OPEN
CHESS DIARY 221-240
22 July 2003 - 25 February 2004
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240. 25 February 2004: Game of the year 2003
My game of the year 2003 seems, until now, to have gone unnoticed. It was
played in August in the North Ural Cup, a women's tournament in Krasnoturinsk,
Russia. It is far from the best game of the year, but it has an abundance of
what makes chess such an irresistable game: brilliant inspirations,
incredible escapes, idiotic blunders, (a least eleven times a half point is
given away) bizarre situations - it is a ferocious and passionate fight from
beginning to end.
Needless to say, I used a brain prothesis for my remarks.
Matveeva (2475) - Skripchenko (2485), Krasnoturinsk (5), 13 August 2003
1.d4 d5 2.c4 e6 3.Nc3 Be7 4.cxd5 exd5 5.Bf4 Nf6 6.e3 O-O 7.Nf3 c6 8.Bd3 a5
9.Qc2 Na6 10.a3 Nc7 11.h3 Re8 12.O-O Ne6 13.Bh2 Nf8 14.Rab1 Be6 15.Rfc1 Rc8
16.Qb3 b5 17.Ne2 a4 18.Qc2 Bd6 19.Bxd6 Qxd6 20.Ne5 (see diagram)
20...c5
The complications begin with what boils down to a pawn sacrifice.
21.Nc3
21.Bxb5 cxd4 22.Qxa4 Qxe5 23.Bxe8 Rxe8 24.Qxd4 Qb8 is not very promising.
21...b4 22.Nb5 Qb8 23.Qxa4 c4 24.Be2 b3 25.Qb4 Bf5 26.Ra1 Re6 27.a4 N6d7
(see diagram)
28.Nxc4
A little too drastic, perhaps. White gets four pawns for a Knight - but
leaves her King without much defense.
28...dxc4 29.Bxc4 Rg6 30.Qxb3 (see diagram)
30...Rxc4!
A defensive sacrifice for the attack.
31.Qxc4 Bxh3 (see diagram)
Black is winning now. After 32.Qd5 there follows Nb6! 33.Qf3 Bxg2 34.Qxg2 Rxg2+
35.Kxg2 Nxa4 and 32.Kf1 Rxg2 33.Qc7 Qa8 also leaves White with a very difficult
defense.
32.Qc7
This doesn't help either.
32...Qa8! 33.Kf1 Qxg2+ 34.Ke2 Bg4+ 35.Kd3 Bf5+ 36.Ke2 Bg4+ 37.Kd3 Qxf2 38.Qf4 Qe2+
39.Kc3
(see diagram)
39...Be6
39...Nb6 was immediately decisive, e.g. 40.e4 Rf6 41.Qc7 Rf3+ 42.Kb4 Qd2+
and Black wins.
40.b3 Nb6 41.Nc7 Rg2
Good enough, but 41...Nd5+ 42.Nxd5 Bxd5 and the computer move 41...Bc4!
were quicker wins.
42.Kb4 (see diagram)
42...Qd2+
Or 42...Bxb3
43.Kb5 Bd7+ 44.Kxb6 (see diagram)
44...Qb4+?
With 44...Rg6+ 45.Kc5 (45.Kb7 Qa5) Rc6+, followed by Rxc1,
Black could win immediately.
45.Ka7?
45.Nb5 was a draw: 45...Rg6+ 46.Kb7 Qa5 47.Nd6 Qd5+
48.Ka7 Qa5+ (Rxd6 49.Qe5!) 49.Kb7 Qd5+ with a repetition.
Now, 45...Qa5+ was again an easy win: 46.Kb8 Rg6 and White is defenseless.
45...Rg6?
(see diagram)
46.a5!
a5 and b6 being guarded, it is suddenly White who is winning:
46...Bc8 47.Qe4 and everything is defended while Black cannot create any
dangerous threats.
46...Bc6 (see diagram)
47.a6?
A terrible blunder - 47.Rxc6 Rxc6 48.Nd5 would have won on the spot. Now Black
could have forced a peculiar draw: 47...Nd7
48.Nd5 Qf8 (not 48...Bxd5 49.Rc8+ Nf8 50.Qb8 Qe7+ 51.Qc7 and after the forced Queen's
exchange, the white queenside pawns decide) 49.Rxc6 Rxc6 50.Kb7 Qc8+ 51.Ka7 Qd8
52.Kb7 Qc8+ with an unsusual repetition of moves.
47...Bg2?
Giving White good winning chances again. (see diagram)
48.e4?
Threatened with mate in one, White panics. The unlikely 48.Na8 would have kept
an advantage; the computer gives 48...Rxa6+ 49.Kxa6 Bxa8 50.Rc7 or 48...Bxa8
49.Kxa8 Qb5 50.Rc7! and White wins in both cases. Even 48.Nb5 (Qxb5 49.Qb8) would
have kept some advantage, but now Black is winning again.
48...Qb6+?
48...Qxd4+ would have mated quickly, but this doesn't throw away the win.
49.Ka8
Completing his journey across the whole board.
49...Rd6 (see diagram)
50.Qxd6
White has no choice.
50...Bxe4+?
Simply 50...Qxd6 is much better but this doesn't spoil the win.
51.Qd5 Bxd5+ 52.Nxd5 Qd8+ 53.Kb7 Qxd5+ 54.Rc6
King moves are better; now Black has an easier task.
54...Ne6 55.a7 (see diagram)
55...Nd8+?
It is hard not to play such a move in time trouble, but it throws away
half a point again. Winning was 55...Qb5+ 56.Rb6 Qd7+
57.Ka6 Nc7+ 58.Kb7 (58.Ka5 Qd5+ 59.Ka4 Qxd4+) 58...Nd5+
59.Ka6 Nxb6 60.Kxb6 Qxd4+ etc.
56.Kc7 Qxc6+ 57.Kxd8 (see diagram)
57...Qa8+(?)
A losing attempt. Black should have taken the perpetual along
the 6th rank with Qd6+ etc. But with a Queen against a Rook, she probably
thought there should be a win.
58.Kc7 (see diagram)
58...h5?
Bad counting? There was a not too difficult draw with Kf8
59.d5 Ke8 Black then threatens Qd8+ followed by a perpetual, and after
60.Re1+ Kf8 White has nothing better than the repetition 61.Ra1 Ke8 etc.
59.b4?
Why not run with the frontmost pawn?
Winning was 59.d5 Kf8 60.d6 Ke8 61.d7+ Ke7 62.Re1+ etc.
Now it's a draw again.
59...h4
Here, 60.d5 does not win anymore:
60...h3 61.b5 (61.d6 probably even loses:
61...h2 62.d7 h1Q 63.Rxh1 Qxa7+ etc.) 61...h2 62.b6 h1Q 63.Rxh1 Qxd5 and the Queen
is freed to give her perpetual.
60.b5 (see diagram)
60...h3?
Black's panic is palpable: 'I'm a Queen ahead, but how
on earth do I draw?'
With Kf8, preparing that check on d8, that was still possible:
61.b6 Ke7 62.b7 (62.Re1+ Kf6 63.Ra1 Ke7 etc.) 62...Qd8+ etc.
Now Black is lost and stays lost.
61.b6 h2 62.b7 h1Q 63.b8Q+
Apart from everything else in this crazily rich game (King's chase,
Steel King, long-term Queen's sacrifice, mutual polygamy), it also sports a
double excelsior: two pawns march to their promotion squares in
five straight steps - be it that the white pawn started from b3.
63...Kh7 64.Rxh1+ Qxh1 65.a8Q Qh2+ 66.Kd7 Qh3+
67.Ke7 Qe6+ 68.Kf8
and Black resigned.
An incredibly exciting, truly fantastic game. Matveeva shared first place
in this tournament and Skripchenko shared last (had she won, they
would have tied in the middle) but that is hardly worth mentioning compared
to the work of art they created together in this game.
The whole game (without annotations)
can be played over in Palview.
239. 11 February 2004: The mystery of Visvanath Apte
In my old files, I found this 30-year old clipping from the Dutch Pravda.
Under the heading CHESS PRODIGY IN INDIA, it says:
A six year old child in India has managed to solve eighty complicated
chess problems within an hour. The boy, Visvanath Apte, learned chess a few
years ago, and in the first junior competition in which he took part, he won
240 out of 340 games. Some time ago, the prodigy even beat New Delhi's number
four.
Anand - you immediately think, even if everything is just off the mark. Anand is from Chennai (the old Madras), not New Delhi, his name was badly mistreated, and, having been born in December 1969, he was only four when this story was published. And a junior competition where a 6-year old plays 340 games...?
Anand's biography says that he learned chess at six - after this story was in the paper. However, things seem so strangely beyond coincidence, especially that mangled Viswanathan Anand, that you wonder if there might have been some pre-Anand. Nothing was ever heard again from Visvanath Apte. Not one game in the databases. Did he exist at all?
Anyone to shed some light on this little mystery?
238. 31 January 2004: Debunking Carlsen - Ernst
13-year old Magnus Carlsen's win against Ernst in the Corus-C tournament
is already famous - it will be reprinted many times, especially if Carlsen
fulfills the high hopes we have for him. Not only was it the decisive
game in group C (before that round,
Carlsen and Ernst shared first place with 9 out of 11);
it was also a dazzling sacrificial attack, ending in a rare epaulettes mate.
Even more amazing: he'd played the initial
sacrifice before.
Carlsen - Ernst, Corus-C, 24 January 2004
1.e4 c6 2.d4 d5 3.Nc3 dxe4 4.Nxe4 Bf5 5.Ng3 Bg6 6.h4 h6 7.Nf3 Nd7 8.h5 Bh7
9.Bd3 Bxd3 10.Qxd3 e6 11.Bf4 Ngf6 12.O-O-O Be7 13.Ne4 Qa5 14.Kb1 O-O 15.Nxf6+ Nxf6 16.Ne5 Rad8 17.Qe2 c5
18.Ng6 fxg6 Pretty naïve when in similar situations,
(as you should know when you play this line), that Knight had been left
alone, even by Anand (against Bologan, Dortmund 2003.)
