Originals from Pat a Mat 28 - second part

Comment on original for solving:
All helpmates have two moves today. 694 and ...

Solution:
1.Kg3 0-0-0 2.Kg2 Vg1#
1.Kh4 Va5 2.Sg4 Sf2#
1.Kh5 Va6 2.Vg4 Vh6#

Three model mates. (JL)

Comment on original for solving:
... 695 shouldn't make troubles for solvers, intended themes are simple too.

Solution:
a) 1.f5 Da6+ 2.Df6 Vg8#

b) 1.Sg5 Va6+ 2.Jf6 Dc2#

Reciprocal changes of function Ra8-Qa2, Qg7-Sh7 and model pin mates. (JL)

Comment on original for solving:
Tichomir Hernadi comes back after longer absence and 696 surely isn't bad comeback.

Solution:
a) 1.Vxd1 d8D 2.Kxf1 Dxd1#

b) 1.Sxc5 f8S 2.Kxf2 Sxc5#

Phenix theme well done - White promotes to the piece taken by Black. (JL)

Comment on original for solving:
Judge of h# 1999 can at last compete in our magazine, 697 didn't get any distinction in Netanya during congress, a bit of surprise for me.

Solution:
a) 1.Sxb6 dxe8S 2.Jf5 Sgxf7#

b) 1.Vxg8 dxe8J 2.Je5 Vxd6#

Zilahi + promotions - judge of TT in Israel hasn't enough understanding for this blend. (author).

Comment on original for solving:
698 was sent to us by French grandmaster and despite the only phase it is interesting piece.

Solution:
1.Va3! hr. 2.Dxg4+ Jxg4#
1...Jd2, Sc2, Je3, Sb3
2.Jc2+ De3+, Jab3+ Dd2+
Sxc2, Jxe3, Sxb3, Jxd2#

Cycle of 1st and 2nd black moves with antibattery use and dual avoidance in every variation. (JL)

Comment on original for solving:
Finnish twins 699 have duals in one by-variation, but main variations are OK.

Solution:
a) 1.f3 f6 2.c8S Kb8 3.h8D Kc7 4.Df8! Kb8 5.Dd8 Ka8 6.Dxb6! V,Sxb6#

b) 1.h8J f3 2.Jg6 f4 3.Jxf4 f5 4.Je6 f4 5.c8J Kb8 6.Jxb6! V,Sxb6#

Prudential play of white with use of twice two promotions, mostly underpromotions. It is interesting to see how a kind of minimalistic twin affects the play. (JL)

Comment on original for solving:
Very airy construction of 700 hints a lot of checks, ...

Solution:
1.Dh4+ Kg8 2.Dh7+ Kf8 3.Dg7+ Ke8 4.Ve7+ Kd8 5.Sb6+ Kc8 6.Dg4+ Se6 7.Dc4+ Sxc4#

All moves checks, they lead to model mate. What about creating the second variation? (JL)

Comment on original for solving:
... on the other hand, in 701 we have peaceful play culminating after black promotions. One must wonder whether it is correct...

Solution:
1.Sc8 f4 2.Da7 f3 3.Ja5 c4 4.De3 c5 5.Sh3 c6 6.Kf5 f6 7.Sxf6 c3 8.Va1 c4 9.Vg1 c5 10.Jd1 f2 11.Vg6 f1J! 12.Dd3+ cxd3 13.Je3+ Jxe3#,
11...f1S! 12.Sg2+ Sxg2 13.De4+ Sxe4#.

It is hard to believe there is no cook, the manoeuvres of white pieces from a-file are suspisious... (JL)

And really (finally), Olivier Ronat cooks this one (English notation):
1.Rxd6+! cxd6 2.Kxf7 f4 3.Qd8/Qc7 f3 4.Be1 Kd4 5.Ke6 d5 6.Qe7 dxc4 7.Bc8 Ke3 8.Kf5+ Kd4 9.Kf4 Kd5 10.Ke3 c3 11.Rd4+ cxd4+ 12.Kd3 c5 13.Bd7/Qf6 c4# or 3.Bc8 f3 4.Be1 Kd4 5.Ke6 d5 6.Qc7/Qd8 dxc4 7.Qe7 Ke3 8.Kf5+ Kd4 and further as before...

Well done, Olivier!