# No.1080 (AO)

 No.1080  Aleksey Oganesjan (Russia) Original Problems, Julia’s Fairies – 2016 (I): January – June    →Previous ; →Next ; →List 2016(I) Please send your original fairy problems to: julia@juliasfairies.com

 No.1080 Aleksey Oganesjan Russia original – 26.05.2016 Dedicated to Unto Heinonen Solution: (click to show/hide) white Kh1 Se8 Bc5 Pb7h7 black Kf7 Sg8e2 Pa2f2e6 hs#2.5                                         (5+6) 1...f2-f1=B 2.b7-b8=R a2-a1=Q 3.h7-h8=S + Qa1*h8 # { (C+ by Popeye 4.75)} For the first time in hs# without fairy pieces and conditions: AUW is realized in single solution form in minimal quantity of moves – 2,5! (hs#2 with AUW in single solution form is impossible even theoretically). A scheme of the problem is similar to longer hs#3 and hs#4 by Unto Heinonen to which I dedicate this problem. (Author)

### 16 Responses to No.1080 (AO)

1. Bartel Erich says:
2. Bartel Erich says:

in my AUW-collection there are more than 100 problems of this genre.
Really this one by Alexej is the first with only 2.5 moves.
Ten years ago I published some articles about this genre:

Die Allumwandlung mit 2 weißen und 2 schwarzen Bauern im Hilfszwingmatt:
Problemkiste (163) II 2006

Die Allumwandlung mit 2 weißen und 2 schwarzen Bauern im Hilfszwingmatt – eine Nachlese:
Problemkiste (164) IV 2006:

Die Allumwandlung mit 2 weißen und 2 schwarzen Bauern im Hilfszwingmatt – 2.Nachlese:
Problemkiste (165) VI 2006:

http://pdb.dieschwalbe.de/search.jsp?expression=PROBID=%27P1188804%27

this was the first without fairy condition, but 3 moves, and

http://pdb.dieschwalbe.de/search.jsp?expression=PROBID=%27P1194066%27

was the only one with 2.5 moves, but with Circe condition.

3. seetharaman says:

Very nice! Surprising that it was not discovered earlier!

4. Luce Sebastien says:

Nice ! Difficult to be shorter !

5. Georgy Evseev says:

Well, I think this was not discovered ealrlier because noone was specifically searching for it.

It is also unfortunate that author has not tried to find the most lightweight position. It has taken me only about half an hour to find 9 pieces version.

6. S K Balasubramanian says:

I don’t know whether Georgy meant one of the following versions with 9 pieces or a different one:

Version 1: HS#2.5
Wh: Kb1 Rb2 Pb7 Pd7 Bd5
Bl: Kb8 Sc8 Pd2 Pe2
(1…d1=B 2.d8=Q e1=R 3.b*c8=S+ Bb3 #)

Version 2: HS#2.5
Wh: Kc1 Rc2 Pa7 Pc7
Bl: Kc8 Sb8 Pe2 Pf2
add wPe7 or bRd8
(1…e1=B 2.a8=Q f1=R 3.c*b8=S+ Bc3 #)

I wonder whether it is possible with less than 9 pieces!

• Georgy Evseev says:

In fact, I have two 9 pieces versions.

The second of them is exactly your second version (up to shift and mirroring – I had black pp a2 and b2).

The first of them has opposite promotions (White BL and black QS)

• seetharaman says:

Surely this is a different scheme and not an economic setting of Aleksey’s idea. But, definitely a nice problem !

7. Aleksey Oganesjan says:

Dear chess friends!
Yes, apparently your 9-pieces-versions are the most ecomonic in this case! Excellent!
But they has another scheme (compare to my No 1080) – so, of course, you can publish one of these version as your independent problem.

8. Georgy Evseev says:

A short table for hs#2,5 AUW without fairy pieces and conditions. Promotions are shown Black/White.

1. QR/BS – 15 pieces, (difficult, try yourself!)
2. QB/RS – difficult!
3. QS/RB – 9 pieces, Kc1 Rc2 Pa7c7e7 – Kc8 Sb8 Pe2f2
4. RB/QS – 9 pieces, Kh1 Rb7h6 Pd7h7 – Kg7 Rg6 Pb2f2
5. RS/QB – 9 pieces, Kh1 Rh7 Pb7f7 – Kc7 Bb6 Pd6b2f2
6. BS/QR – difficult!

• Georgy Evseev says:

2. QB/RS – 13 pieces 🙂

• Georgy Evseev says:

And 6. BS/QR – also 15 pieces

• seetharaman says:

Wow! A complete set…. Please publish!

• Georgy Evseev says:

Well, for me it is much more technical exercises than the problems, especially when position is not optimal. (That is caused my initial reply.)

Here is remaining constructions (one has to find alternative idea how to define black moves’ order, and then avoid any arising hindrances):

QR/BS – Ke2 Qd1 Rb8h7 Bc1 Pa7e7 – Kd7 Ra2 Bb2 Pb6e6g4c2d2

QB/RS – Kc3 Qa4 Rf7b1 Sc1 Pa7c7c4 – Kb7 Rg7 Bh8 Pb2c2

BS/QR – Kc2 Rg7d1 Sc1 Pb7e7 – Kd7 Qh1 Re4b1 Bf5 Pc6d2e2

9. Bartel Erich says:
10. Bartel Erich says: