june 8 2010

 

A small anthology of the last European championship (which, if it had only been on the second day, would have been won by our GM Michel). With a diagram of each of the 6 events. You have already suffered on the h#2 a month ago. Note that the #4 of the event had already been submitted for the 2009 German championship. They must have been in need of help, as Guy does not compose #4. I add a few provisions for the summer : a nice #2 from Emperor Marjan, a curious restored #3, a h#5 which may require some time and a superb s#6 (solution on request).

Exclusively for our readers, the future first prize for an excellent magazine in the reflex category. For once, I give the detailed solution, which will not be found in the said magazine, as the editor obviously did not understand the problem !

For fairy positions, see the report of the master-tabellion.

Merry beaches and see you, God willing, at the end of September.

AV

 


for this last lesson of the season, there was place for fairy composition.
The Master had made a selection of which he has the secret and of which here are the different compositions presented.

For the definition of the rules and fairy tales, I refer you to the reference links:
http://christian.poisson.free.fr/problemesis/condus.html
 http://christian.poisson.free.fr/problemesis/pieces.html

For the starting up, a rather old problem on the scale of the fairy composition (1960) .

1 - #2 with Nightrider and Grasshopper

Nil G.G. VAN DIJK
Probleemblad, 1960
1st HM

white Ph7h4g3 Kg8 Rh1 Gg4e6d5b8 Na4c6c8f5 black Bf1d4 Kh3 Qb2 Ph2 Sf3d2 Re1e2
#2(13+9)
🨟g4, e6, d5, b8: Grasshoppers
🨢a4, c6, c8, f5: Nightriders
 
very nice imbroglio of lines and diagonals

A small diversion to Chinese pieces
 

2 - h#3 with chinese pieces

Unto HEINONEN
Probleemblad, 1987

white Pe7 Kg8 black VAa4f4 LEa1 MAc6e6e4 Kd5 Pb6d6c4e3g3 PAc7b4
h#3(2+14)
4 solutions
🨻c7, b4: Pao
🨼a4, f4: Vao
🨓c6, e6, e4: Mao
🨺a1: Leo

  Very pleasant to search with a coffin construction made of different materials
 
Now here is a magnificent triplet.

3 - #2 with chinese piece

Franz PACHL
Jubilé K. Gandev_60
Shakmatna Misl, 2007
1st Prize

white VAg2 WNe2 Kd8 Pf4 Sc1 WPc7h6 black Bf1 BNh2 Kc5 Pa7 Se3 BPa1
h#2(7+6)
🨻c7, h6, a1: Pao
🨼g2: Vao
🨷e2, h2: Nao
b) ♘c1-->c2
c) ♘c1-->c3

 

A superb work with a beautiful purity of mat

To expand the bestiary, we will take a tour of the Chameleons.

4 - h#2 Andernach

Volker GULKE
The Problemist, 2002
1st-2nd Prize

white Pa4e4h4 Kb7 Rg4 black Pd2f6 Ka5 black Chameleon Qg5 Rh1 Sf5 Be1
h#2 Andernach(5+7)
4 solutions
There are 4 black "chameleon" pieces and 4 solutions, it's up to you !

A little break with a work by the great Michel which is much more affordable at the vulgum pecus, and which provides more aesthetic sensations than some other compositions of the dypterian sodomizers.
 

5 - #2 Circe

Michel CAILLAUD
Jubilé C. Lytton-70
The Problemist, 2009
2nd HM

white Ba3 Kh4 Qf8 Pd3g5e5 Sc6d6 Rg3d7 black Pg7h3 Sf7 Rh2 Kf4
#2 Circe(10+5)
3 black corrections, sorry for the lack ! What a talent !

