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|(61) Posted by Arno Tungler [Tuesday, Nov 19, 2019 12:10]; edited by Arno Tungler [19-11-19]|
For the 15 captures with all black units needed I have this first idea but am quite sure that a more subtle version should be possible. Please check and improve!
AT after Charles Higgie
(= 7+16 )
1.B×b3+ R(P)xa6! 2.Rxa6+ P(R)xa6! 3.Rxa6+ Ba5 4.Rxa5+ Sa3 5.Rxa3+ Kb1 6.Ba2+ Ka1 7.Bxc4+ Kb1 8.Ba2+ Ka1 9.Bxd5+ Kb1 10.Ba2+ Ka1 11.Bxe6+ Kb1 12.Ba2+ Ka1 13.Bxf7+ Kb1 14.Ba2+ Ka1 15.Bg8+ Kb1 16.Bxh7+ g6 17.B×g6+ Sf5 18.B×f5+ Re4 19.B×e4+ Bd3 20.B×d3+ Qc2+ 21.B×c2‡
Also yet with 7 white units...
EDIT: Now I see that you can just remove the wQ in that position...
|(62) Posted by Siegfried Hornecker [Tuesday, Nov 19, 2019 12:51]; edited by Siegfried Hornecker [19-11-19]|
Just out of curiosity: What happens if Black never interposes? Wouldn't Bf1 then attack Ra6 at some point, making the entire problem unsolvable (and also Rb6 to b4 would be prevented for that reason)?
I think we need to swap a whole lot of pieces then to correct it, unless I am wrong. Just in case, here is my proposed correction.
(= 8+16 )
#21 by Arno Tüngler
(proposed correction if wrong by me)
1.B:b3+ Ra3 (1.-Kb1? 2.Qa1/2 mate) 2.Q:a4+ N:a4 3.R:a4+ b:a4 4.R:a4+ Ba3 5.R:a3+, etc.
I noticed afterwards that just Rb6-b5 should correct the issue.
(= 7+16 )
#21 by Arno Tüngler
(proposer correction if wrong by me)
1.B:b3+ b:a6 2.R:a6+ R/Ba5 (2.-Kb1? 3.Ra1 mate) 3.R:a5+ B/R:a5 4.R:a5+ Na3 5.R:a3+, etc.
EDIT: I just noticed that Black has to interpose at least with the knight. So Arno's position is probably correct already.
|(63) Posted by Olaf Jenkner [Tuesday, Nov 19, 2019 13:21]|
Gustav solves both versions within one second. Solutions exist.
|(64) Posted by Arno Tungler [Tuesday, Nov 19, 2019 13:37]; edited by Arno Tungler [19-11-19]|
Siegfried's hint helps to economize the wQ... Please check again and improve!
SH, AT after Charles Higgie
(= 6+16 )
1.B×b3+ Ba5! 2.Rxa5+ Ra4! 3.Rxa4+ bxa4 4.Rxa4+ Sa3 5.Rxa3+ Kb1 6.Ba2+ Ka1 7.Bxc4+ Kb1 8.Ba2+ Ka1 9.Bxd5+ Kb1 10.Ba2+ Ka1 11.Bxe6+ Kb1 12.Ba2+ Ka1 13.Bxf7+ Kb1 14.Ba2+ Ka1 15.Bg8+ Kb1 16.Bxh7+ g6 17.B×g6+ Sf5 18.B×f5+ Re4 19.B×e4+ Bd3 20.B×d3+ Qc2+ 21.B×c2‡
EDIT: See also my edit to post 61 above!
|(65) Posted by Olaf Jenkner [Tuesday, Nov 19, 2019 18:49]|
The last version is C+, by the way a nice model mate.
|(66) Posted by Arno Tungler [Tuesday, Nov 19, 2019 19:18]|
Very nice, and thank you again, Olaf!
Is even the beneath possible??
