Fujitsu CupThe final game of this year's Fujitsu Cup,
jointly sponsored by Fujitsu Ltd. and The Yomiuri Shimbun, was held
on Aug. 7 in Tokyo. This year's finalists were two previous winners
of this cup, Yoo Chang Hyuk of South Korea and Ma Xiaochun of China.
Both had won this cup once before (Yoo in 1993; Ma in 1995). It was
clear that both players were determined to win, and no one was sure
until the very end who would be the victor.
Ma (playing white) staked out more than 60 points of territory at
the top with move 128 (1 in Diagram 1). This was more than
Yoo had on the right side and the lower left corner. But Yoo then
attacked the white stones on the left side forcing them to make eyes
with the sequence to White 17. Next, he staked out territory in the
middle with Black 18, forcing White to defend his own territory with
19. Black then played 20 at the bottom. White had no choice but to
engulf the marked black stone with 21, so Black was able to link up
with his two stones at the bottom with 22, thereby cutting off the
four marked white stones at lower left. Since these white stones
could no longer avoid capture, the game was close.
A hard-fought endgame ensued and the game ended after 244 moves,
with Black emerging the victor by the merest of margins-half a
point.
This was Yoo's third international title. Besides his earlier
victory in the 1993 Fujitsu Cup, he also won the $400,000 Ing
Changki Cup in 1996, and has been runner-up in five other
international tournaments. Over the last three years, he seems to
have been in a slump, but with his victory over Ma, who is
considered to be the world's second best player behind the great Lee
Chang Ho, Yoo has proven himself to be back in form and one of the
top players in the world. In South Korea, only Lee Chang Ho and Cho
Hoon Hyun are ranked above him.
For their efforts, Yoo received 20 million yen and Ma 7 million
yen. Satoru Kobayashi received 4 million yen for his victory over
Cho Chikun in the game that determined third place.
Meijin challengerThe nine player round-robin tournament to
decide the challenger to the Meijin title holder, Cho Chikun, ended
on Aug. 5 with three players having identical scores of six wins to
two losses. According to the rules of this tournament, the two
highest ranked players, Koichi Kobayashi and Norimoto Yoda, met in a
playoff to decide who would be the challenger. Both are considered
to be strong contenders. Kobayashi represents the old guard, while
Yoda, who is only 33, is expected to be one of the leading players
when Kobayashi and Cho pass their primes. Kobayashi held the Meijin
title for seven consecutive years (from 1988 to 1994) and has played
Cho in numerous best-of-seven title matches, while Yoda has faced
Cho only once, in the Yomiuri-sponsored 1998 Kisei title match. In
the hard-fought playoff, Yoda, playing Black, eked out a win by half
a point.
This title match is expected to be a milestone for both Yoda and
Cho. If Cho loses, it could signal the end of his dominance, while
if Yoda wins, it could be the prelude to his own emergence as
Japan's top player. After having lost his Honinbo title last month
to a relatively unknown player, Cho will be out to prove that he is
still the top player in Japan. Yoda, on the other hand, will want to
show that he is not just an ordinary player who wins a few minor
titles, but cannot meet the challenge for the big ones. For these
reasons, the upcoming title match promises to feature some
hard-fought games.
A capturing raceIn the game in Diagram 1, Black 20
was a powerful move that enabled Yoo to equalize the balance of
territories. When White protected the right side with 21, Black
linked up with 22. A capturing race was now set up between the four
marked white stones and the three black stones on the right.
However, Black's three stones had four liberties (points A in
Diagram 2) to White's three liberties (points B). Therefore,
Black wins this capturing race. For example, if White plays first,
he might exchange 1 for 2 in Diagram 3, then start to fill up
the liberties of the black stones with White 3. Black fills one of
White's liberties with 4, and the race to capture continues until
Black captures the four white stones with 8.
Suppose White answered Black 20 in Diagram 1 by playing at
1 in Diagram 4, In that case, Black would link up with 2.
Again, the marked white stones on the left have only three liberties
(points A) to Black's four.
White loses the capturing race as shown in Diagram 5, but
in this sequence he loses six stones instead of the four lost in
Diagram 3. Therefore, White defended the bigger territory on
the right with 21 in Diagram 1.
The placement of Black 20 in Diagram 1 was a skillful
tactical move that wrested the territory at the bottom from White.
However, the timing of this move was crucial. If Black had played it
too soon, without first having exchanged 18 for 19, White would have
gotten too much territory at the top. Black 18 acted as a stumbling
block, preventing White from expanding his territory at the top and
giving Black some territory in the center.
This game was a cliff-hanger which could have gone either way and
was a fitting conclusion to this year's world championship.
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