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Problem 270: Pierre Tritten - Fairy (Anti Circe Clone)

tritten(02.11.2013) Another Anti Circe Clone original by Pierre Tritten from France. This time with captureless play.





270


1.Rb2 Rf4 2.Bc2 Bb4#,  1.Kc4 Bb4 2.Rb5 Rf4#

1.Bc2 Ra1 2.Kb2 Bd4#,  1.Bf3 Rf2 2.Ba8 Rc2#


Control of BK rebirth squares in order to avoid capture of white piece: on the first three solutions, BK rebirth square is guarded by white officer, on the last one BK rebirth square is occupied by BB.

Interchange of white moves between the two firsts solutions

Model mates

Diagonal-orthogonal echo

(Author)

Anti Circe Clone: A capturing unit changes into a piece of the same kind as the captured unit (without changing colour) and is then reborn on the game array square of the new piece. If the rebirth square is occupied, the capture is not allowed.  

 

Comments  

 
0 #1 Kenneth Solja 2013-11-03 17:59
Sorry to say, but first I thought that this is some kind of joke because the solutions works without any fairy condition. In addition I feel that the fairy condition is here to drop the cooks away.
However I know that Pierre can do much, much better problems than this one.
This is from beginning to the end the opinion of one person.
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0 #2 Nikola Predrag 2013-11-03 18:35
Well Kenneth, this problem provokes the question of what is true/false and complex/banal in chess composition.
Actually, I had learned something from it but I suppressed my wish to discuss it here.
Disscussing about some particular problem could look as an attack on the author and his work, although I never have such intention. I am just curious about some general principles which are present or missing in some particular problem.

What is the use or utilization of some fairy element? And more important, how the play is motivated by some fairy element?
So, which answers to these questions gives this problem?
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+1 #3 Pierre Tritten 2013-11-03 20:18
Well, first of all, it's always good to have comments on problems, even negative ones ;-)
I constructed this one using the condition as a starting point and not to avoid cooks!
I just wanted to show final positions where BK cannot capture because cannot rebirth. Maybe it's not enough to make it good...
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0 #4 Nikola Predrag 2013-11-04 02:26
In the fighting genres, the play is naturally motivated by the potential opponent's attacks and defences. There, even for the simpler ideas, a composer must find/invent some motivations for the play both of White and Black. In cooperative genres, a correct problem could be achieved by simple guards and blocks. It is very easy and that is why so many correct helpmates are beeing published nowadays. But a good helpmate should actually present much more complex motivations than a direct problem and that's not so easy.
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0 #5 Nikola Predrag 2013-11-04 02:27
Many themes which require rather imaginative motivations in direct problems, can be achieved in helpmates with trivial tools. But without the resistance of the other side, the essence of such a theme is lost (e.g. there can't be "Umnov theme" in h#, only a very simple ornamental "Umnov effect"). Motivations for the play is what brings the complexity and beauty to a problem, not the superficial ornaments. Someone can say "I like the simple ornaments", that's OK but presenting the simple ornaments is almost surely not original. And we publish problems as "Originals".
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0 #6 Nikola Predrag 2013-11-04 02:28
If some fairy element is present in a problem, it must be utilized in some way. A good problem should use it for the main idea/theme. A simple utilization could be fun but it is actually not enough for a serious composition. A specific fairy element should give a specific motivation for a (specific) play (what is a big difference if a flight is guarded by a Camel instead by a Knight - motivation is the same).
In this particular problem, in the solutions 2.Bb4# and 2.Rf4#, there is no use of, nor motivation by the fairy condition. The only fairy effect is completely accidental since wR attacks f8 all the time. A try 1.Bc2 Re1? 2.Kd2 Bb4+ 3.Kxe1(brRh8)! indicates that the correct echo-play 1...Ra1 2.Kb2 Bd4# is motivated by the attack of h8 by wB. But here it also looks accidental. However, this is an interesting idea for a more convincing realization.
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0 #7 Nikola Predrag 2013-11-04 02:30
The fourth solution 2.Rc2#, utilizes the condition for the guard of b2,d2,(c2) because wRc2 can't be captured.
Now, this "immunity" of wR looks as a specific effect, if we look from the orthodox point of view. But looking from the "AntiCirce-Clone" point of view, it is just an ordinary mating move. Actually, all apparently specific features become perfectly normal and ordinary when we are inside some fairy condition. So what could make a content of a fairy problem? Exactly the same as in orthodox problems-the motivation for the play. To achieve the immunity of wRc2, a8 must be blocked. And not by one of the black Rooks, only bB can do it properly. This 4th solution is not very complex but is interesting enough (a couple of such solutions with possibly enriched white play could make a decent content).
So, 2.Rc2# utilizes the condition after the play which is motivated by the necessary (antidualistic) block of a8.
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+1 #8 Pierre Tritten 2013-11-04 08:30
Kenneth's comment gave me a hint of what could be a better rendering of the idea:

