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MatPlus.Net Forum Selfmates The Slaughterhouse
 
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(41) Posted by Olaf Jenkner [Friday, Nov 15, 2019 18:34]

Yes, no solution after 1. Bxf4+ Bg6 2. Qxg6+ Sg3!
 
   
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(42) Posted by Mark Kirtley [Saturday, Nov 16, 2019 02:25]

@ ichai
Right! Thanks so much for spotting this error.

@ Olaf
Thanks for verifying the error.

Is the following a correction? (I've only reversed the roles of the Rf4 and the Bd8.)

(= 3+16 )


S=18

Intention (same as in posting #39, except that now QxR at g5 instead of QxB):
1.Bxf4+ 2.Qx(B)g6+ 3.Qx(R)g5+ 4.Qx(S)g3+ 5.Qxh3+ 6.Bh2+ 7.Bxe5+ 8.Bh2+ 9.Bxd6+ 10.Bh2+ 11.Bb8+ 12.Bxa7+ 13.Bx(P)b6+ 14.Bx(S)c6+ 15.Bx(P)d4+ 16.Bx(P)e3+ 17.Bx(P)f2+ 18.Qg2+ R/Kxg2=.
 
   
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(43) Posted by Rewan Demontay (Real Name: James Malcom) [Saturday, Nov 16, 2019 04:54]

An idea I had awhile ago was for the longest possible windmill stalemate against all 16 Black pieces.

This is my best work so far on it.

=28

(= 5+16 )


There is an ideal solution; 1. Rh6+ Qh7 2. Rxh7+ Kg8 3. Rxg7+ Kh8 4. Ng6+ Nxg6 5. Rxg6+ Kh7 6. Rg7+ Kh8 7. Rxg5+ Kh7 8. Rg7+ Kh8 9. Rxg4+ Kh7 10. Rg7+ Kh8 11. Rxg3+ Kh7 12. Rg7+ Kh8 13. Rxg2+ Kh7 14. Rg7+ Kh8 15. Rxg1+ Kh7 16. Rg7+ Kh8 17. Rxe7+ Kg8 18. Rg7+ Kh8 19. Rxd7+ Kg8 20. Rg7+ Kh8 21. Rxc7+ Kg8 22. Rg7+ Kh8 23. Rxb7+ Kg8 24. Rg7+ Kh8 25. Rxa7+ Kg8 26. Rg7+ Kh8 27. Bxb2 a1=Q 28. Bxa1

But this works just as well as unfortunately: 3. Rxg7+ Kh8 4. Rxe7+ Kg8 5. Rg7+ Kh8 6. Rxd7+ Kg8 7. Rg7+ Kh8 8. Rxc7+ Kg8 9. Rg7+ Kh8 10. Rxb7+ Kg8 11. Rg7+ Kh8 12. Rxa7+ Kg8 13. Rg7+ Kh8 14. Ng6+ Nxg6 15. Rxg6+ Kh7 16. Rg7+ Kh8 17. Rxg5+ Kh7 18. Rg7+ Kh8 19. Rxg4+ Kh7 20. Rg7+ Kh8 21. Rxg3+ Kh7 22. Rg7+ Kh8 23. Rxg2+ Kh7 24. Rg7+ Kh8 25. Rxg1+ Kh7 26. Rg7+ Kh8 27. Bxb2 a1=Q 28. Bxa1

And that's just what I know about for cooks in my problem.

Feel free to improve upon it.
 
 
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(44) Posted by Arno Tungler [Saturday, Nov 16, 2019 07:02]

@ Mark
Thanks for that nice idea.

@ Olaf
Thanks for all the testing and for Gustav. You already deserve a co-authorship for that alone...

Here is a version of Mark's idea one move nearer to the slaughterhouse ideal of zero non-capturing moves. Correct? And how to get even closer?

(= 3+16 )

S=17

Intention:
1.Bxf4+ Bg6! 2.Qxg6+ Rg5! 3.Qxg5+ Sg4! 4.Qxg4+ Sg3 5.Qxg3+ Kh1 6.Qxh3+ Kg1 7.Bh2+ Kh1 8.Bxe5+ Kg1 9.Bh2+ Kh1 10.Bb8+ Kg1 11.Bxa7+ b6 12.Bxb6+ Bc5 13.Bxc5+ d4 14.Bxd4+ e3 15.Bxe3+ f2+ 16.Bxf2+ Rxf2 17.Qg2+ R/Kxg2=
 
   
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(45) Posted by Olaf Jenkner [Saturday, Nov 16, 2019 11:55]

@ Mark
Your s=18 might be correct.

@ Rewan
Your =28 is stalemate in 27. You don't need 26. Rg7+. Nevertheless there are a lot of duals in this problem.

