No.1080 (AO)

Aleksey Oganesjan


Original Problems, Julia’s Fairies – 2016 (I): January – June

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No.1080 Aleksey Oganesjan

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)

16 Responses to No.1080 (AO)

  1. 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:

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

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

  2. Seetharamanseetharaman says:

    Very nice! Surprising that it was not discovered earlier!

  3. Luce Sebastien says:

    Nice ! Difficult to be shorter !

  4. Georgy EvseevGeorgy 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.

  5. S.K.BalasubramanianS 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 EvseevGeorgy 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)

      • Seetharamanseetharaman says:

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

  6. 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.

  7. Georgy EvseevGeorgy 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 EvseevGeorgy Evseev says:

      2. QB/RS – 13 pieces 🙂

      • Georgy EvseevGeorgy Evseev says:

        And 6. BS/QR – also 15 pieces

        • Seetharamanseetharaman says:

          Wow! A complete set…. Please publish!

          • Georgy EvseevGeorgy 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

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