TFG9 White at the Top

Black plays first  

(White 2,4,6 elsewhere)

As noted elsewhere, if Black plays first, White has four liberties. This means that if Black could make four liberties on the left in sente, Black could start here and capture White - in theory.

What it really means is that if Black makes five liberties on the left, White can't afford to continue there and is forced to come back to the center top and play there. What happens if White does?

White plays first 1a  

• White 1 creates five liberties.
• Blacdk 2 reduces White back to four. If White continues along the top, so does Black. So ...
• White 3 tries to cut Black off, but ...
• Black cuts in turn with 4 and at the same time reduces White to three liberties.
• If White 5, then Black 6. White still has only four liberties (the sequence a, b, c, d captures White). White can also play next at b (Black captures at a) to make a ko. Black needs only three plays to capture White, ignoring the ko threat).

White plays first 1b  

If White tries here, extending the liberties of the cutting stone, B stops White cold with , reducing White to two liberties (either of the points a next puts White in atari).

From this it is clear that above does not help White. What next?

White plays first 2a  

• White tries this next.
• Black tries this .
• threatens to live, so ...
• Black is forced to answer at to prevent White playing here.
• White lives with . (Note that of course if this situation arose Black would not choose to play but would instead attach on the outside of 1 to force White to live smaller.)

Note that this case raises the important question of what happens if both sides live. Assuming that White came back to the center after not being able to capture Black on the left, who wins the game if Black lives on the left while White lives in the center?

But is this the best that Black can do?

White plays first 2b  

• White plays 1.
• B plays 2 to prevent White expanding to the right. White's shape is too small to live so the question becomes how many liberties does White have?
• White connects at 3, threatening to live.
• Black plays 4 to stop White.
• White plays 5 and has two outside liberties and six inside for a total of eight!

White's internal liberties 1  

(2 elsewhere) The count is B + 2, W - 1 = Net +1

W's internal liberties 2  

(2, 4 elsewhere) The count is B + 3, W -1 = Net +2 with +1 in the previous diagram so total Net +3

W's internal liberties 3  

(2, 4 elsewhere) Final count is +3 here with +3 in the previous diagrams for a total of six internal plays required for Black to capture White.

White plays first 2c  

Note that Black can't switch in diagram 2b to here. White has an outside liberty at a, and can throw in at and after Black captures at , White traps two black stones with .

Oops!!!! This is not correct. When Black plays , White will not capture at 6 but instead will extend at . After White b, Black c it is ko but White has fewer liberties than in 2b (total of five?). This changes the conclusions below. I'll have to revise them later. -- Dave

So at this point White has only one eye at the top but has expanded from four liberties to eight liberties in gote, if Black plays next on the left; or seven in sente (if Black plays next at the top and lets White play next on the left). What does this mean for the status of the black stones on the left? Remember that White played tenuki on the left in order to play at the top.

• If Black has two eyes on the left then White is dead.
• If Black can expand to more than eight liberties on the left then White is dead.
• If Black cannot make more than seven liberties on the left then White wins the fight and captures Black.