Julia's Fairies

Original Problems (119)

Original Problems (page 119)

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

Go to →List of Problems ; Page 118 ;  Page 120

No.184 – h==1.5 by Diyan Kostadinov – I’m quoting the author here: “It is not something special, but can attract your attention to the new type of Circe presented by the Romanian composers some months ago!” (JV)


Memory Circe (sent by the problem’s author): Any capture triggers the rebirth of the piece previously captured, on the same square where its capture occurred (if it is not occupied), regardless the color or the type of the unit just captured. The reborn piece resumes its normal course. If the square where the capture occurred is occupied, then the captured unit is annihilated. A unit can exist in memory before the actual play starts. (the condition is not programmed yet)

KoBul Kings  – When a piece (not a pawn) of his own side is captured, a King transforms into a Royal piece of the same type as the captured one. When the King is in the form of any Royal piece and there is a capture of one of the pawns of his own side, he becomes a normal King again. Сaptures are illegal if their result is self-cheсk because of the transformation of the Кings according to KoBul rules. Castling is allowed only if the KoBul King is on his initial square in the form of a normal King and if he has not already moved; however he may already have been transformed. In the case of capture by a King in AntiCirce he is reborn on his initial square and may castle. If the capture is by a King which is in the form of some Royal piece, he is reborn on the initial square of that piece.

Madrasi Rex Inclusiv: A piece (King included) is paralysed if it is threatened by a piece of the same kind.

No.184 Diyan Kostadinov
h==1.5                               (3+0+1)  C-
b/c/d) Pf3>f5/e6/f6
White KoBul King
Madrasi Rex Inclusive
Memory Circe
1 piece in memory:
a/b/c/d) nQ/nR/nB/nS
Neutral King
Solutions: (click to show/hide)

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

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Diyan Kostadinov
December 12, 2012 12:52

The Memory Circe is very interesting new fairy condition, but unfortunately till to the moment it can not be computer check. I believe that the problem is OK.

Dominique Forlot
Dominique Forlot
December 16, 2012 16:06

In Memory Circe The square of circe for the piece in memory must be unique… and we find here several squares to reborn for each piece.

For exemple
a) 1…d8Q+ 2.nK:d8(nK=nrQ,+nQh2) nrQd6 = = is solution to !

something like this position below for exemple try to make unique every solution! What do you think about this proposition :

Diyan Kostadinov (v)
Fen : 8/1nQBKPR1/2p3rp/N2P2P1/3rR1b1/6d1/8/8 neutral= King

White : Qc7 Rg7 Re4 Bd7 Nb7 Pf7 Pd5 Pg5
Black : Qg3 Rg6 Rd4 Bg4 Na5 Pc6 Ph6
Neutral : Ke7

H==1,5 (8+6+1)
KoBul Kings
Madrasi Rex Inclusiv
Circe Memory

Solutions :
{ F = bishop in memory ( Bd8 unique )
D = Queen in mémory ( Qe6 unique )
T = Rook in memory ( Td6 unique )
C = Knigth in memory ( Cf4 unique )
1…f8=Q+ 2.Kn×f8 (+Qe6) Qf6 ==
1…f8=R 2.Kn×f8 (+Rd6) Rf6 ==
1…f8=B+ 2.Kn×f8 (+Bd8) Be7 ==
1…f8=N 2.Kn×f8 (+Nf4) Ne6 ==

KoBul Kings is realy a very fairy new idea.


Dominique Forlot

Diyan Kostadinov
December 16, 2012 16:30

Oh, Domenique, I just upload part of your note here (please see the comment down), where I explain why I think that my problem is correct.
About your version – I still not look it, but I saw that you noted “KoBul Kings” – there should be “White Neutral King” as a condition, because of specific fairy motives: please look the explanations published under the diagram of the problem above.
Thanks again for your interest!

Diyan Kostadinov
December 16, 2012 16:23

Today I received a message in facebook about this problem n184. Here is part of it:

“… I’m sorry but I think there is unfortunately a demolition for this problem.
Here is what asks me question:
In Memory Circe The square of circe for the piece in memory must be unique (last capture) and here we have to be afraid of multiple solutions.
For example: in solution a) 2.nK:d8(nK=nrQ,+nQf4) nrQd6 is correct but 2.nK:d8(nK=nrQ,+nQa6) nrQd6 so! as well as for c6,e6,g6,h6,ç5,e5,b4,a3,g3 and h2 (All these squares lead d6 and note than d7,b6,f6,a5,g5 aren’t allowed because of the “auto-check” if the queen reborn on these squares )
a) 1…d8Q+ 2.nK:d8(nK=nrQ,+nQf4) nrQd6 =
1…d8Q+ 2.nK:d8(nK=nrQ,+nQ[f4,a6,c6,e6,g6,h6,ç5,e5,b4,a3,g3,h2] nrQd6 = = are solutions
And the pawn f3 is superfluous, no? …”

Actually the pawn f3 is important because in the stipulation “= =” both sides (white and black) should be stalemated. So if the nQ in Memory reborn on another square instead of f4 in the final position the wPf3 have possible move, so this is not double stalemate.
In all solutions both white pawns (f3 and d5) should be blocked and both pieces (nRoyal and neutral piece in memory) should be paralised. So I still believe that the problem is correct.

Dominique Forlot
Dominique Forlot
December 16, 2012 21:40

Hi, sorry Diyan, I had missed the condition was in the form of twins : I missed the line: b/c/d) Pf3>f5/e6/f6 :o(
And your problem is absolutely correct!

I had interpreted the problem as having four solutions (without twin)

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