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MatPlus.Net Forum Competitions Illegal Cluster ... Have fun !
 
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(21) Posted by [Saturday, Dec 15, 2012 09:31]

@Mario: Thank you for posting Dittmann's IC from 1976! The solution, as it is shown in PDB (see P1068553) is: +bKb1, +bQb5, +bBe1, +bBh8, +bPd3, +bPf2. The apparently retro-play is: 1.e2-e1=B+ Kc3-d2 2.e4xd4 e.p.+ d2-d4 3.e5-e4+ Kb3xPc3 4.b4xc4 e.p.+ c2-c4. But this leads to an illegal position of bK in SW corner. Without, for example, wPb2, all becomes legal.
 
   
(Read Only)pid=9288
(22) Posted by Mario Richter [Tuesday, Dec 18, 2012 15:51]

I had the same thoughts about the (IMHO unfortunate) mixture of two meanings of "Illegal Cluster" as Juraj.
 QUOTE 
Does anybody by chance know about retro problem with stipulation "IC?" where the position would be provided and it would be solver's task to prove that the position is IC.

I do not know of any, but the Dittmann-IC I gave, might well serve as an example: even if one knows the intended solution, it is not easy to see, why this is indeed an IC - one has to retract a long sequence of moves before the illegality shows up.
Besides this there are some joke problems with the sipulatioon #n, where the solver has to realize that the position is illegal and that removing any of the pieces leads to a legal position which allows to fullfill the original stipulation.
 
   
(Read Only)pid=9320
(23) Posted by Mario Richter [Tuesday, Dec 18, 2012 16:00]

As it turns out, all of the entries for Seetharaman's contest are only IC positions, none of them is an IC problem (using Juraj's terminology).

1. Pietro Pitton (Italy)
-Add 15 men. White: 4 pawns & 1 knight. Black: 8 pawns and 2 rooks.
Add ,, ,
Intended solution:
(= 6+11 )

Alternative setting:
(= 6+11 )


2. Themistoklis Argirakopoulos (Greece)
-Add 15 men: 8 White pawns, 1 WB, 5 black pawns, 1 black bishop.
Add ,,,
Intended solution:
(= 10+7 )

Alternative setting:
(= 10+7 )


3. Mr. Mario Richter, (Germany)
-Add 13 men: 6 white rooks, 1 white bishop, Black bishop and 5 black pawns.
Add ,, ,
Intended solution:
(= 8+7 )

Alternative setting (by Paul Raican):
(= 8+7 )


4. Mr. Vlaicu Crisan (Romania)
- Add 18 pieces: Add 8 white Pawns, 1 white Rook, 1 white Bishop, 7 black Pawns and
1 black Rook
Add ,,,,

Intended solution:
(= 11+9 )

Alternative setting (by Themistoklis Argirakopoulos):
(= 11+9 )

Another alternative setting:
(= 11+9 )


5. Mr. Alexandr Levit (Russia)
-Add 7 white pawns and five white pieces.
Add and five white pieces
Intended solution:
(= 13+1 )

Alternative setting:
(= 13+1 )


6. Mr. Thomas Brand (Germany).
Add 2 bishops and 6 pawns of same colour to an illegal cluster.
Add of same colour to an illegal cluster.
Intended solution:
(= 9+1 )

Alternative setting:
(= 1+9 )

 
   
(Read Only)pid=9321
(24) Posted by Arno Tungler [Tuesday, Dec 18, 2012 16:35]

Probably No. 6 can "survive" with the addition: 2 solutions!
 
   
(Read Only)pid=9322
(25) Posted by Sarah Hornecker [Tuesday, Dec 18, 2012 18:20]; edited by Sarah Hornecker [12-12-18]

I don't think so, since the bP e6 could also be on e7 and it would be an illegal cluster.
EDIT: Oops, see below, Bojan is right.

Also:
(= 9+1 )

The pawns took 15 times, so there was no 16th piece that the bishop f7 would have had to take to give this check. Lots of similar things work if you add black pieces.
 
   
(Read Only)pid=9323
(26) Posted by Bojan Basic [Tuesday, Dec 18, 2012 18:28]

@Siegfried:

With the bPe6 on e7 the position after removing bPd7 is still illegal, because of illegal check from bBc8.
 
 
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MatPlus.Net Forum Competitions Illegal Cluster ... Have fun !