30 March 2012

Striptease

It was in 2011, I don't remember when exactly, that I saw the following chess problem. It is a highly remarkable demonstration of the so-called striptease theme: in each further twin one more piece is removed.
Udo Marks
The Problemist Supplement, 11/2005
[1b4bn/3p4/1p1P4/BP1p1P2/2kPpP2/pR2P3/KpP5/1B6]
  see below(12+9)  

a) diagram, #10: 1. Kxa3! Bxd6+ (1. Bxb6? Bxd6! [2. - Nf7? 3. Kxa3!] 2. Ba5 Nf7! 3. b6 Nd8 4. ?) 2. Ka4 (3. Rc3#) bxa5 3. b6 Bf7 (3. - Bxf4 4. b7 Bxe3 5. b8=Q Bd2 6. c3+ Bxc3 7. Qc7+ Kxd4 8. Qxc3#) 4. Ba2 b1=Q 5. Bxb1 Be5 (5. - Bc5 6. b7 Bxd4 7. b8=Q Bxe3 8. Qb5+ Kxd4 9. c3#) 6.fxe5 Be8 7. Ra3 (8. Ba2#) d6+ 8. Kxa5 Ba4 9. Kxa4 ~ 10. Ba2#
b) -Nh8, #9: 1. Bxb6! Bxd6 2. Ba5 Bc5 3. b6 Bxd4 (3. - Bxb6 4. Rxb6 ~ 5. Rb4+ Kc3 6. Ra4#) 4. exd4 Kxd4 5. b7 e3 6. c3+! Kc5 7. b8=Q (8. Qc7#) d4 8. Qc7+ Kd5 9. Rb5#
c) -Pc2 in b), #7: 1.Be1! (2. Rb4#) Bxd6 2. Bc2 b1=Q+ 3. Rxb1 Bf7 (3. - Bc5 4. Bd1 Bxd4 5. Be2+ Kc5 6. Bb4#) 4. Bd1 (5. Be2#) Bh5 5. Bxh5 (6. Be2#) Kd3 6. Rb3+ Kc2/Kc4 7. Rc3#/Be2#
d) -Bg8 in c), #6: 1. Be1! (2. Rb4#) Bxd6 2. Bc2 Bc5 3.Bd1 b1=Q+ 4. Rxb1 (5. Bd2#) Bxd4/Kd3 (4. - Bb4 5. Be2#) 5. Be2+/Rb3+ Kc5/Kc4 6. Bb4#/Be2#
e) -Bb8 in d), #5: 1. Kxa3! bxa5 2. b6 a4 3. b7 axb3 4. b8=Q Kc3 5. Qb4#
f) -Pa3 in e), #4: 1. Kxb2! bxa5 2. Kc2 a4 3. Rb2 a3 4. Ba2#
g) -Pb2 in f), #3: 1. Bc2! bxa5 2. Bd1 a4 3. Be2#
h) -Pb6 in g), #2: 1. Rb4+! Kc3 2. Ra4#
i) -Pe4 in h), #1: 1. Bd3#!
j) -Pd5 in i), #3: 1. e4! Kxd4 2. Bb6+ Kc4 3. Bd3#
k) -Pf4 in j), #4: 1. Bb6! Kd5 2. Bc5 Kc4 3. Bd3+ Kd5 4. e4#
l) -Pd4 in k), #5: 1. Be4! Kc5 2. Bc7 Kc4 3. Ka3 Kc5 4. Rc3+ Kxb5 5. Bd3#

Really impressive!

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