No.1684
János Csák & Béla Majoros
(Hungary)


(v) 1684.1 János Csák & Béla Majoros (Hungary)


Warm welcome to Béla!

Original Fairy problems
JF-R2021-22:
01.01.2021 - 31.12.2022


No. 1684 János Csák &
Béla Majoros

Hungary
original - 25.01.2022

white Pg2g3b7 Kh5 black Pd2d3g7 Ke3 Bh8a8
h#2            b) bBa8<-->bBh8            4+6
legality / shadow solution?

Solution: (click to show/hide)


No. 1684.1 János Csák &
Béla Majoros

Hungary
version of No.1684 - 28.02.2022

white Pg2g3f7a7 Kh5 black Pd2d3f4g7 Ke3 Bh8g8
h#2            b) bBg8<-->bBh8            5+7
legality / shadow solution?

Solution: (click to show/hide)


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Andrew Buchanan
January 26, 2022 16:53

I think this is a very good chess joke, with a brilliantly original conceit. In a joke problem, the solver must go beyond the stipulation to imagine what the composers might be getting at. A joke may or may not be fairy. This one probably is because the mechanism you supply. The switch of two identical units is a wonderful idea. However I think that your solution needs to include a story for why it’s Bh8 and not Pg7 which is the shadow. Of course an illegal position may be explained in many ways (including just saying it’s illegal so what) but that doesn’t make it fairy.

Béa Majoros
Béa Majoros
January 30, 2022 16:39

We are very glad you like our chess problem!
Yes, it has a funny but serious foundation.
The idea is, how can it be justified to swap two pieces of the same color and the same quality (such as two black Bishop) in a problem?
We found out: illegal (shadov) piece, illegal (shadov) solutions.
The bPg7 is legal by default, making h8bB illegal.
The question is, if we exchange an illegal and legal piecet (of the same color and the same quality), will our legal-illegal nature also be reversed?

Dmitri Turevski
Dmitri Turevski
January 30, 2022 21:07
Reply to  Béa Majoros

There are definitions for legal move and legal position, but what is exactly a legal piece? And why bPg7 is legal by default?

Béla Majoros
Béla Majoros
February 1, 2022 11:51

In my opinion, the illegal piece could not get to the position with a regular step from its original position. A piece in its original position is not considered illegal.
 
A)  (8/8/8/8/8/8/PPPPPPPP/6B1) g1wB is illegal,
B) (8/8/8/8/8/8/PPPPPPP1/6B1) g1wB is legal but promoted,
C) (8/8/8/8/8/8/P1PPPPP1/6B1) g1wB is legal.

Dmitri Turevski
Dmitri Turevski
February 1, 2022 13:09
Reply to  Béla Majoros

But we do not know that the bPg7 is the original bPg7!
As the given position is not a result of series of legal moves from the starting position, the bPg7 could have come from h7 with an illegal move h7-g7 – that is the point.

shankar ram
shankar ram
February 1, 2022 18:22

I’ll try to summarise the intended logic of the composers:
a) BBh8 is obviously illegal. A solution involving an illegal piece is not allowed. So, 1.g5 bxa8Q 2.Bd4 Qf3# is not allowed and only 1.Ke4 b8=Q 2.Bd5 Qf4# works.
b) The illegal BB is now sitting on a8, therefore according to the above argument, 1.Ke4 b8=Q 2.Bd5 Qf4# is now not allowed. But now 1.g5 bxa8Q 2.Bd4 Qf3# works as the legal BB is now on h8!
So, it’s a combination of retro logic and “something is true because I say it is” logic.
As Andrew indicated, we should consider it as a joke problem and leave it at that. Instead of tying ourselves into knots!

TheoGiakatis
TheoGiakatis
February 2, 2022 01:59
Reply to  shankar ram

I agree that the logic of the composers is what you describe but the problem of the problem is what Sergey mentions in the post just below. In the twin (a) everything is fine and I just point out that “A solution involving an illegal piece is not allowed”. But in the twin (b) there is the obvious contradiction: while the solution 1.Ke4 b8=Q 2.Bd5 Qf4#  is considered unacceptable due to the participation of the illegal bΒa8 (2.Ba8-d5) the other solution 1.g5 bxa8Q 2.Bd4 Qf3# is considered acceptable even though the illegal bΒa8 is still involved in the game as is captured (1…bxa8Q) and allows the promotion of the wQ on a8.

