30 September 2011

Oompah-pah!

Obviously, this blog gained more and more popularity since July. And this week, there have been even more pageviews due to a mention on MatPlus.Net. Thanks a lot for the acknowledgement!



It's still Oktoberfest time, therefore I continue to present problems that puzzle you even more after having drunk some Maß (1 liter stein) of beer. Cheers!


  1Bruno Ebner  
Die Schwalbe 02/1972
[8/2p5/2Q5/rpNK4/pk1N4/p1pR4/1p1p4/2r5]
  #1(5+10)  

  2S.C. Herz  
Jugendschach 12/1979
[1Rb5/Kp6/8/8/8/5pR1/6Pk/6qB]
  #1(5+5)  

Solutions
1The position is illegal. It requires more captures by black pawns than white pieces missing. Remove one white piece:
-wNd4: Rd3-d4#
-wQc6: Nd4-c6#
-wNc5: Qc6-c5#
-wRd3: Nc5-d3#
This discloses a cyclical pattern. Very nice!
2Once again, German-speaking people have a little bit more fun. A closer look at the name of the composer tells them already it's a joke (Scherz) problem.
White is in check, so it's him to move: 1. Ka8 Qa1#. But the position of wBh1 is not legal. Rotating the board by 180 degrees results in a legal position, however it does not allow a checkmate in one move.
The explanation is that the kit manager is a negligent guy. The chessboard was torn apart in the middle (between rank 4 and 5) and wrongly put together — upper and lower part have to be swapped. Then, there is 1. Rh7#.

23 September 2011

O'zapft is!

It started last Saturday — the world's largest fair, the famous Munich Oktoberfest. Traditionally, at noon, there was the twelve gun salute and the tapping of the first keg of Oktoberfest beer by the incumbent Mayor of Munich crying "O'zapft is!" afterwards.

I found a chess problem (diagram 1) which is consistent with this subject. It has the slogan "At the Oktoberfest". For a particular reason, I give the original stipulation in German. The translation is something like "White checkmates by making two moves in a row according to the rules of chess". This composer is known for his humorous chess problems and I picked another one for you (diagram 2). Have fun!


  1Hieronymus Fischer  
unknown source, 1879
[8/8/8/4R3/2pk4/8/2KN4/8]
Weiß muss 2 Züge auf einmal machen ohne gegen die Schachgesetze zu verstoßen und setzt so zugleich matt

  2Hieronymus Fischer  
Humor im Schach, 1904
[3n4/n1N2n2/6Bn/3p4/nBpk2pN/1n4Qn/2n1R3/7K]
  #1(7+12)  

Solutions
1This is an Oktoberfest joke which is spoilt by the translation, sorry. White makes one move from the beer-mug, that is takes a sip, and then makes one move on the board to checkmate: 1. Nf3#. The joke lies in the ambiguity of the German word Zug. Of course, if White had to make two legal moves in a row on the board, he could not deliver checkmate.
2There are eight black knights and three black pawns on the board. Therefore, one of these pieces has to be removed:
-Na4: 1. Qc3#
-Na7: 1. Nb5#
-Nb3: 1. Rd2#
-Nc2: 1. Qe3#
-Nd8: 1. Ne6#
-Nf7: 1. Qe5#
-Nh3: 1. Qf4#
-Nh6: 1. Nf5#
-Pc4: 1. Qd3#
-Pd5: 1. Re4#
-Pg4: 1. Nf3#

16 September 2011

Sherlock Holmes travels to Rotterdam

Sherlock Holmes and Dr. Watson were travelling by train from Basel to Rotterdam for the problemists' congress. They had not been in Germany long when Watson spied a slip of paper on the floor. Written on it was: "Ka8 Bg5,h7 Kd8 Be7 h#2 1.Ke8". Watson took out his pocket chess set and soon said "There's something amiss here. 1.Ke8 is wrong; the correct solution is 1.Kc8 Bf4 2.Bd8 Bb7#. (see diagram 1) Holmes said nothing.

