I'm a beginner and have been playing some other beginners and some online computer systems. In one game against COSUMI the final board state was as follows (I've added the 1,2,3)


Now I've been trying to understand why the lower left black stones are dead. I should mention that I am understanding Go by thinking about the New Zealand Rules (which seem by far the most elegant) but I do appreciate that the rule choice might make a different -- COSUMI assumes Japanese rules. The New Zealand rules are here: http://senseis.xmp.net/?NewZealandRules

I've sort of convinced myself that the black stones are indeed dead, but it seems to require just the right moves by white. Is the following analysis correct?
A. If white plays at 1 then black can play at 2, and black's next move can be either north or west of 2 to complete two eyes -- so white fails to kill them.
B. Starting again, if white goes at 3, then black must play at 2, but white can't play east-of-2 since that would cycle the game to the earlier position, hence black can secure it.
C. If instead white plays at 2 then black must just pass, and then
i) if white then follows at 3, then black plays 1, and black will subsequently be able to make two eyes
ii) if white instead follows at 1, then black must play 3, white goes to 2 this does then look hopeless for black Is this indeed white's one and only correct approach?

Is that right?

According to the page you linked, New Zealand rules actually do allow this. You can repeat a previous position if the player to move is different: "the resulting board position does not repeat the whole board position as it was after any of that player's previous moves".

However, this sequence doesn't achieve anything for white.


Correct!

In a situation such as this it is clear that if B plays first then he can only play at 1 or 3 and in doing so he put himself into atari. W plays the other of those two points and captures. So B cannot play any move here to his benefit and you only need to consider W's plays.

What you need to understand is the basic dead shapes. If W first connects at 2 and then follows at 1 he puts himself into atari but he has created a bulky five. When the five stones are captured he can play back in at the critical point of 2 and start filling all over again - there is nothing B can do to prevent it. But next time around it will only take a bulky four to almost fill the space before B has to capture. After that it only takes 3 to almost fill, etc.

The basic idea is that you need to know the short list of basic killing shapes. A good start is http://senseis.xmp.net/?BulkyFive. Once you know that shape then you can be sure that making any of them inside a group will kill that group. The only condition is that the surrounding stones must be alive and not capturable because it actually takes 12 moves to go through the process of capturing if you start out with a bulky five. If the surrounding stones don't have 12 liberties then you could be in trouble.

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Still officially AGA 5d but I play so irregularly these days that I am probably only 3d or 4d over the board (but hopefully still 5d in terms of knowledge, theory and the ability to contribute).

Thanks for the explanation and the link guys.

I think to kill bulky five one needs 8 moves, not 12. You need to take into account the responses by black (in this example) when he tries to live by killing the stones inside the big eye.

Maybe you are correct. I did it from memory. Are you counting the moves in the five before it is fill (4 extra) or after? Regardless of the number, and I will concede I may have misremembered, the point is that it takes many outside liberties to be safe.

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Still officially AGA 5d but I play so irregularly these days that I am probably only 3d or 4d over the board (but hopefully still 5d in terms of knowledge, theory and the ability to contribute).

He's correct, 12 is for a 6 space big eye in a capturing race (possibly where your memory was tricked up), 8 is for a bulky five or crossed five. Counting from the first move white plays on the shape's critical point.

It's 8 from empty bulky five. The calculation is - 4 stones w, 1 stone b to capture, 3 stones w, 1 b, 2 w, 1 b, 2 w. We get 11 w vs 3 b, so b has only 8 extra moves to fill in outside liberties. And indeed, for 6-space nakade we need to add 5w 1b in the beginning of the sequence, which results in 12 net moves for white to capture.