29 March 2011

The Cheshire Cat

Originally, I wanted to write something about Alice's Adventures in Wonderland, more precisely about chapter six of the book. But midway through thinking about it, I (partially) changed my mind.

John Nunn Once again, I present some helpmates, this time four compositions by the famous British mathematician, FIDE GM (Grandmaster of FIDE for over-the-board play) and WFCC GM (Grandmaster of FIDE for chess solving) John Nunn (photo courtesy of ChessBase). He was the third person ever to gain both titles, the others being Jonathan Mestel and Ram Soffer. Since 2008, there is a fourth, namely Bojan Vučković.


  1John Nunn  
The Problemist 11/2005
[8/8/8/8/2pk4/p7/p2Kpp2/R4r2]
  h#4*(2+7)  

  2John Nunn  
The Problemist 03/2005
[8/8/4p1R1/4r3/rk2q3/n7/5PP1/5K2]
  h#4,5(4+6)  


  3John Nunn  
The Problemist 03/2005
4th Honourable Mention
[8/4q3/3b4/1p1k4/1p6/1P1K1p2/nPBR1P2/8]
  h#3      2.1...(6+7)  

  4John Nunn  
The Problemist 07/2006
3rd Prize
[b7/6p1/8/4NPpp/4k1rp/1R6/P1p5/b6K]
  h#3      2.1...(5+9)  

Solutions:
1 *1. - Kxe2 2. Kc3 Rxf1 3. Kb2 Kd1 4. Ka1 Kc2#
1. e1=N Rd1 2. Nf3+ Kc2+ 3. Ke3 Rd4 4. Ke2 Re4#
Totally different play in the set compared to the solution.
2 1. - Rh6 2. Ra5 Rh2 3. Qxg2+ Ke2 4. Qxf2+ Kd3 5. Qc5 Rb2#
I immediately liked that maneuver of the white rook.
3 I) 1. Be5 Rd1 2. Qc5 Kd2 3. Kd4 Ke1#
II) 1. Bf4 Bb1 2. Qe5 Kc2+ 3. Ke4 Kd1#
Perfect analogy.
4 I) 1. Kd4+ Rb7 2. Kc3 Nf3 3. Rc4 Rb3#
II) 1. Kxf5+ Nc6 2. Rd4 Rf3+ 3. Kg4 Ne5#
Switchback of rook and knight respectively delivers a model mate.

26 March 2011

On the traces of Giegold (1)

I have always been fascinated by tricky chess problems with hidden maneuvers, piece sacrifices, etc. Therefore, the works of problemists like Sam Loyd and Fritz Giegold had quite some influence on my composing style and i created several nice puzzles surely reminding solvers of them. Every now and then, I'll show you some of my works. Here are the first two:

  1Gerson Berlinger  
32er 3,4/1990
[1N6/3p4/1P1P4/3P4/4Pk1B/5P2/4K3/R7]
  #4(9+2)  

  2Gerson Berlinger  
32er 11,12/1989
[8/8/8/8/pPk5/K1b5/p3p3/3n4]
  h=6(2+6)  

Solutions:
1 1. Ra8! Ke5 2. Ke3 Kxd6 3. Nxd7 Kxd7 4. Rd8#
The key move only makes sense, and only then you'll actually consider it, until you've (fore)seen the whole course of action. Additionally, those quiet White moves add to the difficulty of the problem.
2 1. axb3 e.p. Ka4 2. Kd3 Kb5 3. Kc2 Kc4 4. Kb1 Kd3 5. Ba1 Kxe2 6. b2 Kxd1=
Admittedly, help(stale)mates with retro content are nothing special and not really unusual. Still, you may not be aware immediately that there is such a possibility. Especially in this case, where you have to capture the only white piece (apart from the king) in order to achieve the aim.

23 March 2011

Black & White

Chess problems where one side has an overwhelming majority might not seem to be very interesting. How attractive are the following two to you?