@ Arno
Your s=17 might be correct.
 
   
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(46) Posted by Mark Kirtley [Sunday, Nov 17, 2019 02:33]

@ Arno
Your S=17 setting is a nice improvement!

@ Olaf
Thanks for the testing.
 
   
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(47) Posted by Arno Tungler [Sunday, Nov 17, 2019 10:39]

And back again to the selfmate…
As Jacques stated, additional white force "open a very large number of possible settings."
Anyhow, I would like to show this idea with the maximum of possible captures by White - 14!
Probably this justifies the additional white blocking bishop?

JM, JR & AT
(= 4+16 )

s#21

1.B×e4+ Bf4! 2.Qxf4+ Sf3 3.Qxf3+ Kg1 4.Qxh1+ Kf2 5.Qf3+ Kg1 6.Qxg3+ Kf1 7.Bg2+ Kg1 8.B×d5+ Kf1 9.Bg2+ Kg1 10.B×c6+ Kf1 11.Bg2+ Kg1 12.B×b7+ Kf1 13.Bg2+ Kg1 14.B×a8+ Kf1 15.Bg2+ Kg1 16.Bb7+ Kf1 17.B×a6+ b5 18.B×b5+ c4 19.B×c4+ d3 20.B×d3+ e2+ 21.B×e2+ Q×e2‡
 
   
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(48) Posted by Olaf Jenkner [Sunday, Nov 17, 2019 10:53]

Congratulations!
I tested (but not C+) and found no cook.
In my opinion the maximum of possible captures by White even justifies promoted pieces as long as no better setting is found.
 
   
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(49) Posted by Michael McDowell [Sunday, Nov 17, 2019 13:05]; edited by Michael McDowell [19-11-17]

You may be interested in this, which shows capture of 15 black pieces in a directmate:

Charles Higgie
Gameknot 2016

(= 7+16 )

Mate in 22

1.Rh8+ Rg8 2.Rxg8+ Rf8 3.Rxf8+ Be8 4.Rxe8+ Sc8 5.Rxc8+ Ka7 6.Bb8+ Ka8 7.Bxd6+ Ka7 8.Bb8+ Ka8 9.Bxe5+ Ka7 10.Bb8+ Ka8 11.Bxf4+ Ka7 12.Bb8+ Ka8 13.Bxg3+ Ka7 14.Bb8+ Ka8 15.Bxh2+ Ka7 16.Rxa6+ Sxa6 17.Bxg1+ Be3 18.Bxe3+ d4 19.Bxd4+ Sc5 20.Bxc5+ b6 21.Bxb6+ Kb7 22.Sd6# (16…bxa6 17.Bb8+ Kb7 18.Sd6+ Ka8 19.Bc7+ Ka7 20.Bb6# or 17…Ka8 18.Bc7+ Kb7 19.Sd6+ Kc6 20.Bb8#)

Higgie is a player who lives in the north of England, though I see from the FIDE database that he is listed as Welsh. The above wasn’t quite his original setting – he had a promoted piece because he couldn’t find a way of getting rid of the a-pawn while maintaining legality, but I made a suggestion that worked. Strangely he just seemed content to know that the idea could be realised. He showed no interest whatever in problem chess.
 
   
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(50) Posted by Olaf Jenkner [Sunday, Nov 17, 2019 15:39]; edited by Olaf Jenkner [19-11-17]

Higgie's problem is C+.
The solution of the last two problems can be seen via autoplay in the PDB:
http://pdb.dieschwalbe.de/search.jsp?expression=probid=%27P1369409%27or%20probid=%27P1369412%27
 
   
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(51) Posted by Rewan Demontay (Real Name: James Malcom) [Sunday, Nov 17, 2019 19:20]

Perhaps my s#17 from post 5 should be added in as the forefather?

Also, the s#27 from earlier in this thread would also be a great addition too.
 
   
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(52) Posted by Mark Kirtley [Sunday, Nov 17, 2019 23:39]; edited by Mark Kirtley [19-11-18]

@ Arno
Whoa! The new s#21 is a delight to play through, and understand why the bR and bS in the corners are necessary to capture. Fruitful idea, shifting the position one unit to the left!

Charles Higgie's #22 is also a delight, but in a sense it doesn't really count as 15 captures of black units, in terms of captures that influence the solution. Please coach me if I'm wrong, but the presence of the black pawns at a6 and h2 has no bearing on the solution, as White has to visit those squares anyway.
 
   
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(53) Posted by Olaf Jenkner [Sunday, Nov 17, 2019 23:56]

Yes, there is the same solution without these two pieces. So it can hardly considered a record. As we know, Higgie is not interested in problem chess, so he doesn't care about such things.
 
