Problem 958: Francesco Simoni - Helpmate |
(29.12.2018) A nice helpmate from Francesco Simoni. Three solution helpmates have a charm of their own. The unpin selfblock play is especially interesting in 958.
958. Francesco Simoni (Italy)
H#2 3 Sols. (6+9)
1.Qa1 Sg4 2.Qe5 Sxh6#
1.Ba4 Bf3 2.Re5 Bg4# 1.Re8 Rg7 2.Re5 Rxf7# Three direct white unpins in B1. Three different selfblocks on the same square in B2. (Author)
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Comments
But Lois/Kapros work has better unity of the Black's play: all three pieces which performs direct unpinning block the same square.
Yes I agree. At the same time I thank you for the opportunity to deepen the analysis of the two problems.
In the problem by Lois and Kapros:
1.Ra6 is also guard abandonment of f7.
Strategy is linear; the bQ and the bB don't have other choices to block the square e6.
In 958:
the bB has a try: 1.Bb5?
the bQ has tries: 1.Qb2? 1.Qe2?
This is debatable since f7 is not guarded in the initial position.
Any way, I think it's not important for thematic content. In my view, the main aim of departure effect of 1.Ra6 is direct unpinning.
I think you are but that's ok
Quote:
Since the main goal of Black's strategy is to self-block the same square near the BK with three different pieces in two moves, I believe that the departure effect in B1 (whatever it is) is of secondary importance. Moreover, for this strategy the pins of white pieces does not need at all!. For instanse, take a look at the following Meredith showing another departure effect in B1 - unguard of mating square pdb.dieschwalbe.de/.../
As you can see, all three thematic white pieces have full mobility in the initial position which is more valuable in respect of solution difficulty.
BTW the version at #7 is the best exploitation of the known scheme thanks to the greater economy.
In my problem the pin of the Rg6 is unfortunately redundant, while the pins of the Bd5 and the Sf2 are necessary to force their unpins in B1. All this, however, may appear a bit accidental and therefore not entirely satisfactory. For this reason I did not reported dual avoidances in the solution of my problem.
Bristol (bicolor, r-R, impure, 2, 1) – Random addition can be neglected
I don't like 1.Rc8 Re8, that is the wR moves in the square left by the bR. In my opinion this weakens the unpin.
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