Problem 270: Pierre Tritten - Fairy (Anti Circe Clone) |
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. |
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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.
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?
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...
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.
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.
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‡
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?
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]#
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 ;)
prntscr.com/21xqlq
By the way - you can see both conditions (Circe Clone and Anti Circe Clone) together in one problem in my post n.12 above.
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.
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.
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 :)
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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