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Problem 958: Francesco Simoni - Helpmate
francesco.simoni(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)
958
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)
 
 

Comments  

 
0 #1 Vitaly Medintsev 2018-12-29 18:22
Can be compared to pdb.dieschwalbe.de/.../ with slightly different set of thematic pieces
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+1 #2 Francesco Simoni 2018-12-29 20:11
Quoting Vitaly Medintsev:
Can be compared to pdb.dieschwalbe.de/.../ with slightly different set of thematic pieces
Thanks Vitaly. I already know the problem by Lois and Kapros. In my opinion my problem is more complex, also because the three white thematic pieces are of different types.
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+1 #3 Vitaly Medintsev 2018-12-30 09:31
I agree in respect of complexity, Francesco.
But Lois/Kapros work has better unity of the Black's play: all three pieces which performs direct unpinning block the same square.
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+1 #4 Francesco Simoni 2018-12-30 09:51
Quoting Vitaly Medintsev:
I agree in respect of complexity, Francesco.
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?
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0 #5 Vitaly Medintsev 2018-12-30 17:52
Quote:
1.Ra6 is also guard abandonment of f7.

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.
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0 #6 Francesco Simoni 2018-12-30 18:20
Quoting Vitaly Medintsev:
Quote:
1.Ra6 is also guard abandonment of f7.

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.
Maybe I'm too fussy :-) , but if I composed that problem, the guard abandonment wouldn't like me.
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0 #7 Vitaly Medintsev 2018-12-30 21:31
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Maybe I'm too fussy :-)

I think you are :-) but that's ok
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but if I composed that problem, the guard abandonment wouldn't like me.

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.
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-1 #8 Rodolfo Riva 2018-12-30 23:08
N° 958 is a rough version of the problem quoted at #1: the solution 1.Ba4....2.Re5 is like a hoof and shoe and the modelmates are lost(! ). I don't see guard abandonment after the quoted 1.Ra6 since the rook does not guard f7.
BTW the version at #7 is the best exploitation of the known scheme thanks to the greater economy.
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+1 #9 Francesco Simoni 2018-12-30 23:32
To #7]In the problem by Lois and Kapros the pins of the Sb7 and the Sc2 are redundant. The pin of the Bh4 is good. It forces the move 1.Qb3 to mate 2.Bf4.
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.
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0 #10 Yuri BILOKIN 2019-01-11 10:20
More economical. 4r3/b3Rp2/rB1k1P2/3p1P2/8/q2NK3/8/8 (6+7)
Bristol (bicolor, r-R, impure, 2, 1) – Random addition can be neglected
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0 #11 Francesco Simoni 2019-01-11 12:24
Quoting Yuri BILOKIN:
More economical. 4r3/b3Rp2/rB1k1P2/3p1P2/8/q2NK3/8/8 (6+7)
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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0 #12 Yuri BILOKIN 2019-01-11 19:30
I do not perceive much difference, but I am convinced that its execution is always expensive for the author.
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