Good old times 10


We have seen some time ago some problems from the award of A section of the match between France and Germany. Now here you can find six problems coming from the section B, dedicated to #2 with fairy pieces (from the prescribed list). Each problem had to feature exactly two types and it was required to show black correction with both types of used fairy pieces. This section was judged by Roland Baier and Markus Ott.


Roméo Bedoni
13th Place France - Germany 1983-89

1.GId4! zz
1...CAh~ 2.g8CA#
1...CAg8! 2.fxg8CA#
1...GI4~ 2.f8S#
1...GIf8! 2.gxf8S#
1...GI7~ 2.a8CA#
1...GIa8! 2.bxa8CA#
1...CAc~ 2.b8S#
1...CAb8! 2.axb8S#

The mechanism for the black correction is a fairly simple one. Any move of black leaper unguards promotion square, leading to mating promotion. The move to promotion square corrects, but allows the same checkmate after capture by the other pawn. The scheme is symmetric and thus show the same content twice.









#2 (8+5)
0+2 camel, 2+2 giraffe

Jean-Pierre Boyer
10th Place France - Germany 1983-89

1.LIh1! zz
1...LId2~, f4, R~ 2.Sg5#
1...LIg2! 2.exf5#
1...Sb5~ 2.exf5#
1...Sd6! 2.Re7#
1...MAc6~ 2.Rxe7#
1...MAd8! 2.Sf8#
1...LIb8~ 2.Sf8#
1...LIbf4! 2.Sg5#
1...f~ 2.Rf6#

Cycle of mates after random moves and corrections of four different pieces: LId2, Sb5, MAc6 and LIb8. In other words it is 4-fold cyclic Feldmann 1 theme. The motivation of play is based on guard of g5 by Black, of d5 by White, of e7 by Black and of f8 by Black.









#2 (10+10)
1+1 mao, 2+2 lion

Joachim Brugge
5th Place France - Germany 1983-89

1.Kh5! th. 2.Rg4#
1...Ng2~ 2.Gd4#
1...Nc4! 2.Sd5#
1...Ne3! 2.Gg3#
1...Ge6~ 2.Gc7#
1...Gb6! 2.Qh6#
1...Gxe4! 2.Qxe4#
1...Qe2+ 2.Sxe2#

Both random defences show theme B (opening of white line allowing closing of other white line in mating move) and both of them allow two different corrections each. 1...Nc4! and 1...Gb6! also both close black lines resulting in unguards, other motifs are different. But the overall content is rather rich.









#2 (15+9)
1+1 nightrider, 4+3 grasshopper

Hans Peter Rehm
4th Place France - Germany 1983-89

1.Rd5! th. 2.Re5#
1...VAf5~ 2.Sf6#
1...VAxd7! 2.PAa4#
1...PAd4~ 2.Sc5#
1...PAxd7! 2.VAh7#

Only two corrections, but they have quite original motivation (for 1983). Capturing black vao/pao blocks the possible square for jump of the other Chinese piece.

Compare this e.g. to #2 with Chinese pieces and hoppers that have found it 's way to FIDE Album 2001-2003 as G81.









#2 (9+7)
2+1 pao, 1+2 vao

Jean-Pierre Boyer
2nd Place France - Germany 1983-89

1.Bc1! zz
1...MAe6~ 2.Rd4#
1...MAc5! 2.Sc3#
1...LIc7~ 2.Sc3#
1...LIg7! 2.MAhxf6#
1...MAhf8 2.MAhxf6#
1...MAg5! 2.MAg3#
1...LIg8~ 2.MAg3#
1...LIxd5! 2.Re2#
1...LIb5~ 2.Re2#
1...LIe5! 2.Rd4#
(1...LIbg5+! 2.Mg3#)

Cycle of mates after random moves and corrections of five different pieces: MAe6, LIc7, MAh7, LIg8 and LIb5. In other words it is 5-fold cyclic Feldmann 1 theme, thematically improving the 4-fold cycle.

The motivation of play is based on black guards of d4, c3, f6 and g3 and on blocking/guarding of potential flight d5. It is worth noting that Black uses only five fairy pieces and all of them are thematical.









#2 (14+7)
4+2 mao, 2+3 lion

Hans Peter Rehm
1st Place France - Germany 1983-89

1.Se6! th. 2.Sh4#
1...VA~ 2.Sfd4#
1...VAd5! 2.Qxb1#
1...VAc6! 2.Sed4#
1...PAe~ 2.Sed4#
1...PAed5! 2.VAc2#
(1...PAd4 2.VAd7#
1...bxa6+ 2.Qxe5#
1...Bxe7 2.Sg7#)

Again only two black fairy pieces correct their random moves.

After the key, f6 is guarded by both PAa6 and VAa1, that is why 2.Sed4+ is not mate (both lines are deactivated by the move). Similarly, g4 is guarded by both PAc4 and VAd1, that is why 2.Sfd4+ is not mate. Random moves deactivate lines through d4, but that means knights can leave their lines and go to d4. This is a specific form of Levman defences.

Corrections are based on the antibattery activation of PAd7 to d4. But 1...VAd5 pins both VAd5 and PAe5 in advance, allowing 2.Qxb1#, and 1...PAd5 blocks VAe4, making possible mate over immobilized hurdle. Both mates after corrections are unified by being given on the same diagonal.









#2 (12+9)
3+3 pao, 3+1 vao

Comments to Juraj Lörinc.
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