[Retros] 6-fold pin
rbosch at adelphia.net
Mon Nov 13 20:21:13 EST 2006
In your position I presume a mate would be made by a Rook (or Queen) on e8.
Then both Knights and both Bishops would be able to stop the mate if they
weren't pinned. But the two Rooks wouldn't be able to stop the mate even if
they weren't pinned. So those two pins are gratuitous, while in the 4-pin
game of Nicolas all 4 pins are essential to avoid the mate. So if you
interchanged the Rooks and Knights you might have a more satisfying
Best regards -- Renny Bosch
----- Original Message -----
From: "Christoph Fieberg" <christoph at fieberg.com>
To: <retros at janko.at>
Sent: Sunday, November 12, 2006 2:30 PM
Subject: [Retros] 6-fold pin
> 6-fold pin:
> 8/2B3B1/3n1n2/2RbkbR1/3rpr2/2B1K1B1/8/7Q w - - 0 1
> This to achieve in a SPG would be a real challenge.
>>Date: Sun, 12 Nov 2006 08:23:24 +0100 (CET)
>>From: Nicolas.Dupont at math.univ-lille1.fr
>>To: afretro at yandex.ru, "The Retrograde Analysis Mailing List"
>><retros at janko.at>
>>Andrey asked :
>> > How many pins can be effective in the mate position of an SPG
>> > (regardless
>> > of the mating piece)? Three? Four? Or even more?
>>It's not so complicated to get four...
>>1.Pa2-a4 Ph7-h5 2.Pa4-a5 Ph5-h4 3.Pa5-a6 Ph4-h3
>>4.Pa6xb7 Ph3xg2 5.Ph2-h4 Pa7-a5 6.Ph4-h5 Pa5-a4
>>7.Ph5-h6 Pa4-a3 8.Ph6-h7 Pa3-a2 9.Th1-h6 Pa2xb1=T
>>10.Cg1-h3 Pg2-g1=T 11.Ta1-a3 Tg1-g5 12.Ta3-g3 Ta8-a1
>>13.Ph7xg8=T Cb8-a6 14.Pb7-b8=T Fc8-b7 15.Tb8-c8 Fb7-h1
>>16.Pe2-e4 Pc7-c5 17.Ff1-b5 Pc5-c4 18.Dd1-h5 Tg5-c5
>>19.Pf2-f3 Pe7-e5 20.Th6-e6
>>Obtaining five effective pins needs of course a completely different kind
>>of strategy (the thematic king must probably stand around the center of
>>the board). It should probably exists, although I have strictly no idea of
>>how getting it.
>>Other researches can be made on that 4-pins task. For examples :
>>- fewer moves
>>- no visible promoted piece
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