No.237,238 (MP)

No.237, 238
Mario Parrinello (Italy)
joint with Paul Rãican (Romania)


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

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No.237 – hs#4 by Mario Parrinello – An interesting festival of Locust– family! (JV)

No.237.1 – hs#4 by Mario Parrinello & Paul Rãican – An improved version to No.237 after Paul‘s comments! (JV)

No.238 – h#2 by Mario Parrinello – A nice combination between Grasshopper and its “Contra-brother”! (JV)


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.

Contra Grasshopper (CG): As G, but in reverse: the hurdle must be adjacent to the CG, which may land anywhere on the line beyond.

Locust (L): moves on Q-lines but only by capturing an enemy unit, arriving on the square immediately beyond that unit, which must be vacant.

Rook-Locust (LR): Moves like Locust, but on Rook-lines only.

Bishop-Locust (LB): Moves like Locust, but on Bishop-lines only.

Siren SI / Triton TR / Nereid ND (“marine pieces”): Move as Q/R/B respectively, but capture by hopping over and removing an adverse unit, landing on the next (necessarily empty) square, i.e. they capture like locusts.

No.237 Mario Parrinello
hs#4             2 solutions            (7+9)
Locust f3
Rook-Locust c3
Bishop-Locust d3
Siren b5
Tritons d4, c5
Solution: (click to show/hide)
No.237.1 Mario Parrinello & Paul Rãican
Italy / Romania
Version of No.237, 23.02.2013
hs#4             2 solutions            (6+11)
Rook-Locust c6
Bishop-Locust c5
Locust c4
Siren e7
Triton e6
Solution: (click to show/hide)
No.238 Mario Parrinello
h#2        b) CGh4↔Gh8       (4+5+1)
Neutral Contra Grasshopper d4
Contra Grasshopper h4
Grasshopper h8
Solution: (click to show/hide)

3 Responses to No.237,238 (MP)

  1. Paul Raican says:

    Nr. 237, wTRd4 could be a Pawn, I think:
    White Sa7 LRc6 LBc5 Kc3 Pc2 Pd2 Rd1
    Black Be8 Kd7 SIe7 Pd6 TRe6 Pb5 Lc4 Bc1
    1.d2-d4! Lc4xd4-e4 2.Rd1-d4 SIe7-g7 3.LBc5xd6-e7 + TRe6-d6 4.LBe7xd6-c5 + Le4xd4-c4 #
    Try: 1.d3? Lxd3-e2 2.Rd3 TRe3 3.LRxd6-e6+ SId6 4.TRxd6-c6+ Lxd3-c4#? but 5.LBxe3-f2!

    • Mario Parrinello says:

      The idea was to show a rather difficult combination: creation both of two different white batteries (Rf4/LBb5 or Rg4/LRc5) and two different black ecto-batteries (SIb8/Lof5 or TRg5/Loh5), switchbacks both by the front piece of the white batteries (4.LRxc4-c3 + and 4.LBxc4-d3 +) and of the ecto-batteries (4… Lxg4-f3 # and 4… Lxf4-f3 #) and, as result, the Zilahi.
      I appreciate Paul’s attempt to replace a fairy piece with an orthodox one (namely wTRd4 which I am not very happy of, but I was not able to improve the construction) but his version shows only one solution which in my opinion seems not to be a step forward of this idea.
      Of course I’d like to know Paul’s and readers’ opinion.

  2. Paul Raican says:

    The second version has two solutions and all four switchbacks:
    White Sa7 LRc6 LBc5 Kc3 Pd2 Rd1
    Black Be8 Kd7 SIe7 Pd6 TRe6 Pb5 Lc4 Pa3 Pf3 Pc2 Rf2
    Stipulation Hs#4, 2 solutions, C+
    1.d2-d4 Lc4*d4-e4 2.Rd1-d4 SIe7-g7 3.LBc5*d6-e7 + TRe6-d6 4.LBe7*d6-c5 + Le4*d4-c4 #
    1.d2-d3 Lc4*d3-e2 2.Rd1-d3 TRe6-e3 3.LRc6*d6-e6 + SIe7-d6 4.LRe6*d6-c6 + Le2*d3-c4 #

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