AT after Charles Higgie
(= 4+16 )
1.B×b3+ Ra6! 2.Rxa6+ bxa6 3.Qxa6+ Ba5 4.Qxa5+ Sa3 5.Qxa3+ Kb1 6.Ba2+ Ka1 7.Bxc4+ Kb1 8.Ba2+ Ka1 9.Bxd5+ Kb1 10.Ba2+ Ka1 11.Bxe6+ Kb1 12.Ba2+ Ka1 13.Bxf7+ Kb1 14.Ba2+ Ka1 15.Bg8+ Kb1 16.Bxh7+ g6 17.B×g6+ Sf5 18.B×f5+ Qe4 19.B×e4+ Bd3 20.B×d3+ Rc2 21.B×c2#
Whenever Black plays Rxa2 White mates with Qc1#
Does the black rook protect all other white attempts?
|(67) Posted by Olaf Jenkner [Tuesday, Nov 19, 2019 20:24]|
C+, but no model mate anymore.
|(68) Posted by Mark Kirtley [Wednesday, Nov 20, 2019 06:45]|
Only 4 white units in the #21 (posting #66) surprises me -- well done!
I'm wondering about direct stalemates with 15 black captures. It seems the task is not difficult -- but I wonder how short the problem can be made. The old P1264320 (14 captures) of mine is 56 moves, and the following has 43 moves (15 captures, perhaps unsound?).
(= 3+16 )
Solution (scroll below with the mouse):
(only white moves are given and are all checks except the last move)
1.-42.Rxe7-g7xg6-g7xg5-g7xg3-g7-g5xh5-g5-g7-g3xh3-g3-g7-c7!xc8-c7-g7-g2xh2, Qg6, Rh8, Re8, Qe6, Rc8, Qc6, Rc7-b7xb5-b7xb4-b7xb3-b7xb2-b7xb1-b7-b3xa3, 43.Kxc3=
The setting by James (posting #43) is only 28 moves! -- although as it stands it is unsound. In that setting the battery is not taken down from one corner and re-assembled in another. which really helps in saving moves...
|(69) Posted by Olaf Jenkner [Wednesday, Nov 20, 2019 10:30]; edited by Olaf Jenkner [19-11-20]|
There is a dual in the second move:
1. Rxe7+ Kg8 2. Rg7+! Kh8 3. Rxg6+ Kh7 4. Rg7+ Kh8 ...
White can save three moves playing 15. Qg6+! but there is a dual in the 18th move:
1. Rxe7+ Kg8 2. Qxg6+ Kf8 3. Qf6+ Kg8 4. Rg7+ Kh8 5. Rxg5+ Kh7 6. Rg7+ Kh8 7. Rxg3+ Kh7 8. Rg7+ Kh8 9. Rg5+ Kh7 10. Rxh5+ Kg8 11. Rg5+ Kh7 12. Rg7+ Kh8 13. Rg3+ Kh7 14. Rxh3+ Kg8 15. Qg6+! Kf8 16. Rf3+ Ke7 17. Rf7+ Kd8 18. Qd6+,Df6+ Ke8 19. Qe6+ Kd8 20. Rf8+ Kc7 21. Rxc8+ Kb7 22. Qc6+ Ka7 23. Rc7+ Kb8 24. Rb7+ Ka8 25. Rxb5+ Ka7 26. Rb7+ Ka8 27. Rxb4+ Ka7 28. Rb7+ Ka8 29. Rxb3+ Ka7 30. Rb7+ Ka8 31. Rxb2+ Ka7 32. Rb7+ Ka8 33. Rxb1+ Ka7 34. Rb7+ Ka8 35. Rb3+ Ka7 36. Rxa3+ Kb8 37. Rb3+ Ka7 38. Rb7+ Ka8 39. Kxc3 h1=Q 40. Qxh1=
Note that white captures the promoted pawn in the last move.
|(70) Posted by Jacques Rotenberg [Wednesday, Nov 20, 2019 11:14]|
The point with the capture on g3 in the S#21 is that it is completely passive : it is captured because it is on the way, not because it is needed.
In other words, it is better when white "wants" to capture. Moreover the use of this pawn is very weak, it blocks g3 only during the 1st move.
|(71) Posted by Arno Tungler [Wednesday, Nov 20, 2019 12:12]|
Yes, I understand but the difference is not great as the Bg3 also is just in the way. In your position, if you replace the Bg3 with a bP it has the same number of moves and exact the same solution - only the alternative with 1...Bf4 disappears. So, while your argument has merits it does not really give much and you have to weigh it with the added two moves. So, for me it is 50-50...
|(72) Posted by Jacques Rotenberg [Wednesday, Nov 20, 2019 23:23]|
No, the Bg3 is not in the way, it plays to f4 : this is the thematic defence.
