Cho Chikun loses Honinbo title
After holding the Honinbo title for a record 10 consecutive terms, Cho Chikun finally went down in defeat at the hands
In the crucial sixth game, Cho Chikun was in overtime: he had used up his allotted eight hours and had to make each of his moves within one minute. This situation is normal for Cho in two-day games. He was clearly winning, but the pressure of time and the realization that he would lose the title if he did not win this game probably caused him to make an uncharacteristic blunder.
The final position is shown in Figure 1. A capturing race had developed between the marked black and white stones in the upper right. Cho Chikun (white) could have easily started a ko to capture the six marked black stones, but he descended to White 1. When Black also descended to 2, it was clear that the white stones would die, so Cho Chikun resigned.
12, 18: in 7;     15, 21: in 9
If White had played 1 in Diagram 1, the sequence would have continued to White 5. If Black captured with 6, White would atari with 7 in Diagram 2. Black would play 8, putting the White group into atari, so White would capture the ko with 9. Black would then have to threaten to increase the liberties of his embattled stones with 10 and 16 as ko threats, but White 13 and 19 are huge ko threats that Black cannot ignore. After White takes the ko with 21, Black has no significant ko threats, so he will lose nine stone in the top right when White plays at A. Moreover, Black would now have to play at B to make two eyes for his group on the left.
In spite of this loss, Cho Chikun still holds the top two Japanese titles-the Kisei and the
Answers to last weeks problem
In Problem 1, White has an eye on the second line at the top, so if Black is going to kill the white stones, he has to destroy the whites eye shape below. Black 1 in Diagram 3 fails to accomplish this because White will answer with 2. Although Black has created a false eye at A, White gets his second eye at B, so his stones are alive.
White 2 in Diagram 3 is clearly the key point because if White can play there he gets two eyes. Therefore, Black must play on 1 himself as in Diagram 4. White plays 2 to stop Black 1 from linking up to his stones on the outside, but Black now plays 3, putting a white stone in atari. White will capture with 4, but Black throws in a stone at 5 in Diagram 5. White can capture this stone, but that point is a false eye. White will eventually have to play there, so he is left with only one eye and his stones are dead.
In Problem 2, White has to make a second eye at the top if he is going to live. A placement at 1 in Diagram 6 is the key point. White plays 2 to stop Black from linking up to his stones on the left, but Black pushes in at 3, sacrificing two stones. After White captures with 4, Black throws in a stone at 5 in Diagram 7 leaving White with a false eye at the top. Whites stones are now dead.
You might think that Black 1 in Diagram 8 would force White to connect at A, then Black can play 2 destroying the eye at the top. However, White will answer Black 1with 2. White can now fight a ko. If he is able to win the ko by connecting at A, he will have his second eye to the right of 2. Clearly, it would be better to kill white cleanly by playing the sequence in Diagrams 6 and 7.
In the position in Problem 3, the only move that will kill the white stones is for Black to throw in a stone at 1 in Diagram 9. White cannot play at A because he puts his stones into atari. Therefore, he makes an eye in the corner with 2 in Diagram 10. Next, Black ataris with 3 and 5, and the eye at 1 is now a false one, so White is dead.
If White captures Black 1 with 2 in Diagram 11, Black will play 3. If White next plays A, Black will play B, leaving the white stones with only one eye (at 1). If White B, Black plays A and the eye at 1 is a false one. Either way, Whites stones are dead.