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Problem 793: Igor Agapov - Threemover
igor.agapov(31.12.2016) Nice threemover miniature with AUW on the same square by Igor Agapov.
 
 
 
 
 
 
 
 
 
 
 
793. Igor Agapov (Russia)
793.1
#3                           (3+2)
a) Rd7-d8   b) Rd7-e7 
c) Pb5-c6   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 5-piece-problem in #3.
Compare to yacpdb/351680 (Author).
 

Comments  

 
0 #1 Seetharaman Kalyan 2016-12-31 11:27
Huge task! Pity zeroposition is necessary.
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0 #2 Seetharaman Kalyan 2016-12-31 20:20
S.N.Ravishankar points out that in position (a) after 1.a8Q 2.Kc7 there is dual 2.Kc5,Ka5 & also the threat 2.Qb8+. But I think can be ignored in a task of this magnititude.
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+1 #3 Diyan Kostadinov 2017-01-01 06:30
Well, I think that the zeroposition and duals can be avoided if we combine both problems - by Titarenko and Agapov. In that way we have a good economy - 2 solutions with 4 pieces and two with 5 (included wS) with a perfect play:

white: Kc4, Ra7, Pb7
black: Kc6
b)Ra7-e8, 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)Re7-d8 c)Re7-c7

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#
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+1 #4 Diyan Kostadinov 2017-01-03 16:54
I am not sure which of the two versions is better - the first is with perfect twins and play, but the second is record for economy...
Or both can be published as different problems?
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+1 #5 Igor Agapov 2017-01-03 18:43
Dear Seetharaman and Diyan!

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) Kd7-b8 c) +Pe6-c6 d) +Kb8-c8

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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