Counting a surrounded non-alive carpenter's square

Question

I was playing against Crazy Stone and we ended up with (essentially) this corner situation:

$$W
$$ -------------
$$ | . . 5 X O .
$$ | 4 1 3 X O .
$$ | . 2 , X O .
$$ | X X X X O .
$$ | O O O O O .
$$ | . . . . . .

This certainly taught me (black) a valuable lesson about what I learned is called the "Carpenter's Square" and how I should not let W1 happen.

What I did not understand was the counting of this situation. I would have expected this to be counted under seki rules (as far as I can tell, Crazy Stone uses Japanese scoring), which in my understanding means all stones are alive and the four empty intersections are dame, so this corner is worth zero for either player.

The program however marked the black stones as dead and counted the corner as white territory, making the above diagram worth 22 points for white.

Is my understanding incorrect? Or the computer's? How should this situation be counted?

Answer 1

During the game

$$
$$ -------------
$$ | a b O X O .
$$ | X O O X O .
$$ | . X , X O .
$$ | X X X X O .
$$ | O O O O O .
$$ | . . . . . .

The corner is not settled yet, as both players can choose to:

• either start a ten thousand year ko, white by playing at a and black by playing at b
• or settle for a seki, if black plays at a or white plays at b.

At scoring time

If no move has been added in the corner, I would expect this to be considered a seki under Japanese rules (see example 12).

Under Chinese rules, both players would have to agree on the seki, or keep playing (in which case a or b will eventually be played).