  1Dr. Karl Fabel  
Die Schwalbe 12/1952
 
[7K/8/2N5/6pp/4ppnq/4pkp1/4nprb/5bb1]
  #12(2+15)  

  2Gerhard Paul Latzel  
Schach, 1951
1st Honourable Mention
[6QN/3N3B/3PK3/1PPr4/P1k2B2/2P5/1PP5/8]
  #2(13+2)  

Diagram 1 features a minimal. This is a position where, apart from the king, White has only one piece. This expression was introduced by the German problemist Ado Kraemer in 1924.
The aim is to capture the black queen forcing Black into zugzwang. Obviously, the white king has to do all the work, whereas the knight controls his black adversaries on e2 and g4. He has to avoid the checks Nf6+ and Nh6+ allowing the escape of the black king. Thus, the squares d7, e8, f7, g8, h7 are taboo. On the other hand, he has to move onto a white square once (apart from g6). He has to lose a tempo, as the attempt 1. Kg7 Qh3 2. Kg6 Qh4 3. Kxg5?? shows. The nearest square to accomplish this is e6.
So, we have collected all necessary details. By now, the solution should be easy to understand.
1. Kg7! Qh3 2. Kf8 Qh4 2. - h4? forces Black to move one of his knights on move 3 thus allowing a mate in 4. 3. Ke7 Qh3 4. Kd6 Qh4 5. Ke6! Qh3 6. Ke7 Qh4 7. Kf8 Qh3 8. Kg7 Qh4 9. Kg6 Qh3 10. Kxg5 Qh4 11. Kxh4 Ne~ / Ng~ 12. Nd4 / Ne5#

Looking at diagram 2, you might think this is pretty boring to try to mate in two moves. But it's not such an easy task, though there are so many white pieces and Black has just this rook to defend!
The theme is the different ways of White to answer the checks on d6 and e5. There are four phases. First, we look at the set play (as if it were Black to move):
1. - Rxd6+ 2. Kxd6#
1. - Re5+ 2. Kxe5#
Now, we examine two tries, each invoking zugwang:
1. Nf7?
1. - Rxd6+ 2. Nxd6#
1. - Re5+ 2. Nfxe5#
1. - Rf5!
1. Qg4?
1. - Rxd6+ 2. Bxd6#
1. - Re5+ 2. Bxe5#
1. - Rd4!
Finally, there is the solution:
1. Qc8! (thr. 2. Nb6#)
1. - Rd6+ 2. cxd6#
1. - Re5+ 2. Nxe5#

20 March 2011

Who's Missing

Don't worry, I am not going to write about the album of the same title. Or should I say I'm sorry? It's not that I would not like music by The Who. But I want to write about chess problems (and I will).

A certain type of retros asks you to add one or more pieces in a way yielding a legal position. There are different variations. You might know what and how many units to add. In other cases, you have no further restrictions and must find out yourself what is missing and where. Sometimes, there are additional conditions to be met — do you remember the second problem in the previous post?

The examples I compiled for you are simply asking to add one piece. The first two are not difficult, right? Of course, it's always an aid when one or even both kings are in check. This is similar to the colouring problems. Didn't I also write about them?

  1Andrej Frolkin  
PDB Website 03/2011
[4k3/3P4/8/7B/8/8/8/8]
  Add one piece!(2+1)  

  2Andrej Frolkin  
PDB Website 03/2011
[6r1/8/8/4k3/8/8/7B/B5K1]
  Add one piece!(3+2)  

In diagram 1, the white king is missing and the illegal double check can only be avoided if we add a wKg6. Similarly, in diagram 2, we need a piece on g3. This can only be a white rook, so that the check of the wBa1 is legal. The last move was Rc3xg3+, but we can't say what was captured. This was fairly easy. But I have more to show.

