Originals from Pat a Mat 23 - third part

Comment on original for solving:
213 has changes of variations...

Solution:
1...e1S, e1J 2.Dxc3+, Dc2+ Sxc3#, Jxc2#, 1.De1? t., 1...fxe1S, fxe1J 2.Vxc3+, Sc2+ Sxc3#, Jxc2#, 1...Vh1!, 1.Dd1+! exd1S, exd1J 2.Sc2+, Vxc3+ Jxc2#, Sxc3#

Change Z-32-64 is made possible by non-standard key in a very (for selfmate) economical position.

Comment on original for solving:
... 214 changes of functions...

Solution:
1...Jf1~ 2.Ve4! A, 1.Ve4? A t., 1...Jb8 (Jb4) 2.Sc6 B Jxc6#, 1...Jc7 (Jc5) 2.Je5 Je6#, 1...Dc6!, 1.Sc6! B hr. 2.Se4 Dc5#, 1...Jb8 (Jb4) 2.Ve4 A Jxc6#, 1...Jc7 (Jc5) 2.Se4 Jxe6#

Simple change of functions of moves.

Comment on original for solving:
... 215 is strategical...

Solution:
1.Vf5! t., 1...Dxa7 2.Jf6+ Kxg7 3.Jd5+ Dxa1#, 1...b7~ 2.Jh5+ Kxh7 3.Jf4+ Dxh1#

Strong D-O analogy, Michal can't disclaim influence of his helpmate heart. Look at the following classical work for comparison.

Comment on original for solving:
... 216 has echoed mates...

21.5.1999: I received an e-mail from mr. Rainer Staudte stating that this problem is selfanticipated: it won Comm in Moscow tournament (date?) and was reprinted in Shachmatna Kompozicija Nov-Dec 1998.

Solution:
a) 1.Dg1? g3 2.Dd4 g2+ 3.Kg1 Kg3 4.?, 1.De1! g3 2.De4 (Db4?) g2+ 3.Kg1 Kg3 4.Df5! h3 5.De4 h2#, b) 1.Df1! h3 2.Df6 h2 3.d4 Kh3 4.Df2 g3 5.Df4 g2#

This selfmate doesn't compete in 1998/99 tourney due to selfanticipation.

Comment on original for solving:
... 217 has echoed mates too (it is the correction of No. 234 from PaM 19)...

Solution:
Author intention: 1...b4 2.Df3 Ke1 3.Kc1 b2+ 4.Sxb2 b3 5.Se4 d5 6.Sb1 d4 7.Dg2 d3 8.Df3 d2#, 1.Sf2! b4 2.De1+ Kd3 3.Kb1 b2 4.Dd1+ Kc3 5.Sb3 d5 6.Sc5 d4 7.Sa2 d3, b3 8.Dc2+ dxc2#.

Unfortunately, author had not happy hand during composing this correction as is shown by two solvers:
1.Df3 b4 2.Sf6 Ke1 3.Kc1 b2+ 4.Sxb2 etc. as in set play, 1...Ke1 2.Kc1 b2+ 3.Sxb2 b4 4.Dg2 b3 5.Se4 d5 6.Sd3 d4 7.Sb1 d3 8.Df3 d2#, 2...b4 3.Dg2 b2+ 4.Sxb2 etc. as in set play.(Jozef Holubec) or 1.Se4! d5 2.Sd3 b4 3.Df3 Ke1 4.Kc1 b2+ 5.Sxb2 d4 6.Dg2 b3 7.Sb1 d3 8.Df3 d2# (Milan Svrcek).

Comment on original for solving:
... and finally 218 shows underpromotions.

Solution:
Author intention: 1.h8V+! Kf7 2.Jb6+ Kf6 3.Dd8 axb6 4.Vh7 bxc5 5.e4 fxe4 6.Jg5 e3 7.Jf7 e2 8.Vh5 e1~ 9.Dd4+ cxd4 10.g8J+ Jxg8#

Again Jozef Holubec found cooks, this time shorter than intention: 1.Sd3 Kf7 2.Sxf5 Ke8 3.Sd7+ Kf7 4.Sg4 Ke8 5.h8V+ Kf7 6.Vh7 K~ 7.Jf6+ Kf7 8.Vh5 Kxf6 9.g8J+ Jxg8#.