  1
[K2k4/4b2B/8/6B1/8/8/8/8]
  h#2

A few hours later, when they were already in Holland, they came back from the restaurant-car and Watson found another slip of paper with a chess problem on it: "Kc4 Pa6 Kc8 Pc5,c6,c7 h#3 1.Kb7". "Again there's something wrong," said Watson immediately. (see diagram 2) "1.Kb7 is a move into check and so impossible. Maybe it's another mistake?" And before long he said: "Yes indeed, you can mate by 1.Kb8 a7+ 2.Kb7 Kxc5 3.Ka6 a8=Q#". Curious. What's your view, Holmes?

  2
[2k5/2p5/P1p5/2p5/2K5/8/8/8]
  h#3

Do you have an idea what Holmes answered? Click here to find out.



(After the text in the book Eigenartige Schachprobleme by Werner Keym, page 165. All compositions by Barry P. Barnes, Klüver Memorial Tourney 1990-93, 1st/2nd Prize.)

09 September 2011

Sherlock Holmes in France

Entertaining stories mixed with tricky chess problems is what most of us enjoy. The book Chess Mysteries of Sherlock Holmes by Raymond Smullyan offers no less than fifty of them. It is a classic and a must-have for all chess lovers, especially for fans of retrograde analysis. Smullyan also wrote The Chess Mysteries of the Arabian Knights, but its stories and chess puzzles are nowhere near as good as those of the first book.

The problemist Barry Barnes followed in Smullyan's footsteps composing chess puzzles packaged in stories with Sherlock Holmes and Dr. Watson. Apparently, they were all published in the magazine The Problemist, but I don't know how many of them exist. Most probably, there are more than just the two that I have found.

Today's story was published in 1994 and I'll stick to the original text as close as possible. For those of you who don't know it yet and who want to try to crack the puzzles themselves I changed things a bit by hiding the decisive deductions of our famous detective at first. Have fun!



"YOU CAN PREVENT THE CRIME OF THE CENTURY THIS COMING WEEK BY ALERTING THE POLICE NOW"
DEDICATED TO CJM AND SFBM

  1
[3k4/P1N1N2R/1BPb4/3B1pQ1/1PK1P3/rP3P2/5Pr1/n2R2nq]
  #2

  2
[BnNN4/K7/Pn1PP3/p2p2r1/R2P2k1/R2PP1q1/P5pP/B7]
  h#2

  3
[{FEN}]
  White or Black to play and win

  4
[{FEN}]
  Legal?

£16,000 REWARD


Sherlock Holmes and Dr Watson were in France for the 37th meeting of the FIDE Problem Commission in Belfort. On their arrival on Saturday at the Salle des Fêtes there was a letter waiting for Holmes. He examined the envelope carefully before opening it and taking out a single sheet of paper. "Listen, Watson", said Holmes, "it reads "You can prevent the crime of the century this coming week by alerting the police now." There are 4 chess problems to solve. There is no name or address of the sender. Here, see for yourself. I suggest that we take our cases to our rooms, and that we meet here in 30 minutes".

Half an hour later, the two friends were studying the letter in a bar in a silence only broken by the drumming of Holmes's fingers on the table top. After the time it takes to savour more than one large cognac, Holmes said, "I would value your comments, Watson".

"I do not understand the reference to stopping a crime and telling the police", admitted Watson, "but I think I have solved the problems. And my guess is that the CJM of the dedication is Christopher Jeremy Morse — Sir Jeremy Morse. Again I can only guess that SBFM is one of Sir Jeremy's sons — he has four — and that it is a dedication to father and son. Problem 1 has set play 1... Bxc7/Bxe7/Rxb3/Bg3/Bh2 2.Bf7/Be6/a8=Q/Qg8/Rh8#. It solves most disappointingly by 1.Qxf5! when 2.Qc8/Qd7/Qf8# is irrefutable. The helpmate 2 is a letter problem "FCL" which solves by 1.Qe5 dxe5+ 2.Kf5 Ne7#. But who is "FCL"? I know composers with the initials "FL" — Frank Libby, Frédéric Lazard, and Frithiof Lindgren — but I do not know if their middle names began with "C". Regarding position 3, my guess is that, with or without the move, White always wins in this curiously balanced position. 1.f3 is very strong. Problem 4 has 2 promoted Black bishops and 1 promoted White bishop — and too many captures. The position is illegal. But a £16000 reward is magnificent!" exclaimed Watson.

