16 December 2011

Christmas Quizzes (3)

Welcome to round three. Are you ready? Here we go!

  1E. C. Mortimer  
Chess, April 1952
[rnbqkbnr/ppppp1pp/5p2/8/4P3/8/PPPP1PP1/R1BQKBNR]
  Position after 8,0 moves. Find a game (first moves not unique).

  2unknown author/source  
 
[8/1KP5/8/2p5/1pP5/p7/k7/1R3R2]
  #4(5+4)  

[rnbq1bnr/ppppkppp/8/4R3/4N3/8/1PPPPPPP/2BQKBNR]

3   This puzzle has two parts:
  1. The diagram shows the position after 5. RxPe5#. What were the moves?
  2. Hugh Courtney pointed out that he had found five other sequences of moves wherein the white queen's rook mates the black king in 5 moves. Two of them are quite similar and have the black king on g6. In the other three solutions the king is mated on c6 (twice) and h6, respectively.


  4A. Herbstmann  
Tyovaen Shakki, 1935
1st Prize
[8/8/5P1P/4P3/3p1q2/3bn2K/3k1N2/7Q]
  Win(6+5)  

  5B. J. da Costa Andrade  
unknown source
 
[R7/PPQPPkPP/8/7K/B7/2B5/4r3/8]
  #1(11+2)  

  6Lord Dunsany  
The Times Lit. Supplement, 1922
[3N1rk1/pppp1N1p/5B1P/8/8/RP1P4/rPP3P1/4K2R]
  White to play, can he castle?

  7K. Fabel  
source unknown
[8/8/8/1pNp4/PpbpP3/1bkb4/1N1N4/2K5]
  #3(6+8)  
Solutions:
1There are three ways to make the first three white moves. Apart from that, everything is unique: 1. Nc3 f6 2. h4 Kf7 3. Nh3 Kg6 4. h5+ Kxh5 5. Ng5+ Kxg5 6. e4 Kg6 7. Ne2 Kf7 8. Ng1 Ke8.
21. c8=B! b3 2. Bg4 b2 3. Bd1 Kxb1 4. Bb3#
3These are the six solutions that were given by Courtney. Unfortunately, there are always interchangeable moves. It's still a challenge to find all final positions, isn't it?
a) 1. a3 e5 2. Nc3 Bxa3 3. Ne4 Bf8 4. Ra5 Ke7 5. Rxe5#
b) 1. e4 e6 2. a4 Ke7 3. Ra3 Kf6 4. Qh5 Q,Be7 5. Rf3#
c) 1. e4 e5 2. a4 Ke7 3. Ra3 Kf6 4. Qg4 Q,Be7 5. Rf3#
d) 1. a4 d6 2. a5 Kd7 3. Ra3 Kc6 4. e4 Q,Bd7 5. Rc3#
e) 1. a4 d6 2. a5 Kd7 3. Ra4 Kc6 4. Nc3 Q,Bd7 5. Rc4#
f) 1. e4 f6 2. a4 Kf7 3. Ra3 Kg6 4. Qg4+ Kh6 5. Rh3#
41. Qe1+! Kc2 2. Qc1+ Kb3 3. Qb2+ Kc4 4. Qb4+ Kd5 5. Qd6+ Kc4 6. Qc5+ Kb3 7. Qb4+ Kc2 8. Qb2+ Kxb2 9. Nxd3+. Very nice.
51. d8=N#
6No. If it's White to play, Black's last move must have been 0-0. Therefore, the rook at a2 was promoted from one of the pawns from e7, f7, or g7, any of which would have required either the white king or white rook to have moved. The promotion could not have taken place on a1. Otherwise, the bPe7 would have had to capture four white pieces on black squares — but there are only three available!
71. a5! Bbc2 2. Nb1+ Bxb1 3. Nd1#
1.- Bdc2 2. Nd1+ Bxd1 3. Nb1#
1.- Ba4 2. Nb,cxa4+ (dual) bxa4 3. Nxa4#
This last variation shows why 1. e5? fails: Black plays 1.- Bf5..h7!

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