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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)


Definition:

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
Hungary
original – 26.08.2012
 
ser-h=8                                       (8+6)
Anti-Circe
 
 
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

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seetharaman
seetharaman
8 years ago

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!

János Mikitovics
János Mikitovics
8 years ago

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.

seetharaman
seetharaman
8 years ago

I meant an additional bishop promotion at d1 also.

János Mikitovics
János Mikitovics
8 years ago

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

Solutions:

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.

Requirement:

Testability of the solutions by a computer!

Question:

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

seetharaman
seetharaman
8 years ago

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 !!

János Mikitovics
János Mikitovics
8 years ago

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 =

seetharaman
seetharaman
8 years ago

:((