Problem 793: Igor Agapov  Threemover 
(31.12.2016) Nice threemover miniature with AUW on the same square by Igor Agapov.
793. Igor Agapov (Russia)
#3 (3+2)
a) Rd7d8 b) Rd7e7
c) Pb5c6 d) Pb5 a) 1.a8Q threat 2.Qb8+ Kc6 3.Rd6# b) 1.a8R Kc6 2.Rd8 Kb6 3.Rd6# c) 1.a8S+ Ka6 2.Kc5 Ka5 3.Ra7# d) 1.a8B Ka6 2.Kc5 Ka5 3.Ra7# White AUW on the same square. Task for 5pieceproblem in #3.
Compare to yacpdb/351680 (Author). 
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Comments
white: Kc4, Ra7, Pb7
black: Kc6
b)Ra7e8, c)+Sb5, d)+Se4
a) 1.b8R! (2.Kd4 Kd6 3.Rb6#)
1...Kd6 2.Re8 Kc6 3.Re6#
b) 1.b8Q! Kd7 2.Qc8+ Kd6 3.Re6#
c) 1.b8S+! Kb6 2.Sd7+ Kc6 3.Rc7#
d) 1.b8B! Kb6 2.Sc5 Kc6 3.Ra6#
But also is possible all solutions to be with only 4 pieces  AUW on the same square with only 4 pieces and black Minimal  a task for economy:
white: Kb4, Re7, Pa7
black: Kb6
b)Re7d8 c)Re7c7
a) 1.a8B Ka6 2.Kc5 Ka5 3.Ra7#
1.a8R Kc6 2.Rd8 Kb6 3.Rd6#
b) 1.a8Q Kc7 2.Qb8+ Kc6 3.Rd6#
c) 1.a8S+ Ka6 2.Kc4(5) Ka5 3.Ra7#
Or both can be published as different problems?
1) Russian proverb says: "Trust but verify"
The program Olive has shown a solution in the first twin in the following recording (short notation): 1.a8Q! threat: 2.Qb8 + Kc6 3.Rd6 #, 2 ... Ka6 3.Rd6#. I.e. the program has "considered" that 1...Kc7  does not protect against the threat and it did not consider this move as a defense. I believed it, but ... I have not checked it! Of course, these duals 2.Ka5/Kc5  look very unpleasant. To improve the purity of the solution, we can adjust the solution recording in the first twin:
a) Pb5>a6
1.a8Q a5+ 2.Q:a5+ Kc6 3.Qb5/Qc7# (the rest  without changes)
Let this adjustment will be an alternative for the tourney judge (#3).
2) This idea has bee shown for 4 pieces: I.Agapov, Seven chess notes, in 2016 (November)  yacpdb/416963. But that task can not be considered as a "pure" because the last solution has organic dual on the 2nd move (1.a8S+ Ka6 2.Kc4/Kc5). That is why I have decided to adapt this scheme for 5 pieces.
3) I propose another (new) version of this task  №793.1:
FEN: 8/P2K4/1k2p3/4R3/8/8/8/8
White: Kd7, Re5, Pa7 (3)
Black: Kb6, Pe6 (2) #3
b) Kd7b8 c) +Pe6c6 d) +Kb8c8
a) 1.a8R! Kb7 2.Raa5 Kb6 3.Reb5#
b) 1.a8B! Ka6 2.Kc7 Ka7 3.Ra5#
c) 1.a8Q! c5 2.Qa4 c5 3.Qb5#
d) 1.a8S+! Ka6 2.Rc5! Ka7 3.Ra5#
No duals + usual twins (no zero) = quite a Christmas story ...

Happy New Year 2017!
Igor Agapov
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