# No.985 (DK)

 No.985  Diyan Kostadinov (Bulgaria) No.985.1 Diyan Kostadinov (Bulgaria) & Georgy Evseev (Russia) Original Problems, Julia’s Fairies – 2015 (II): July – December    →Previous ; →List 2015(II) Please send your original fairy problems to: julia@juliasfairies.com

No.985 by Diyan Kostadinov – Fairy pin mates with the white KoBul King. (JV)

Definitions:

Phantom Chess: Any unit except a king may move either normally (from its current square) or as though from its Circe rebirth square (game-array squares, of the same colour in the case of pieces, on the file of capture in the case of pawns, and on the promotion square of the file of capture in the case of fairy pieces) if the latter is vacant.

White KoBul King: When a white piece (not a pawn) is captured, the white 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 a white pawn, he becomes a normal King again. Captures are illegal if they result in self-check by the transformed King.

 No.985 Diyan Kostadinov Bulgaria original – 30.12.2015 Solutions: (click to show/hide) white Kb7 Ba5 Sc5 Rh2 black Ka1 Qd3 Pa7f3g3 hs#3            b) Sc5→c6            (4+5) Phantom Chess White KoBul King a) 1.Ba5-d8 {(Kb8??)} g3-g2 2.Kb7-b8 g2-g1=S! 3.Sc5-b3 + Qd3*b3[b8=rS] # { [4.rS~?? selfcheck by Sg1 from b8; 3.Bb6?? selfcheck by Qb3 from d8]} b) wSc5-->c6 1.Sc6-d8 {(Kc8??)} f3-f2 2.Kb7-c8 f2-f1=B! 3.Ba5-c3 + Qd3*c3[c8=rB] # { [4.rB~?? selfcheck by Bf1 from c8; 3.Sc6?? selfcheck by Qc3 from d8] (a) C+ by Popeye 4.73)} Fairy pin mates with pinned white pieces on d8 and non-standard pin of the Royal piece on b8/c8, change of functions between wB/wS, Meredith. (Author) No.985.1 Diyan Kostadinov & Georgy Evseev Bulgaria / Russia version of No.985 – 30.01.2016 Solutions: (click to show/hide) White Kc7 Qf2 Ba6 Sc6 Pb2 Black Ka2 Ra4 Pa7 Pg5 Pe4 Pe3 Ph3 Pg2 hs#3           2 solutions         (5+8) Phantom Chess White KoBul King 1.Sc6-b8 h3-h2 2.Kc7-c8 h2-h1=B 3.Ba6-c4 + Ra4*c4[c8=rB] # { [4.rB~??, 4.Sc6??]} 1.Ba6-c8 Ra4-d4 2.Kc7-b8 g2-g1=S 3.Sc6-b4 + Rd4*b4[b8=rS] # { [4.rS~??, 4.Bb7??] (C+ by Popeye 4.73)}

### 18 Responses to No.985 (DK)

1. Geoff Foster says:

In each solution a white piece moves to d8, becoming pinned after the wK moves to the top rank (the pinned piece on d8 cannot move because of self-check from the bQ via d8). In (a) the wS is captured and the wK becomes a royal S, so the wK moves to b8 and Black promotes to S on a dark square. This “pins” the white royal S because if it makes a S move from b8 then it is automatically in self-check from the bS via b8. In (b) the wB is captured and the wK becomes a royal B, so the wK moves to c8 and Black promotes to B on a light square. This “pins” the white royal B because if it makes a B move from c8 then it is automatically in self-check from the bB via c8. In each solution Black cannot play 3…Kb1?, because of self-check from the wR via a1.

This is a very nice problem, but it is a pity that the B1 moves are a bit dull. Also, it would be nice to have a Phantom-specific move during the solution, if this could somehow be arranged (although I can’t see how).

2. seetharaman says:

Very Nice effects! Black rook instead of queen will also work, but with black queen the block on d8 is more ‘Phantom’ specific.

• Laco Packa says:

Not so, the black Rook (Ra3 or Re3) works only in the second solution.

• seetharaman says:

Sorry for my oversight. You are right Laco Packa!

3. Geoff Foster says:

An interesting point is that the wR could be on h8. The bK cannot move to a2 or b1 because of self-check from the wR via a1, nor can it move to b2 because of self-check from the wB via c1!

Yes, the wR can be on h2 or h8. I put it on h2 because with wRh8 the square b2 in a) is clearly fairy protected by the wB via c1, but in b) this fairy effect is not needed because the Bc3 protect b2 also in normal way. Usually I don’t like such an imbalance in fairy elements, so I leave the wR on h2 from where it protect a2/b2 squares in orthodox way, so both solutions are fairy equal with only b1 square being fairy protected.
Of course with wRh8 there is possible wPg2 to be used instead of bPg3, so in both solutions only one bP play in both solutions (1…fxg2/ 1…f2) which is nice.
So both versions have their pluses and minuses, and it is matter of taste which one is preferable.