There is nothing wrong with Rfe8 - see below.
19.Qxe6+ Kh8 20.hxg6 (see diagram) 20...Ng8
After Rde8, 21 Rxh6+ gxh6 22.Bxh6 with the
unstoppable Rh1 is winning, but in a game
Smolkov - Rumyantsev, Petersburg 2003 21.Bxh6 proved to be sufficient too:
21...gxh6 22.Qe3 Ng4 23.Rxh6+ Kg7 24.Rh7+ Kxg6 25.Qe4+ Rf5 26.Qxg4+ Bg5
27.Qh5+ and Black resigned.
21.Bxh6 gxh6 22.Rxh6+ Nxh6 23.Qxe7 Nf7
Even this position (see diagram right) had already occurred.
Almagro Llanas - Gustafsson,
Madrid 2003, ended in a draw after 24.Qf6+ Kg8 25.Rh1 Nh6 26.Qe7 Nf7
27.Qf6 Nh6 28.Qe7 Nf7 with a threefold repetition. But at move 27,
White could still have won with 27.Rh5 (or Rh3, or Rh4), e.g. 27...Ng5
28.Rh7 Qe1+ 29.Qxe1 Nxh7 30.gxh7+ etc.
Carlsen - Ernst brilliantly ended: 24.gxf7 Kg7 25.Rd3 Rd6 26.Rg3+ Rg6
27.Qe5+ Kxf7 28.Qf5+ And, being a gentleman, Ernst let Carlsen carry
out the beautiful mate: 28...Rf6 29.Qd7 mate (see diagram left)
Analyses of this fascinating game are all over the internet - agreeing
that after 18...fxg6? Black is lost.
But how much of this did Carlsen see when he played
18.Ng6?
According to Hans Ree, who interviewed him on stage right after the game,
he said that he had seen everything right up to the end, although it was
unclear whether he meant he had calculated this at the board during the
game, or had seen it at home before the game. Elsewhere, he has
said that he didn't even know Ernst played the Caro-Kann, and had prepared
against a different opening. He also did not know the Almagro game, but he
did remember Ng6 from Bologan - Anand.
He certainly knew Ng6 - it has been played in at
least ten tournament games - in fact, he had already played it himself.
Just a few days before the Corus tournament started, the following blitz
game was played on ICC.
Magnus - AndrewMartinIM, ICC 3 0, 30 December 2003
Magnus is Carlsen's ICC handle.
1.e4 c6 2.d4 d5 3.Nc3 dxe4 4.Nxe4 Bf5 5.Ng3 Bg6 6.h4 h6 7.Nf3 Nd7 8.h5 Bh7
9.Bd3 Bxd3 10.Qxd3 Ngf6 11.Bf4 e6 12.O-O-O Be7 13.Kb1 O-O 14.Ne4 Nxe4
15.Qxe4 Nf6 16.Qe2 Qd5 17.Ne5 Rad8 18.f3 Qa5 19.g4 c5 (see diagram)
With the same position als in Carlsen - Ernst, except that the white f- and g-pawns
have advanced a little. That slightly changes the combination.
20.Ng6 Rfe8
If here 20...fxg6 21.Qxe6+ Kh8 22.hxg6 Ng8 23.Bxh6 gxh6
24.Rxh6+ Nxh6 25.Qxe7 Nf7, then 26.gxf7 Kg7 doesn't work, because the
3rd rank is not available, but White can choose from two other wins:
a) 26.Rh1+ Kg7 (or 26...Kg8 27.Rh5 as in b)) 27.Rh7+ Kxg6
28.Qe4+ Kf6 29.g5+ Kxg5 30.Rg7+ Kh5 31.Qg6+ Kh4 32.Qg4 mate
b) Like in Almagro Llanas - Gustafsson:
26.Qf6+ Kg8 27.Rh1 Nh6 28.Qe7 Nf7 29.Rh5! and White wins the Knight,
and the Queen for a Rook, e.g. 29...Qa6 30.gxf7+ Kg7 31.Qe5+ Qf6 32.Rh7+ etc.
The game continued: 21.Nxe7+ Rxe7 22.dxc5 Red7 23.Bd6 Ne8 24.Be5 Qxc5
25.Rxd7 Rxd7 and was drawn after many adventures.
Finally - what would have happened, had Ernst simply played 18...Rfe8?
Amazingly, just two days after his celebrated game, Carlsen had that exact
position again, in a blitz game on ICC.
Magnus - Inspektor, ICC 3 0, 26 January 2004
(First 17 moves as in Carlsen - Ernst)
18.Ng6 Rfe8 19.Nxe7+ Rxe7
20.dxc5 Red7 21.Rxd7 Rxd7 22.Qe5 Nd5 Rd5 - and others - draw, but now
White could have won with 23.c6 and 24.c4 After 23.Be3 Nxe3 24.Qxe3 Rd5
25.c6 bxc6, a drawn ending was reached which Magnus eventually won.
PS 4 February: I've been told that even during the Corus
tournament, Carlsen and Ernst have been playing each other on ICC,
Carlsen possibly without knowing.
I found three earlier (correspondence) games with the triple sacrifice on e6 and h6, all
beginning with the position that also occurred in Bologan - Anand,
and which was reached there after 1.e4 c6 2.d4 d5 3.Nc3 dxe4 4.Nxe4 Bf5 5.Ng3 Bg6
6.h4 h6 7.Nf3 Nd7 8.h5 Bh7 9.Bd3 Bxd3 10.Qxd3 Ngf6 11.Bf4 e6 12.O-O-O Be7
13.Kb1 Qa5 14.Ne5 Rd8 15.Qe2 O-O (see diagram) 16.Ng6 fxg6
(Anand played Rfe8) 17.Qxe6+ Kh8 18.hxg6 Ng8 19.Bxh6 and now:
19...gxh6 20.Rxh6+ Nxh6 21.Qxe7 Nf6 22.g7+ Kg8 23.gxf8Q+ Rxf8 24.Qxb7 Re8
25.Qxc6 Re1 26.Qf3 Qd2 and White resigned (Andonov - Curtacci, cr 1989)
19...gxh6 20.Rxh6+ Nxh6 21.Qxe7 Nf7 22.Rh1+ Kg7 23.Nh5+ Kg8 24.Nf4 Ng5
25.Ne6 Rf7 26.Qxg5 Qxg5 27.gxf7+ Kxf7 28.Nxg5+ Kg6 29.Nh3 Nf6 30.c3 Rxd4 31.cxd4 and
Black resigned (Snape - Herries, cr 1994)
19...Nxh6 20.Rxh6+ gxh6 21.Qxe7 Nf6 22.d5 Rg8 23.Qxf6+ Rg7 24.Nf5 Qc7
25.Nxg7 Qxg7 26.Qxd8+ Qg8 27.Qf6+ Qg7 28.Qd6 and Black resigned
(Ruggeri Laderchi - O'Callaghan, email 1998)
PS 19 February: In an interview for
Chess Today (see issues 1196 and 1197),
made well after the Corus tournament, Kavalek asked Carlsen:
"Was this smashing victory [against Ernst] based on a marvellous imagination or
did it represent homework well done?"
Carlsen answered: "Well, Ernst left my preparation for him at move 1! During the game I knew that 17.Qe2 was theory, but I spent 25 minutes figuring out why. I had seen Ng6 in Bologan-Anand, Dortmund 2003. Whether that was imagination or homework, decide for yourself!"
Strange... still no recollection that he played Ng6 himself before
he played it against Ernst.
237. 20 January 2004: Bareev's missed save
Dennis Monokroussos sends me some wonderful analysis of a moment in yesterday's fascinating Anand - Bareev game at Corus.
Anand - Bareev, Corus (8), 19 January 2004
1.e4 e6 2.d4 d5 3.Nc3 Nf6 4.Bg5 dxe4 5.Nxe4 Be7 6.Bxf6 Bxf6 7.Nf3 O-O
8.Qd2 Nd7 9.O-O-O Be7 10.Bd3 b6 11.h4 Bb7 12.Neg5 Nf6 13.c3 Bxf3 14.gxf3 c5
15.dxc5 Qc7 16.Kb1 bxc5 17.Rdg1 Rfd8 18.Qc2 h6 19.Bh7+ Kf8 20.Nxf7 Kxf7
21.Qg6+ Kf8 22.Qxg7+ Ke8 23.Re1 Rd6 24.Qh8+ Bf8 25.Bg6+ Ke7 26.Rhg1 Rb6
27.Bf5 (see diagram)
Here, Bareev played 27...Kf7 and eventually lost, but Monokroussos,
"with the help of our mutual friend the tin can", suggests 27...Qf4,
attacking the Bf5, and guarding the Nf6. White then has nothing better than:
28.Rg7+ Kd6 29.Rxe6+ Kd5 30.Rd7+ (30.c4+ Qxc4 31.Rd7+ is the same) 30...Nxd7
31.c4+ Qxc4 (see analysis diagram 1 below)
|
|
|
|
| analysis diagram 1 | analysis diagram 2 | analysis diagram 3 | analysis diagram 4 |
The black King seems safe, and White seems lost now, but the wonderful quiet move 32.Qg8, unguarding b2, almost turns the tables. But after 32...Rxb2+ 33.Kxb2, Black has Bg7+! (see analysis diagram 2)
Now, 34.Qxg7? Rb8+ is mate, but White has the amazing 34.Re5++! - a counter-doublecheck and selfpin. Black must then choose between:
a) 34...Kd4 35.Re4+ Kd3+ 36.Re5+ Kd4 37.Re4+ etc., with a highly unusual repetition where three out of four moves are checks. In this line White can even play 36.Qxg7 Rb8+ 37.Ka3 Qc1+ 38.Ka4 Qc2+ with another repetition - it is dangerous for Black to deviate with 38... Nb6+ 39.Kb5 Nc4+ 40.Kc6 Rb6+ 41.Kc7 (see analysis diagram 3) and now White's King has also crossed the board, and White wins.
b) 34...Kxe5 35.Qxc4 Kxf5+ (see analysis diagram 4) with an unclear position where Black appears to be slightly better.