Béla Majoros
Béla Majoros
February 3, 2022 09:18
Reply to  TheoGiakatis

In twin (b), the “obvious” contradiction is the subject of controversy.
We don’t know if we think well.
That’s why the chess problem was made.
In our view, an illegal h8bB (shadov) going to a8 (legal location) will be legal, and a8bB (non-shadow) going to h8 (illegal location) will be legal.

TheoGiakatis
TheoGiakatis
February 5, 2022 15:34
Reply to  Béla Majoros

Dear Bela, if you want to understand my reasoning please follow the next steps.
1) Give the criterion according to which a solution is considered acceptable (real) or unacceptable (shadow).
2) In the twin (b) choose whether bBa8 is legal or illegal.
3) Compare the answers you have given to the two previous steps and decide:
In twin (b) the solution1.Ke4 b8=Q 2.Bd5 Qf4# is acceptable (real) or unacceptable (shadow)?
 In twin (b) the solution 1.g5 bxa8Q 2.Bd4 Qf3# is acceptable (real) or unacceptable (shadow)?
4) Compare the answer you gave in the previous step with your comment “So in both positions, one solution is illegal (‘shadov’) and the other is legal.”
Honestly with good intentions.
Theo

Last edited 3 months ago by TheoGiakatis
TheoGiakatis
TheoGiakatis
February 5, 2022 16:06
Reply to  Béla Majoros

Before you give your answers I will give my own.
1) A solution is considered acceptable (real) if only legal pieces participate in it.
A solution is considered unacceptable (shadow), I will borrow your comment “There is no definition of shadow piece. It behaves like it’s a real piece, but since it is illegal, it does not even exist (shadov), so the solution it would include is also illegal (shadov).”
2) In twin (b) the bBa8 is illegal.
3) In the solution 1.Ke4 b8=Q 2.Bd5 Qf4# the illegal bBa8 obviously participates so it is considered unacceptable (shadow). The solution 1.g5 bxa8Q 2.Bd4 Qf3# includes the illegal bBa8, captured although it does not exist, so this solution is also considered unacceptable (shadow).
4) In twin (b) both solutions are considered unacceptable (shadows) in disagreement with the comment “So in both positions, one solution is illegal (‘shadov ‘) and the other is legal.”



Last edited 3 months ago by TheoGiakatis
Béla Majoros
Béla Majoros
February 10, 2022 18:21
Reply to  TheoGiakatis

Dear Theo, What you have written is completely true, thank you very much for your reply! We are working on the correction. One correction:
6bb/5Pp1/8/7K/8/3pk1P1/3p2P1/8
h#2 (4+6) B: bBg8<->bBh8
A: 1.g5 f8=Q 2.Bd4 Qf3# ? shadow 1.Ke4 f8=Q 2.Bd5 Qf4#
 B: 1.Ke4 f8=Q 2.Bd5 Qf4# ? shadow  1.g5 f8=Q 2.Bd4 Qf3#
Here white’s first steps are the same, it’s a fault, but it doesn’t change the point.
 

TheoGiakatis
TheoGiakatis
February 11, 2022 01:46
Reply to  Béla Majoros

Bela, because I really like your idea, I worked on it and I found the next position:
(a) 7b/6p1/b7/5K2/8/4k3/2QpP3/8
(b) bBh8↔bBa6
Solutions:
(a) 1.g5 Qc6 2.Bd4 Qf3# ? (shadow) but 1.Kd4 Qc7 2.Bc4 Qe5# ! (real)
(b) 1.Kd4 Qc7 2.Bc4 Qe5# ? (shadow) but 1.g5 Qc6 2.Bd4 Qf3# ! (real) 
The promotions have been lost, but in (b) the unwanted capture of the illegal (shadow) bB has also been lost.

