|(1) Posted by seetharaman kalyan [Thursday, Sep 9, 2010 19:52]|
how many flights can a keymove give?
Can the key-move of a directmate twoer give more than 4 flights? I have seen several where the key gives 4 flights. In one especially nice problem the key gave all the four flights for a star-flight (but it had the drawback that black was in stalemate initially). Any problem beating that?
|(2) Posted by Dmitri Turevski [Thursday, Sep 9, 2010 20:30]|
W. Shinkman, Der Westen 1902
(= 5+1 )
|(3) Posted by Paz Einat [Thursday, Sep 9, 2010 23:17]; edited by Paz Einat [10-09-09]|
WinChloe contains a total of 13 #2 with a key granting 5 flights and one problem with a key granting 6 flights...using a checking key. Since this seems to be the record I give it here:
after Otto WURZBURG & William A. SHINKMAN
Le Journal de Genève 1972
(= 10+5 )
‡2 (10+5) C+
1.Sd4+! R~ 2.c×d8=Q‡
|(4) Posted by Siegfried Hornecker [Friday, Sep 10, 2010 00:35]|
Actually, it is probably impossible to show a key that give six flights without check since there is no possible move that would protect the seven necessary fields for a mating move at once (the king's field and six around it).
However, maybe zugzwang can do even this miracle but I don't see how.
|(5) Posted by seetharaman kalyan [Friday, Sep 10, 2010 14:59]; edited by seetharaman kalyan [10-09-10]|
see my next post
|(6) Posted by seetharaman kalyan [Friday, Sep 10, 2010 15:45]|
Thanks for the problems. Actually I was trying a scheme giving five flights, but seeing the ancient problems quoted, will stop work on it!
Actually a vertical Rook+P battery with a pawn promoting to queen capturing a black piece will guard the seven squares. But while it can be a threat, no variations can be worked on it (unless it is dummy black piece!!). However to force a non-checking key giving the six flights will be impossible.
|(7) Posted by Siegfried Hornecker [Friday, Sep 10, 2010 16:07]|
Yes, I also realised in the meantime that it only works like in the diagram, because the front piece of the battery must also give a flight without taking one, so only a knight at this distance works.
|(8) Posted by Hauke Reddmann [Saturday, Sep 11, 2010 15:54]|
Oh, the dummy wouldn't be the problem - in the Fulpius problem,
replace the bishop d8 by a knight and add pawns on b7 c6 f7. See?
Artificial knight dummy. :-)
|(9) Posted by Milan Velimirović (+) [Saturday, Sep 11, 2010 18:36]|
This seems to be a pretty boring constructional task. A real challenge is to grant as much flights as possible and meet them with different mates. Here 5 seems to be the maximum like in the classic with mate on 6 squares by Ua Tane and a more subtle (quiet key!!) masterpiece by Michael McDowell.
Good Companion 1918
(= 12+6 )
The Problemist 1986
(= 12+4 )
|(10) Posted by seetharaman kalyan [Monday, Oct 25, 2010 21:56]; edited by seetharaman kalyan [10-10-25]|
Not even commendations for this wonderful problem by McDowel?
|(11) Posted by Hauke Reddmann [Wednesday, Oct 27, 2010 12:21]|
Everybody still in shock and awe :-)
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