Originals from Pat a Mat 29 - fourth part

Comment on original for solving:
Why 783 isn't placed among orthodox twomovers? Everyone must answer this for himself...

Solution:

a) It is possible to get with one of two black pieces Rb1, Kc1 to the SW corner through 2nd rank at queen's flank without need to move wK - but not both together! White castling is illegal.
1.Jf2? e4!
1.Kf1! hr. 2.Kg2#, 1...Jxc3, Jf4, Vb2 2.Sxe3, Vd1, Dxb2#

b) 1.Kf1? no threat
1.Jf2! hr. 2.0-0, Jd3#, 1...Sd2+, Vxc4, Vxb3 2.Sxd2, Jd3, 0-0#

I liked the most the paradox implied by the uncertainty of preceding play - it is known that wK had to move because of bR or bK, but it is not known, which piece caused it in fact... (JL) Retromotive (as is usual with this kind of problems) is the most interesting. (BM)

Comment on original for solving:
This #2 Circe carries the risk of anticipation.

Solution:

1...Sb6, Vb6 2.d5, Vc1#
1.Vb6? hr. 2.d5, Vc1#, 1...Jxe4(pe2) 2.Df3#, 1...Je6!
1.c6! hr. 2.Dxb4#, 1...Sb6, Vb6 2.Vc1, d5#

Reciprocal change and one Nowotny try. (JL)

Comment on original for solving:
Reto (with the help 4 different kinds of fairy pieces) produced difficult new-strategical theme containing black corrections.

Solution:

1.VAg7? A hr. 2.Sc5# B
1...Sf5~ a, Se6! b, Se4!! c 2.Vxd5 C, Jxf3 D, VAe5# E
1...Sxd7!

1.VAe5? E hr. 2.Vxd5# C
1...Sf5~ a, Se6! b, Se4!! c 2.Jxf3 D, Sc5 B, VAg7# A

Excellent letter theme - Shedej cycle after defences a, b and key-mate reversal after defence c. Its prototype was probably shown by me in 2nd Prize Lipskie Centrum Kultury C 1.9.1998. But... Reto incorporated in admirable manner black corrections! In my opinion top favorit for prize in year tourney. And - thanks for dedication! Moreover, Reto removed some duals in by-variations by following changes (better version): replace all nightrider lions by naos, bpf3 -> g6, bNAg4 -> bRg4. The content doesn't change. (JL)

Comment on original for solving:
Hybrid with various stipulations is simple.

Solution:

a) 1.Je5+ dxe5 2.Da6#

b) 1.Ja5 a1S, a1J 2.Va3, Jc4 Sxb2, Jxb3#

c) 1.a1S 2.Sxb2+ Jxb2#.

Varying play. Every single stipulation is worth nothing here, but it is interesting, how one can get it all together. (JL) It is a little, 100 years ago the composers were able to add h#2 too. (BM)

Comment on original for solving:
Jozef for the first time introduces kentaur king. (Kentaur king moves alternately as knight and king, he has royal function.)

Solution:

1.KKd4 Ja3 2.Ve2 Jac2#

1.Sf4 Jxe5 2.KKg5 J1f3#

Two model mates with different movement possibilities of bK. (JL) Bohus Moravcik and Olivier Ronat cook by 1.KKd4 Ja5, Jxd2 2.Ve2 Jb3, Jf3#.

Comment on original for solving:
My join problem with ex-chief of PCCC arose as a by-product of composing for Problem-Echo 2000 tourney.

Solution:

a) 1.Sg4 Cxc7 2.Vf6 Cg3#

b) 1.Vg4 Txd2 2.Jf4 Te4#

Reciprocal batteries N-G and G-N with model mates. (JL)

Compare to very similar problem by JL or more varying problems by Klaus Wenda and Zvonimir Hernitz.

Comment on original for solving:
Finaly, we have two maximummers - one selfmate with orthodox model mate and ...

Solution:

1.Jd8 Sh5 2.Se6+ Ke8 3.Sg4 Sf7 4.Kf5 Sa2 5.Kg6 Sg8 6.Kh5 Sa2 7.Jf7 Sxf7#

Two switchbacks, of bB and wS, to f7. (JL) Possible second model mate with blocking of g6 and with mating from d1-h5 line needs (with duals, pity) 8 moves, e.g. after 1.Sh8. Doesn't author want to realize it? (BM)

Comment on original for solving:
... one helpmate in Circe with logical structure.

Solution:

1.Da4 Ke7! 2.Dh4+ Kd6 3.Da4 Kc5 4.Dh4 Kb5! 5.Da4+ Kxa4(Dd8) 6.Dd1+ Kb4 7.Dd8 a4! 8.Dd1 a5 9.Dd8 a6 10.Dd1 a7 11.Dd8 a8D+ 12.Da5+ Kb3 13.Da2 Dxa2(Dd8)#

Round trip of pa2, but the return in the form of queen. (author)
To do something with wp, white must first remove bQ. But the play is rather linear, without more interesting possibilities. (JL)