"But no reward for you, dear Watson", replied Holmes. "You have not told me about the crime and what we should tell the police. You have solved the problems — but it was not necessary. The solutions do not matter. They are pure deception! The positions will speak to us! They can tell us the when, where, what and who of the planned crime. I shall explain!"

What did Holmes discover? Click here to find out.


"So, Watson, let us alert the police of the crime. Please ask the telephone operator to connect me with the Quai d'Orsay in Paris. I will speak to Detective Chief Inspector Lestrade". "But is not Inspector Lestrade in London at Scotland Yard?" queried Watson. "Not that Lestrade" laughed Holmes — "his French cousin, Inspector Marcel Lestrade".

(original text: SHERLOCK HOLMES EN FRANCE by Barry Barnes, The Problemist, September 1994)

02 September 2011

A + B = C

Today, I present some awarded problems from the Sherlock Holmes Tourney that Barry P. Barnes initiated several years ago in The Problemist. He demanded a combined position (C) that is composed of positions (A) and (B). All positions have a different solution in the same number of moves, (C) does not introduce new pieces or conditions. But it turned out that not all composers fully understood what was expected. Anyway, there were some very interesting entries and here are a few for you.

S. J. G. Taylor
Sherlock Holmes Tourney, 1994
1st Prize
  (A)
[r3k3/b1p5/P1p5/3N1N2/8/8/8/4K3]
  h#2(4+5)  
  (B)
[4k2r/6p1/6n1/5N2/8/8/8/R3K3]
  h#2(3+4)  
  (C)
[r3k2r/b1p3p1/P1p3n1/3N1N2/8/8/8/R3K3]
  h#2(5+8)  

Economic composition spiced with white and black castling options.
(A)1. 0-0-0 Nf6 2. Bb8 Ne7#
(B)1. 0-0 Ra7 2. Nh8 Rg7#
(C)1. Kd8 0-0-0 2. Re8 Sb6#


Attila Benedek
Sherlock Holmes Tourney, 1994
2nd Prize
  (A)
[8/1b5K/3p4/8/3pp1B1/3rk3/N2n4/2r5]
  h#2(3+8)  
  (B)
[1b1n4/1bk4K/8/3P2q1/6B1/8/N2n4/8]
  h#2(4+6)  
  (C)
[1b6/1b5K/2p2kP1/4p3/6B1/8/N4p2/8]
  h#2(4+6)  
  (D)
[1b1n4/1bk4K/2pp1kP1/3Pp1q1/3pp1B1/3rk3/N2n1p2/2r5]
  h#2(4+16)  

A funny idea. In each of the three diagrams the black king is on a different square. Thus, the combined position (D) has three black kings that are to be checkmated (such problems with more than one king are filed in the category Rex Multiplex).
(A)1. Rf1 Nc3 2. Rf4 Nd1#
(B)1. Nc4 Nc3 2. Nb6 Nb5#
(C)1. Bd6 Nc3 2. Be7 Ne4#
(D)1. cxd5 Nb4 2. Rc6 Nxd5#


R. C. Nascimento
Sherlock Holmes Tourney, 1994
2nd Honourable Mention
  (A)
[8/8/3p4/1p6/1p4BK/3Pk3/4P2B/8]
  h#2(5+4)  
  (B)
[8/8/8/3R3P/6BK/N2Pk2P/3p1p2/4b3]
  s#2(7+4)  
  (C)
[8/8/8/8/7K/4k2P/5p2/3b3R]
  h#3(3+3)  
  (D)
[8/8/3p4/1p1R3P/1p4BK/N2Pk2P/3pPp1B/3bb2R]
  #3(10+8)  

Another ingenious work. Merging two helpmates and a selfmate results in a mate problem!
(A)1. Kd4 Bf3 2. Kc5 Bg1#
(B)1. Nc4+ Kf4 2. Bd1 f1=~#
(C)1. Ke2 Rh2 2. Kf1 Kg3 3. Be2 Rh1#
(D)1. Rf1! (thr. 2. Kg5 3. Bf4#) Bb3 2. Kg5 Bxd5 3. Nc2#