Fairy chess composition

# No.675 (RJM)

 No.675 René J. Millour(France) Warm welcome to René in Original problems section of JF! Original Problems, Julia’s Fairies – 2014 (III): September – December →Previous ; →Next ; →List 2014(III) Please send your original fairy problems to: julia@juliasfairies.com

No.675 by René J. Millour – 8 Invisible pieces and author’s presentation of the Invisible system in his problem! Welcome to René’s first publication on JF!

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

Definition sent by the author:

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.

 No.675 René J. MillourFranceoriginal – 28.12.2014 Solution: (click to show/hide) White Qa6 Ra8c5 Pe5e6h7 Black Ke8 h==13                                  (6+1) C-8+0 Invisibles! {Animation is not possible for this problem. (JV)} 1.Ke7! I--! 2.Ke8! e7 3.Kd7 Rd5 4.Kc6 Qa3! 5.Kb6 Qh3 6.Ka5! Qc8 7.Ka6! Rd7 8.Kb5 I--! 9.Kc5! I--! 10.Kd5 h8R 11.Ke6 Rg8! 12.Kf7 Re8 13.Kxe8 e6 ==. The stipulated 8 Is are completely identified!The 3 I-moves are also exactly identified!Promotion to B and R+tempo move!bK spectacular long rundlauf!Given position = miniature! (Author) See below detailed explanation by author ->

Show detailed explanation by author: (click to show/hide)

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Paul Rãican
December 29, 2014 14:52

Not at all a trivial problem, as usually with René’s work. A fine gift in Christmas tree. Thank you, René!

Vlaicu Crisan
December 30, 2014 20:16

Chapeau bas, René! Does anyone know which is the record of number of Invisible identities revealed in a chess composition?

Mike Neumeier
March 24, 2015 20:31

Up to now I have seen only four Invisibles problems. Two on JF, this one
No.675, No.739 and the two I just composed (not published). Guessing 8 is a record the world over. Just one or two I’s are hard to compose…no program to solve for you. Awesome problem, René!!

Mike Neumeier
March 25, 2015 02:50

I wonder if somebody has done something like this already, though: 8/8/8/8/7R/2P2P1s/5K2/5R1k, =1, 10 Black Invisibles. Solution (if I can get away with it): 1.Rxh3=. True, not very interesting play, but 10 Invisibles!

Mike Neumeier
March 25, 2015 06:26

Amending that to 11 Black Invisibles (remove Black King). Same solution.

Mike Neumeier
March 25, 2015 06:36

11 is cooked. Returning to the 10 problem. Sorry.

René J. Millour
March 25, 2015 12:06

Mike,

In 8/8/8/8/7R/2P2P1s/5K2/5R1k, I can imagine 8 Ps on c4,5,6,7 f4,5,6,7 / S on g1 / B on h2. In this precise case, 1.Rxh3 is stalemate. But this kind of “solution” is not at all a solution. As stated in the definition: “all must be deduced, not simply asserted”. 1.Rxh3 does not prove that there are Ps on these precise squares (on the board there are perhaps 8 Ps promoted to Q), and does not prove that there is a S on g1 rather than a R, and a B on h2 rather than a S. You cannot simply “construct”, your solution must DEMONSTRATE and ASCERTAIN the stipulated mate/stalemate.

At http://ubp.org.br/wccc2009/composing/Sake_Tourney.pdf (pages 3-5), have a look on my problem, which reveals 10 Invisibles in only 4 single moves.

Best wishes!
René

Mike Neumeier
March 25, 2015 13:31

Thanks René, then I’d have to make it a retro problem, I suppose: “Add 10 Black units, then =1.” Would that be a legitimate way to pose this problem? (Sadly, removing “Invisibles” from the wording.) Julia, my mate-in-2 would be invalidated for Invisibles, as well, yet there again it could be retro “Add two White units for a forced mate-in-2” although I’d want to change e-Rook to bP for economy’s sake. The helpmate non-twin I think asserts through play, so, still Invisibles.

Mike Neumeier
March 25, 2015 13:42

Also, thanks for the record of 10 updpdf. I will read the pdf.

Mike Neumeier
April 3, 2015 14:17

I get a ‘page not found’ error when trying to look at this pdf.

Mike Neumeier
April 4, 2015 12:51