4
$$ ------------------
$$ | O . O X . . . .
$$ | . O O X . . . . 
$$ | O O O X . . . .
$$ | X X X X . . . .
$$ | . . . X . . . .
$$ | . . . X . . . .
$$ | X X X X . . . .
$$ | . . . . . . . .

How is this position scored, because white and black cannot be captured. Who controls that territory?

1
  • note that black still can be captured :-)
    – Tomas
    Jul 31, 2021 at 10:04

2 Answers 2

5

Summary

It does not matter if a group is “surrounded” (which is hard to define); all that matters for the score is enclosed vacant points.

The rules

Vacant points only connected to stones of one player count as territory for that player; if “area scoring” is in use, points occupied by a player also count for them, but this seldom alters the result(!). Stones which players agree are captured are removed before counting like this and count as territory points for the the capturer but are ignored in area scoring.

Some traditions recognise several special cases, which you can find in or via the article on Scoring in Sensei’s Library. See there also for how the rules are applied in practice.

Your case

$$ ------------------
$$ | O w O X . . . .
$$ | w O O X . . . . 
$$ | O O O X . . . .
$$ | X X X X . . . .
$$ | b b b X . . . .
$$ | b b b X . . . .
$$ | X X X X . . . .
$$ | . . . . . . . .

White has 2 points of territory marked w and 7 stones played, making 9 area points. Black has 6 points of territory marked b and 13 stones making 19 area points.

The remaining unoccupied intersections theoretically count for Black if there are no other white stones on the rest of the board but in practice White would still be able to make more points somewhere else. If there are white stones, then neither player has surrounded the remaining intersections, so these do not fall under the scoring rules and count as neutral.

Surrounding live groups makes no difference

Note that the rules are not interested in whether one group surrounds another: the white group in your example may be surrounded by black stones, but since it cannot be captured, it earns White points.

It is just as well it does not matter, as it is not always clear who has surrounded whom: consider two opposing lines of stones from top to bottom.

$$ Who surrounds whom?
$$ -------
$$ |..OX.|
$$ |O.OX.|
$$ |.O,X.|
$$ |.OX.X|
$$ |.OX..|
$$ -------

It is better not to think of “surrounding”, i.e. going all round the “outside”, to capture a group. What really matters is whether all the strings (connected chains) in the group have liberties (adjacent vacant points). This approach lets you play using logical rules on an arbitrary finite graph, where “outside” may be meaningless.

A note on efficiency

Incidentally, since Black has played more stones than White in your example, it is sensible to compare how efficiently they have made their territory: for White this is ²₇ ≈ 0·29 and for Black ⁶₁₃ ≈ 0·46, so Black is doing roughly 1¹₂ times as well as White, even though their territory is 3 times the size.

2
  • why the odd formatting in the efficiency calculation? It seems to me you've just calculated 2/7 = 0.29 and 6/13 = 0.46, so why mess around with superscript/subscript glyphs and the centered dot instead of the more usual division line and regular decimal point?
    – ilkkachu
    Nov 14, 2021 at 14:33
  • @ilkkachu: I am sorry if you found my formatting hard to read. I grew up with centered dots for decimal points, and definitely prefer the look of them and of fractions formatted like that, which is the same as in “1¹₂”, which is common on road signs in Britain — I find them less cluttered and more readable than “2/7” and “6/31”, so it does not feel like messing around to me. If we had mathematical markup in this community I might have used that.
    – PJTraill
    Nov 16, 2021 at 22:13
2

Depends on the ruleset.

White is alive on its own, enclosing two points.

Black however, might suffer from a ko threat and might seki.

$$ -----------------
$$ | O . O X O . . .
$$ | . O O X O . . . 
$$ | O O O X O . . .
$$ | X X X X O . . .
$$ | 6 2 4 X O . . .
$$ | . 5 . X O . . .
$$ | X X X X O . . 3
$$ | O O O O O . . .

Here, 2 is the ko-threat, 3 is taking the ko and 4 is w following its ko-threat.

So, if the game ends without the the ko threat, assuming Japanese rules, w has 2 points while b has 6.

With Chinese rules, however, if the ko threat is not played, w has 9 points while b has 19 points.

2
  • You have a good point about the ko threat — though it can only happen if White gets something like all those extra stones on the board, which are not there in the question! As to the score, one needs to bear in mind that White has made 2 or 9 points with 7 stones, while Black has made 6 or 19 with 13 stones, so in either case Black has played more efficiently — and efficiency is vital in Go!
    – PJTraill
    Nov 16, 2021 at 22:31
  • Counting the points requires isolating the counted areas, otherwise w gets 2 points and b gets 342, i.e., rest of the board. You can't comment on efficiency from such a small part of the board. For all I know, b traded lower right corner for upper left corner and w invaded to reduce the corner, or w passed 7 times.
    – ck1987pd
    Nov 17, 2021 at 8:07

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