After 27...Kf7, the game continued: 28.Bg6+ Ke7 29.Bc2 Kf7 30.Rg6 Qf4 31.Reg1 e5 32.Rg7+ Ke6 33.R1g6 Rab8 34.Qg8+ Kd6 35.Rxf6+ Qxf6 36.Rg6 Kc7 37.Rxf6 Rxf6 38.Qh7+ Kb6 39.Be4 Rd6 40.h5 a6 41.Qf7 Rd2 42.a3 Rd1+ 43.Kc2 Rd6 44.b4 cxb4 45.axb4 Rdd8 46.Qe6+ Rd6 47.Qc4 Rf6 48.Qd5 and Black resigned.
PS: It turns out Anand saw this line during the game.
See
John Henderson’s report at TWIC. I can't begin to describe my awe.
It's interesting to see Anand's assessment of
diagram 4: "Black has a small material advantage, but it will be a
draw due to the king never being able to find shelter from the many
queen checks."
236. 10 January 2004: Some properties of the en passant capture
A reader told me that he had given his opponent a discovered check by removing an enemy piece from the checking line, and asked me if that was perhaps unique. At first, it even sounded impossible - you can only remove an enemy piece by capturing it. But in that case, it cannot be a discovered check. Of course, he had done it with an en passant capture - apart from castling, the only move that clears two squares and two lines at once.
In games, such a discovered check is rare, but it
does occur, even in one opening line: 1.e4 Nf6 2.e5 Nd5 3.d4 d6
4.Nf3 dxe5 5.Nxe5 Nd7 6.Nxf7 Kxf7 7.Qh5+ Ke6 8.c4 N5f6 9.d5+ Kd6
10.Bf4+ Qf7 and Qf5 are the normal moves 10...e5 (see
diagram) and now 11.dxe6+. White removes Black's pawn from
the checking line, giving a discovered check which involves three instead of
the usual two pieces. In Popelka - Hlavac, Chechoslovakia 1988, White
didn't have much compensation after 11...Kxe6 12.Qf3 Kf7, and lost
without much ado.
Had the white Queen been on d1, this would have
been a unique double check, without either of the checking pieces
making a move themselves.
Van der Heiden - Chess Challenger Voice, Amsterdam 1980
1.d4 Nf6 2.c4 d5 3.cxd5 Nxd5 4.Nf3 e6 5.e4 Nf6 6.Nc3 Bb4 7.Bd3 O-O
8.e5 Nd5 9.Bxh7+ Kxh7 10.Ng5+ Kg6 11.h4 Nxc3 12.bxc3 Bxc3+ 13.Kf1 Bxa1
14.h5+ Kh6 15.Qg4 Qxd4 16.Nxe6+ g5 (see diagram) 17.hxg6 mate.
The double-checking pieces Rh1 and Bc1 remain in their
places, but note how the clearance of the g-file also plays its part -
without Qg4 covering g6, it wouldn't be mate. In fact, Van der Heiden
didn't really play but composed this game, just to hear his newly
acquired Voice speak its longest possible sentence: "Pawn takes pawn en
passant check and mate I lose."
The horizontal version of this discovered check is even stranger.
In Dlugy - Bologan, New York 1993, White had just played 57.f2-f4+,
and Black countered with the discovered cross-check 57...exf3+,
removing two pieces at once from the checking line. Black won quickly
after 58.Kxf3 Kf5 etc.
When a discovered check is possible on a doubly occupied line, a
pin must also be possible.
White to play
Plaskett - Parker
Eastbourne 1991
With 48.f4, a strange phenomenon occurred: the pinned pawn g4
remained pinned even if White interfered the pinning line. In fact,
the move f2-f4 both unpins the Pg4 because g3 is now possible,
and pins it, as gxf3 e.p. is impossible. (This is the solution to
the riddle in item 234 - see the PS there.)
There followed: 48...h5 49.Ra5 f6 50.Ra1 g3
51.h3 51.hxg3+ should be winning easily, but Plaskett missed Black's
devilish idea. 51...f5 52.Kf3 g2 53.Kxg2 Rxa7 54.Rxa7 stalemate.
235. 8 January 2004: Perpetual at 7
In the Garayazli 60-JT endgame study composing tourney, organized by the city of Sumgait in Azerbaijan, one of the commendations went to this study.
White to play and draw
M. Iskenderov
Commendation, Garayazli 60-JT, 2003
1.Rf7+ Now, Kh8 2.Bf6+ loses. 1...Kg8 2.Rf8++! Kxf8 Kg7 or Kh7 3.Rf7+ would lead to a repetition, but now White has an amazing perpetual: 3.Bh6+ Ke7 4.Bg5+ Kf8 5.Bh6+ etc.
A bit light for an award perhaps, but in EG 151 which I received today, I read that
Misreddin Iskenderov was born 14 January 1995 - so he was 8 years old when
he composed this. That might very well make him the youngest published composer ever. (PS 26 August 2007: Susan Polgar composed a two-mover at 4, although it is not clear when it was first published.)
The position is very natural and in fact, Iskenderov found his idea over the
board, in a game he played in the Sumgait under-10 championship he won a year ago,
when he was still 7.
Iskenderov - Babazade, Sumgait, 3 January 2003
1.e4 c5 2.Nf3 d6 3.d4 cxd4 4.Nxd4 Nf6 5.Nc3 g6 6.Be3 Bg7 7.Be2 O-O
8.O-O Nc6 9.f4 Re8 10.Bb5 Bd7 11.Nxc6 bxc6 12.Bc4 Ng4 13.Bd4 e6 14.Bxg7 Ne3
15.Qd4 Nxc2 16.Qf6 Nxa1 17.Rxa1 Rb8 18.Qd4 e5 19.fxe5 c5 20.Bf6 cxd4
21.Bxd8 dxc3 22.Bg5 cxb2 23.Rf1 b1Q 24.Bxf7+ Kg7 25.Bc4 (see diagram)
Almost any move would win - but not 25...Qxe4 26.Rf7+ Kg8
26...Kh8 loses here, too: 27.Bf6+ Kg8 28.Rg7+ and 29.Rg8 mate) 27.Rf8++!
Quite something to see, when you're 7. 27...Kxf8 28.Bh6+ Ke7 29.Bg5+ draw
234. 4 January 2004: Pinning riddle
I wonder how obvious the solution is to this little riddle: How can you play a move which unpins an enemy piece, and pins it? (This is about (un)pins against the King only.)
Solution in a few days. It has happened in games, so there's no hanky-panky.
PS 10 January: Several readers rightly guessed I meant a situation
like in the problem below.
Mate in 2
O. Sommerfeldt
Danske Skakopgaver, 1902
1.d4! threatens 2.Qf2 mate. It keeps Pe4 pinned (exd3+ is not possible)
but also unpins it, as Black now has: 1...e3+ But that blocks
e3, allowing 2.d5 mate.
For an example of this pin/unpin from a game, and some
other properties of the en passant capture, see item 236.
It turned out to be hard to formulate my riddle unambiguously without giving
away too much - initially, I even left out the King-stipulation. But with
relative pins (against other pieces than the King, or even squares),
pinning could be generalized too far.
Still, a few nice pinning sophisms turned up.
Hauke Reddmann had a counter-riddle: can the King be pinned?
It can be pinned relatively by a pawn; when the Ke3 moves here,
e2-e4 can follow, with material gain.
And in a selfmate, it could be said that the white King is partially
pinned in a situation like this - it cannot leave the h-file, as Black
would be mated.
This situation is often called a half-pin - but that is the
problemist's term for the situation in the diagram below. More
appropriate is partial pin; this black Rook can move vertically,
but not horizontally. (One might even wonder if it is pinned at all,
as the pin does not deprive it of any move. However, imagine
a wKc7 and wBh1, and the Rook turns out to defend against Bxb7 mate.)
Halfpins are a popular theme in problems, but happen rarely in games.
White to play
Fischer - Robatsch
Olympiad, Varna 1962
After 13.g4, Black did not play Nxg4 because of the half-pin 14.Rdg1
In that case, the Bg7 and Ng4 could both capture on h6, but would thereby
expose the other to the pin by Rg1, allowing Qxh6.
After 13...Nf8 Robatsch lost quickly anyway:
14.gxh5 Ne6 15.Rdg1 Kh8 16.Bxg7+ Nxg7 17.Qh6 Rg8 18.Rg5 Qd8 19.Rhg1 Nf5
20.Bxf5 and Black resigned.
233. 31 December 2003: Cortlever's inspiration
In the study-composing tournament for his 50th birthday, Jan Timman gave a minor award to a pawn-ending with an unusual material imbalance; White has one pawn, Black has them all. Still, White can win.
White to play and win
D. Novomesky and L. Siran
4th Honourable Mention, Jan Timman 50-JT, 2002
1.c7 e2 2.c8Q e1Q With a Queen's ending of seven against
zero pawns - the zero pawns win. 3.Qc4+ Qb4 4.Qc6+ Qb5 If b5,
then 5.Qc2+ and mate. 5.Qxe4+ Qb4 6.Qd3! And for the first time,
the crucial Zugzwang has arisen. Qb3 mate must remain guarded;
Qb5 would be met by Qa3 mate, and b5 by Qc2+ and mate. Black can only
make pawn moves. 6...a6 7.Qd7+ Qb5 8.Qd4+ Qb4 9.Qd3 But White
can repeat the Zugzwang. 9...g4 And he can also repeat it now.
