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MatPlus.Net Forum Fairies Twins vs. multiple solutions
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|(1) Posted by Bojan Basic [Friday, Dec 31, 2021 15:03]|
Twins vs. multiple solutions
I am wondering which answer you would prefer to the question from the title in the case of this problem.
4th HM, Fairy Section Tzuica 2021
(= 8+8 )
locust f2, moarider-lion d8
1...Qg3-e3 2.Lf2*e3-d4 + Rf5-f6 3.Ld4*f6-g7 Se6-f4 =
1...Rf5-f7 2.Lf2*f7-f8 + Se6-c5 3.Lf8*c5-b4 Qg3-f4 =
1...Se6-d4 2.Lf2*d4-c5 + Qg3-c7 3.Lc5*c7-c8 Rf5-f4 =
Cyclic Zilahi, where each of the three thematic black pieces (which are the only black officers on the board) is once captured in the first White’s move, once in the second one, and once delivers stalemate. The final moves are Pelle moves on the same square and along the same pinning line.
I also had the following version.
(= 8+7 )
locust a8, moarider-lion e2
b) La8→c8; c) La8→e8
a) 1...Qb7-a6 2.La8*a6-a5 + Rc5-c3 3.La5*c3-d2 Sd4-c6 =
b) 1...Rc5-c4 2.Lc8*c4-c3 + Sd4-f3 3.Lc3*f3-g3 Qb7-c6 =
c) 1...Sd4-b5 2.Le8*b5-a4 + Qb7*e4 3.La4*e4-f4 Rc5-c6 =
With such a homogenous twinning, I was wondering whether the price of an additional plugging pawn is justified to reach the twinless version. But my impression was that the multi-solutions form here really puts an icing on the cake, and thus that was the version I settled on.
Actually, this is not the end of the story, since after the tourney deadline I found the following version.
(= 7+7 )
locust h8, moarider-lion b6
1...Sd5-f6 2.Lh8*f6-e5 + Re4-e2 3.Le5*e2-e1 Bg3-f4 =
1...Re4-e8 2.Lh8*e8-d8 + Bg3-b8 3.Ld8*b8-a8 Sd5-f4 =
1...Bg3-e5 2.Lh8*e5-d4 + Sd5*b4 3.Ld4*b4-a4 Re4-f4 =
If I had found this version on time, it would certainly be my preferred version. (Though there is a tiny drawback in it: namely, in the second solution, wL is already immobilized before the final move, while this is not the case in the first and the third solution; but I would say that this is nitpicking.) But assuming that this version does not exist (or if somebody considers it inferior to the other ones), which one would you choose from the previous two?
All the best greetings for the year 2022, and let us hope that it does not turn out as 2020-too! :D
|(2) Posted by Nikola Predrag [Friday, Dec 31, 2021 19:10]|
Zero-position is usually avoided and even forbidden in most tourneys. However, if the twinning mechanism is well structured with a clear principle, it may bring in its own merits, better than some lousy standard twinning.
An example of a 'zero-switch' position:
original version of 3rd Prize Benkö-90 JT 2018/19
(= 3+13 )
HAPPY NEW YEAR!
a) bPd6-->b2: 1.Rf4 Kb1!(Kd1?) 2.Kf5 Re6 3.Qg4 R1e5#
b) bPd7-->a2: 1.Rf5 Kd1!(Kc1?) 2.Kf6 Re7 3.Qg5 R1e6#
c) bSc8-->b2: 1.Rf6 Kc1!(Kb1?) 2.Kf7 Re8 3.Qg6 R1e7#
Cycle of right & wrong wK's flights for self-unpin through the phases.
Together with the flight disabled by the respective twin, there's a full cycle of changed effects on the 3 squares (illegal??/right!/wrong?)
|(3) Posted by Kostas Prentos [Friday, Dec 31, 2021 22:08]|
Here are two of my problems, in which I was happy with twins instead of two solutions:
Sake, Batumi 2013
(= 4+8 )
h#2 b) Kd7->c7 C+
a) 1.Sb4 [1.Sd4?] Se2 2.Rh6 Sd4#
b) 1.Sd4 [1.Sb4?] Bf7 2.Bg2 Rd3#
Here the twin was out of necessity, as I could not find a sound version with 2 solutions without computer help (at the time of the tourney, face-to-face was not programmed, yet). Note the dual avoidance that separates the two solutions, in relation to the position of the wK.