   
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(54) Posted by Arno Tungler [Monday, Nov 18, 2019 04:25]

@Olaf
Thanks again for testing and posting!
@J. Malcolm
Will try to get all the correct problems of this thread to the PDB. Somewhere on the site I already saw your first name but now seemingly cannot find it! If you do not want to post it here you can send me a private message so that the full name can be added for the database.
@Mark & Olaf
That was my first thought on the Higgie #22 and I also identified those two pawns. Can Gustav confirm that it would be C+ without those? Then it really should not count as a record and we have a challenge - get a correct version. Additional goal to reduce move numbers and white force...

A simple idea for one black pawn would be to place the wR to h8, the wB to b8, and the bPh2 to c7 with 1.Bxc7+ as key. For the bPa6 it is not so easy...
 
   
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(55) Posted by Jacques Rotenberg [Monday, Nov 18, 2019 12:25]; edited by Jacques Rotenberg [19-11-18]

about the s#21.
It remains some small flaws
1) the Bh6 and the pawn g3 are somewhat artificial : the "natural" place for this bishop would be g3, and then you remain with a pawn to capture. (if you want to get your 14 captures...)
2) the rook b7 also could be a pawn, but then you remain with a rook to capture...

so perhaps slightly better is this :

(= 4+16 )
s#24

1.B×e4+ Bf4 2.Q×f4+ Sf3 3.Q×f3+ Kg1 4.Q×h1+ Kf2 5.Qf3+ Kg1 6.Qg3+ Kf1 7.Q×h3+ Kf2! 8.Qf3+! (8.Qh2+?) Kg1 9.Qg3+ Kf1 10.Bg2+ Kg1 11.B×d5+ Kf1 12.Bg2+ Kg1 13.B×c6+ Kf1 14.Bg2+ Kg1 15.B×b7+ Kf1 16.Bg2+ Kg1 17.B×a8+ Kf1 18.Bg2+ Kg1 19.Bb7+ Kf1 20.B×a6+ Rb5 21.B×b5+ c4 22.B×c4+ d3 23.B×d3+ e2+ 24.B×e2+ Q×e2‡

The answer for the underused Rb7 is a bit formal.
It is worth noticing the nice potential third battery (8.Qh2+?)
 
   
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(56) Posted by Rewan Demontay (Real Name: James Malcom) [Monday, Nov 18, 2019 13:49]

@Arno You can see my full first name now. :D
 
   
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(57) Posted by Olaf Jenkner [Monday, Nov 18, 2019 16:04]

It is cooked in 23:
1.B×e4+ Bf4 2.Q×f4+ Sf3 3.Q×f3+ Kg1 4.Q×h1+ Kf2 5.Qh2+ !
 
   
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(58) Posted by Jacques Rotenberg [Monday, Nov 18, 2019 16:18]; edited by Jacques Rotenberg [19-11-18]

ok, but it is not really a cook, the solution is simply shorter (even, nicely shorter)

So it is a S#23

And thank you very much for your quick answer!!

so, the solution :

1.B×e4+ Bf4 2.Q×f4+ Sf3 3.Q×f3+ Kg1 4.Q×h1+ Kf2 5.Qh2+! (5.Qf3+?) Kf1 6.Q×h3+ Kf2! 7.Qf3+! (7.Qh2+?) Kg1 8.Qg3+ Kf1 9.Bg2+ Kg1 10.B×d5+ Kf1 11.Bg2+ Kg1 12.B×c6+ Kf1 13.Bg2+ Kg1 14.B×b7+ Kf1 15.Bg2+ Kg1 16.B×a8+ Kf1 17.Bg2+ Kg1 18.Bb7+ Kf1 19.B×a6+ Rb5 20.B×b5+ c4 21.B×c4+ d3 22.B×d3+ e2+ 23.B×e2+ Q×e2‡
 
   
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(59) Posted by Olaf Jenkner [Monday, Nov 18, 2019 17:07]

Now it is OK, with some shuffeling of the first five moves.
 
   
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(60) Posted by Arno Tungler [Tuesday, Nov 19, 2019 11:50]

@Jacques:
Thank you for your fresh ideas for the 14 captures! I like especially the shift of the units on the b-column. I am not so convinced whether the bBg3 is better there than a bP as it adds another 2 (non-capturing) moves and also leads to different possibilities in the first black move (if 1... Sf3 2.Qxf3+ Kg1 3.Qxh1+ Kf2 4.Qf3+ Kg1 5.Qxg3+ Kf1 6.Qxh3+ etc.). Probably it is a matter of taste - my version is "straight forward" without any black alternatives and shorter.
 
   
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MatPlus.Net Forum Selfmates The Slaughterhouse