The rest is evident. 10.Qd7+ Qb5
11.Qxg4+ Qb4 12.Qd7+ Qb5 13.Qd4+ Qb4 14.Qd3 g5 15.Qd7+ Qb5 16.Qd4+ Qb4
17.Qd3 g6 18.Qd7+ Qb5 19.Qd4+ Qb4 20.Qd3 g4 21.Qd7+ Qb5 22.Qxg4+ Qb4
23.Qd7+ Qb5 24.Qd4+ Qb4 25.Qd3 g5 26.Qd7+ Qb5 27.Qd4+ Qb4 28.Qd3 g4
29.Qd7+ Qb5 30.Qxg4+ Qb4 31.Qd7+ Qb5 32.Qd4+ Qb4 33.Qd3 and after
Zugzwang number 8, Black will finally be mated.
Of course, this particular Zugzwang wasn't new, or the study would have obtained
a higher reward. Still, it is the lengthiest elaboration of a 60-year old
idea, which was first shown by Nico Cortlever (1915 - 1995),
a Dutch top player for many years, and a prominent endgame composer.
White to play and win
N. Cortlever, 1941
1.Qe7+ Qg5 2.Qe4+ Qg4 3.Qe3 and White wins.
That this little gem should bear Cortlever's name is
uncontested, but the exact source is not clear. According to the Van der
Heijden database, it was first published in 1941, in something called
Spirit. In other sources, it is 'from a game played in Holland in
1954 or 1955'. I first saw it in the booklet Schachmatt
by Kurt Richter, where he says it is 'from a game played in Holland in
1941, as told by E. Cortlever.' According to Harrie Grondijs, author of
the book Charged Moves and Progressions about Cortlever's studies,
an earlier edition of this booklet had appeared in 1942 (with an
introduction dated 1941), as Das Matt, subtitled 'Eine Plauderei
über den Mattangriff im Schach' and as a 'Sonderdruck für das Oberkommando
der Wehrmacht, Abteilung Inland'.
Grondijs also quotes Selman, who wrote in 1952:
'According to N. Cortlever [this position] occurred in a game played in 1941
in the Netherlands.' A game then - but by Cortlever himself? When I
published the position in 1978 in Schaakbulletin, with the strange
caption 'After Cortlever, Netherlands 1954', Cortlever phoned me to correct that, saying
something like: 'I am not very proud of this, but in 1941, I played in the
international tournament in Munich. When analyzing a game there, that
position came up.'
Stupidly, I didn't ask him which game - or I
forgot. But even if Cortlever said contradictory things to Selman and me
(a game from Holland; analysis from Munich) - the position is from 1941.
Cortlever's little gem proved to be fruitful - soon interesting elaborations
were shown.
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White to play and win
A. Ojanen Schackvärlden 1943 1.Qc4+ Qb4 2.Qc6+ Qb5 3.Qxe4+ Qb4 4.Qd3 g6 5.Qd7+ Qb5 6.Qd4+ Qb4 7.Qd3 g5 8.Qd7+ Qb5 9.Qd4+ Qb4 10.Qd3 g4 11.Qd7+ Qb5 12.Qxg4+ Qb4 13.Qd7+ Qb5 14.Qd4+ Qb4 15.Qd3 and White wins. |
White to play and win
K. Arnstam Schackvärlden 1943 1.b6 cxb6 2.dxc6 f3 3.c7 fxe2 4.c8=Q e1=Q 5.Qc4+ Qb4 6.Qc6+ Qb5
7.Qxe4+ Qb4 8.Qd3 and White wins.
|
White to play and win
H. Rinck Chess 1944 1.Qb5+ Kh4 2.g5 Bxg5 3.Rxg5 Qxg5 4.Qb4+ Qg4 5.Qe7+ Qg5 6.Qe4+
Qg4 7.Qe3 and White wins.
|
Ojanen was the first to show the repetitive pawngrabbing potential of the Zugzwang (in the Czech Sach, also in 1943, Prokeš published almost the same position, only with Qe2 at c7 and Pg7 at g6.) Arnstam, in the same magazine as Ojanen, but in a later issue, showed how the Zugzwang can arise from a pawn endgame, and the great Rinck added some not-too-impressive tactical foreplay. (PS 2 January: It is also cooked, as the Kuwait Wizard, Bader Al-Hajiri, had already spotted in 1987. White mates in 4 with 1.Kg3 h4+ (hxg4 2.Qb5+ Be5+ 3.Qxe5 mate) 2.Kf3 h3 3.Qf4+ Kh4 4.Qh6 mate.)
Over the years, composers added new effects.
White to play and win
D. Petrov, 1946
Apart from Kxb4, Black threatens mate by Qa8+ Therefore: 1.Ba5! bxa5
Not 1...Qa8 2.Qd3+ Ka4 3.Qc4+ Kxa5 4.Qa2+ 2.Qd3+ Ka4 3.Ka2
Threatening to win the Queen. 3...Qb7 4.Qc4+ Qb4 5.Nb6+! cxb6 6.Qd3
and now the old story: 6...g6 7.Qd7+ Qb5 8.Qd4+ Qb4 9.Qd3 g5 10.Qd7+ Qb5 11.Qd4+ Qb4
12.Qd3 g4 13.Qd7+ Qb5 14.Qxg4+ Qb4 15.Qd7+ Qb5 16.Qd4+ Qb4 17.Qd3 and White
wins.
An elegant feature is that both black pawns reach
their blocking squares during the solution.
White to play and win
A. Mamedov
'64', 1979
After the introduction 1.h4+ Kxh4 2.Qf4+ Qg4 3.Qf6+ Qg5 4.Qd4+ Qg4 5.Qxd8+ Qg5
6.Qd4+ Qg4 7.Qe3 the Zugzwang even turns out to work when Black has a Bishop.
It must keep guarding f7 (otherwise something like 7...Bd7 8.Qf2+ Kg5 9.Nf7+ Kg6
10.Ne5+) and 7...Qg7 is not possible in view of 8.Qh3+ Kg5 9.Qg3+ Kf6 10.Qc3+ So:
7...Bg6, but now: 8.Qe7+ Qg5 9.Nxg6+ hxg6 (Kg4 10.Qe4+) 10.Qe4+
Qg4 11.Qe3 and White wins.
White to play and win
M. Matous
Ceskoslovensky Sach, 1999
1.Qb2+ Kf5 (1...Kd6 2.Qb4+ Ke5 3.Qc3+ Qd4 4.Re2+ Kd5 5.Rd2) 2.Rf2+ Kg5 3.Qc1+ Kh4 4.Rf4+ Bg4 5.Qe1+ Kg5 6.Qe3! A silent move which restricts Black to 6...Kh4 But now, the very surprising: 7.Rxg4+! Qxg4+ Otherwise, Black loses his Queen or is mated. 8.Kh2 and White milks the Zugzwang again: 8...b6 9.Qe7+ Qg5 10.Qe4+ Qg4 11.Qe3 b5 12.Qe7+ Qg5 13.Qe4+ Qg4 14.Qe3 b4 15.Qe7+ Qg5 16.Qxb4+ Qg4 17.Qe7+ Qg5 18.Qe4+ Qg4 19.Qe3 and wins.
If what Cortlever told me is true, and a Munich 1941 game inspired him, then which one was it? In his own fifteen games, there were no Queen's endings. There were five in the whole tournament, but only one which might possibly have been his inspiration.
Mross - Alekhine
Munich 1941
This position arose after 37...Qxc3 For almost 70 moves, Alekhine
doggedly tried to win this last round game, even if he could not catch up
with Stoltz, who was already the sole tournament winner, 1½ points ahead of him.
There followed: 38.Qb7 Qe3 39.Qa8 Qd4 40.Qb7 f5 41.Qc8 f6 42.Qc7+ Kh6
43.Qc1+ Kh5 44.Qc8 Qe5+
45.Kh1 Kh4 46.Qc4+ f4 47.Qc6 Kg5 48.Qf3 f5 49.Qb7 Qd6 50.Qf3 Qd4 51.Qb7 Qe5
52.Qa8 Qc7 53.Qd5 Kh4 54.Qf3 Qb8 55.Qf2+ Kg5 56.Qc5 Kh5 57.Qe7 Qg8 58.Qb7 Kh6
59.Qe7 Kh5 60.Qe2+ Kh6 61.Qe7 g5 62.Qf6+ Qg6 63.Qh8+ Qh7 64.Qf6+ Kh5 65.Qe5 Qf7
66.Qd6 Qg6 67.Qe5 Qf7 68.Qd6 Qb3 69.Qe5 Qd1+ 70.Kh2 Qd7 71.Qe2+ Kg6 72.Qe5 Qc6
73.Qe2 Qe4 74.Qd1 Qe5 75.Kh1 Kh6 76.Qf3 Qe4 77.Qd1 Kg6 78.Kh2 Kf6 79.Qd6+ Qe6
80.Qd8+ Kg6 81.Qd3 Qe5 82.Qf3 Qc7 83.Kh1 Kf6 84.Qh5 Qg7 85.Qf3 Qe7 86.Qd1 Qe6
87.Qd4+ Kf7 88.Kh2 Kg6 89.Qd3 Kh6 90.Qf3 Qe5 91.Kh1 Kg6 92.Kh2 Qd6 93.Kh1 Kf6
94.Qe2 Qe5 95.Qf3 Qe1+ 96.Kh2 Ke5 97.Qh5 Qg3+ 98.Kh1 Kd4 99.Qh8+ Kd3 100.Qe5 Kd2
101.Qxf5 Ke2 102.Qc2+ Kf1 103.Qc4+ Ke1 104.Qc1+ Ke2 draw.
But even if White has these h- and g-pawns here, I don't really
see how that could lead to Cortlever's ending.
An intriguing footnote is that Kurt Richter, who was the first to publish the
position, was also one of the participants in Munich. If Cortlever told him
('mitgeteilt von') about it, then that conversation must have taken place during
that tournament. But apparently, he didn't say what inspired him. Did Richter
forget? Was it a Dutch game after all - was Cortlever mistaken when he told me
a Munich game had been his inspiration? Do I remember him wrongly?
Where did his gem come from?
You can play over
the studies and the game online in Palview.
