2
$$ ------------------
$$ | 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 at 10:04
3

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.

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