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August 1999's Go Answer of the Month

Dia 1 - Stacey v Macfadyen.

Black has 13 prisoners, White has 7, assume 5½ komi.
The position is available as an Ishi format file.

This position is taken from a game published in an old BGJ - I'll reveal which game later on. According to the commentary, the result of the game was unexpected. Be that as it may, this is an exercise in counting and in semeai.

Your starter for ten - what is live, what is dead?
The bonus question - who is winning and by how much?

Dia 2 - Identify the groups.

Lets work our way around the groups:

• A: This has no obvious eyes and no obvious way to run. A problem group.
• B: This has one eye (2 points with a prisoner on one of them) and all the other potential eyes falsified by white.
• C: This has two eyes at n3 and at n1. Black can't do anything about them, EG:
  • If Black captures k1 with j1, white can atari m2 and n3 with m3, and whilst black could play m2 to convert l1/m1/n1/m2 into one clump, it is still captured for one eye. n3 being the second eye.
  • If Black plays at m3 to spoil the above, an atari at o3 sets up either a capture of two clumps (two eyes) or, should Black connect, the capture of a non-nakade clump.
• D: The stone can cut this group off from stone E. So it needs eyes on the right. Recent analysis (tjh, sgb & gotools) is that the best Black can do is to make two eyes here in ko.
  Dia 3 - Ko.
  
• E including the five stones f7..f11: appear to be connected to K.
• F, G, H, I, J, K, L: all look to be alive.

Therefore we have some form of capturing race between groups A & B, and a life and death problem on group D.

Dia 4 - Liberties.

You are referred to the liberty counting tables for terminology and "formulae".

Group A has no eyes and no prospect of obtaining any. It cannot expand its liberties either. Group B has a small eye and again cannot expand its liberties.
This makes it a type 3 'eye v no eye' fight - to the death - where the favourite is Black who gets all the inside liberties.
White effectively has 7 outside liberties . Black effectively has 4 outside liberties . The status of 'j4' may be complicated by a possible ko. X, Y and Z are effectively inside liberties.
With the score being White 7 to Black 4+3, we have a 'first to play wins the race' situation. It is Black's turn, so he should win the semeai.

Therefore White should not respond and leave group A dead and play to capture group D, a group of similar size to group A. Think of it as an exchange. Should White fail to isolate and kill group D then he is way adrift in the score.

Now let's count the game, noting that the current 13 prisoners captured by black equal the 7 prisoners plus 5½ komi that white has. So effectively the result is that on the board. (My counting is not good enough to worry about the odd ½ point!

White: 10 top left; 2 top middle; 12 top right; 6 right side; 7 bottom; 46 in the midle due to group D. Total = 83.
Black: 80 left side and group A; 6 bottom right; 10 surrounding top right corner. Total = 96.
So I have black winning by 13.
Tim Hunt did a similar calculation and had each group within 2 points of my estimates and yet his total was more than 10 different! Counting is not trivial.

The commentary agrees that Black is winning the semeai, but remarks that had White managed to start the semeai, it would have been a ko.

In the actual game from BGJ 67 page 6, Black was confused about the semeai and wasted moves on it ensuring that White group A died. However in the process Black's D group did get cut off and died. This permitted White to regain the lead and win by 3.5 points.

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