232. 24 December 2003: The wrong side in a Steinitz Gambit
A reader, Andreas Fecke, sent me an interesting game he once lost. It is an unhappy echo to the famous (fake; see item 72) game Steel - NN, where the white King reached a7 in a Steinitz Gambit and won.
Fecke - Mackenberg, 45', Lippstadt 1989
1.e4 e5 2.f4 Nc6 3.Nc3 exf4 4.d4 Qh4+ 5.Ke2 d5 6.exd5 Qe7+ 7.Kf2 Qh4+
8.g3 fxg3+ 9.Kg2 Bd6 10.dxc6 In 1883, Steinitz lost to Chigorin with 10.Qe1+ Nce7
11.hxg3 Dxd4
10...gxh2 and to Englisch with 11.Qf3 hxg1Q+ 12.Kxg1 Qxd4+
11.Nge2 Nf6 12.Be3 Much better Qd3 12...Ng4 13.Qd2 Qh3+
A nice Queen's sacrifice which forces mate.
14.Kxh3 Nxe3+ 15.Kh4 g5+ 16.Kh5 Bg4+ 17.Kxg5 Rg8+ 18.Kh6 Bf8+(?) (18...Nf5+
19.Kxh7 Rg7+ 20.Kh8 O-O-O mate is easier) 19.Kxh7 Rg7+ 20.Kh8 O-O-O 21.cxb7+ Kb8 22.Qxe3
Bf5 23.Qh6 Bg6 24.Qxg7 After Qh4 it takes a little longer. 24...Be7+
25.Qg8 Bf6 mate. (See diagram.)
231. 19 December 2003: The Strong Queen
Accidentally made a surprising discovery. I thought it was generally accepted that two Rooks are slightly stronger than a Queen. However, in the 4287 pure endgames of Queen + pawns vs. 2 Rooks + pawns that I found in my database, the Queen scores 55,7 %, and the Rooks only 44.3 %
I also thought that generally, three minor pieces would be at least as strong as the Queen in such endgames. But here, the Queen scores 62,8 % and the pieces only 37,2 %
230. 9 December 2003: The Bishop's Pair
Today's Chess Today had a very nice pair of Bishop's moves.
Jakab - Cernousek, Attila Schneider Memorial, Budapest 1 december 2003
1.d4 d5 2.c4 e6 3.Nc3 c5 4.cxd5 cxd4 5.Qa4+ Bd7 6.Qxd4 exd5 7.Qxd5 Nc6
8.Bg5 Nf6 9.Qd2 h6 10.Bh4 g5 11.Bg3 Qa5 12.e3 O-O-O 13.Bc4 (see diagram)
13...Bh3! 14.Qxd8+ Nxd8 15.gxh3 and now the other one: 15...Ba3
and White resigned.
This made me think of a double double Bishop's mate by Gerard Welling.
Welling - Wrobel, Mondorf Open 1982
1.d4 d5 2.e4 dxe4 3.Nc3 Nf6 4.f3 exf3 5.Nxf3 c6 6.Bc4 Bf5 7.Ne5 e6 8.O-O Bxc2
9.Nxf7 Bxd1 10.Nxd8 Kxd8 11.Rxd1 Nd5 12.Re1 Bb4 13.Rxe6 Nxc3 14.bxc3 Bxc3
15.Bg5+ Kd7 16.Re7+ Kd6 17.Rd1 h6 18.Re6+ Kd7 19.Bf4 Here, Welling hoped
for 19...Bb4 20.d5 cxd5 21.Bxd5 Nc6 22.Bc4+ Kc8 (see diagram right)
23.Rxc6+ bxc6 24.Ba6 mate. But after 19...Re8 20.Rd6+ Ke7 21.Rd3 Bb4 22.Re3+ Kf8,
he still got his chance: 23.Rf6+ (see diagram left) and Black resigned.
229. 19 November 2003: World Champion
It's interesting that today's New York Times / Associated Press write: "World chess champion Garry Kasparov tied his computerized opponent X3D Fritz in a final match Tuesday, leaving the four-game series in a draw."
They are right - everything has been in vain.
228. 24 October 2003: Latest ultimate blunder
Thanks to Chess Today (a must for any chess follower) for showing the latest example of this wonderful theme.
Korchnoi - Van der Stricht, European Team Championship, Plovdiv
1.c4 e6 2.e4 d5 3.exd5 exd5 4.d4 Bb4+ 5.Nc3 Nf6 6.Bd3 O-O 7.Nf3 Re8+ 8.Be3 Bg4
9.O-O Nbd7 10.Qb3 Bxc3 11.bxc3 Bxf3 12.gxf3 Nb6 13.c5 Nc4 14.Rae1 Qc8
15.Kg2 b6 16.cxb6 axb6 17.Bc1 Qa6 18.Bb1 h6 19.Rg1 Qc8 20.Qd1 Qd7 21.Kh1 Rxe1
22.Qxe1 Re8 23.Qf1 Kh8 24.Bf4 Rg8 25.Qg2 c6 26.Bc1 Qe6 27.Bd3 b5 28.Qg3 Ne8
29.Bf4 Ned6 30.Qh4 Re8 31.Qh5 Qf6 32.Rg4 Re1+ 33.Kg2 Re8 34.Bg3 Rg8 35.Be5 Qe6
and here Korchnoi saw a stroke to end it all: 36.Rg6 (see diagram).
His strong Belgian opponent (rating 2465) saw the idea too, and resigned. Instead,
he could have won with 36...Nxe5 37.Rxe6 Nxd3 and the threat Nf4+ decides - 38.Rxh6+
fails to gxh6+ with check.
For more examples, see items 12, 150, 182, 186, 188 and 190 in this Diary,
and The ultimate blunder,
elsewhere on this site.
227. 2 October 2003: A 99 year old record
In his latest column in the Washington Post, Lubos Kavalek showed a fantastic recent game. My remarks are based on his.
Studnicka - Cernousek, ProMoPro Open, Prague September 2003
1.d4 d5 2.c4 c6 3.Nf3 Nf6 4.Nc3 e6 5.Bg5 dxc4 6.e4 b5 7.e5 h6 8.Bh4 g5
9.Nxg5 hxg5 10.Bxg5 Nbd7 11.g3 Bb7 12.Bg2 Qb6 13.exf6 c5 14.d5 b4 15.O-O
bxc3 16.dxe6 Ne5 17.Re1 Bxg2 18.Rxe5 Qc6 19.Qd7+ Qxd7 20.exd7+ Kxd7 21.Kxg2 cxb2
22.Rb1 Bd6 23.Re3 Rab8 24.Rc3 Rb4 25.Bd2 The first new move in the game. Black
is already winning. 25...Be5 26.Rc2 Rhb8 27.Bxb4 Rxb4 28.a3
(see diagram left)
28...c3 Missing the win, which was there with 28...Rb3 29.Rxc4 Rxa3 30.Rxc5 Ra1 31.Rxe5 Rxb1 32.Rb5 a5 33.h4 a4 'and Black wins the pawn race.'
29.axb4 cxb4 30.Kf3 a5 No better is 30...b3 31.Rxc3 Bxc3 32.Ke4 Kc6
33.Kd3 Bxf6 34.Kc4 and White wins 31.Ke4 a4 Or 31...Bxf6 32.Kd5 a4 33.Kc4 a3
34.Kb3 etc. 32.Kxe5 a3 (see diagram right)
33.Rd1+ Kc6 34.Kd4 a2
35.Rcc1 Kb5 36.Kd3 Ka4 37.h4 This loses. The win was still there with 37.Kc2 bxc1Q+
(Ka3 38.Rb1!) 38.Rxc1 Ka3 39.Re1! b3+ 40.Kxc3 b2 41.Re8 b1N+ 42.Kc2 Nc3 43.Ra8+ Na4
44.Rxa4+ Kxa4 45.Kb2 37...Kb3 38.h5 c2 (see diagram) and the extremely
rare phenomenon of three adjacent pawns on the 7th rank has occurred.
Kavalek calls them 'the Bourdonnais monsters', after a famous game McDonnell - La
Bourdonnais, 1834. White is helpless. 39.Rh1 b1Q 40.h6 a1Q 41.h7 Qxc1 42.Rxc1 Qxc1
43.h8Q Qd1+ and White resigned.
Kavalek does not mention this himself, but this game is highly reminiscent of the
following classic masterpiece.
Gufeld - Kavalek, Student Olympiad, Marianske Lazne 1962
1.e4 e5 2.Nf3 Nc6 3.Bb5 Bc5 4.c3 f5 5.d4 fxe4 6.Ng5 Bb6 7.d5 e3 8.Ne4 Qh4
9.Qf3 Nf6 10.Nxf6+ gxf6 11.dxc6 exf2+ 12.Kd1 dxc6 13.Be2 Be6 14.Qh5+
Qxh5 15.Bxh5+ Ke7 16.b3 Bd5 17.Ba3+ Ke6 18.Bg4+ f5 19.Bh3 Rhg8 20.Nd2
Bxg2 21.Bxg2 Rxg2 22.Rf1 Rd8 23.Ke2 Rxd2+
24.Kxd2 e4 25.Bf8 f4 26.b4 Rg5 27.Bc5 Rxc5 28.bxc5 Bxc5 (see diagram)
29.Rab1 f3 30.Rb4 Kf5 31.Rd4 Bxd4 32.cxd4 Kf4 and White resigned.
But the record in this type of thing is already 99 years old.
