| No.237, 238 Mario Parrinello (Italy) No.237.1 joint with Paul Rãican (Romania) |
Original Problems, Julia’s Fairies – 2013 (I): January – April →Previous ; →Next ; →List 2013(I) Please send your original fairy problems to: julia@juliasfairies.com |
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)
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.
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.
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No.237 Mario Parrinello
Italy
original-23.01.2013
 hs#4 2 solutions (7+9)
Locust f3 Rook-Locust c3 Bishop-Locust d3 Siren b5 Tritons d4, c5 Solution: (click to show/hide)
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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)
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No.238 Mario Parrinello
Italy
original-23.01.2013
 h#2 b) CGh4↔Gh8 (4+5+1)
Neutral Contra Grasshopper d4 Contra Grasshopper h4 Grasshopper h8 Solution: (click to show/hide)
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Nr. 237, wTRd4 could be a Pawn, I think:
[/img]
White Sa7 LRc6 LBc5 Kc3 Pc2 Pd2 Rd1
Black Be8 Kd7 SIe7 Pd6 TRe6 Pb5 Lc4 Bc1
[img
hs#4
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!
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.
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 #