I intentionally chose to have twins instead of two solutions in the next example:
5th FIDE World Cup 2017
(= 5+9 )
hs#3,5 b) Qe6->e4 C+
a) 1...Qh6 2.Rg6 f6 3.Sa6+ Kb6 4.Rxf6+ Qxf6#
b) 1...Qh1 2.Bg2 f3 3.Sd7+ Kd5 4.Bxf3+ Qxf3#
It was possible to have two solutions with bQe6->c6, but it would limit the content shown by some tries in the published version:
1...Qf5?, 1...Qe7?, 1...Qe2?, that aim to control f6 or f3, in order to answer the check with 4...Qxf6# or 4...Qxf3#, respectively. They all fail because 4.Rxf6+ is answered by 4...Qe6! (or Qd6!) and 4.Bxf3+ by 4...Qe4!. For the same reason, the solution of each twin is unique and cannot be interchanged: In twin (a), 2.Bg2? does not work; likewise, 2.Rg6? in twin (b). The Queen must hide behind the white piece.
The judge (Petko Petkov) wrote the following relevant comment in his award: According to the author`s comments, it is better to have twins than two solutions (with a black Qc6). His main motives are some tries such as in position "A" the moves 1...Qf5?, 1...Qe7?, 1...Qe2? etc. Of course, this view seems somewhat debatable. On one hand, with Qc6 there will be 2 solutions or twins with eventual material economy. But on the other hand, with Qc6 the mobility of the white figures is too limited and that is not good. Therefore, I accept the version with twins better on this basis!
So, my view is that the two solutions are preferable to twins, but sometimes it makes sense to have twins instead of multiple solutions. This is clearly debatable. Many composers would go for multiple solutions even at the cost of a heavier diagram, or even unused white material in each solution. It would be interesting to see what other composers think about my choice of twins in diagram B above.
Happy New Year to all! Stay safe!
|(4) Posted by seetharaman kalyan [Saturday, Jan 1, 2022 08:24]|
I think twins are preferable if it is not obvious why the other solution won't work after the twin. This type of subtle twinning adds interest to the problem. Otherwise multi- solution.
|(5) Posted by Hauke Reddmann [Saturday, Jan 1, 2022 10:43]|
The default argument is that twins are easier to construct
than multiple solutions.
|(6) Posted by Vlaicu Crisan [Saturday, Jan 1, 2022 21:04]|
To twin or not to twin, that's the question. The answer depends on the context.
The default choice is not so obvious, as one can imagine. It is not always about the classic dilemma material economy versus unified presentation.
In this particular case, I think the first version is better. Why?
Just because shifting the position of the thematic Locust in the twins slightly suggests "cheating" or making it look simple.
Having to build all three solutions from the same position reveals a more elaborate conception.
But there are also many cases when twinning should be preferred instead of multi-solutions.
In an article published in The Problemist Supplement a long time ago :), you can find 6+1 examples where authors preferred the twins to multi-solutions.
|(7) Posted by Joost de Heer [Sunday, Jan 2, 2022 09:37]|
Suppose you have a h# with solution distribution 0.2.1.1 and 126.96.36.199 (so a double set play, and two solutions after black's first move). What form is better? Diagram with set play and a solution that forks after the first black move, or a twin where both parts have the same solution distribution with two white starting moves?
|(8) Posted by Dmitri Turevski [Sunday, Jan 2, 2022 13:37]|
I'd say that generally "0.2.1.1 + 188.8.131.52" is superior to twins, for the reason that Vlaicu mentioned: it's more challenging to achieve and it has more thematic integrity (a single thematic piece that makes a move versus two unconnected thematic pieces replaced in twins).
|(9) Posted by Georgy Evseev [Sunday, Jan 2, 2022 21:01]|
Currently, there is a tendency to forget that twins are _different_ positions.
Historically, the idea of "technical" twins was to keep intact the rule "one position - one solution".
One may look at helpmate section of AF14-44 - it is blatantly obvious there.
So, from my point of view, one should only use "technical" twins if there is the definite advantage of some sort over the multiple solutions presentation.
(Twins are technical, if twinning itself does not contribute to the contents or artistic value of problem.)
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MatPlus.Net Forum Fairies Twins vs. multiple solutions