Another compositional genius approaches the Zebra and the equistopper.
Here's the thing :
 

6 - #2 Transmuted Kings with Zèbra and Equistoppers

Hans Peter REHM
(The Problemist, 2005
1st Prize

white QFh6f4 Pd7g7a2 Kc8 Ze4 Sb3 black Pe5b7f3e2d2c3b4 Sa8 Bh2 Ke6
#2 Transmuted Kings(8+10)
🨷e4: Zebra
🩈h6, f4: Equistopper

 

Here below is the definition of the Equistopper from an English site
"The Equistopper (invented by Peter Harris and named by Brian Stephenson at Pitlochry, see The Problemist May 2004, p. 361) is the opposite of the Equihopper; it moves to any square that is exactly halfway towards any other unit of either colour, which must (naturally) be an even number of rows (ranks and files) away orthogonally. In J1, similarly to F1, we see how a Transmuting K, stripped of all powers except that of the attacking ES, is unable to escape the attack, and the backstop ESd2, separated from the Ks by an odd number of rows, cannot relieve it either. In J2 a Fersk”onig is chosen to be stalemated; e2 cannot be wK on account of many cooks, and surprisingly, a3 is the only square for the ES – all the units are placed with precision. J3 shows a neatly-stalemated Royal ES on an Anchor Ring, on which we note that if the ES can go to one square, it can also go to the squares 4 rows distant either way, corresponding to the choice of hurdles for an Equihopper going to the same square."
The preliminary question is : why is d8=N doesn't mate ? (to help you a little : it's a double-check !)

  Un petit détour par le Triton

7 - h#3 with Pao and Triton

Chris FEATHER
The Problemist, 2009
3rd Place

white Kd1 PAd4 TRa4 black Pd2d3e3e5c5 Ke4 Rf3 Sd5 Bh3
h#3(3+9)
b) 🨋a4-->h4
🨋a4: Triton
🨵d4: Pao

  due to the absence of a strike, here is a problem in Functionary Chess

8 - #2 - Functionary Chess

Jim DUCAK
The Problemist, 2005
3rd HM

white Bc8e7 Kh8 Qf3 Pe3 Sb5b4f7 Rc2h4 black Qb8 Pg3g5 Rb3g4 Ba2g1h7 Ke5
#2 Functionaries(10+9)
Something more complicated  

9 - series s#15 - AntiCirce - Madrasi

Unto HEINONEN
The Problemist, 2010
1st Prize

white Pc5e5a7g7d3 Ka5 black Kd4 Pb2b4b6f6 Bd1 Rh1 Sd6
ser-s#15 AntiCirce Madrasi(6+8)

 

h#2 Mars Circe 5 (!) solutions

Frantisek SABOL
15è T.T. Chess Composition
Microweb, 2004 - 1st Prize

white Ba8 Kh4 Qc7 Pe7 Sg8 Rd5 black Bg3 Kh3 Qh2g1 Pf2a2b4g5h5 Se3e1 Rg2a1
h#2 Mars Circe 5sol.(6+13)
5 solutions

 

The Master managed to discreetly put a position

11 - s#11 - AntiCirce with Leo et Moa

Klaus WENDA
Die Schwalbe, 2006
1st Prize

white LEg7 Kf8 Pf7f3c5a2b2c2 Sc3 Bc4f2 Rb1 MOe2 black Pd2 Sh1 Ba1 Kd1
s#11 AntiCirce(13+4)
🨷e2: Moa
🨴g7: Leo

 
Hoping that you will not have abused strong drinks to try to understand the statements and solve the problems, it remains for me to wish you a good holiday.
The holiday homework is in the electronic version of the Mestre to which I wish a well-deserved rest.

Happy reading to all.
Yours sincerely

Le greffier 

Master's clarifications



I can confirm that the first two diagrams are correct. For the remaining part, I haven't seen any anomaly, but I have to trust the master greffier, not keeping any duplicate of the fairy ones. Let's also mention the nice key in problem 5.

Furthermore, his remark on the "sodomizers of diptera" proves how much he tries to make the atmosphere of the course palpable for those who could not attend. An in vivo report, so to speak. Lined with a real and undoubtedly temporary optimism about the "absence of strike".

I must however add a position that I had chosen to end this walk among the madmen with dignity.

Kh8, Qf6, Pg7 & h7 / Kh1, Ra8, Be5. It seems that White only have to give up. Yet they can save themselves. They play in a traditional circle with a very old wooden chessboard and such old pieces...

Have a good rest.

AV

Add a comment

Anti-spam