Fairy chess composition
 No.739 Tadashi Wakashima (Japan) No.739.1 Tadashi Wakashima (Japan) & René J. Millour (France) Warm welcome to Tadashi in Original problems section of JF! Original Problems, Julia’s Fairies – 2015 (I): January – June    →Previous ; →Next ; →List 2015(I) Please send your original fairy problems to: julia@juliasfairies.com

No.739 by Tadashi Wakashima – A new interesting example of the Invisible(s) piece, that was a theme of 9th Japanese Sake TT, 2009. (JV)

The Invisibles are not programmed, the problem is C-, and no animation is possible here. I keep the same abbreviation “I” as used in 9th Japanese Sake Tourney. Please don’t confuse with Imitator! (JV)

Definitions:

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): Chinese piece operating along Rook lines: moves as Rook, but captures only by hopping over a hurdle to any square beyond.

 No.739 Tadashi Wakashima Japan original – 04.03.2015 Solutions: (click to show/hide) white   ke1 bd1 rg2 pa2 black ka1 sg1 paa4 h#2           2 solutions         (4+3)  С- 2 White Invisibles PAO a4 {(no animation for Invisibles for now (JV)) } 1st solution: 1.PAh4! Ba4! 2.PAxa4! Ixg1# (2.PAxa4 implies that wI stands somewhere on the b4-g4 line. This capturing switchback also means that B1 move 1.PAh4 is actually a capture: 1.PAxwIh4. Thus, whereabouts of 2 wIs are almost revealed. W2 move 2…Ixg1 is possible only when wIg4=PA and the move is actually 2…PAxg1, which is a checkmate because no more wIs remain on the board. wId4=B and 2…Bxg1 is impossible because Bd4 will give check to bK.) Theme: Seemingly non-capture move and switchback by Pao (Invisible- specific strategy). 2nd solution: 1.Sf3+ Bxf3 2.PAg4! Kg1# (2…Kg1 is actually 2.0-0 with wIh1=R on the board. Another wI is revealed from the moves 2.PAg4! Kg1! because 2…Kg1 is legal only when there is wI on g3. Thus, 2.0-0 is checkmate because no more wIs remain on the board. Try 1.Sf3+ Bxf3 2.PAf4? Kg1#? is refuted by 3.Kb2! [e.g. wIf2=P].) Theme: Anti-selfcheck moves 2.PAg4 Kg1 (Invisible-specific strategy). Castling. No.739 René J. Millour & Tadashi Wakashima France / Japan version of No.739 – 27.08.2015 Solutions: (click to show/hide) white   ke1 rg2 bf2 black ka1 paa7 bg1 h#2           2 solutions         (3+3)  С- 2 White Invisibles PAO a7 {(no animation for Invisibles for now (JV)) } (1) 1.PAh7! Ba7 2.PAxa7! Ixg1# (2) 1.Bxf2+ Rxf2 2.PAf7! Kg1#

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Nikola Predrag
March 5, 2015 12:34

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?

March 5, 2015 13:26

Non-capturing 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.

seetharaman
March 5, 2015 13:35

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#

March 5, 2015 14:03

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)?

seetharaman
March 5, 2015 14:28

Got it! Thanks.

Nikola Predrag
March 5, 2015 16:04

Thanks, the difference between capturing and non-capturing 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 🙂

March 5, 2015 16:55

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.

paul
July 26, 2015 21:55

I found the article here: http://blog.zaq.ne.jp/propara/article/34/

Mike Neumeier
March 20, 2015 14:39

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.

Mike Neumeier
March 20, 2015 14:46

*to be able to effect

Mike Neumeier
March 20, 2015 15:17

Stronger perhaps is adding a check (+) to Black’s first, 1.PAb4+, determining position of invisibl wQ on c3 in the diagram.

Mike Neumeier
March 20, 2015 15:21

Oops. A cannon does not play checks diagonally. Forget adding a check symbol.

Mike Neumeier
March 20, 2015 18:14

Which makes me think of this alternative: 1.PAe4+ Ic3 2.PAb4 Ixb4#. This should prove the white invisibles were e3Q and b2R.

Georgy Evseev
March 21, 2015 09:32

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.

