No.739 No.739.1 Теплое приветствие Tadashi в разделе Оригинальных задач JF! 
Оригинальные задачи, Julia’s Fairies – 2015 (I): январь – июнь →Предыдущая ; →Следующая ; →Список 2015(I) Пожалуйста, присылайте свои оригинальные сказочные задачи на: julia@juliasfairies.com 
No.739 – Tadashi Wakashima – Новый интересный пример фигуры Invisible(s), которая была темой 9^{ого} Japanese Sake TT, 2009. (JV)
Условие Invisibles не запрограммировано, задача C, и анимация здесь невозможна. Я сохраняю аббревиатуру “I” как она использовалась в 9^{ом} Japanese Sake Tourney. Пожалуйста, не путайте с Имитатором! (JV)
Определения:
Invisibles(I): Beside ordinary units, a given number of Invisible pieces are on the board, their identity and whereabouts being not known. Moves are played so as to determine partially or totally the identity and location of the Is, in order to finally achieve with certainty the stipulated aim. An I may play a quiet move denoted 1.I– (a priori we don’t know which I moved to what square), or may capture a visible unit (1.Ixd5) to show its arrival square. An I remains I until both its identity and whereabouts are determined, in this case it is revealed and becomes an ordinary piece. The notation expresses what we know, and only what we know, all must be deduced, not simply asserted through the notation.
PAO (PA): ПАО. Китайская фигура, играющая по линиям ладьи: без взятия ходит как ладья, а забирает только перепрыгиванием через препятствие на любое последующее поле.
No.739 Tadashi Wakashima 
Solutions: (click to show/hide) 
white ke1 bd1 rg2 pa2
black ka1 sg1 paa4
h#2 2 solutions (4+3) С 



No.739 René J. Millour & 
Solutions: (click to show/hide) 
white ke1 rg2 bf2
black ka1 paa7 bg1
h#2 2 solutions (3+3) С 

Looks very nice and tricky for the inexperienced like me.
A stupid question, why not:
1.PAxa2 Ic5 2.PA(x)h2 Ic1# ?
Black moves reveal the initial positions of both wIs and white moves reveal wQ?
Noncapturing move by an Invisible is denoted simply as I– and its destination square cannot be specified without proof.
Therefore, you cannot write 1…Ic5 and the sequence 1.PAxa2 I– 2.Pa(x)h2 I– does not necessarily end in checkmate, of course.
It is of course a possible solution. The one way to rule it out is of course to say that the Invisible can only the kind of pieces on board.
Even then, I don’t see why there cant be a invisible WR on a3 with W1 move being 1…Rc3 and mate 2….Rc1# Seems black can also play 2.Sxh3 Rc1#
Again, you cannot specify an Invisible without proof.
How can you prove wIa3 is some specified piece when you cannot see its movement (it’s Invisible, after all)?
Got it! Thanks.
Thanks, the difference between capturing and noncapturing moves by I is indeed stated in the definition.
However, drawing a solver’s attention to the important consequence of that difference would be a help to the solvers who deal with the Invisibles for the first time.
For instance:
“…An I may play a quiet move denoted 1.I– (a priori we don’t know which I moved to what square), or may show its arrival square by capturing a visible unit (1.Ixd5)…”
But reading the rules carefully might also be a part of solving 🙂
Those who are not familiar with Invisibles are advised to consult the result of the 9th Japanese Sake Tourney:
http://ubp.org.br/wccc2009/composing/Sake_Tourney.pdf
I hope this will help.
I found the article here: http://blog.zaq.ne.jp/propara/article/34/
What about this solution?: 1.PAb4 Ib3 2.PAxb2 Ixb2# implying a white Q must have played to b3 yo be able to rffect a mate next move. Any of white P, R, S or PA were sitting on b2 in the diagram, else black’s second move would not be possible.
*to be able to effect
Stronger perhaps is adding a check (+) to Black’s first, 1.PAb4+, determining position of invisibl wQ on c3 in the diagram.
Oops. A cannon does not play checks diagonally. Forget adding a check symbol.
Which makes me think of this alternative: 1.PAe4+ Ic3 2.PAb4 Ixb4#. This should prove the white invisibles were e3Q and b2R.
AFAIU, there is no declared check or mate with invisibles.
So this your solution is interpreted as
1.Pae4 ? 2.Pab4 ?xb4 and there is no way to identify any piece from it.
Right. I understand what can be written (deduced) now. I accidentally responded outside of this thread yesterday, explaining I had accepted the first solution by the author as valid, but had a doubt lingering regarding his castling solution. Since then, I have composed a direct #2, no fairy pieces, with two white invisibles, which I will submit to Julia for possible publication. I find the genre interesting, with close ties to retro analysis, a field in which I never composed, just solved. More grist for the mill at least, if not a valid one.
The definition of Invisibles says what is initially known and what might become known from the play.
It requires some “discipline” to comprehend the definition fully, especially about what IS NOT revealed by the play.
So, you can’t say: “It’s a mating move by the Invisible, therefore the Invisible must be a Queen.”
First you must prove the Invisible was a Queen and therefore it’s a mate.
