Original Problems (pages 61-70)

Original Problems (page 61)

Original fairy problems published during 2012 will participate in the informal tourney JF-2012

The site is mostly about fairies, but h# and s# are also welcomed for publication! Please send your problems to my e-mail: julia@juliasfairies.com

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I’m happy to see 2 Hungarian authors with a joint problem! This time I’ve a pleasure to welcome Tibor Érsek! And I’m thankful to János Mikitovics for his activity!

No.117 – ser-h=8 by Tibor Érsek & János Mikitovics  –  An interesting work of the both Hungarian national grandmasters! (JV)


Anti-Circe: After a capture the capturing piece (Ks included) must immediately be removed to its game array square (necessarily vacant, else the capture is illegal). Captures on the rebirth square are allowed. Game array squares are determined as in Circe. antiCirce Cheylan: As antiCirce Calvet except that captures on the rebirth square are not allowed.

You can сlick on “Solutions” to show or hide the solutions!


No.117 Tibor Érsek  & János Mikitovics
original – 26.08.2012
ser-h=8                                       (8+6)
Solution: (click to show/hide)
Promotions – rundlauf – self-pins (Authors)

The diagrams are made on WinChloe and its Echecs font is used for Logo design

7 Responses to Original Problems (pages 61-70)

  1. Seetharamanseetharaman says:

    Very nice. The rook coming back to a1 and three bishop promotions are quite attractive. The authors obviously noticed the possibility of adding another bishop promotion at d1!! I Forcing the move order will of course be very difficult!

  2. Janos MikitovicsJános Mikitovics says:

    Thank you very much for your appreciative words!
    You’re right! Forcing the move order is very difficult, because Black’s moves are interchangeable many times.

    We think that between the bishop promotions (d1B and b1B) is not an essential qualitative difference.

  3. Janos MikitovicsJános Mikitovics says:

    Of course your idea is very good!
    We didn’t think of the fourth bishop promotion.

    Please look at this schema:

    White Rb8 Rg8 Qc7 Kd3 Pe2
    Black Bc8 Ke8 Bf8 Rh8 Pa2 Pb2 Pd2
    ser-h=5 / AntiCirce


    1.d2-d1=B 2.Bd1-c2 3.Rh8-h1 4.Rh1-a1 5.Bc2-b1 Kd3-c2 =
    1.d2-d1=B 2.Rh8-h1 3.Bd1-c2 4.Rh1-a1 5.Bc2-b1 Kd3-c2 =

    1.Rh8-h1 2.Rh1-a1 3.d2-d1=B 4.Bd1-c2 5.Bc2-b1 Kd3-c2 =
    1.Rh8-h1 2.d2-d1=B 3.Bd1-c2 4.Rh1-a1 5.Bc2-b1 Kd3-c2 =

    Visible, the solutions are possible also without b1B because the d1B blocks the b2-pawn on b1.


    Testability of the solutions by a computer!


    How would it be possible to extort the move b1B and to stymie the interchange of Black’s moves?

  4. Seetharamanseetharaman says:

    My thought was to have 9.d2-d1B and 9….Kc2= extending your solution by another move. (Of course after guarding the seventh row differently) After all 9….Kc2 will stop both b1 and d1 bishops (with WPe2 in place). I could think no way to stop interchange of last two moves, which was what I commented. Back to the experts !!

  5. Janos MikitovicsJános Mikitovics says:

    As illustration of last two moves is an example here with the interchange of moves:

    White Rb8 Rg8 Ra7 Sc6 Kd3 Pe2
    Black Bc8 Ke8 Bf8 Pa2 Pb2 Pd2 Ra1
    The last two moves from ser-h=9 / AntiCirce
    8.b2-b1=B 9.d2-d1=B Kd3-c2 =
    8.d2-d1=B 9.b2-b1=B Kd3-c2 =

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