10 January 2011

Forsberg twins

One of the most often reproduced helpmates ever is shown in the following diagram. Five different pieces on a6 lead to five different model mates. Moreover, the composer achieved this with just five pieces. Outstanding!

Henry Forsberg
Revista Romana de Sah 1935
[8/8/q7/8/1R4K1/k2N4/8/8]
  h#2(3+2)  
a) Diagram
b) bRa6
c) bBa6
d) bNa6
e) bPa6

Such a twinning by replacing a piece by any other piece on the same square is no trivial task, especially if you want to have a classic Forsberg twin with all five orthodox pieces. Here are three more problems, the last one featuring some grasshoppers.


Lajos Bukovinszky
Romanian tourney, 1962
[8/5Q2/1K6/4p2N/4k2P/8/6p1/6B1]
  h#2(5+3)  
a) Diagram
b) wRf7
c) wBf7
d) wNf7
e) wPf7

Christer Jonsson & Kenneth Solja
Springaren 12/2010
[4r3/1P6/4n3/1b3N2/1p2k2p/1P5P/2Pb4/r1nK4]
  h#2(6+9)  
a) Diagram
b) wNb3
c) wBb3
d) wRb3
e) wQb3

Heinrich Bernleitner
feenschach 09-10/1991
[*2Q7/1*2q6/1p6/4k1K1/1p5*2q/2R5/1b*2q1n3/*2q7]
  h#3(3+9)  
a) Diagram
b) bQb7
c) bRb7
d) bBb7
e) bNb7

Forsberg
a) 1. Qf6 Nc5 2. Qb2 Ra4#
b) 1. Rb6 Rb1 2. Rb3 Ra1#
c) 1. Bc4 Ne1 2. Ba2 Nc2#
d) 1. Nc5 Nc1 2. Na4 Rb3#
e) 1. a5 Rb3+ 2. Ka4 Nc5#

Bukovinszky
a) 1. Kd3 Qf1+ 2. Ke4 Sf6#
b) 1. Kd5 Nf4+ 2. Kd6 Bc5#
c) 1. Kf3 Bb3 2. e4 Bd1#
d) 1. Kd5 Kb5 2. e4 Sf4#
e) 1. Kf5 Nf4 2. Kf6 f8=Q#

Jonsson & Solja
a) 1. Kd5 b8=Q 2.Kc5 Qd6#
b) 1. Kd5 Kxd2 2.Kc4 Ne3#
c) 1. Bf4 Ke1 2.Kf3 Bd5#
d) 1. Nf4 Rd3 2.Re5 Nd6#
e) 1. Bc4 c3 2.Kd3 Qc2#

Bernleitner
a) 1. Ba3 Ga2 2.Gd4 Rc6 3.Gd5 Re6#
b) 1. Qc6 Gd5 2.Qd6 Gd7 3.Nd4 Re3#
c) 1. Re7 Rd3 2.Ge4 Gf3 3.Re6 Rd5#
d) 1. Bd5 Ge4 2.Gd4 Ge1 3.Ke4 Re3#
e) 1. Nd6 Rc7 2.Gc8 Gd8 3.Bd4 Re7#

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