Lee - Shoosmith, London 1904
1.d4 d5 2.Nf3 e6 3.e3 Bd6 4.Bd3 f5 5.c4 c6 6.Nbd2 Qf6 7.Qc2 Nd7 8.b3
Nh6 9.Bb2 Nf7 10.Ng1 Qg6 11.g3 Nf6 12.f4 Ne4 13.Ngf3 Qh6 14.Bxe4 fxe4
15.Ne5 Bxe5 16.dxe5 Qh3 17.Ba3 Bd7 18.cxd5 exd5 19.Qc5 Qe6 20.h3 b6
21.Qc2 a5 22.Bb2 h5 23.Nb1 Nh6 24.Nc3 Nf5 25.Qf2 b5 26.Ne2 c5 27.Rc1 Rc8
28.Rg1 Nh6 29.Rh1 Nf5 30.Rg1 Nh6 31.Rh1 Rf8 32.Rc2 g6 33.Kd2 b4 34.Kc1 a4
35.Kb1 Bb5 36.Nc1 Bd3 37.Nxd3 exd3 38.Rd2 c4 39.Bd4 Qa6 40.Qf1 Ra8 41.g4 Rf7
42.gxh5 axb3 43.hxg6 bxa2+ 44.Ka1 Qxg6 45.Rg1 Qe6 46.Qg2 b3 47.Qg6 Qxg6
48.Rxg6 Nf5 49.e6 Rf8 50.e7 Nxd4 51.exf8Q+ Kxf8 52.exd4 c3 53.Rgg2 Re8
54.Rd1 d2 55.Rgg1 Re2 56.h4 c2 57.Rc1 b2+ (see diagram)
58.Kxb2 d1Q 59.Kc3 Qd2+ 60.Kb2 a1Q+ 61.Kxa1 Qc3+ and White resigned.
226. 1 September 2003: A new chapter in the Babson saga
In the August issue of the German problem magazine Die Schwalbe, there was, under the title First ascent with oxygen, the problem below by Peter Hoffmann.
His name, together with the four white Bishops and three white Knights
in the initial position, and the white and black pawns ready to promote,
whispers the mythical word: Babson. Hoffmann is, after Drumare and
Yarosh, the Dritte im Bunde in the
Babson saga.
If Drumare is the tragic hero of the Babson Task and Yarosh is its genius,
then Hoffmann is its Stakhanov, achieving it more often than anyone else.
But this is a special Babson.
Mate in 4
Peter Hoffmann
Die Schwalbe, August 2003
As tries like 1.hxg8Q Bxg8; 1.h8Q+ Rxh8 and 1.Qxe2 dxe1Q (more or
less) obviously fail, and Bishop checks might be a nuisance, we
quickly consider 1.Nxe6
The main threat is then 2.hxg8Q followed by 3.Qf7
mate. As Rook moves are defeated by 2.Qxe2 or 2.h8Q+ or, in case of Rxe8,
by 2.Rxe8, and 1...dxc1Q or dxe1Q do not stop the threat,
that leaves 1...d1Q. Now 2.hxg8Q Qd4+? 3.c4! would work
fine, but 2...Qd7+! 3.Bxd7 is stalemate. 2.hxg8R has the same drawback,
and after 2.hxg8N+ Kxe6, White is not in time. But 2.hxg8B! is
good, the main point being that 2...Qd7+ 3.Bxd7 isn't stalemate
now, and 3...Kxg6 4.Rxh6 is mate. The threat 3.c4+ also cannot be
stopped by 2...Qd4+ because 3.c4 Qxb2 4.Bxb2 is mate. A
nice side variation is 2...Qxc1 3.Rxg5! followed by 4.Rf5 or Qh8 mate.
But after 1...d1R(!), both 2.hxg8Q and
2.hxg8B fail to 2...Rd4+ 3.c4 stalemate. 2.hxg8R does not threaten
enough, but as the Rd1 does not guard e2, White now has 2.hxg8N+!
Kxe6 3.Qxe2+ Kd5 4.Qe5 mate.
Obviously, the selfparalysis 1...d1B(!)
also defeats the threats; promotions to Q and B are now stalemate,
while 2.hxg8N+ lacks the necessary follow-up Qxe2. But here, White has
2.hxg8R! Kxe6 3.Rd8 Kf6 4.Rd6 mate.
Finally, Black has 1...d1N(!), threatening
a nasty check. 2.hxg8N+ Kxe6 3.Qxe2+ Ne3 is no good, but 2.hxg8Q!
works: 2...Nxb2+ 3.Kb5 (or Bxb2 - an insignificant dual) and
4.Qf7 mate.
The idea of this problem is that Black's four different promotions on his
first move, must be answered by White's four different promotions,
and that these promotions do not form echos, as in the classic Babson Task,
but a cycle, the scheme of which is: QB-BR-RN-NQ.
"It seems," Hoffmann wrote to me, "that the full
echo, that is the normal Babson, is the easiest - which of course sounds
comical when speaking of the Babson." In Die Schwalbe, he adds:
"Because the motivations of the Rook and Bishop promotions (setting up
and preventing stalemate) are not related to each other in a closed system
anymore, more stalemating situations are required." He quotes his
correspondent Ulrich Auhagen: "If the regular Babson is the Everest, then
achieving the cyclic Babson must be compared to the ascent of a 10.000
meter mountain, without oxygen" - comparing the help of supplementary
oxygen to the help of promoted pieces.
But whether or not Hoffmann or somebody else will be able to do away
with this help, there remain a few other missions to be accomplished,
with or without oxygen, before we can really say that in this new sense,
the Babson Task has been completely mastered.
In all, 24 schemes are possible.
One, QQ-RR-BB-NN, is the full echo - the standard Babson Task.
6 schemes are half reciprocal: QQ-RR-BN-NB; QQ-RN-BB-NR; QQ-RB-BR-NN; QN-RR-BB-NQ; QB-RR-BQ-NN and QR-RQ-BB-NN.
Then there are 8 schemes that are 1/4 echos and 3/4 cyclic: QQ-RB-BN-NR; QQ-RN-BR-NB; QB-RR-BN-NQ QN-RR-BQ-NB; QN-RQ-BB-NR; QR-RN-BB-NQ QR-RB-BQ-NN and QB-RQ-BR-NN
Three schemes would show crossed reciprocal pairs: QR-RQ-BN-NB; QB-RN-BQ-NR and QN-RB-BR-NQ - this last one would be the inverted Babson.
Finally, 6 schemes are fully cyclic: QR-RB-BN-NQ (the standard sequence of the pieces); QR-RN-BQ-NB; QB-RN-BR-NQ (the one Hoffmann made); QB-RQ-BN-NR; QN-RB-BQ-NR and QN-RQ-BR-NB.
2 down, 22 to go.
Met dank aan Joost de Heer.
PS 17 December:
Mate in 4
Peter Hoffmann
Die Schwalbe, December 2003
In this month's Die Schwalbe, Hoffmann publishes a new version of his cyclic Babson Task, with one promoted man less. There are now only two promoted Bishops; the total number of pieces is down from 26 to 24 and there are no duals in the main variations. The key, which captures a Queen, is worse - but the reader decide whether this version is an improvement.
After 1.exf7, the main variations are:
1...d1Q 2.fxg8B! Qd7+ (Qd4+ 3.c4 Qxb2 4.Bxb2 mate; Qxc1 3.Qxc1 e1Q 4.Qxf4 mate)
3.Rxd7 Kxg6 4.Rd6 mate
1...d1B 2.fxg8R! Ke6 3.Re7+ Kd6 (Kf6 4.c4 mate) 4.Rd8 mate
1...d1R 2.fxg8N+ Ke6 3.Qxe2+ Re5 (Kd6 4.Qe7 mate) 4.Qxe5 mate
1...d1N 2.fxg8Q! Nxb2+ 3.Bxb2 c1Q 4.Qxf7 mate
225. 27 August 2003: The Great Annotated Empty Moving Boxes Robbery
Fischer's letter that he sent in January from Tokyo to the The Chief of Police of the Budapest Police Department, is heartbreakingly funny, on a par with I was tortured at the Pasadena Jailhouse.
In 1998, Fischer rented an apartment in Budapest
via one Janos Rigo. For security, he had the wooden front door changed
for a heavy steel one with a new lock, immediately nullifying this effort
by giving a copy of the key to Rigo, so Rigo's sister could clean while
Fischer was away. Strangely and suspiciously, this cleaning lady seemed
to double (she looked a lot like Rigo) as the piano player / singer in a
restaurant Fischer used to frequent.
Early 1999, Fischer received from the USA "about
31 very old boxes containing some of my very old belongings." A little
later, he moved the contents elsewhere, keeping the boxes in the apartment,
and telling Rigo that they should under no circumstances be thrown away,
because they had "vitally important" annotations on them by himself and
his late mother "and are therefore of very great value to collectors."
Nevertheless some time later in 1999, "all of these
approximately 31 annotated empty boxes had completely disappeared from the apt.!
I immediately asked Mr. Rigo what the hell had become of these boxes and demanded
their return." Rigo thought his sister had probably thrown them out, but Fischer
did not accept "this fanciful story" and accused him of stealing the boxes,
hoping "to profit enormously from them". Or perhaps "he was put up to it by
others and there is a much broader conspiracy involved..."
In fact a few years earlier, in 1996, Rigo had already
proven his untrustworthiness on a trip Fischer and he made from Budapest to Vienna
in Rigo's car. In Fischer's luggage, there were prototypes of his
chess clock and his Fischerandom shuffling device. "And it just "happened"
that Mr. Rigo knew this." When after some shopping they wanted the car again,
it had "disappeared". The police couldn't help, and Rigo and Fischer
had to go back to Budapest by train. "At this stage of the game Mr. Rigo
carefully avoided talking about what was really on his mind: extorting money
from me for the return of my clock and shuffler!"
When Fischer threatened Rigo to go to the Budapest
police about this, the car "magically reappeared (with my luggage and
the clock and the shuffler intact) in the huge fenced off Vienna city parking
lot for cars that the city had towed away for parking violations." This was
two weeks after the disappearance, "more than enough time for someone or some
people to study the clock and the shuffler at their leisure..." (And to this
"gangster", this "sick, greedy thieving bastard", this "con-artist and predator"
he gives the key to the brand-new heavy steel door of his apartment?)
