Originals from Pat a Mat 30 - third part

Comment on original for solving:
Polish selfmate has one phase, ...

Solution:

1.Ve2! t.
1...c2, c6, Sxe7, Dxe2
2.Jc3+, Jxc5+, Sxd5+, Dxd3+
2...Vxc3, Vxc5, Sxd5, Dxd3#

Unpinning of black pieces, twice by interference, twice by capture. (JL)

Comment on original for solving:
.. in s#3 893 we have model mates. Isn't it possible to create more of them than here?

Solution:

1.Dc5! t.
1...Jg3 2.Ve4+ Jxe4 3.De5+ fxe5#
1...f5 2.Sxf5 Jg3 3.Je2+ Jxe2#

Two model mates. (JL)

Comment on original for solving:
Similar question is valid also for 894 by latest International Slovak Champion in solving, also the originality is doubtful.

Solution:

1.Se3! t.
1...h6 2.Df5 h5 3.Sg1 h4 4.Vf2 h3 5.Vg2+ hxg2#
1...h5 2.Sg5 h4 3.Jh2 h3 4.Vg1+ Kf2 5.Vg2+ hxg2#

Key allows 2 moves for bp, at the end two model mates. (author) Originality is questionable in this type of problems... (JL)

Comment on original for solving:
Difficult selfmate 895 was the only problem that wasn't fully solved by any participant of mentioned championship within time limit. At home, in armchair everyone should have more chances. The result is worth the effort.

Solution:

1.Jf7+? Sxf7+! 2.Dxh8
1.Dd4~? hr. 2.Jf7+, 1...Kxe5!
1.Jf5+? gxf5!

1.Sh2! hr. 2.Jf5+ gxf5 3.Jd3+ f4 (Jg3) 4.Df6+ Se6+ 5.Df8+ Vxf8#
1...Jxg3 2.Sxg3! hr. 3.Dc5+! Kxc5 4.Sf2+ Kd6 5.Jf7+ Sxf7#
2...Jd3 3.Dxb4+! Jxb4 4.Jc4+ Kc5 5.Sd6+ Sxd6#
(2...Je2? 3.Jd3+ and as in threat after 1st move)

Queen sacrifices,lively play of Se5. Probably hard for solving. (author) As I already wrote with original, its difficulty was proved by International open championships of Slovakia in solving. From composing point of view I am fascinated by author's ability to motivate the threats and black counterplay. (JL)

Comment on original for solving:
In former Soviet Union the ice is finally broken - fairy chess is developing very quickly and it brought us 4 originals out of 6 present. 896 shows reciprocal change.

Solution:

1...Vb5, Vxb6 2.Dh1# (Db1?), Db1# (Dh1?)

1.c6! hr. 2.Sc5#
1...Vb5, Vxb6 2.Db1# (Dh1?), Dh1# (Db1?)

Reciprocal change with dual avoidance in every variation. (JL)

Comment on original for solving:
Authors of 897 originally sent it as an entry to 4th TT CCM, but as it was not thematical, it has no chance. I think it has it in PaM despite only one phase , but thanks to intensity.

Solution:

1.c3! hr. 2.Dc2#
1...Kc4, Kc5, Kd3, Kd5, Ke3, Ke4, Cd3
2.Db5, Dd5, c4, De5, Df4, Cg6, Dc8#

Before key bK has 6 flights, the key gives d3 and takes c3. The variations consequently use all fairy elements and although it can be hard to say that there is some special concrete theme here, this problem is excellent example of almost infinite possibilities of combining more fairy elements. (JL)

Comment on original for solving:
Here solvers must look for accurate promotions.

Solution:

a) 1.a1D gxf7 2.Dxb2 f8S 3.Df2 Sh6#

b) 1.a1J gxh7 2.Jb3 h8V 3.Jd2 Vh3#

AUW, black promoted piece cannot interfere with mating check due to selfcheck. Joost de Heer has pointed that Sa7 is absolutely unneccessary. (JL)

Comment on original for solving:
Again miniature h#2 Circe from Lithuania.

Solution:

1...f8D 2.Jd2 Df1#

1.Jd2 d8D 2.Jf1 Dd2#

Analogy of set and real mates. (JL)

Comment on original for solving:
Final position of 900 has many common elements with finales of a problem by Jaroslav Lunacek that was awarded in competition PaM 1991-1992 Rest - see diagram 611 in PaM 28.

Solution:

1.e1V Je2 2.f1J Jc1 3.dxc1S(Jg1) c8D 4.Jf3+ Dc2=

Again AUW. (JL)

Comment on original for solving:
And finally, we have also Sergej Smotrov with his standard quality.

Solution:

1.Sd2+? Vxd2+!

1.Te6+! Kg4 2.Jh6+ Kh5 3.Jf7+ Kg4 4.Vg8+ Kh5 5.Tg7+ Kg4 6.Te8+ Kh5 7.Vh8+ Kg4 8.Vf2+ Kg3 9.Vxh2+! K~ 10.Vf2+ Kg3 11.Vg8+ Kh4 12.Vf6+ Kh5 13.Tg7+ Kg4 14.Te6+ Kh5 15.Vh8+ Kg4 16.Jh6+ Kh5 17.Jg8+ Kg4 18.Tc2+ Kg5 19.Sd2+! cxd2#

Everything is clear - Rh2 must be removed. (JL)