Last edited 3 months ago by TheoGiakatis
Béla Majoros
Béla Majoros
February 15, 2022 10:15
Reply to  TheoGiakatis

Theo,
Thank you very much for your efforts!
We found the following solution to fix the error:
János Csák-Béla Majoros
Julia’s Fairies, 2022.
correction
6bb/P4Pp1/8/7K/5p2/3pk1P1/3p2P1/8
h#2 (5+7)
b) bBg8<–>bBh8

a)

1.g5 a8=Q 2.Bd4 Qf3# shadow
1.Ke4 f8=Q 2.Bd5 Qxf4# legal
b)
1.Ke4 f8=Q 2.Bd5 Qxf4# shadow
1.g5 a8=Q 2.Bd4 Qf3# legal
That is what we want to keep.
 
 

 

shankar ram
shankar ram
February 3, 2022 10:30
Reply to  TheoGiakatis

One more assertion: “Illegal pieces cannot move, capture or check, but they can be captured“! 🙂

Last edited 3 months ago by shankar ram
TheoGiakatis
TheoGiakatis
February 4, 2022 01:24
Reply to  shankar ram

I thing “Illegal pieces cannot move, capture or check, but they can be captured, but only on first white move and only on square a8” is more accurate. (Smile)

Last edited 3 months ago by TheoGiakatis
Béla Majoros
Béla Majoros
February 3, 2022 08:54
Reply to  shankar ram

Perfect wording!
We think so too.
Just an add-on:
 Of course, it is arguable that if an illegal piece (h8bB) is swapped with a legal piece of the same color and quality, it was also illegal-legal to swap places (bBa8 and bBh8 bBa8 and bBh8).
We found no answer.
That was the purpose of publishing the chess problem.
 If there will be a decision, it is worth publishing the problem.
If there is no decision, we will have a good time, so everyone can look at our idea …

Béla Majoros
Béla Majoros
February 3, 2022 08:32

Illegal move cannot be assumed!
Illegal moves must be proven!

Andrew Buchanan
Andrew Buchanan
February 9, 2022 05:13

It’s curious to me that while this problem has provoked a lot of friendly discussion, which presumably people enjoyed, there are still very few upvotes to say “thanks for provoking this dialogue”. Most jokes actually follow a known pattern, but this one goes outside the rules of chess logic. It makes no sense and that’s the point.
Rules and conventions are great, and they promote creativity both in the following and in the breaking. I showed this problem in the monthly international zoom call, and the intended solution was found quickly – I think most of the upvotes came from that one audience. It’s not the same idea as Caillaud, where his inspired joke transposition is to avoid the stern e.p. convention.
There are some superbly constructed problems in this site with some brilliant architectures. But I think we deserve to be challenged sometimes by trickster problems like this.

Last edited 3 months ago by Andrew Buchanan
Georgy Evseev
Georgy Evseev
February 9, 2022 13:46

This problem has given me the idea of “Illegal anti-cluster”, where illegal position can be legalized by removal only one specific piece. An attempt for such problem:
W:Ke1 – B:Kb1. White to move. Add white RBSpppp to receive an illegal anti-cluster. (may be cooked in various ways, of course)

Sergey Shumeiko
Sergey Shumeiko
January 26, 2022 23:25

The composition is a joke, but even if we follow your logic, then in gemini b) both solutions are shadow: b) 1.g5 bxa8Q??? It is impossible to take a shadow.
The position can be shortened: 2b4b/3P2p1/8/8/ 4pk1K/4p3/8/8

Béa Majoros
Béa Majoros
January 30, 2022 16:40

In position A), field a8 stands for a8 (legal) bB and field h8 stands for h8 (illegal) bB.
in position B) h8 (illegal) bB is in field a8 and a8 (legal) bB is in field h8.
So in both positions, one solution is illegal (‘shadov’) and the other is legal.
The job offered is remarkable.

