|(1) Posted by Neal Turner [Tuesday, Apr 30, 2019 18:55]|
I see we have the latest Wenigsteiner-Jahrespreis - (http://www.wenigsteiner.de/).
In the pre-selection we have 32 problems - 31 of which ask the solver to put maximum 4 pieces on the board, and one asks him to put 8(!) pieces on the board.
Guess which came in 1st place.
But you say: The board is empty at the start with a piece count (0+0) - Wenigsteiner!
But by that logic every problem is a Wenigsteiner - if the solver is using a board and pieces he starts with an empty board (Wenigsteiner!) sets up the position and then tries to solve it.
The only difference is that that normally the composer is telling you where to put the pieces, while in René's problem you have to work it out for yourself.
Now I'm sure this is an amazing problem (if only I could understand it) - but is it a Wenigsteiner?
|(2) Posted by Eric Huber [Tuesday, Apr 30, 2019 21:59]|
The rules of the Wenigsteiner-Jahrespreis tourney can be read here, in German: http://www.wenigsteiner.de/Rules.htm
We may not agree with the provided definition, but the rules are quite clear. The problems that take part in this tourney must have at most 4 pieces in the diagram position. The diagram counts, the stipulation doesn't.
"Zugelassen sind Wenigsteiner aller Art, orthodox und heterodox, mit beliebigen Steinen und Zusatzbedingungen, auch Studien, Retroanalyse-Stücke, Schachmathematisches usw., sofern die Maximalzahl von 4 Steinen in der Diagrammstellung nicht überschritten ist."
|(3) Posted by Joost de Heer [Wednesday, May 1, 2019 10:05]|
Is a Sentinelles composition with 2 kings in which 16 pawns are created a Wenigsteiner?
Is a Proca retractor with 2 kings in which 10 pieces are uncaptured a Wenigsteiner?
(And nitpick: #9 and #17 are also 'add more than 4 pieces' compositions)
|(4) Posted by seetharaman kalyan [Wednesday, May 1, 2019 15:05]|
They all qualify but I would not select them for awards.
|(5) Posted by Neal Turner [Wednesday, May 1, 2019 15:30]|
From Google Translate: '...provided that the maximum number of 4 stones in the diagram position is not exceeded.'
My point is that the empty board isn't a 'position'.
With the problem in question the solver is being asked to produce a position - with 8 'stones'.
|(6) Posted by Vlaicu Crisan [Wednesday, May 1, 2019 19:12]|
This stipulation is also encountered in Illegal Clusters. Solving this kind of problems is generally like detective stories. Creating (...and solving!) good problems of this type requires out-of-the-box thinking. Some judges like these challenges, others don't - but that's another story.
I think empty boards can be initial positions even for traditional stipulations. Look for instance at PROBID=P1205982 from PDB. :-)
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