Even if there were a white Ib3 (after 1st Wmove ) in your “solution”, it could have been wR. So, 2…Ib3xb2 would be possible without mate.
But already 1…Ib3 is not allowed by the rules which allow merely 1…I~ (with unknown arrival square) or, 1…Ixb3. Only a capturing move reveals the arrival of I.
Well, at least we know there is a # (mate) in the future. It is a given.
We don’t know that for sure, despite the stipulation, the problem could be incorrect, insoluble.
Hum… For sure the composer is asking the solver to develop a sequence such that invisibles will be revealed in such a way that a checkmate is possible. But you can’t use the fact that a checkmate exists to gain information on the invisibles.
This is very clear if you look at the last move in the second solution, which is 00. This move forces an invisible to be wRh1 and the other to stand on g3 in order to make castling legal.
Thus the checkmate follows “a posteriori”, you can’t use this fact “a priori”. In a certain sense such a problem is not purely helped, as the white invisibles are defending against the goal (they are saying “I refuse to checkmate black unless you are able to force me to do it”).
Crossed messages Nikola, I answered to Mike.
Well done, I really like it.
With such problems I always wonder about soundness as rules are difficult enough and it is not easy to avoid tricky cooks. Fortunately we use approach “innocent until proven guilty”.
Wait a minute! Something is wrong about the way the problem is posed. In a normal, visible h#2 with two solutions, all units begin in their same locations in the diagram in both solution lines. This invisible problem is a twin, each part featuring different starting locations for the invisibles. Should not the diagram be underscored with ‘A) Diagram B) I?–>? & I?–>?’? I feel like I cannot accept the castling solution B), otherwise.
No, I don’t think the problem is posed incorrectly. The thing is that invisibles are very strange animals, much like electrons (and other particles) in quantum physics. They are not situated in any particular place until they have been detected there – until then, they are everywhere they possibly could be. (Forgive me, you quantum physicists out there, this most likely was not a scientifically correct description.) So it is quite normal for one solution to detect an invisible in one place, and for another solution to detect it in a completely different place.
I still think there should be a way to describe under the diagram how to say it’s a true twosolution (single demonstrable origin, or at least some neighborhood of it) as opposed to a twin (parts not of the same origin) to be revealed. I say neighborhood because I see moves like Ixa1 not necessarily revealing exact original square, such as in a solution where an invisible line unit could have materialized from any number of ranks further up the afile to make its capture. In a way, these invisible problems are proofs that such origins exist.
Why in second solution black cannot play 2.PAxa2 instead of 2.PAg4?
1.Sf3+ Bxf3 2.2.PAxa2 Kg1#
I think you’re right!
2.PAxa2 seems to be a dual/cook. The move implies that the second white invisible is standing on a3 (as opposed to g3 in the author’s solution), which implies that no white invisible is standing on the first row, which implies that 00 is sure to be a mate.
Congratulations, Georgy! You found a different origin (a3) for an Invisible! Author maybe will fix it for us, removing one of the two castling solutions, which unfortunately look too much alike to keep this as a triplet problem.
Just checking with the rules experts, is Black allowed to play 1.Kb2 (selfcheck in the visible realm), which establishes at least one Invisible is on the second rank between bK and wR?
Not being certified expert, I take the liberty to answer “yes”.
Thanks, Georgy, for your opinion on that. We need all the detection powers that can be used against the Invisibles!
I think the not so aesthetic Pa2 prevents 1.PAb4 Bb3 2.PAb1 (wI captured on b1) Ra2#. It is perhaps possible to remove it.
Suggestion to author: White(3) Ke1 Rg2 Sd2, Black(3) Kc1 Sg1 PAc4, 2 white Invisibles.
1st solution: 1.PAh4 Se4! 2.PAxe4 Ixg1#
Try : Not 1…Sc4? Here too, the switchback 2.PAxc4 is present, but as a trap! Instead of f4g4 when the S is captured on e4, the hurdle is on g4f4e4d4 when the S is captured on c4. Consequently, 2…Ixg1 is possibly the nonmating 1…Bd4xg1!
2nd solution: 1.Sf3+ Sxf3 2.PAg4! Kg1#
Try: Now if 1.PAb4 Re2 (no I on f2e2) 2.PAb1 (blocking b1) Sb3+ (this move proves that, thanks to wI on d1, the wK is not in check from PAb1) …but no mate: Black answers 3.Kb2+!!, because Rg2 does not guard b2 with the second wI on c2!
It is also a question of taste. The PAmoves are here perhaps less spectacular because shorter. However, I think all the effects mentioned in 739 are present and, instead of “wP, 7 pieces and no corresponding try by blocking b1”, here “no wP, only 6 pieces and a specific try by blocking b1”!
The Invisibles are a not so easy condition. We are never sure to have seen everything. Perhaps you will find flaws in what I say.
Hoping author accepts this correction. Looks sound to me.
The initial set of data does not give one exact position and even the final position might not be exactly revealed although the stipulation is fulfilled.
The set of data changes after each move, continuously determining the set of POSSIBLE positions in a certain moment.
A player may make any move which is legal in at least one of the possible positions in that moment.