Sergey Shumeiko
Sergey Shumeiko
February 2, 2022 20:31
Reply to  Béa Majoros

Béa, there are three versions of this problem:
№1
Retro. Board-orientation problem.
8/8/1K1kp3/8/8/8/1p4P1/b6b (diagram with discolored margins).
Paint the board. h#2 (2+5)
Solution:
b6b/1P4p1/8/8/8/3pk1K1/8/8  1.Ke3-e4 b7-b8=Q 2.Ba8-d5 Qb8-f4 # (?) Illegal position.
7b/6P1/8/2p5/2k5/8/2K3p1/7b 1.Kc4-b4 g7-g8=Q  2.Kb4-a3 Qg8-b3 #. Legal position 
№2
Сoloring problem.
7b/6p1/8/2kb4/4kp2/8/8/8
h#2 b) f4->f3 (0+6)
Which shapes are repainted?
Solution:
7b/6P1/8/2Kb4/4kp2/8/8/8
a)     1.Ke4-e5 g7-g8=Q 2.Bh8-f6 Qg8*d5 #
b)     1.Bd5-e6 g7*h8=Q 2.Be6-f5 Qh8-d4 # 
№3
2b5/3P4/8/6bK/4pk2/4p3/8/8
h#2  2.1.. (2+5)
1.Kf4-f5 d7-d8=Q 2.Bc8-e6 Qd8*g5 #
1.Bg5-f6 d7*c8=Q 2.Bf6-e5 Qc8-g4 #
Simplified Chumakov theme (2). (In one phase, a black piece blocks a square, while another black piece is captured; in another phase, vice versa).

Béla Majoros
Béla Majoros
February 3, 2022 09:42

Sergey, thanks for the trouble!
Our chess problem is a little different.
There are two solutions in each position, only one of which is good.
Of course, only if our idea is confirmed.

Georgy Evseev
Georgy Evseev
January 27, 2022 22:17

There is no definition of shadow piece.
How and when is it nominated?
What are its abilities?

Béla Majoros
Béla Majoros
January 30, 2022 16:41
Reply to  Georgy Evseev

There is no definition of shadow piece.
 It behaves like it’s a real piece, but since it is illegal, it does not even exist (shadov), so the solution it would include is also illegal (shadov).

Georgy Evseev
Georgy Evseev
January 31, 2022 09:21
Reply to  Béla Majoros

Under this assumption either both bishops are shadow and the problem has no solution, or only pg7 is shadow and there is a number of cooks.

Béla Majoros
Béla Majoros
February 1, 2022 11:59
Reply to  Georgy Evseev

Pg7 can’t be a shadow, it’s in its original place! The point is, if we replace an illegal piece with an equally legal piece (by exchange), it will be legal. And the originally legal piece will be illegal because the illegal piece has been put in place.

Bojan Bašić
Bojan Bašić
January 28, 2022 17:00

Here is a different problem with the same twinning mechanism.

Michel Caillaud
Special HM
10th TT Chess Composition Microweb C 30.9.2002

h#2 (8+13)
b) f4 ↔ g5

MIchel.gif
Last edited 3 months ago by Bojan Bašić
Béla Majoros
Béla Majoros
January 30, 2022 16:43
Reply to  Bojan Bašić

Yes, this problem is also great.
But the e5bB is not an illegal piece, but a promoted piece.

Seetharaman
Seetharaman
January 28, 2022 21:12

Sorry. However hard i think i could not understand why the twin is a different position. 🤔

Nikola Predrag
Nikola Predrag
January 28, 2022 22:01
Reply to  Seetharaman

Twin position could be different if the 2 wP’s exchange their initial game array position.

Last edited 3 months ago by Nikola Predrag
Georgy Evseev
Georgy Evseev
January 29, 2022 11:31
Reply to  Seetharaman

See comments to this problem in the award: https://www.jurajlorinc.com/chess/ann10tt.htm

Béla Majoros
Béla Majoros
January 30, 2022 16:45
Reply to  Georgy Evseev

The link cannot be opened!

shankar ram
shankar ram
January 30, 2022 20:34
Reply to  Béla Majoros

Here’s the judge’s comments:
a) 1.Rg3 Sxg3+ 2.Kf3 0-0#
b) 1.gxf3 e.p. Rxg1 2.Rf4 Sg3#

In a) last move cannot be f2-f4 : this would imply sBe5 promoted on e1 or c1 (via d2) and castling would be illegal. So the last move must be g3xf4, and it can be demonstrated that wPf4 is wPg2 and wPg5 is wPf2 (the missing sBc8 must be captured on white square by f3xg4, so the sequence f2xg3xf4 is illegal).

