# Gokyo Shumyo, Section 1, Problem 54 / Solution

Main line

After , Black must capture two stones: shortage of liberties prevents him from playing at a. White captures at b to isolate Black's one-eyed group.

In spite of the me ari me nashi proverb (Black destroys an eye at c), White will win this race to capture (semeai) because she has many more outside liberties.

Continuation

With respect to the main line, Black has captured two stones with , and White one stone with , and the captured stones are removed.

We'll assume a black stone at . And we'll also assume -. White's outside group has 7 liberties (squared). Black's inside group has the shared liberties, the approach liberties (circled) and the liberty at , 6 in total.

Variation

Another option for Black is to play first, creating a ko and threatening to make an eye. White calmly connects. If we exchange a and b, Black has four liberties: the two circled ones, one in the eye and the one in the ko, which White will play as an atari.

Compared to this, White has eight liberties, plus one in the ko, and she may start. So, Black will only be able to capture White if he ignores five of her ko threats while she never ignores any of his. This is a five move approach ko. Technically, White is not alive. Practically speaking, she is.

The original diagram does not contain , so White either escapes or prevents the approach ko, and she is technically alive.

Variation at

White captures the three stones in the corner.

Gokyo Shumyo, Section 1, Problem 54 / Solution last edited by hnishy on January 15, 2023 - 09:46