Lee Chang Ho, undisputed world go championThere are four
annual tournaments that bill themselves as world championships:
Samsung Cup, LG Cup, Fujitsu Cup, and around Tong Yang Securities
Cup. Every year, top players from around the world head for South
Korea or Japan to compete for hundreds of thousands of dollars in
prize money.
These are all elimination tournaments, so the odds are quite high
against any one player making it to all four finals. However, over
the past year, South Korean superstar Lee Chang Ho not only managed
to get into the finals of all four tournaments, he also managed to
win three of them, and is on the verge of triumphing in the fourth.
Lee is now 23 years old, but he has been a force to be reckoned
with in the go world for the last seven years. Since he was 16, he
has dominated all the South Korean titles, and he has also taken his
share of international titles. However, it will be a major
achievement should he make a clean sweep of all the world cup
championships this year.
Lee's odyssey began last April when he reached the best-of-five
final of the 9th Tong Yang Securities Cup. His opponent was
compatriot Yoo Chang Hyuk, a regular contender in the finals of
these international tournaments. Lee won the match 3-1, making it
the fourth time he has won this tournament.
Then during the spring and autumn, Lee won his way to the final
round of the 11th Fujitsu Cup. There, he faced another superstar,
top Chinese player Chang Hao, who is only 22 years old. After a
titanic struggle worthy of these two young geniuses, Lee managed to
win the game by a mere 11/2 points. This was
the second time for Lee to win this championship.
In the best-of-five Samsung Cup final, held in early February,
Lee faced Ma Xiaochun, China's No. 2 player. The match was closely
contested with both players splitting the first four games, but Lee
managed to eke out a 21/2 point win in the
fifth and final game. This was the second year in a row that Lee has
won this most prestigious of the world championships, which has a
first prize of more than $300,000.
Lee and Ma are facing each other again in the 3rd LG Cup. This is
also a best-of-five final. The first two games were held on March 2
and 3, with Lee winning both of them. A two-game lead in a
best-of-five match is a difficult handicap for anyone to overcome
against Lee, so it seems quite likely that he will prevail in this
tournament as well, making a clean sweep of all the world
championship go tournaments over the last year.
Lee will no doubt be busy over the next year or so. China founded
a new international tournament at the end of past year called the
Spring Orchid Cup. The first two rounds were held in late December
and Lee made it to the quarterfinals, which will be held sometime
this month.
Answers to last week's problems Problem 14. This is
another example of a bent three-point eye space. The key point for
the life or death of the white stones is at Black 1 in Diagram
1. If Black plays here, White cannot make two eyes. On the other
hand, if White is able to play at this key point in Diagram
2, his stones are alive because he can make two eyes: at A and
B.
Problem 15. The white
stones form a square four-point eye space and they cannot make two
eyes, even if it is White's turn to move. For example, if White
plays 1 (Diagram 3) , his stones will form a three-point eye
space similar to the one in Problem 14. It is now Black's
turn, so he can play at 2 and White cannot make two eyes. Note that
the position is symmetrical, so wherever White plays his first move,
he will always end up with a three-point eye space.
Problem 16.
In this straight four-point eye space, the white stones have enough
space to make two eyes without White doing anything. If Black plays
1 (Diagram 4), White plays 2, and he has two eyes: one on the
right and one on the left. If Black plays at 2 instead, White will
defend at 1. Note that White must answer Black 1. If he ignores
Black's move, Black will play 3 in Diagram 5 and White will
be unable to make two eyes.
Problem 17. The white
stones here form a bent four-point eye space. As in Problem
16, they are alive as they stand. If Black plays 1 in Diagram
6, White gets two eyes by playing at 2. If Black 1 at 2, White
will play at 1. Again, White must answer Black 1 with 2. If Black
can play at both 1 and 2, the White stones will be unable to make
two eyes and they will die.
Problem 18. In this
pyramidal four-point eye space, White 1 in Diagram 7 is the
key point; White has made three eyes. If Black gets to play 1 in
Diagram 8, White is reduced to one big eye space and cannot
make two eyes.
Problem 19.
The white stones form a five-point eye space. White must play 1 (or
A) in Diagram 9. Either way, he has two eyes. If Black gets
to play on the key point with 1 in Diagram 10, White cannot
make two eyes and his stones will die. If White plays 2, Black will
play 3.
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