  3A. Jarosch  
Schachmatnaja komposizija 79,
2007
[8/1p1p4/8/1p6/p1p1p3/Rpp5/2K5/k7]
  Add one piece!(2+9)  

  4A. Jarosch  
Die Schwalbe 04/2007
 
[8/p1ppp3/4p3/4P3/3PP3/1K1P4/P1PPk3/1bn5]
Add one piece and resolve the position! (8+8)

Looking at the third problem, we see that the black pawns captured all missing white men, so we have to add a black piece. Obviously, this has to be on a2. We can't put a queen or a rook there, that would give an illegal check. The bishop from c8 never left this square and all black pawns are on the board. This leads to the conclusion that we have to add a black knight on a2.
Next one. In no. 4, we immediately see that the last move was b2xc1=N+, but we do not know (yet) what was captured. The bishop on b1 is a promoted piece. The black pawns captured seven times, as one came from g7. In order to accomplish this, the white pawns on d3, d4, e4 and e5 had to come from e2 to h2 and thus captured all missing eight black pieces. So, we know we have to add a white piece. The white pawn from b2 was captured right there by the black pawn that promoted into the bishop b1. After that there was only the last capture on c1. So, only the white bishop could be captured on c1. From this, we see that the white rook a1 never could leave, either wBc1 or bBb1 blocked it.
In short: Add wRa1. Last move was b2xLc1=S+. Before, the moves b3-b2 (this is the pawn from b7), b2-b1=B and bPc3xwPb2 were played (exactly in that order, but maybe with intermediate moves).

This fourth one was already a little bit tricky, but not too difficult. In the last one, no king is in check.

  5Alain Brobecker  
Jubilé TLG-50 2010
1st Commendation
[8/1p5p/6pP/p4pPP/1k4pP/7p/KPP3P1/8]
  Add one piece!(8+8)  

The white pawns in diagram 5 captured eight times, so the piece to add has to be white. One of the taken pieces was the promoted black c-pawn which in turn captured once to get to the d-file. The black pawns on the board already captured six times. Hence, only the white a-pawn can be added. It never left his file, as there are no black pieces left that he could have captured. Likewise, the black a-pawn went straight to a5. Therefore, the white pawn stands on a4 (not a3 due to an illegal check).

17 March 2011

More fun with twomovers

Once again, I present two twomovers.

  1Sam Loyd  
New York Mail and Express, 1889
[8/1P1PpP1K/1P2k3/1P6/7P/3NBN2/3R3P/4QR2]
  #2(14+2)  

  2W. Pauly  
Pittsburgh Gazette Times, 1912
[8/2P3p1/1p4P1/1P2B3/5pK1/3R1P2/8/8]
  #2 (add black king)(7+3)  
4 solutions

Solutions:
1 1. Qe2! Kd~ / Kf~ 2. Nf4 / Nd4# Typical Loyd!
2 I) +bKb1: 1. c8=Q! Ka2 2. Qc2#
II) +bKe7: 1. c8=R! Ke6 2. Re8#
III) +bKa7: 1. c8=B! Ka8 2. Ra3#
IV) +bKe8: 1. c8=N! Kf8 2. Rd8#
Another funny way to employ the Allumwandlung.

14 March 2011

Light fare?

Time for some funny chess problems. I hope those two twomovers animate you to make an attempt at solving them. Just two moves. Come on, let's go!

  1Knud Hannemann  
Skakbladet 01/1922
[2K5/2PP3R/5k1B/8/1N3P2/5PP1/1P2P3/1B6]
  #2(12+1)  
a) Diagram
b) Rotate position 90 degrees
c) Rotate position 180 degrees
d) Rotate position 270 degrees

  2Tim Krabbé  
Chess Curiosities, 1985
[8/8/8/5B2/2p5/2B2kPp/2P2P1P/5K1R]
  #2(8+3)  
a) Diagram
b) after key move