(Moving of an Invisible piece is defined by the rules of Invisibles.)
The solution changes the set of data to create the final set of possible positions, such that in EACH ONE of them, the stipulation is fulfilled.
1st solution of no.739 ends in a unique position, although it’s not known which piece was captured on h4.
The end of 2nd solution doesn’t determine the nature of Ig3.
Georgy’s solution doesn’t determine the nature of Ia3.
But in any of the possible final positions, Black is mated.
So, if my “explanation” is correct, 1.Kb2 is legal.
Thanks, Nikola. Your explanation also confirms my belief that there is a distinction between twins (No. 739 here) and nontwin multisolution problems (I sent one to Julia) wherein we may use that all solutions are simultaneous. Solving first one can reap Iinformation available to the next solution line.
You’re welcome Mike, I’ve actually tried to clarify the condition to myself, in the first place.
It’s up to the more experienced to say how much correct and applicable is my “approach”.
However, No.739 has 2 (+1 cook) solutions. There are no twins, there’s only one initial set of data for all phases.
The data change differently in different phases, just as in any ordinary multisolutions helpmate 🙂
As far as I understand, Nikola’s explanation is correct (and 1.Kb2 really IS a legal move).
There’s an interesting distinction between Invisibles and Kriegspiel, where the exact whereabouts (and sometimes nature) of some pieces are also unknown. If White plays Ra1a8 in Kriegspiel, but Black has one of his “invisible” pieces on a4, that move is illegal and White will have to try some other move (there is an Umpire to inform him of the illegality). But if Whte plays the same move Ra1a8 with Invisibles, he proves that there is no Invisible on the afile.
So in Kriegspiel, the “invisible” pieces are situated somewhere definite but unknown (and the solution must cater for every possible placement), whereas in Invisibles, they can be in any possible (and we can influence what places are possible).
All of this is reminiscent of the difference between partial retrograde analysis (Retro Variations) and Retro Strategy. In the former, we must take care of every possible set of move rights (withiin some limitations), and in the latter, we can influence the move rights during the play (typically, White castles and thereby proves that Black may not castle).
I apologize for this very late reply. Somehow I missed the thread after some point and didn’t know what was going on in the discussion….
Yes, as Georgy pointed out, this problem of mine is cooked. Thanks, Georgy, and René too, for your valuable comment and version.
My correction is as follows.
White: Ke1 Rg2 VAOe2 (3)
Black: Ka1 Sg1 PAOb6 (3)
H#2 2 solutions
2 White Invisibles
1st solution: 1.PAOh6! VAOa6! 2.PAOxa6! Ixg1#
(Switchback is missing here, but with PAOa6 instead b6 there is no mate because wI=Bb6 and 2…Bxg1!)
2nd solution: 1.Sxe2 Rxe2 2.PAOe6! Kg1#
I hope it is sound this time. Any “human” testing is welcome.
For me, one of the peculiar attractions of Invisibles is that you must deal with them without the aid of computertesting. Even simple positions like this one have many pitfalls. My oversight will give us a good lesson, I hope.
Why does the second solution work? I think the black Invisible could be a black Vao e3, so perhaps 3.I~~ would parry the check.
There is no black Invisible.
Remark – Simply moving like a B in 1st solution, simply captured in 2nd solution, isn’t VAOe2 a questionable new fairy piece?
Another suggestion – White(3) Ke1 Rg2 Be4, Black(3) Ka1 Sg1 PAOa8, 2 white Is.
1.PAOh8 Ba8 2.PAOxa8 Ixg1#, 1.Sf3+ Bxf3 2.PAOg8 Kg1#.
PAOswitchback on a8.
Tries – 1.PAOa7 Rb2 2.PAOa2 Rb1+ but 3.Kxb1 (wId3)!, 1.PAOb8 Bd5 2.PAOb1 Ra2+ but 3.Kxa2 (wIc4)!, 1.Sf3+ Bxf3 2.PAOf8 Kg1+ but 3.Kb2 (wIf2)!
wBe4 is an inspired setting. Yes, I agree your version is much better than mine. Thanks, René!
1.Se2/Sh3 Ixa8 2.Sd4/Sf4/Sg5 Kg1#
Attempt: why not restart from the initial position of Tadashi
and avoid the cook 2.Ra2# with a bBg8 instead of wPa2
the latter causing many problems (I hope this the only pur pose of wPa2…) Of course it´s ugly
but the problem deserves to be saved!
Sorry, it´a bVaog8 shich is needed, not a bBg8,
The Vao guards a2 to prevent the cook
1.Paob4 Bb3 2.Paob1 Ra2
In “August 25, 2015 at 22:30”, for me VAOg8, of no use in the solution, is not acceptable (see remark “August 20, 2015 at 07:44”).
With only 6 pieces, the suggestion “March 25, 2015 at 15:48” is still valid as far as I know.
Here another 6piecer. White(3) Ke1 Rg2 Bf2, Black(3) Ka1 Bg1 PAOa7, 2 wIs.
1.PAOh7 Ba7 2.PAOxa7 Ixg1#, 1.Bxf2+ Rxf2 2.PAOf7 Kg1#.