|(1) Posted by Hauke Reddmann [Sunday, May 13, 2012 18:26]|
Yet another free-for-all construction task
Ka5 Bc7 Pb7 - Ka7
White can #1 (now guess how :-) but if he chooses not to
(what do I know, maybe he is into S&M :-), he will
need 25 moves starting with b8S.
Improve "25". Also positions with less than 25 moves but with
dual-free play are welcome.
|(2) Posted by Georgy Evseev [Sunday, May 13, 2012 19:29]|
Kf8 Qh2 Sf6e1 - Kh8 Qh7
|(3) Posted by seetharaman kalyan [Sunday, May 13, 2012 23:07]|
In this position can white win at all, if he does not mate immediately? e.g... 1.Qe5? Qf7+ ! 1.Qb8? Qc7! 2.Qa8? Qc8+! (2.Qe8? Qd8!)
|(4) Posted by Jacques Rotenberg [Monday, May 14, 2012 00:29]; edited by Jacques Rotenberg [12-05-14]|
Also fast search :
Ka2, Rb1, Be3 / Kc3, Nc2, Nc4
1.Rb3# or 1.Bc5 gives mate in 210
Thanks to Nalimov tables all other white moves draw
see http://www.k4it.de/index.php?topic=egtb&lang=en for those who don't know that yet.
|(5) Posted by Arno Tungler [Monday, May 14, 2012 04:19]|
But are those dual-free...?
|(6) Posted by Jacques Rotenberg [Monday, May 14, 2012 08:26]|
ok this is another question...
|(7) Posted by Georgy Evseev [Monday, May 14, 2012 09:02]; edited by Georgy Evseev [12-05-14]|
I was also thinking about adapting some long moremover with multiple win of tempo to the stipulation in question. It may lead to dual-free or "almost dual-free" result, though I expect that on this path it will be impossible to reach really big lengths.
Kf6 Ra7 Sh2 - Ke8 Sd8 Sf8
1.Re7# or win on 232nd move after 1.Sg4
|(8) Posted by Hauke Reddmann [Monday, May 14, 2012 11:55]|
I checked, in the "main" variant (B plays the most logical looking
move, W always the fastest) it's dual free (in fact, riddled
with "only moves" that keep the win) until move 198
(Kd5 Rf1 Sf8 - Kg5 Sf7 Sf5) where Ra1/Rb1 wins fastest. But since
the rook goes via rank 7 to g7 in both cases, I checked a bit
farther until the pittoresque Kd4 Ra4 Se4 - Kh4 Sg4 Sf4 arises
where 1...Se6/Se2+ gives an echo (both #186).
Then I gave up :-)
|(9) Posted by Georgy Evseev [Tuesday, May 15, 2012 05:51]|
For dual-free line the following problem may be used (taken from WinChloe database, solution is in French notation)
The Problemist 1980 (v)
1.Te1! Ce4 2.Txe4 Cc5 3.Te1 Cd3 4.Tb1 a2 5.Ta1 f2 6.Td1 Cb2 7.Tc1 Cd3 8.Ta1 a5 9.Td1 Cb2 10.Tc1 Cd3 11.Ta1 a4 12.Td1 Cb2 13.Tc1 Cd3 14.Ta1 a3 15.Td1 Cb2 16.Tc1 Cd3 17.Ta1 a6 18.Td1 Cb2 19.Tc1 Cd3 20.Ta1 a5 21.Td1 Cb2 22.Tc1 Cd3 23.Ta1 a4 24.Td1 Cb2 25.Tc1 Cd3 26.Ta1 d6 27.Td1 Cb2 28.Tc1 Cd3 29.Ta1 d5 30.Td1 Cb2 31.Tc1 Cd3 32.Ta1 d4 33.Td1 Cb2 34.Tc1 Cd3 35.Ta1 f6 36.Td1 Cb2 37.Tc1 Cd3 38.Ta1 f5 39.Td1 Cb2 40.Tc1 Cd3 41.Ta1 f4 42.Td1 Cb2 43.Tc1 Cd3 44.Ta1 f3 45.Td1 Cb2 46.Tc1 Cd3 47.Ta1 h6 48.Td1 Cb2 49.Tc1 Cd3 50.Ta1 h5 51.Td1 Cb2 52.Tc1 Cd3 53.Ta1 h4 54.Td1 Cb2 55.Tc1 Cd3 56.Ta1 h3 57.Td1 Cb2 58.Tc1 a1=D 59.Txa1 Cd3 60.Tb1 a2 61.Ta1 a3 62.Td1 a1=D 63.Txa1 a2 64.Td1 Cb2 65.Tc1 a1=D 66.Txa1 Cd3 67.Tb1 Ce1 68.Rxf2 d3 69.Txe1‡
Move white rook to a1 and one may choose between 1.Kf2# and the solution shown above.
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