Solutions:
1 a) 1. d8=Q+! Ke6 2. Qe7#
b) 1. b8=R! Kf4 2. Rf8#
c) 1. d8=B! Kd4 2. Bf6#
d) 1. f8=N! Kd5 2. Bb7#
What an interesting way to show an Allumwandlung. I guess it's already pretty hard to compose a chess problem with four different solutions by rotating the board, not to mention to include a theme like the Allumwandlung. Just great!
2 a) 1. Ke1 Kg2 2. Be4#
b) 1. 0-0! Ke2 2. Bg4#
I really hope you tried to solve it, maybe even without setting up the pieces on a real chess board. I am curious: How long did it take you to find the solution for b)? Or did you give up in total despair? It is quite a psychological barrier to see or even consider the castling. You just "moved" the white king from f1 to e1, so castling seems to be forbidden. But ... the position of b) has no history, there is no link to a).

11 March 2011

Losing chess

The previous post already was about suicidal tendencies. This one continues in that direction though in a slightly different way. Losing chess is also known under many other names, e.g. antichess, giveaway chess or suicide chess. This chess variant is quite popular, as it is easy to understand:
  • Capturing is compulsory. In case there are different choices to capture a piece, a player may choose which piece to capture.
  • There is no check or checkmate. Therefore, the king plays no special role and can be taken as any other piece. This also means that pawns may also promote to kings.
  • Castling is not allowed.
  • A player wins by losing all his pieces. According to the International Rules, stalemate is a win for the stalemated player.
  • The game is drawn like in the orthodox game. Additionally, it's a draw when a win is impossible, e.g. if a dark-squared bishop and a light-squared bishop are the only pieces remaining.

On the internet, you'll find lot of links to software, sites to play it online, opening theory and even endgame databases. I'll give you a few addresses:

Moreover, there are chess problems with that condition. Here are some for you.

  1John Niemann  
Schachmatt, 1947
  
[8/8/nk6/8/P7/5Q2/8/8]
  White wins(2+2)  
  Losing chess

  2J. Iglesias  
France-Echecs, 2006
Special Prize
[rn1qkbnr/p1ppppp1/1p5p/8/8/4P3/PPPP1PPP/RNBQ1BNR]
  SPG in 4,5 moves(15+15)  
  3 solutions
  Losing chess

This first problem is not as easy as it may look at first glance: 1. a5! Kxa5 2. Qh1 Not 2. Qg2? Ka4 3. Qh1 Kb3 and Black wins. 2. - Ka4 3. Qg2 Ka5 4. Qf3 and Black will have to capture the white queen with his next move. The attempts 1. Qh3? or 1. Qd1? are refuted by Kc5!

The solutions to the second diagram are
I) 1.e3 b6 2.Ke2 Ba6 3.Qe1 Bxe2 4.Qxe2 h6 5.Qd1
II) 1.e3 b6 2.Ke2 Ba6 3.Kd3 Bxd3 4.Bxd3 h6 5.Bf1
III) 1.e3 b6 2.Ke2 Bb7 3.Kf3 Bxf3 4.Nxf3 h6 5.Ng1


Vincent van der Bilt
Internet (?), 1997
[1n6/1kp5/pq6/2pp2p1/pp4pP/3b4/r2b2P1/nr6]
Black to move and win (2+16)
Losing chess
  