Active captures and passive captures are two different worlds. Probably a task of passive captures would be more difficult to build, but every single one here is an active one, it is not such an accessory detail.
About the length, yes, shorter is better (in principle), but here you have already 7 no capturing moves, so two more is not that a difference.
Also very small things :
- the play is a little bit more interesting with two places for mistakes :
5.Qh2+! (5.Qf3+?) and 7.Qf3+! (7.Qh2+?)
- I like 7.Qh2+? that gives immediately an efficient battery Kf2, Bg2, Qh2 but the queen controls e2.
To say it shortly, in the meanwhile, I like better the S#23
|(73) Posted by Mark Kirtley [Thursday, Nov 21, 2019 01:09]|
Thanks so much for finding the duals in the =43.
Here's perhaps a correction but now with rather uninteresting play:
(= 3+16 )
1.-34.(all checks)Rxe7-g7xg5-g7xg4-g7xg3-g7xg2-g7xg1-g7-g5xh5, Qg6, Rh8, Rf8, Qf6, Rc8, Qc6, Rc7-b7xb5-b7xb4-b7xb3-b7xb2-b7xb1-b7-b3xa3, 35.Kxd4 c2 36.Kxc2= (4.Rxh5+? 5.Rg5+...14.Rh1+ Rh3+!)
Can the number of non-capturing moves be reduced? Or can an ideal (stale)mate be made with only 3 white units starting in the diagram?
|(74) Posted by Arno Tungler [Thursday, Nov 21, 2019 01:29]|
I understand all of your arguments. So, the 21- mover is the first setting (already in PDB and reprinted on Superproblem site) and the s#23 is a valid version with small additional merits. I will add it to PDB also. From this treat I would also add s#27 (Post 33: JM + JR + AT), the s=17 (Post 44 MK + JM + JR + AT) and the #21 (Post 66: As the whole matrix as well as the additional helping ideas developed from this threat - compare i.e. post 28 - I feel the authors should also be JM + JR + AT after CH). All agree?
|(75) Posted by ichai [Thursday, Nov 21, 2019 02:01]; edited by ichai [19-11-21]|
|(76) Posted by Jacques Rotenberg [Thursday, Nov 21, 2019 03:09]|
Thank you Arno, I agree, but not for the s=17 and not for the #21, I have no part there.
Btw, it is an opportunity to underline a kind of paradox.
An original shown here is considered by many as « published », though you might hesitate on the position, and on the authors too
So ... is it really « published » ?
|(77) Posted by Mark Kirtley [Thursday, Nov 21, 2019 08:18]|
Thanks for the offer of including my name on the s=17, but that would make me blush. I don't feel I contributed enough.
|(78) Posted by Arno Tungler [Thursday, Nov 21, 2019 08:47]|
The Codex says: Article 20 – Definition of Publication
(1) Publication of a chess composition consists of communicating it to the public, whether in permanent form (e.g. a document or a recording medium) or transient form (e.g. on a demonstration board or through an electronic medium).
(2) For the purposes of this Article, "communicating to the public" means enabling an unrestricted number of people to have the opportunity of access to a chess composition by
(a) presenting it in permanent form, or
(b) showing or using it in a lecture or solving tournament which falls within the categories listed in Annex I, or
(c) showing it in transient form through a generally accessible medium (e.g. an electronic network).
Footnote 22 says: Including the Internet and electronic mailing lists, but not e-mail. (see https://www.wfcc.ch/1999-2012/codex/#c5)
So, MatPlus.net is "a generally accessible medium" and if we publish a problem here, it is deemed published whether we like it or not...
The forum of chessproblems.ca is mostly a closed environment and when we discuss positions there they are not yet "published". That makes it easier to identify the positions that finally will go to the public. Usually I prefer that way but with this not very ambitious task it is not so important anyhow.
|(79) Posted by Arno Tungler [Thursday, Nov 21, 2019 08:52]|
Please give your consent as really you were the initiator here! Nobody else had the s= idea. And then you are also one of my co-authors...
|(80) Posted by Jacques Rotenberg [Thursday, Nov 21, 2019 12:41]|
"...it is deemed published whether we like it or not..."
in any case it sounds strange. Laws that we are generating ourselves should be done to protect us, not to coerce us with no reason.
it raises some questions
for example :
why should the codex force somebody to sign a problem, if he does not want ?
why should the codex forbid a problem shown here to participate in any tournament ?
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