Rigo, Fischer thinks, must have gotten the idea to steal
the boxes because he was "emboldened (...) by the publicity
surrounding the mega-robbery in about late 1998 or early 1999 of all of my cash,
valuables and other belongings (including my gold and silver coin collections)
in my storage room at the Bekins moving and storage company in Pasadena,
California USA", committed by the US government and helpers. "Said robbery
in terms of monetary value was probably one of the biggest if not the biggest
robbery in the history of the United States." But while the boxes robbery
"is an infinitesimally small and utterly trivial crime by comparison," it is
still a crime - and a serious crime for which Rigo must be arrested and brought
to justice. Even if more than three and a half years have gone by, in which Fischer
has vainly waited for handwritten declarations from Rigo and his sister about
the fate of the boxes.
Vintage Fischer - except that one word is shockingly missing - Jew.
One thing I never thought Bobby Fischer would
be capable of, is hypocrisy. He has always spoken his mind, no matter
how offensive people might think that was. In approximately 21 radio
interviews he gave over the last few years, there were hardly ten
consecutive words in which he wasn't persecuted by a world-wide conspiracy
of dirty, hooknosed, circumcised Jews who were bastards, criminals, parasites,
liars, thieves and murderers. Bob Ellsworth, allegedly involved in the
mega-robbery, was "a dirty secret Jew, worthy of death" - and now, when
Fischer needs the Budapest police to help him get back his treasures,
the Jewish conspiracy had evaporated and the criminal Janos Rigo is a
straight-nosed gentile?
I have never been so disappointed in Bobby Fischer.
Two more points.
I agree it shouldn't matter to the Budapest Chief of Police, but Fischer could have mentioned Janos Rigo is an International Master and chess organizer, and was for years Fischer's main sparring partner in Fischerandom chess.
In this letter, Fischer again calls himself "the World Chess Champion", and mentions
"successfully defending my chess crown against Mr. Boris Spassky in a $5
million title rematch [in] 1992."
I wonder about his logic. "World Chess Champion" is a
legally unprotected title; anyone may call himself that, and more and more
do. But if Fischer defended his chess crown in 1992 in a title rematch,
he can only mean the crown and the title he won in Reykjavik in 1972. That,
however, was the title owned and bestowed by FIDE, and for FIDE to bestow on
others - which it did in 1975. FIDE giveth, FIDE taketh. If Fischer won his crown
in 1972, that means he doesn't have it now. If he has it now, that means he didn't
win it in 1972.
In fact, Fischer has also used this second, equally
illogical claim. At the press conference before the 1992 match against
Spassky, he made an interesting remark to an Icelandic journalist, who had
innocently used the phrase: "Iceland, where you became World Champion," in
his question.
Fischer then said: "I really didn't become a World
Champion in Iceland, I was the World Champion long before Iceland."
Fischer probably meant that "long before" 1972,
he was the strongest player in the world, but even if that were true
(his score against Spassky before 1972 was +0 =2 -3), it is a
different interpretation of "Chess World Champion" than is generally
accepted.
Titles and qualities are not the same - but in
neither way can Fischer call himself World Champion. FIDE took
his FIDE-title in 1975 and if you don't play, you can't call yourself the
strongest.
224. 21 August 2003: When two are doing the same, it might just be the opposite
The old chess maxim "When two are doing the same, it isn't always the same", was very nicely illustrated in the last Dutch championship.
White to play
Van Wely - Delemarre, ch Netherlands, Leeuwarden 2003
White is a pawn up, but he cannot do very much with it; the position
is more or less equal. With 22.Qg4, he threatened Bxf6. There
followed 22...Kh8 23.Kh1 This mysterious copycat King's move is
a brilliant trap. Not only isn't White really copying Black, he is in fact
doing the opposite: Kh8 exposed Black to back-rank dangers, while White,
with Kh1, defended his back rank. Black didn't notice that.
23...Bc7? Wanting to pick up the pawn - and falling for the trap.
He should have kept blocking; 23...g5 was playable. 24.d6!
Black must capture; after Bb8 25.Bg3 the extra pawn is alive and kicking.
24...Rxd6 After 24...Bxd6 Van Wely's develish idea is revealed:
25.Qxd7! Qxd7 26.Rxd6 and because after Qxd6 27.Rxd6 White's back rank is
guarded by Ng1, the weakness of Black's bank rank costs him a piece. But
25.Rxd6 Bxd6 26.Qe6 Bxf3 (26...Rc6 27.Ne5) 27.gxf3 Qa8
(27...Rc6 28.Bg3) 28.Qxd6 Qxf3+ 29.Kg1 had the same effect;
White won easily with his extra piece.
223. 14 August 2003: Machine vs. King of Claustrophilia
A new "Man vs. Machine" (you should see me with my dishwasher) Challenge has
been announced, this time the First Official World Chess Championship in total
virtual reality" no less. From 11-18 November, Kasparov will play four games
with a version of Fritz in New York for a million dollars.
But even if it beats Kasparov 4-0, machine cannot
be said to "play chess" until it makes two final little steps.
If given this position (Petrosyan - Bronstein, Candidates Amsterdam 1956;
the preceding moves were
1.c4 Nf6 2.Nc3 g6 3.g3 Bg7 4.Bg2 O-O 5.Nf3 c5 6.O-O Nc6 7.d4 d6 8.dxc5 dxc5
9.Be3 Nd7 10.Qc1 Nd4 11.Rd1 e5 12.Bh6 Qa5 13.Bxg7 Kxg7 14.Kh1 Rb8 15.Nd2 a6
16.e3 Ne6 17.a4 h5 18.h4 f5 19.Nd5 Kh7 20.b3 Rf7 21.Nf3 Qd8 22.Qc3 Qh8 23.e4 fxe4
24.Nd2 Qg7 25.Nxe4 Kh8 26.Rd2 Rf8 27.a5 Nd4 28.b4 cxb4 29.Qxb4 Nf5
30.Rad1 Nd4 31.Re1 Nc6 32.Qa3 Nd4 33.Rb2 Nc6 34.Reb1 Nd4 35.Qd6 Nf5)
a million times, machine should at least once play 36.Ng5 like Petrosyan.
Machine should, within a million hours, be able to find the draw
in this position from 1951 by the King of Claustrophilia, J. Hašek.
1.Ng4+ hxg4 2.d4+ K- 3.Rh1 Qh8 4.Ke1 Qa8 5.Kf1 Qa6+ 6.Kg1
and the Queen must go back or else it is stalemate.
PS 17 August: An apology to machine is in its place. I must have
been behind a little, not being aware of the giant strides forward it has
made.
For one thing, machine now seems to be capable of
making incomprehensible human blunders, too. Chess programmer
Gian-Carlo Pascutto tells me his Deep Sjeng has a level where
it simulates human blunders. "It would effortlessly produce a move like
Ng5 within a million tries." Showing praise where praise is due, he
gives an example by another program from a recent big computer tournament.
Nullmover - XiniX, Leiden 2003
1.e4 e6 2.d4 d5 3.Nc3 dxe4 4.Nxe4 Qd5 5.Nc3 Qd8 6.Nf3 Nf6
7.Bc4 Be7 8.0-0 0-0 9.Qe2 Nc6 10.Rd1 Nd5 11.Bb5 Bd7 12.Ne4 a6
13.Bxc6 Bxc6 14.Ne5 Ba4 15.b3 f5 16.bxa4 fxe4 17.Bd2 Nf6
18.Bg5 Nd5 19.Bxe7 Qxe7 20.Qxe4 Nc3 21.Qxb7 Nxd1 22.Rxd1 Qd6
23.Nc6 Qf4 24.Rf1 Kh8 25.Qb3 Rf6 26.c3 Qd6 27.Ne5 Raf8
28.Nd3 Rh6 29.g3 Qd5 30.h4 Qc6 31.Ne5 Qd6 32.a5 Rhf6 33.a4 Kg8
34.f4 Rd8 35.Qc4 Rb8 36.Rd1 h6 37.Re1 Rb2 38.Nd3 Rd2 39.h5 Kh7
40.Re5 Kh8 41.Rc5 Rf5 42.Ne5 Kh7 43.Qf1 Qd8 44.Qe1 Rb2
45.Qe4 Qb8 46.Nd3 Rd2 47.Nb4 Kh8 48.Rc6 Qa8 49.Qxe6 Rxh5
50.Qe4 Qg8 51.Rxc7 Rh3 52.g4 Rg3+ 53.Kf1 Rd1+ 54.Ke2 Rb1
55.f5 Rbg1 56.Qc6 R1g2+ 57.Kf1 (see diagram)
"XiniX is a strong program that probably would hold its own against
Petrosyan," writes Pascutto, yet here, it played 57...Qc4+ and
resigned after 58.Qxc4
Pascutto says this was not caused by a
bug, but by a "flawed knowledge pattern" which led machine to
think White would be mated. Very human - although I do not see
the mate man might have mistakenly seen here.
The other stride forward I hadn't been aware of is that, as Pascutto
and Dennis Monokroussos wrote, Shredder 7.04 has no
problem at all finding 1.Ng4+ and 3.Rh1 in the Hašek position. I am
truly amazed and awed by this.
To quote Monokroussos: "I suspect that all
computer chess Turing-type tests can be passed (these two can, at any rate)
or will be passable in the foreseeable future; the lesson I draw
from this is that the Turing test is an inadequate test for intelligence."
I agree that will be man's next and probably best
defense, but we don't yet need it - here is a new Turing test (can you
tell man and machine apart from their answers to questions) for chess.
(See diagram left.)