This is a retro! White's last move was not h3-h4, otherwise White would have had to capture the pg4. Also, g3xh4 was not possible, for all black pieces are still on the board. Thus, the last move had to be h2-h4. You know what that means, right? Yes, Black is allowed to capture en passant.
1. - g4xh3 e.p.! 2. gxh3 g4 3. hxg4 Lf5 4.gxf5 Qe6 5.fxe6 Nd7 6.exd7 Ka8!
Not 6. - Kc8? 7. dxc8=B! and Black will always keep his dark-squared bishop.
Also not 6. - Ka7? 7. d8=Q! Rh1 8. Qxd5 Bg5 9. Qxa2! Rh3 10. Qxa4! and now Pa6 is protected by the black king.
Now, White has the choice:
a) 7. d8=K b3! 8. Kxc7 Kb7 9. Kxb7 Ba5 10. Kxa6 Bb6 11. Kxb6 Nc2 12. Kxc5 Rc1 13. Kxd5 Nd4 14. Kxd4 Rc4 15. Kxc4 a3 16. Kxb3 Rd2 17. Kxa3 Ra2 18. Kxa2 or
b) 7. d8=B Ka7 8. Bxc7 Kb6 9. Bxb6 Bf4 10. Bxc5 Re1 11. Bxb4 Be5 12.Bxe1 a5 13. Bxa5 Bc3 14. Bxc3 Rg2 15. Bxa1 d4 16. Bxd4 Rb2 17. Bxb2 a3 18. Bxa3 or
c) 7. d8=N Kb7 8. Nxb7 Rf1 9. Nxc5 a3 10. Nxa6 b3 11. Nxb4 Be3 12. Nxd5 Rd2 13. Nxe3 a2 14. Nxf1 Nc2 15. Nxd2 Nb4 16. Nxb3 a1=K 17. Nxa1 Nc2 18. Nxc2 or
d) 7. d8=R Ka7 8. Rxd5 Be1 9. Rxc5 Nb3 10. Rxc7 Na5 11. Rxa7 Bf2 12. Rxa6 Re2 13. Rxa5 Rh1 14. Rxa4 Bd4 15. Rxb4 Rg1 16. Rxd4 Rg4 17. Rxg4 Re4 18. Rxe4 or
e) 7. d8=Q! Rh1! 8. Qxa8 b3! 9. Qxa6 Ba5 10. Qxa5 Rg2 11. Qxa4 Nc2 12. Qxb3 Rg6 13. Qxc2 etc.
Of course, not all Black moves are unique. Anyway, the retro idea together with the Allumwandlung is quite nice.

08 March 2011

Mate me!

A selfmate is a chess problem in which White forces Black to deliver checkmate within a specified number of moves against his will. I think you'll like the following two examples.

  1Jan A. Rusek  
Thémes-64, 1969
3rd Prize
[8/3Pp3/3p1P2/2P5/7P/2Q1PB1K/4q2P/3k4]
  s#4(9+4)  

  2Mark Kirtley  
The Problemist, 1986
1st Prize
[3Q4/p7/PP3p2/R4P2/1pP2p1P/N3BN2/kr2BK1P/4R3]
  s#8(14+6)  

Diagram 1 shows a Pickaninny combined with an Allumwandlung:
1. cxd6! e6 2. d8=R e5 3. Te8 e4 4. Txe4
1. - e5 2. d8=Q e4 3. Qa8 exf3 4. Qxf3
1. - exd6 2. d8=B d5 3. Lb6 d4 4. Lxd4
1. - exf6 2. d8=N f5 3. Ne6 f4 4. Nxf4

and the pawn is stopped just in time forcing Black to play Qxf3#.

The second problem features a severe case of homesickness: 1. Nb1+ Kb3 2. Qd1+ Rc2 3. Bc1 axb6 4. Ra1 b5 5. Rh1 bxc4 6. Ke1 c3 7. Ng1 f3 8. Bf1 f2#. All white pieces return home!

05 March 2011

Chameleon echoes

With respect to chess problems, the term echo is used to describe various types of repeated effects. Primarily it refers to a specific visual idea, namely the recurrence of a mating configuration. When the same mating position occurs several times (mirrored, rotated or shifted), we call that echo mates. A special type is the chameleon echo which means the black king is mated on different coloured squares. I chose three different miniatures with that feature. Enjoy!

