Another explanation on Chinese komi

    Keywords: Rules

These are two earlier rec,games.go postings of mine on "Chinese komi", very lightly edited. Charles Matthews


7 October 2000

There is a hidden parity effect here, mathematically speaking. It isn't hard to see it at work with some concrete numbers. Let's suppose that there is no seki to worry about. Then area scores will add up to 361 (an odd number),

eg

  • Black 181 White 180
  • Black 182 White 179
  • Black 183 White 178
  • Black 184 White 177
  • Black 185 White 176

with differences 1, 3, 5, 7, 9 and always odd.

The simple way to give "komi" with area counting is to say for example Black must score 184 to win, White 178 to win.

Let's look at territory (Japanese-style) scores assuming Black played 121 times and White 120 (say), net of captives. These would give you corresponding pre-komi territory scores

  • Black 60 White 60
  • Black 61 White 59
  • Black 62 White 58
  • Black 63 White 57
  • Black 64 White 56

and so on, with difference stepping up by two each time. Then 5.5 komi to White is enough to make 58 a winning score for White, and 63 would be a winning score for Black. Either of 6.5 or 7.5 komi makes 57 the winning score for White, 64 the winning score for Black. The point is how to translate this back into Black's target number 185. Of course you can do all this with a little algebra, but probably most people find a few actual numbers just as convincing.


8 October 2000

I think it's clearer not to worry about translating scores between area and territory counting, but to concentrate on the result of the game. Basic equation, since numerical data seems to throw some people:

C(B) - C(W) = J(B) - J(W) + (P(B) - P(W))

where C(B) and C(W) are the area (Chinese-style) counts for Black and White J(B) and J(W) are the territory plus prisoners (Japanese-style) counts. P(B) and P(W) are the number of plays made in the game by Black and White.

Winning condition C(B) - C(W) >= 2n + 1 for some value of n = 0, 1, 2, 3, ... .

Normally P(B) = P(W) or P(W) + 1, because players don't pass until the game is over (fill what would be dame for Japanese alternately). We know that C(B) - C(W) is odd because C(B) + C(W) is odd, provided there is no seki. So then P(B) = P(W) whenever the difference of Japanese scores is odd, and P(B) = P(W) + 1 whenever the difference is even.

Summary: 5.5 komi means Black wins is equivalent to J(B) - J(C) >= 6. If J(B) - J(C) = 6 we have P(B) - P(W) = 1, so 5.5 komi equates to C(B) - C(W) >= 7, or Black needs 184 to win. 6.5 komi means Black wins is equivalent to J(B) - J(C) >= 7. If J(B) - J(C) = 7 we have P(B) - P(C) = 0, so again this is C(B) - C(W) >= 7.


Another explanation on Chinese komi last edited by rubilia on April 4, 2004 - 03:02
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