White : Kb3 Rf1 Ba1
Black : Ke3 Bb7 Sg2 Se1 Pe4

h‡2 (3+5)
b)anticirce clone

a)
1.Sd3 Re1+ 2.Kd2 Bc3‡
b)
1.Ba6 Rf2 2.Bd3 Bd4‡
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+1 #9 Nikola Predrag 2013-11-04 14:59
I would say that it's an interesting idea of antidualistic guard/non-guard of wB on c3/d4.
Well, I would say it if I would understand how in b), 2...Bd4 can be a check while e1 is occupied. And if somehow this is indeed a checkmate, why bS can't block d3 in b)? Actually, now I have noticed that I did not understand the published original. Why one of bR's can't block a8 to enable 2...Rc2#? I wrongly supposed that 3.Ra8-e8 would block the rebirth square which should actually be e1.
I can't use Popeye for checking Anticirce Clone (or I don't know how), so what do I missunderstand in the definition?
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+1 #10 Kenneth Solja 2013-11-04 18:54
Well, Nikola has explained a lot. I'll say things straight like I feel.Sometimes this hurt more, but what I'm trying to say here is that Pierre is much better composer than what this problem shows.
If I had composed this problem I would not have published it at all. There is too weak link between the problem and the fairy condition, only one solution needs the fairy condition.
Pierre's another problem (above here) is intresting idea really, but this also should been with 2+2 solutions.
I can't check AntiCirceClone problems but I have made(not composed)some small problems like this (after Tritten):

White : Kb3 Rf1 Ba1
Black : Ke3 Sg2 Se1 Pe4
h#2, 2 solutions, circeclone
1. Sd3 Te1+ 2. Kd2 Lxc3[+bLf8]#
1. Sc2 Kxc3[+bbc7] 2. Sxa1[+wSg1] Txf3[+bTa8]#
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0 #11 Diyan Kostadinov 2013-11-04 20:04
Hi all - here are two links where you can see the versions by Pierre and Kenneth with diagrams and solution:

For the version of Pierre Tritten (with AntiCirceClone)click here:
prntscr.com/21x41u

For the version of Kenneth Solja (with CirceClone) click here:
prntscr.com/21x85a

By the way - Kenneth, the 1st solution in your version is also orthodox ;)
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0 #12 Diyan Kostadinov 2013-11-04 20:41
Actually it is possible both versions to be combined and now there are 4 solutions: 2x2 AntiCirceClone and CirceClone. Look here:
prntscr.com/21xqlq
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+1 #13 Pierre Tritten 2013-11-04 21:04
@Nikola: Bd4 is a check because it doesn't transform to King when capturing. Its rebirth square is c1, so BS can't block on d3. Same for my first problem where WR rebirth square is a1 and not e1.
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0 #14 Nikola Predrag 2013-11-04 22:14
I have to beg for the explanation. How can in "Anticirce clone" bK be checked on any square by any piece when square e1 is occupied? The piece which would capture bK should be reborn on e1 as wK and the occupied rebirth square forbids the capture which means that there's no check.
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0 #15 Diyan Kostadinov 2013-11-04 22:52
Nikola, I approving the Pierre's comment too late and probably you not saw his answer... In Circe Clone and Anti Circe Clone it is not allowed the pieces to transform into Kings after the capture, so they returns on its original squares without transformations in such cases.
By the way - you can see both conditions (Circe Clone and Anti Circe Clone) together in one problem in my post n.12 above.
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0 #16 Nikola Predrag 2013-11-05 00:06
Thanks Pierre and Diyan! The explanation of what is check and what is not should be included in the definition. And if by any chance now I've understood the condition, you have still not explained it well.
It seems that R/B after capturing the King, would BE TRANSFORMED-into the Royal R/B! Otherwise, the attack of the rebirth square would be irrelevant.

I don't think that this transformation follows the logic of other Anticirce clone transformations but it adds some dynamics to the condition (rebirth squares depend on the checking piece). This could be nicely used for the thematic play.
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0 #17 Nikola Predrag 2013-11-05 01:07
Of course, I have not understood anything!
If a capture of bK would result with a selfcheck to a white Royal piece, it would also be irrelevant, because bK would be hypothetically captured first.

What does in Pierre's original prevent 1.Rh8 Rf2 2.Ra8 2.Rc2# ?
I give up.
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0 #18 Nikola Predrag 2013-11-05 01:30
I have to apologize for my stupidity. The potential rebirth squares would be blocked in 3rd black move, after the check.

As a bit of excuse for my mistakes and as a warning to all, I'll mention my recent discussions about "Beidmatt" and "Gegenmatt". You can see how this affects a poor human brain :)
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+1 #19 Kenneth Solja 2013-11-05 17:29
Well done Diyan! I like your AntiCirceClone+CirceClone problem ..
I know that in my problem the first solution was orthodox, I would say it filling. I'm still trying to find idea how to get analogical solutions with CirceClone because I don't know how to check AntiCirceClone
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0 #20 Diyan Kostadinov 2013-11-07 14:45
Nikola, don't give up ;) If you still don't understand Anti Circe Clone and Circe Clone conditions please look problem n.273 - there I include more full definitions.
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