White to play
As I do not have Shredder, I asked Pascutto and Monokroussos to feed it this position; it turns out it does not find the easy draw with 1.Qxc7+ Kxc7 2.b4
Note: I took the position, with a minor change, and
colors reversed, from Alexander - Menchik, Margate 1937
1.e4 e6 2.d4 d5 3.Nc3 Nf6 4.Bg5 Be7 5.e5 Nfd7 6.Bxe7 Qxe7 7.f4 O-O 8.Nf3 c5
9.Bd3 f6 10.dxc5 Nxc5 11.O-O f5 12.Nb5 Bd7 13.Nd6 Nc6 14.Qd2 Be8 15.b4 Nxd3
16.cxd3 a6 17.Qc3 Qc7 18.Qc5 b6 19.Qc3 Qb8 20.Nd4 Nxd4 21.Qxd4 Ba4 22.Rac1 Ra7
23.Qe3 Rd8 24.Rc3 Rc7 25.Rfc1 Rdd7 26.Qd4 h6 27.h3 Kh7 28.Kh2 Re7 29.Qe3 Bb5
30.Nxb5 axb5 31.Rc6 Qa7 32.Qf2 Rxc6 33.Rxc6 Qa4 34.Qd2 Qa7 35.Qf2 Qa3 36.Qd2 Qa7
37.d4 Kg8 38.Qc2 Kf7 39.Qb3 Qa6 40.a3 Ra7 41.Qg3 Qxa3 42.Rc7+ Ke8 43.Qg6+ Kd8
44.Rxa7 Qxa7 45.Qxe6 Qd7 46.Qxd7+ Kxd7 (see diagram) 47.g4 fxg4
Now, 48.Kg3 or f5 would be winning. But with 48.hxg4 White, blocking
g4, prepared Black's fortress which Menchik completed with 48...g5
Alexander immediately recognized the fortress was impenetrable now, and
agreed to a draw.
Finally, as Pascutto remarks: "An interesting question is perhaps why
in heavens this Kasparov-Fritz event is called an 'Official World
Chess Championship', when neither participant has any claim to any
world title. Of course this sort of nonsense has become commonplace in
the chess world, but the fact that an official 'independent academic'
organisation like ICGA (the International Computer Games
Association - TK) condones this, I cannot digest."
222. 23 July 2003: The Richter riddle
In his Gambit Cartel column of today at The Chess Cafe, Tim McGrew gives that strange game where Chigorin played the Damiano, blundered his queen, missed an obvious mating combination twice, and was granted a draw in a losing position.
Schiffers - Chigorin, 13th match game, St.Petersburg 1897
1.e4 e5 2.Nf3 f6 3.Nxe5 Qe7 4.Nf3 d5 5.d3 dxe4 6.dxe4 Qxe4+ 7.Be2 Nc6
8.O-O Bd7 9.Nc3 Qg6 10.Ne5 Nxe5 11.Bh5 O-O-O 12.Bxg6 hxg6 13.Qe2 Bd6 14.Ne4
Nf3+ 15.gxf3 Bxh2+ 16.Kg2 Bh3+ 17.Kh1 Be5 18.Kg1 Bh2+ 19.Kh1 Be5 20.Qe1 Bg4+
21.Kg1 Bxf3 22.Ng3 Ne7 23.Qe3 Bc6 24.Qxa7? (see diagram) Allowing
a forced mate: 24...Rh1+! 25.Nxh1 Bh2+ 26.Kxh2 Rh8+ 27.Kg3 Nf5+ and Rh4 mate.
24...b6? 25.Be3? Allowing it again 25...Nf5? And missing it again.
26.f4 Nxg3 27.fxe5 Rh1+ 28.Kf2 Rh2+ 29.Kxg3 Rdh8 30.Qa6+ Kb8 31.Bxb6 Rg2+
32.Kf4 Rh4+ 33.Ke3 Rh3+ 34.Kf4 Rh4+ and a draw was agreed - but White
could have won with
35.Ke3 Rh3+ 36.Kd4 Rd2+ 37.Qd3 fxe5+ 38.Kxe5 Rdxd3 39.Bxc7+ Kxc7 40.cxd3 etc.
There is a strange footnote to this game.
This facsimile is from Deutsches Wochenschach, 1911. The text below the diagram
on the right is: White played Be3, whereupon Black announced a mate in 5: 1...Rh1+
2.Nxh1 Bh2+ 3.Kxh2 Rh8+ 4.Kg3 Nf5+ 5.Kg4 Rh4 mate.
That this must have been plagiarism, a naive hoax, is
suggested by the other Spielstellung. There, it says: Black had played
Re2, and White made a last attempt with 1.Rf2. And indeed, there followed gxf2, when
of course Qxg5+ forced stalemate. With 1...g2+ 2.Rxg2 Qf3 3.h3 Kh4 or 3.h4 g4
the win could have been forced easily.
A very weak Spielstellung, if only because (1.Rf2) Rxf2
2.Qxf2 Qxh2+ also wins. Somebody who thinks that such a stupid trap is worth
showing, cannot have announced a mate that Chigorin had missed. He must have seen
that game somewhere (he even copied the introduction with Be3) and have
wanted to attach his name to it.
The strange thing however, is the name of the hoaxer.
Kurt Richter (1900-1969) was a strong master, a romantic who played
very imaginative games, and published innumerable books and articles on tactics.
But in October 1911, he was still a 10-year old boy in Berlin. And when he later
published the Chigorin game, he did not mention having played the missed
combination himself. The hoaxer was a false Kurt Richter, a premature apparition
of the name that was to become a synonym for combinational beauty in chess.
PS 19 August: Browsing around 1911 in Deutsches Wochenschach,
Harold van der Heijden found several pre-1911 mentions of one Richter from Berlin,
sometimes with Kurt as his first name, sometimes without a first name. In 1910,
he played a match against Von Bardeleben, losing with +1 =2 -3. In 1911, a
few readers noted the Spielstellung was 'almost identical' with Schiffers -
Chigorin. In issue number 36, from 6 September 1908, there is a remark saying
"Herr Richter" was Lasker's second in the World Championship match against Tarrasch,
played earlier that year in Düsseldorf. In January 1912, Kurt Richter is among a
few gentlemen who have founded a chess club in Triëst. After that, his name
does not reappear in the magazine.
Strange: what business did a Kurt Richter from
Berlin have in a town like Triëst? How could someone with such a serious
function as the World Champion's second, do something as frivolous as
plagiarize a well-known combination in a widely read chess magazine?
221. 22 July 2003: R = N + B
Richard Evans pointed out a "truly amazing relationship" between
Rook, Knight and Bishop. Counting all the moves a Rook could ever make,
you get 896 (64x14). Doing the same for Knight and Bishop, you get 336 and
560, respectively. And 336 + 560 = 896.
Fascinating indeed and new to me - but I quickly found
that in the bible for this kind of thing, Schach und Zahl (Chess and numbers)
(Bonsdorff, Fabel & Riihimaa, 1971) this relation had already been observed.
And generalized. So here is a nice problem: for which other n x n board(s)
is it true that R = N + B?
I will give the answer in a few days.
PS 28 July: Four readers, Olaf Teschke, Pekka Karjalainen, Ulrich
Schimke and Rein Halbersma, took up this challenge and all found
the correct answer: R = N + B also (and only) for the 1 x 1 (0 = 0 + 0)
and the 3 x 3 (36 = 16 + 20) boards. Using different reasonings, they all
arrived at the formulae already given in Schach und Zahl:
R = 2*n^2*(n-1)
N = 8*(n-2)*(n-1)
B = 2*n*(2n-1)*(n-1)/3
Of course, S&Z also gives the formulae for the other pieces:
K = 4*(2n-1)*(n-1)
Q = 2*n*(5n-1)*(n-1)/3 (= R + B)
P = (3n-4)*(n-1) (for n=4 and greater)
Olaf Teschke wonders whether S&Z also gives the formulae for n x m boards, but it doesn't. It does point to an unpublished study by one Dr. E. von Fabrizi, titled Brett und Figur im Banne der Dualität (1945) for a generalization 'in the broadest possible sense', which however 'requires a lot of mathematical understanding'.
Ulrich Schimke mentions having sent his formulae years ago to the On-Line Encyclopedia of Integer Sequences, giving the linked page as an example.
Pekka Karjalainen remarks that on a 0 x 0 board, the Knight is granted 16 moves by the formula and the King 4, adding: "One could perhaps create some very theoretical chess compositions without a board, but that is beyond my skill."
PS 7 August: Ulrich Schimke and David Bevan sent me
generalized formulae for m x n boards. They are:
K = 8*m*n - 6*(m+n) + 4 (Valid for m,n >= 1)
The formula for R = N + B on non-square boards, then, becomes:
David Bevan takes this a few steps further, for one thing also giving the formula
for a (p,q)-leaper (the Knight is a 2,1 leaper) on a m x n board:
Bevan also gives a few boards with special relationships of this kind:
Q = R + B (Valid for n >= m)
R = (n+m-2)*n*m
B = (6*m^2*n - 2*m^3 - 6*m*n + 2*m) / 3 (Valid for n >= m)
N = 8*m*n - 12*(m+n) + 16 (Valid for m,n >= 2)
Bevan also gives a formula for the Pawn. On a board with f files
and r ranks, it is:
P = 3*f*r - 5*f - 2*r + 4 (Valid for r >= 4)
m*n*(n-m-8) + 2/3*m^3 + 34/3*m + 12*n - 16 = 0 (if n >= m and m,n >=2)
As Schimke says, it is clear that if m is large enough,
for instance greater than 20, the left part of the equality is positive,
which means there is a finite number of boards where R = N + B.
These are (apart from the square boards 8 x 8, 3 x 3 and 1 x 1):
3 x 4 (R = 60; B = 32, N = 28)
4 x 6 (R = 192; B = 104; N = 88)
L = 4*m*n - 2*q*(m+n) (Valid for p=0)
L = 4*(m-q)*(n-q) (Valid for p=q)
L = 8*(m*n+p*q) - 4*(m+n)*(p+q) (Valid for q>p>0)
On 4 x 7 (files given first); 6 x 5; 3 x 17 and
16 x 4 boards, K = N + P
Not only on the normal board of 8 x 8, but also on 20 x 6
and 6 x 41 boards, B = K + P
On the 8 x 106 board, B = K + N
On the 5 x 9 board, B = K
On any 6*s x 6*s board, B = s*K
And finally, to get back to the equation everything started with: on the
8 x 8 board it is not only true that R = N + B, but also that R = K + N + P
© Tim Krabbé, 2003, 2004
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