Julia's Fairies

No.962
Aleksey Oganesjan
(Russia)

Original Problems, Julia’s Fairies – 2015 (II): July – December

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No.962 by Aleksey Oganesjan – Underpromotions and bicolor Bristol in HS# Meredith without fairy pieces. (JV)

No.962.1 by Aleksey Oganesjan – An improved version of No.962, after comment by N.Predrag. (JV)

No.962 Aleksey Oganesjan Russia original – 27.11.2015	Solutions: (click to show/hide)
White Kg8 Qa2 Rh5 Ra1 Bg1 Ph7 Pa4 Black Ka5 Be5 Pa7 Pa6 Pb2 hs#3,5 zero (7+5) a) Bg1→g7; b) Kg8→g2	a) wBg1-->g7 1...b2-b1=R {b1Q?} 2.Kg8-h8 Rb1-b8 + 3.Qa2-g8 Rb8-a8 4.Qg8-b8 Ra8b8 # b) wKg8-->g2 1...b2-b1=B {b1Q?} 2.Kg2-h1 Bb1-e4 + 3.Qa2-g2 Be4-a8 4.Qg2-b7 Ba8b7 # { (C+ by Popeye 4.73)} Meredith with underpromotions and model mates. Bicolor Bristol and Pelle theme. Wide play of pieces of both sides. [Zero-position has not a solution, so to achieve a goal, white King and Bishop must connect: either Bishop approaches King for a) or, vice versa, King approaches Bishop for b)] (Author)

No.962.1 Aleksey Oganesjan Russia version of No.962 – 03.12.2015 Dedicated to Nikola Predrag	Solutions: (click to show/hide)
White Kg2 Qa2 Rh5 Ph7 Pg6 Pa3 Black Ka5 Be5 Pa7 Pg7 Pa6 Pa4 Pd2 hs#3,5 zero (6+7) a) Kg2→g8; b) Pg7→h2	a) wKg2-->g8 1...d2-d1=R 2.Kg8-h8 Rd1-d8 + 3.Qa2-g8 Rd8-a8 4.Qg8-b8 Ra8b8 # b) bPg7-->h2 1...d2-d1=B 2.Kg2-h1 Bd1-f3 + 3.Qa2-g2 Bf3-a8 4.Qg2-b7 Ba8b7 # { (C+ by Popeye 4.73)} Underpromotions and model mates. Bicolor Bristol and Pelle theme. Wide play of pieces of both sides. [Zero-position has not a solution, so to achieve a goal, white King and black Pawn g7 must connect: either King approaches Pawn for a) or, vice versa, Pawn approaches King for b)] (Author)

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seetharaman

November 30, 2015 21:57

Very nice! Is it better to have the black pawn b2 at d2?

Aleksey Oganesjan

December 1, 2015 08:32

No, it’s not better, because if bPb2->d2 then cooks in a) will appear:
1…Kb4/Kb6 2.a5 Kb5 3.Bf8 d1Q 4.Qd5+ Qxd5#
(and three more cooks, with another order of moves)

Nikola Predrag

December 1, 2015 16:55

bPd2 would be better because in a), 3….Rd8-a8 would be a critical Bristol-move across b8. Rb8-a8 is not a Bristol-move.

And wB has a function of a Dummy which is too expensive for a helpselfmate. bP g7/h2 would be much cheaper than wB g7/g1.

Also, wRa1 is only a static guard of a4 which could be blocked by bP which could be immobilized by wPa3.

December 1, 2015 23:12

Reply to Nikola Predrag

hm…. bph2/g7 is an obvious choice, but how to stop h2 pawn for the ZZ finale?

December 1, 2015 19:24

You are right of course. I tested with expensive BSa4!

December 1, 2015 19:55

Reply to seetharaman

Yes, you are right – the version with bSa4 and bPb2->d2 is correct, and I saw it during composing… But in this case the mates are not model (in view of this bSa4). Also, according a principle of economy, I prefer the published version with wPa4.

December 2, 2015 00:17

Zero-position, just as the original:
White Ph7 Pg6 Rh5 Pa3 Qa2 Kg2
Black Pa7 Pg7 Pa6 Ka5 Be5 Pa4 Pd2
Hs#3.5; Zero Move g2 g8; Twin Move g7 h2