5. seetharaman says:

The black pawn promotions are part of the solution, while guarding is part of construction. So I would have preferred the version with same black pawn promoting in both solutions on different squares. But then it is individual preference.

Hm… Probably really the problem is in better form with wRh2 to h8, wPg2, -bPg3 so my initial choice should be changed. But what will follow – is the problem will participate as a 2015 original (because the initial original was published in 2015) or for 2016, because this little change of the scheme practically will be formal published in 2016?
If there is a choise – I prefere to be as 2015 entry, because I have very small number composed problems last year and was hurry to publish it before the end of 2015.
Some opinions?

• Julia says:

If we were talking about the tournament the problem participates in, I’d suggest to leave a version in JF-2015/II too.
But here the question is about a date of publication. I doubt it would be legal to put 31.12.2015 for a version as well. Or?

7. Georgy Evseev says:

I think everyone already knows that I very much dislike twins when they may be avoided.

In original Diyan’s problem there is a colour parity issue which makes twinning seemingly unavoidable. From the other side, to move for the twin a crucial thematic piece is also not a good idea.

And yes, it is really possible to avoid twins, though there are some other issues))).

Georgy, of course when the twins can be avoided this is always better, but in problem this looks impossible – wS can not play on d8 and b3 without twin. Probably if some fairy piece is used (Rose maybe) this can be done somehow, but there the “Phantom reborn” square also will be different, so I am not sure is it possible. I prefere the problem to be only with orthodox pieces.
So I don’t understand your: “And yes, it is really possible to avoid twins”. Am I miss something!?

The question here was another – the original was published on 31.12. with wRh2 and bPg3, but later I change my thoughts (also with help of the other composers’ opinions) that the problem is better to be with wR on h8 and wPg2 instead of bPg3, so the question is – can this version participate in the 2015 tourney instead the published original version or should be for 2016…
Personally I prefere it to be for 2015, because I have very few compositions last year.

9. Nikola Predrag says:

Julia and the relevant judge may decide to which informal tourney the version belongs.
But I think they can’t decide to which WCCI or FA period it belongs.
A possibly antedated publication of the version might be noticed by the WFCC subcommittees and their word is decisive.
(Personally, I would accept your choice in this particular case.)

10. Georgy Evseev says:

Diyan,

As I have written, there are some issues))).

Still the following version is possible, with interference instead of obstruction. It is even with three solutions, though third one is slightly different.

It was very difficult to get this version working, so I was not able to write my comment earlier.

White : Kc7 Qg3 Bh6g4 Sc6 Pa6f4f2
Black : Ka2 Ra4 Pa7e7e4h3g2

hs‡3 (8+7)
Phantom
Roi KoBul blanc

1.Bc8 Rd4 2.Kb8 g1=S 3.Sb4+ Rxb4‡
1.Sb8 h2 2.Kc8 h1=B 3.Bc4+ Rxc4‡
1.Bf8 Ra5 2.Kd8 g1=Q 3.Qd5+ Rxd5‡

It’s nice that you found a way to include also Q-variation, but actually this version is not step ahead. In my opinion the Q-variation is very different – there is no double pin mate. Also the S and B-variations are with reciprocal change of functions between the wS/B which also make the Q-variation more different. I prefere version where all thematic elements are same in all variations even if they are “only” two.

But is it possible to be created 3 solutions proble (with twins or without) where the three white peces B/S/Q cyclically change their function!? This will be fantastic, but probably impossible…

12. Georgy Evseev says:

The third solution may be removed by moving Bh6 to g5.

Concerning cyclic change of functions, I’ll think about it))).

Georgy, I realized that 3 solutions with cycled change of functions between the white S/B/Q is not so promising idea, because the wS and B will be idle piece in two of the solutions. So I tried to improve your suggested version and now one wB is removed and 2 pieces are saved. Please look:

8/p1K5/B1S5/6p1/r3p3/4p2p/kP3Qp1/8
HS#3 2 solutions Phantom, White KoBul King

1.Sb8 h2 2.Kc8 h1B 3.Bc4+ Rxc4(c8=rB)# [4.rB~??, 4.Sc6??]
1.Bc8 Rd4 2.Kb8 g1S 3.Sb4+ Rxb4(b8=rS)# [4.rS~??, 4.Bb7??]

I tried to make in Meredith, but probably will be not possible. What you think? Probably we can publish it as a join composition if this is the best possible version.

• Georgy Evseev says:

Looks OK for me. At least, obvious attempts to improve do not work.

And an additional question, not tied to this problem. Is it possible to computer test KobulKings problem, when king(s) are not in “king” phase in diagram position?