No.222 (VA&GB)

No.222 
Valerio Agostini &
Gabriele Brunori

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Original Problems, Julia’s Fairies – 2013 (I): January – April

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Please send your original fairy problems to: julia@juliasfairies.com


No.222 – h#2,5 by Valerio Agostini & Gabriele Brunori – An elegant Grimshaw interpretation! (JV)

No.222.1 – h#2 by Valerio Agostini – A new version of No.222 composed after comments! (JV)


Definitions:

Grasshopper(G): Moves along Q-lines over another unit of either color to the square immediately beyond that unit. A capture may be made on arrival, but the hurdle is not affected.

PAO(PA): Moves as Rook, but captures only by hopping over a hurdle to any square beyond.

VAO(VA): Moves as Bishop, but captures only by hopping over a hurdle to any square beyond.

Lion(Li): Moves along Q-lines over another unit of either color to any square beyond that unit. A capture may be made on arrival, but the hurdle is not affected.


No.222 Valerio Agostini & Gabriele Brunori
Italy
original-10.01.2013
 
222-h#2,5-va-gb
h#2,5             2 solutions             (7+5)
Grasshopper e8
Pao h8
Vao e7
 
 
Solutions: (click to show/hide)
 
No.222.1 Valerio Agostini
Version of No.222
original-27.01.2013
 
222.1-h#2-va
h#2             2 solutions             (3+4)
Lion a4
 
 
Solutions: (click to show/hide)
 

7 Responses to No.222 (VA&GB)

  1. Seetharamanseetharaman says:

    Elementary grimshaw shown many times in orthodox helpmates. Three hoppers and no Anti-battery mates !

    • Nikola Predrag says:

      Yes, move bPg3->h3, add bPc5, remove wPs d4&f4 and replace w PA/VA by w R/B.

      • Dmitri TurevskiDmitri Turevski says:

        It’s a pity 2.Bf1 is a pure hideaway, but 2.Re1 is also a selfblock. Perhaps it could be rearranged? Scheme:
        3(Q2)1(Q2)2
        h#2 2.1…
        1.Bxf1 Bc1 2.Rd3 Rc7#
        1.Rxd1 Rf2 2.Bd3 Bg7#
        Can this be turned into h#2.5 with a single grasshopper?

        • Seetharamanseetharaman says:

          Very nice. I think but for cooks h#2.5 would be very nice!

          • Nikola Predrag says:

            Indeed very nice. Still I don’t like the static Grasshoppers, the play is not fairy. The accidental captures help to get rid of superfluous white material, cleverly done but perhaps not fully justifying these fairy pieces. White Knights on a5&c2 would do the job of Grasshoppers and Pawns.

  2. JuliaJulia says:

    Yesterday I’ve published a new version of problem No.222 – No.222.1 by Valerio Agostini. Valerio has sent it to me with some comments in Italian asking to quote him in English if possible. So, the main things from Valerio are: He is very happy that his problems usually produce some discussions, it is very interesting and helpful to him. Valerio sends his gratitude to all commentators! Valerio also writes that he’s agree with Seetharaman only partly, as Grimshaw was shown thousands of times in orthodox composition, but his idea was to show it exactly in fairy chess using anti-battery.
    After comments Valerio has simplified the position, has got a miniature – and now he’s waiting what will be said about such an idea??

    • Nikola Predrag says:

      Battery/antibattery mates are results of the departure/arrival effects of a FRONT piece/hurdle, not of the arrival effect of a REAR piece. That was the first remark by Seetharaman. Therefore 222 is not a genuine fairy problem.

      Nevertheless, the reciprocal functions of thematic white pieces is good but it is lost in the version, which is also not the most economical. First, wLI could be replaced by wG in 1st solution. So wKa5 and wGb4 would be more economical. Second, the power of wQ is not used enough and bPe5 is added. A rough example:

      White Qb8 Kb4 Gd3; Black Re7 Bc6 Kd4; h#2; 2 sol
      1.Bf3 Gg3 2.Re4 Qd6#; 1.Re3 Gf3 2.Be4 Qh8#

      Direct & indirect antibattery with reciprocal mate/guard function in next example, still do not justify enough the presence of Chameleon piece:

      White chameleon Qa4; White LIh7 Kb4; Black Re8 Bh6 Kd3
      h#2.5 2 sol
      1…LIh4 2.Bd2+ Kb3 3.Re3 cQf4=cS#; 1…Kc5 2.Re2 Kd5 3.Be3 cQe4=cS#

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