Model mates in fairy directmovers 1

1.Rb5! th. 2.Ra5#
1...PRc3 2.PRc4#
1...PRc4 2.PRxc4#
1...PRxb1 2.PRc4#
1...PRxb4! 2.PRxb4#
1...PRxb3! 2.PRb2#
1...Sc6 2.PRc2#
1...Sxb5 2.PRc2#
1...Sxb3! 2.PRb2#

The quartet of white pieces on b-file guards all bK flights, thus white princess is free to mate. The threat employs guard of b4 by princess for non-model mate, however all variation mates are given by princess. Defences by PRd2 unguard c4, but there are two stronger defences, corrections of a kind. Similarly, defences by Sd4 unguard c2 and there is one stronger defences. Altogether there are 3 different model mates shown.

1.Nb8! th. 2.Ngc6 th. 3.Ne5,Nb4#
1...Sf1 2.Qd4+ Ke2 3.Qd2#
1...c1=~ 2.Qe4+ Kd2 3.Qe2#

Good key introduces a threat with temporaty doubling of two white nightriders, but more importantly sets two white nightriders to a symmetric position with regard to line between d- and e- files. Two defences block f1 and c1 respectively and thus lead to a pair of symmetric variations with set-up and firing of white queen battery and model mates with double checks. It is a pity there is no third model mate.

1.Qf1+?
1...Kg4 2.Qf3+ Kh4 3.Nf8#
1...Ke4 2.Kc2 th. 3.Qd3#
1...Kg6 2.Qf7+ Kh6 3.Nb3#
1...Ke6!

1.Nf2! th. 2.Qg4#
1...Gd4 2.Qf7+ Kg5 3.Nf3#
1...Kg6 2.Qg8+ Kf5 3.Qg4#
1...Kg5 2.Qg4+ Kh6 3.Qg8#

Not only two model mates in the solution, but two further model mates in the checking try. Note that the key gives flight besides creating short threat.

1.g3! zz
1...rWd5 2.Rb6 rWxc5 3.Rd6 rWb5/rWc4 4.Rd5/Rc6#
      2...rWe5 3.Rd6 rWxf5/rWe4 4.Rd5/Re6#
1...rWxf5 2.Rb6 rWe5 3.Rxf6 rWd5/rWe4 4.Rf5/Re6#
      2...rWg5 3.Rxf6 rWxh5/rWg4 4.Rf5/Rg6#
1...rWe4 2.Rb2 th. 3.Re2#
      2...rWe3 3.Rd2 rWf3/rWe4 4.Rd3/Re2#
      2...rWe5 3.Rd2 rWxf5/rWe4 4.Rd5/Re2#

Unbelievable scheme shows forking play leading to 12 subvariations. All mates are model and only a few of them repeat. Although there is no deep strategy involved, the record character of the result can be inspiring.

a) 1.Nf4! Kg1 2.Nd5 Kh1 3.Sc3 Kg1 4.Se2+ Kh1 5.Nb4#

b) 1.Kf2! Kh2 2.Nb6 Kh1 3.Nd5 Kh2 4.Nd4+ Kh1 5.Ne7#

Very economical presentation of the key difference between knight and nightrider. While nightrider is strictly stronger piece than knight, it can be too strong sometimes. That is why the solution of a) does not work in b) - 1.Nf4? stalemate!

1.Qe1? zz 1...Kc4 2.Qd2 Kb3 3.Qc3+ Kxa4 4.Qb2 Ka5 5.Qb5#
      2...Kc5 3.Kc7 Kc4 4.Nf5 Kc5 5.Qd4#
1...Kd4 2.Nf2 Kc5,Kc4 3.Qb4+ Kd5 4.Ke7 Kc6/Ke5 5.Qb5/Qe4#
      2...Kd5 3.Qe4+ Kc5 4.Nh3 Kxd6 5.Qd5#
            3...Kxd6 4.Qf5+ Kc6 5.Qb5#
1...Kc2!

1.Qf4! zz
1...Ke2 2.Nb2 Ke1 3.Qh2 Kf1 4.Nf4 Ke1 5.Qe2#
1...Kc2 2.Qd4 Kb3 3.Qc3+ Kxa4 4.Qb2 Ka5 5.Qb5#

And once again Jevgenij Bogdanov. Actually, he has produced a lot of similar problems during period 2000-2002, publishing them in obscure places. I had the pleasure (?) to judge some of them for FIDE Album 1998-2000 and indeed one of them managed to get in. Perhaps we will come to it.