Mike Neumeier
March 21, 2015 22:45

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.

Nikola Predrag
March 20, 2015 15:41

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 W-move ) 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.

Mike Neumeier
March 20, 2015 20:58

Well, at least we know there is a # (mate) in the future. It is a given.

Nikola Predrag
March 20, 2015 21:29

We don’t know that for sure, despite the stipulation, the problem could be incorrect, insoluble.

dupont
March 20, 2015 22:01

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 0-0. 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”).

dupont
March 20, 2015 22:03

Crossed messages Nikola, I answered to Mike.

March 21, 2015 12:35

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”.

Mike Neumeier
March 22, 2015 09:37

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.

Kjell Widlert
March 22, 2015 11:47

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.

Mike Neumeier
March 23, 2015 03:13

I still think there should be a way to describe under the diagram how to say it’s a true two-solution (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 a-file to make its capture. In a way, these invisible problems are proofs that such origins exist.

Georgy Evseev
March 22, 2015 10:53

Why in second solution black cannot play 2.PAxa2 instead of 2.PAg4?

1.Sf3+ Bxf3 2.2.PAxa2 Kg1#

Kjell Widlert
March 22, 2015 11:37

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 0-0 is sure to be a mate.

Mike Neumeier
March 23, 2015 03:30

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.

Mike Neumeier
March 25, 2015 05:54

Just checking with the rules experts, is Black allowed to play 1.Kb2 (self-check in the visible realm), which establishes at least one Invisible is on the second rank between bK and wR?

Georgy Evseev
March 25, 2015 13:21

Not being certified expert, I take the liberty to answer “yes”.

Mike Neumeier
March 25, 2015 14:08

Thanks, Georgy, for your opinion on that. We need all the detection powers that can be used against the Invisibles!

René J. Millour
March 25, 2015 15:48

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 f4-g4 when the S is captured on e4, the hurdle is on g4-f4-e4-d4 when the S is captured on c4. Consequently, 2…Ixg1 is possibly the non-mating 1…Bd4xg1!

2nd solution: 1.Sf3+ Sxf3 2.PAg4! Kg1#

Try: Now if 1.PAb4 Re2 (no I on f2-e2) 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 PA-moves 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.

Mike Neumeier
June 5, 2015 13:55

Hoping author accepts this correction. Looks sound to me.

Nikola Predrag
March 25, 2015 16:49

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.

Mike Neumeier
March 25, 2015 19:21

Thanks, Nikola. Your explanation also confirms my belief that there is a distinction between twins (No. 739 here) and non-twin multi-solution problems (I sent one to Julia) wherein we may use that all solutions are simultaneous. Solving first one can reap I-information available to the next solution line.

Nikola Predrag
March 25, 2015 20:18

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 multi-solutions helpmate 🙂

Kjell Widlert
March 27, 2015 00:30

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 Ra1-a8 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 Ra1-a8 with Invisibles, he proves that there is no Invisible on the a-file.
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).

August 18, 2015 22:12

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 computer-testing. Even simple positions like this one have many pitfalls. My oversight will give us a good lesson, I hope.

Vlaicu Crisan
August 19, 2015 21:10

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.

August 19, 2015 21:52

There is no black Invisible.

René J. Millour
August 20, 2015 07:44

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#.

PAO-switchback 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)!

August 20, 2015 19:12

wBe4 is an inspired setting. Yes, I agree your version is much better than mine. Thanks, René!

Vlaicu Crisan
August 24, 2015 07:54

1.Se2/Sh3 Ixa8 2.Sd4/Sf4/Sg5 Kg1#

Maryan Kerhuel
August 25, 2015 21:31

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!

Maryan Kerhuel
August 25, 2015 22:30

Sorry, it´a bVaog8 shich is needed, not a bBg8,
The Vao guards a2 to prevent the cook
1.Paob4 Bb3 2.Paob1 Ra2

Millour
August 27, 2015 09:00

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 6-piecer. White(3) Ke1 Rg2 Bf2, Black(3) Ka1 Bg1 PAOa7, 2 wIs.
1.PAOh7 Ba7 2.PAOxa7 Ixg1#, 1.Bxf2+ Rxf2 2.PAOf7 Kg1#.

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