filled them in on intuition.)

Anyone an idea? This could be handy in Handicap for smaller board sizes.

thx

xela: 9H=140K? (2007-11-03 10:03) [#4013]

I think the relationship is supposed to be non-linear--the first few handicap stones are equivalent to around 12 points of komi each, but a nine stone handicap is worth about 140 points. Sorry, I can't remember a source for this assertion; I have a vague memory that there were some pro vs pro handicap games to test this sort of thing.

X
71.192.11.206: Re: 9H=140K? (2007-11-03 15:33) [#4018]

Bob McGuigan: I have the game record of one pro-pro handicap game with a 9 stone handicap, Minami Yoshimi 6p (W) vs. Suzuki Etsuo 7p (B), played around 1967. Black won by 124 points. It is only one game so it is hard to draw any conclusions about the correct komi but it seems safe to conclude that 90 points is too low.

71.192.11.206: ((no subject)) (2007-11-04 01:32) [#4022]

Bob McGuigan: I went through my library of go material and found thirteen pro vs. pro handicap games played all the way through to a final counting. The results and handicaps were: H2 B+18, H3 B+35, H3 B+38, H3 B+41, H4 B+42, H4 B+53, H5 B+68 H5 B+58, H5 B+67, H7 B+92, H9 B+124. A scatter plot of these results shows reasonable linearity. The correlation between number of handicap stones and winning margin is quite strong at 0.95. Of course this is a non-random sample so the statistics aren't very useful.

X
MrTenuki: Re: ((no subject)) (2007-11-04 00:49) [#4023]

What's interesting about your non-random sample is that extrapolating the linear fit of your data gives a result of y = 6.28 for x = 1 (essentially black playing first and giving no komi).

In other words, this small sample (by coincidence) weakly supports the case for a 6.5-point komi. I say "weakly" not only because it's a small sample, but also because extrapolating to x=0 (white playing first, no komi) gives y = -8.26, which would lead to the conclusion that the perfect komi is about (6.28 + 8.26) / 2 = 7.27 points.