|(1) Posted by Siegfried Hornecker [Monday, Mar 21, 2011 16:18]|
h#2 with AUW
(= 5+4 )
harmonie 105 (iii.2011)
1.Sxf8 exf8S 2.a1R h8Q mate
With the last move (gxf8B+) we have an AUW here. Is the task possible without retro or virtual play?
|(2) Posted by Kostas Prentos [Monday, Mar 21, 2011 19:24]|
A single solution AUW is most probably impossible in a helpmate in 2.
It was done in 2.5 moves for the first time:
Branko Koludrovic, Mat 1984
(= 10+13 )
1…bxc8=D 2.d1=L Dxe3 3.c1=T exd8=S#
and more economically:
Michael McDowell, The Problemist 1996
(= 4+5 )
1…gxf8=L 2.c1=T h8=S 3.Tc8 bxc8=D#
The first time this was done in 3 moves, according to my research:
Branko Koludrovic, Mat 1983, 1st comm.
(= 4+16 )
1.c1=L a8=T 2.Lg5 gxh8=D 3.f1=S Dxh1#
and more economically:
Martin Hoffmann, after Dirk Borst & Harald Haverkorn, Die Schwalbe 1992
(= 4+5 )
1.b1=T h8=S 2.Tb7 axb7 3.c1=L bxc8=D#
|(3) Posted by Frank Richter [Monday, Mar 21, 2011 20:35]|
I think, it is impossible.
Some years ago I spent a lot of time to the theme "single-phase h#2 with 4 promotions". As result I wrote a small article (feenschach 04-06/2003, Nr. 151). Even using promoted force and white king in check the "best" result was the following terrible position with one (unique) cook:
feenschach 151/2003, p. 64
(= 9+14 )
1.h:g1T h:g8S 2.b:c1L d:e8D#
1.K:h7 d:e8D+ 2.Dg7 Dg6#
Who is able to do this better?
At least I could compose two problems showing sDtS (P1074760) resp. tTsD (P1074761) promotions.
And P0521591 shows AUW in 2,5 moves too.
|(4) Posted by Siegfried Hornecker [Monday, Mar 21, 2011 23:17]|
I did not want to quote your full article, or else I would have added this problem as well here. :-)
|(5) Posted by Jacques Rotenberg [Monday, Mar 21, 2011 23:24]|
with a bQg8 you can take the bPh3 off the board
|(6) Posted by seetharaman kalyan [Tuesday, Mar 22, 2011 15:19]|
BQg8 will also remove the cook.
|(7) Posted by Frank Richter [Tuesday, Mar 22, 2011 15:34]|
No, it remains the same.
|(8) Posted by seetharaman kalyan [Tuesday, Mar 22, 2011 18:52]|
Right :( Noticed after posting my comment. Could not remove it !
|(9) Posted by Jacques Rotenberg [Tuesday, Mar 22, 2011 20:04]|
You were right to publish this even with the 2nd sol, the matrix and the construction are crafty
|(10) Posted by Frank Richter [Tuesday, Mar 22, 2011 20:23]|
Thanks. It seems that in difference to the 100$ theme it is impossible to find a correct realization even working with bK in chess and/or promoted pieces.
There are only very few matrixes allowing a "only promotion h#2".
The main difficulty is in my opinion to determine the black moves, especially their order.
|(11) Posted by Kevin Begley [Wednesday, Mar 23, 2011 23:18]; edited by Kevin Begley [11-03-23]|
Besides the question of whether achieving AUW is possible in h#2, w/o retro content, Olaf's fine problem begs 3 questions:
1) can additional retro-content be added, to achieve an AUW in fewer (forward) moves?,
2) can the retro-content be more precise (the last move is, after all, g7x?f8=B+)?, and
3) can anybody combine the AUW with an AUC (where AUC = all units captured/uncaptured)?
I'd most enjoy an answer to the last question.
|(12) Posted by Frank Richter [Thursday, Mar 24, 2011 17:14]|
because I got already some new positions for part 2 of the article, I'd like to ask not to publish here any results. Please send me your diagrams for publication in next harmonie issue, thanks!
|(13) Posted by Hauke Reddmann [Friday, Mar 25, 2011 14:31]|
@Frank: Bh1+Nh4 instead of Qh1?
(Just a first glance idea - I haven't studied where all the
problems will pop up in this construction, but I'm aware that
there will be brickloads :-)
|(14) Posted by Frank Richter [Friday, Mar 25, 2011 16:07]|
You get the AUW-solution, but also with 1.Bg2, 1.Sg2 and 1.Sg6 ...
|(15) Posted by Hauke Reddmann [Wednesday, Mar 30, 2011 10:36]|
I already suspected such. (I could have checked with
Popeye first but h#, even 2 moves, are rather slow.
Yes, my PC is a Zuse :-)
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