Bo Lindgren
Tidskrift för Schack, 1944
[8/8/8/5p1B/7B/6K1/8/7k]
  #4(3+2)  

Stanislaw A. Kirilitschenko
Le Minotaure 06/1971
[K7/1p6/1k6/p2P4/8/8/N7/8]
  h#3      2.1...(3+3)  

Frank Müller
Sachmatija 04-06/2004
[8/3p4/8/1B6/R2Q4/1k6/3K4/2B5]
  s#7(5+2)  

Lindgren
1. Kf2! Kh2 2. Bg3+ Kh3 3. Kf3 f4 4. Bg4#
1. - f4 2. Bf3+ Kh2 3. Bg2 f3 4. Bg3#
This is an exact chameleon echo. The mating positions match each other completely (taking into account every piece on the board).

Kirilitschenko
I) 1. Kb5 Nc3+ 2. Ka6 Nb5 3. b6 Nc7#
II) 1. a4 Nb4 2. Ka5 Ka7 3. b5 Nc6#

Müller
1. Kd1! d6 2. Qa1 d5 3. Bc4+ dxc4 4. Qa2+ Kc3 5. Qb2+ Kd3 6. Qe5 c3 7. Qe1 c2#
1. - d5 2. Ba3 Ka2 3. Kc1 Kb3 4. Qa1 d4 5. Qb1+ Kc3 6. Be2 d3 7. Bd1 d2#

02 March 2011

Colours

Today, we'll have a look at colouring problems. No, we won't paint chess pieces in blue, red or any other colour — except black and white and even then only virtually. This kind of chess problems has been invented by Albert H. Kniest. In 1964, he published an article on the topic in Diagramme und Figuren. In a colouring problem, you have a position where you don't know (yet) of which colour the pieces are. You only know that they are either black or white and that the position is legal.

Gideon Husserl
feenschach 1986
[6KQ/8/6RK/8/8/8/8/8]
Colour the pieces
a) Diagram
b) Rg6 → g7
  Let's have a look at an example (see diagram). Obviously, in a), one of the kings is in double check by the rook and the queen. Otherwise, both kings were in check simultaneously and that is illegal. The double check can only be explained by the promotion g7xh8=Q+. Therefore, Kh6, Qh8 and Rg6 are white, Kg8 is Black.

In b), only the queen gives check to one of the kings. There is just one last move which leads to this position, namely h7-h8=Q+. From there, we see that Kg8 and Rg7 (and Qh8) have to be white, Kh6 is black.

That was no big deal, right? The next problem should be a little bit more challenging.

Gideon Husserl
Israel Ring Tourney 1966-1971
Prize
[B1KRN3/PP6/2QK1R2/8/8/8/8/8]
Colour the pieces. Last move?
  Here, one of the two kings is in double check by Rd8 and Qc6. As we already have seen in the previous problem, the only possible explanation is a promotion: c7xd8=R+. So, we conclude that Rd8, Qc6 and Kd6 are white and Kc8 is black. Furthermore, to exclude illegal checks, Ne8 and Rf6 have to be white and Pb7 black. As there is a black pawn on b7, the Ba8 has to be white. Pa7 is also white to allow the promotion into wBa8. The last move was c7xNd8=R+ because only a Nd8 gives Black the opportunity to retract a move (Ne6-/xd8).

Still not thrilling enough? Good, here's something to keep you busy for quite some time.

Henrik Juel
Thema Danicum 07/2004
 
[8/3PPPP1/4P3/3PPP1P/4P3/K2PPK1P/1P2P1P1/8]
Colour the pieces

Dmitri Baibikow
Schachmatnaja komposizija 12/2008
2nd Prize
[2K1K3/PPPPPPPP/1PP1P3/PPPP4/1P6/8/8/8]
Colour the pieces. Last two half-moves?

Did you actually solve at least one of the problems? Congratulations. If not, I hope you tried at least. Anyway, here are the solutions.
Juel: Kf3 and Pd3,d5,d7,e4,e7,f5,g2,h3 are white, the rest black.
Baibikow: Ke8 and Pa5,b6,c5,c6,d7,e6,e7,f7 are white, the rest black. Last moves were d6-d7+ Kb8-c8.