Given a) solution and the twin, in b) wPf4 is wPf2 and wPg5 is wPg2; as already seen, g3xf4 cannot be the last move if wPf4 is wPf2, so this leave only f2-f4 making castling illegal.

Laut F.I.D.E.-Kodex darf der e.p.-Schlag nur dann erfolgen, wenn nachgewiesen werden kann, dass der dazugehörige Bauerndoppelschritt unmittelbar voranging. Dies ist hier zunächst nicht der Fall, weshalb in a) nur jene Lösung ausführbar ist, die „zufällig“ die weiße Rochade enthält. Mit der Ausführung der Rochade steht aber gleichzeitig fest, daß nur Bg3xf4 der letzte Zug gewesen sein kann, weil eine Rücknahme des Doppelschritts den sUW-Le5 von g1 aussperrt und dieser somit nicht hätte entstehen können, ohne das Rochaderecht durch ein Schachgebot zunichte zu machen. Genau hier setzt nun die witzige Zwillingsbildung an: Angesichts der Vertauschung zweier gleichfarbiger Bauern kommt ein Löser, der dem Autor eine vernünftige Absicht zutraut, nicht darum herum, in b) dem Bf4 das andere Ursprungsfeld f2 zu unterstellen, damit den Doppelschritt als einzig möglichen letzten Zug anzunehmen und die e.p.-Lösung zu spielen. Die Notation des Zwillings allein funktioniert dabei ähnlich verbindlich wie das bei Scherzaufgaben häufig verwendete Ausrufungszeichen (“!”) und erweist sich hier als ganz und gar unübliche Methode, voneinander abhängige Sonderzüge zu inszenieren. Natürlich ist dieses Vorgehen nicht ganz seriös geschweige denn exakt und findet daher sicher nicht jedermanns Zustimmung. Dem versuche ich mit dem „speziellen“ Charakter der Auszeichnung Rechnung zu tragen.

According to F.I.D.E. codex the e.p. capture may take place only if it can be proven that the pertinent pawn double step happened in the immediately preceding move. Initially this is not the case here, so in a) only that solution is executable, which “coincidentally” contains the white castling. With the execution of the castling it is certain however that Pg3xf4 must have been the last move, because an inversion of the double step would cut the obtrusive bBe5 from g1 and its genesis would not have been possible without destroying W’s right to castle by a pawn check from the second row. This is exactly where the funny twinning mechanism takes effect: Exposed to the interchange of two equivalent pawns a solver willing to take for granted a reasonable author’s intention can’t help subordinating the other square of origin (f2) to the Pf4 in b), assuming the double step as the only possible last move and playing the e.p. solution. Here the mere notation of the twin is equivalent to the exclamation mark frequently used with joke problems (“!”) and proves a totally uncommon way to demonstrate the potential dependency of special moves. Sure this procedure is not completely respectable let alone accurate and therefore will not be given everyone’s consent. I try to take this into account with the “special” character of the distinction.

Last edited 3 months ago by shankar ram
Béla Majoros
Béla Majoros
February 1, 2022 11:01
Reply to  shankar ram

 
Thank you very much for your detailed reply!

Joost de Heer
Joost de Heer
January 30, 2022 13:29
Reply to  Seetharaman

wPf4 and wPg5 originally came from f2 and g2.
In one twin, wPf4 is the pawn from f2 and wPg5 is the pawn from g2
In the other, wPf4 is the pawn from g2 and wPg5 is the pawn from f2
See the detailed explanation in Shankar Ram’s post for the exact details.

Last edited 3 months ago by Joost de Heer
Béla Majoros
Béla Majoros
February 1, 2022 11:06
Reply to  Joost de Heer

Thanks for your reply!