Note: In these solutions, those moves that
were required are given in bold type. All else is given for the
sake of completeness and for the interest of solvers. The symbol
'S' is used for knight and threats are given in parentheses.
1. Joel Fridlizius
4th Prize, Sporten , 1893
Mate in 2
Show solution
1.Sc3! (2.Se6#) (5)
1...Kxc3 2.Sb3#
1...bxc5 2.Qg7#
1...Qxc3 2.Qd6#
1...Qxb5 2.Sxb5#
1...Sxc3 2.e3#
1...Bxe2 2.Sxe2#
1...Bb3 2.Sxb3#
2. Josef Holzman
Sachove listy , 1900
Mate in 3
Show solution
1.exd5! (1)
(2.Sc6 & 3.b4#) (1)
1...Kxd4 2.Qxd3+ Kc5 3.Sa6# (1)
1...Kxb4 2.Sd2+ Kc5 3.Sb3# (1)
Ka5 3.Sb3#
1...Rxd5 2.Sa6+ Kxd4 3.Qg4# (1)
3. Franz Pachl & Hermann Weissauer
4th Prize, Sachsische Zeitung , 2004
Mate in 3
Show solution
1.g6! (1)
(2.Sf6+ Kxf5 3.Bd7#) (1)
1...Bxf4 2.Re5+ Bxe5+ 3.Qxe5# (1)
1...Rxe3 2.Bd3+ Rxd3 3.Qxd3# (1)
1...Qxf4 2.Qg2+ Qf3 3.Qxf3# (1)
4. Valentin F Rudenko
3rd Place, Dniepropetrovsk Thessaloniki, 2004-2005
Mate in 4
Show solution
1.Qh3? 1...Bf8!
1.Qh5! (1)
(2.Sf4+ exf4 3.Qd5+ Kxe3 4.Qd2# (½)
Bd4 4.Qxd4#
Kxe3 3.Sd5+ Kd3 4.Qxf3#) (½)
1...Be2 2.Qf5 & 3.Rd4+ exd4 4.e5# (1)
1...Se2 2.Qxf3 & 3.Sf4+ Sxf4 4.exf4# (1)
exf4 4.exf4#
1...Sc3 2.Rxc3+ Kxe4 3.Qg4+ Kxd5 4.Rc5# (½)
1...Sxb4 2.Sxb4+ Kxe3 3.Sd5+ Kd3 4.Qxf3# (½)
5. Enrico Paoli
54 Studi Scacchistici (1947-1957) , 1959
White to play and win
Show solution
1.h7 (i) (½)
1...Be5+ 2.d4! (ii) (1)
2...Bxd4+ 3.Ka2 (1)
3...Nc3+ 4.Kb3! (1)
4...Nd5 (iii) 5.Kc4 (1)
5...Ke4 6.f3+ 1-0 (½)
(i) 1.d4?? Kf6 2.Ka2 Nc3+ 3.Kb3 Nb5 4.Kb4 Nc7 5.Ka5 Ba7 6.d5 Nxd5 7.Kb5 Bxf2 8.Kc6 Nc3 0-1; 1.Ka2? Kf6 2.Kb3 Nxf2 3.d4 Ne4 4.Kc4 Ba7 5.Kd5 Nc3+ 6.Kc4 Ne2 7.d5 Nd4 8.d6 Nc6 9.Kd5 Nb8 =
(ii) 2.Ka2? Nc3+ 3.Kb3 Nb5 4.Kc4 Na7 5.Kc5 Ke6 6.Kb6 Bd4+ 7.Kb7 Nb5 =
(iii) 4...Nb5 5.Kc4 1-0
6. Mikhail Marandiuk
3rd Prize, Chepizhni-70 JT, 2004
Selfmate in 3
Show solution
1.Rg4! (1)
(2.Se3+ Qxe3 3.Qe6+ Qxe6#) (1)
1...Bg1 2.Bd3+ Qxd3 3.Qd5+ Qxd5# (1)
1...Qxa4 2.Rxf4+ Bxf4 3.Qxb5+ Qxb5# (1)
1...Bxg4 2.Rxb4+ Qxb4 3.Qc5+ Qxc5# (1)
7. Camillo Gamnitzer
feenschach , 2005
Selfmate in 4
Show solution
1.Bc8? 1...fxe2! or 1...bxa3!
1.Be4? 1...a1=Q+!
1.Bd5? 1...Bxg5!
1.Ba8! (1)
(2.Qd7+ Kxc5 3.Qc6+ Kd4 4.Sxf3+ Sxf3#) (1)
1...fxe2 2.Bf3 (3.Sc6+ Kc4 4.Bxe2+ Sd3#) (1)
Bxg5 3.Se4+ Kd5 4.Sxg5+ Sxf3# (1)
1...bxa3 2.Rg4+ hxg4 3.Qa4+ Kxc5 4.Sd3+ Sxd3# (1)
8. Rupert J Wood
feenschach , 2005
Helpmate in 6; 2 solutions
Show solution
1.h5 Kg5 2.Rb1 Kf6 3.g1=R Kxf7 4.Rg7+ Ke8 5.Rc7 Sf7 6.Rb8 Sd6# (2½)
1.Kd8 Kg4 2.g1=Q+ Kf5 3.Qg4+ Kf6 4.Qc8 Ke5 5.Re1+ Kd6 6.Re8 Sxf7# (2½)
This problem was previously used in the 2006-2007 Postal Round, a fact mentioned by
just one solver. I wonder how many others noticed? I clearly hadn't added the 2006-2007
problems to my list of used problems, but that lapse has now been corrected.
