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| (1) Posted by Hauke Reddmann [Friday, Nov 19, 2021 22:35] |
Longest series mover with...
Most sx# let one party make n moves, then the
other fulfils the stipulation. This one is a
bit different: Only White makes n moves, then
it shall be mate. (Dawsons "Revolver", for example.)
How long can it get? Not all to long, I guess,
even *with* promoted pieces or even an illegal
position.
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| (2) Posted by Joost de Heer [Saturday, Nov 20, 2021 11:39] |
 (= 6+15 )
Arno Tüngler, Strategems 2013
ser-#154
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| (3) Posted by Olaf Jenkner [Saturday, Nov 20, 2021 19:15] |
https://pdb.dieschwalbe.de/P1278400
It is C+ in less than one second.
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| (4) Posted by Joost de Heer [Saturday, Nov 20, 2021 19:23] |
An explanation on this series-mover: https://www.scienceblogs.com/evolutionblog/2015/06/21/sunday-chess-problem-45
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| (5) Posted by seetharaman kalyan [Sunday, Nov 21, 2021 04:54] |
Amazing problem with an ingenious mechanism.
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| (6) Posted by Hauke Reddmann [Sunday, Nov 21, 2021 08:56] |
Yup. I would have thought of "something with bishops
and a king" myself, but the repeated unpins, and
especially things like Ke8 blocking the road and
the bishop must take the long way around...and he
must let Bd3 live, and an underpromotion crowning it...brill.
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| (7) Posted by Olaf Jenkner [Sunday, Nov 21, 2021 14:01] |
In sunday chess problem is written:
Great stuff! Of course, this is well beyond anything that can be computer tested, so cooks are always a worry.
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| (8) Posted by Arno Tungler [Sunday, Nov 21, 2021 15:46] |
Yes, I also felt that this was quite funny! Anyhow, there are still other long records that are not so easy to test for computers, for example with the “auto-stalemate” goal. Meaning that after the indicated number of moves White has a position where he is stalemate if it’s his turn. My current records without (https://pdb.dieschwalbe.de/P1333509) and with promoted force (https://pdb.dieschwalbe.de/P1273550) remain unconfirmed up to now. And even the for human testers quite clear series helpstalemate overall record (https://pdb.dieschwalbe.de/P1307677) took 40 minutes by a strong program.
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| (9) Posted by Olaf Jenkner [Sunday, Nov 21, 2021 18:28] |
I tested the last 167 moves of P1333509 and full length P1273550 assuming that white must not promote.
Both have a unique solution.
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| (10) Posted by Hauke Reddmann [Sunday, Nov 21, 2021 19:53] |
P.S.: I asked on CSE: Starting position, what is the shortest serial sequence
where a) White stalemates Black b) White stalemates himself?
Surely this already has been tried...
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| (11) Posted by Arno Tungler [Monday, Nov 22, 2021 10:21] |
I got a Ser-=19 with 1. d4 2. d5 3. d6 4. dxc7 5. cxb8=R 6. Rxc8 7. Rxa8 8. Rxa7 9. Rxb7 10. Rxd7 11. Bg5 12. Bxe7 13. Bxd8 14. Rxf7 15. Rxg7 16. Rxh7 17. Rxh8 18. Rxg8 19. Qd5 =
Is there a possibility to reduce to 18?
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| (12) Posted by Ulrich Voigt [Monday, Nov 22, 2021 11:03] |
deleted
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| (13) Posted by Arno Tungler [Monday, Nov 22, 2021 11:50] |
And the shortest so far auto stalemate is a Ser-!=56:
1.d4 2.Sc3 3.Sd5 4.Sxe7 5.Sxg8 6.Sf3 7.Sh4 8.Sg6 9.Sxh8 10.b4 11.e4 12.e5 13.Bd3 14.Bxh7 15.Qg4 16.Qxg7 17.h4 18.Rh3 19.Rg3 20.Rg6 21.a4 22.Ra3 23.Rf3 24.Rxf7 25.f4 26.f5 27.f6 28.Ke2 29.Kf3 30.Kg4 31.Kh5 32.Kh6 33.h5 34.g4 35.g5 36.b5 37.Ba3 38.Bxf8 39.e6 40.e7 41.b6 42.bxc7 43.cxd8=B 44.a5 45.a6 46.axb7 47.bxa8=R 48.Rxb8 49.Rb7 50.Rxd7 51.c4 52.c5 53.c6 54.c7 55.d5 56.d6!=
This should certainly be shortened with another final position…
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| (14) Posted by Arno Tungler [Monday, Nov 22, 2021 15:03] |
Three moves shorter…
1.Sc3 2.Sd5 3.Sxe7 4.Sxg8 5.Sf3 6.Sh4 7.Sg6 8.Sxh8 9.e4 10.b4 11.b5 12.d4 13.d5 14.Ba3 15.Bxf8 16.b6 17.bxa7 18.axb8=B 19.a4 20.a5 21.a6 22.a7 23.d6 24.dxc7 25.cxd8=B 26.c4 27.c5 28.c6 29.c7 30.Qg4 31.Qxg7 32.Ke2 33.Ke3 34.Kf4 35.Kg5 36.Kh6 37.e5 38.e6 39.e7 40.Bd3 41.Bxh7 42.Ra6 43.Rg6 44.h4 45.h5 46.Rh4 47.Rf4 48.Rxf7 49.f4 50.f5 51.f6 52.g4 53.g5!=
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| (15) Posted by Arno Tungler [Monday, Nov 22, 2021 15:41] |
P.S.: The so far longest series help stalemate from the starting position (obviously with Black to move…) is from a Duckburg resident:
Ernst Eygenloob, Entenhausen, after Hemmo Axt & Markus Ott, dedicated to Ulrich Ring, feenschach I-IX 1992
Ser-h=43
1.Sf6 2. Se4 3.Sxd2 4.Sxf1 5.Sg3 6.Sxe2 7.Sxg1 8.d5 9.Kd7 10.Kc6 11.Kb5 12.Ka4 13.a5 14.b5 15.Sc6 16.Sb4 17.e5 18.e4 19.e3 20.e2 21.Qf6 22.Qxb2 23.Qd4 24.c5 25.c4 26.c3 27.h5 28.h4 29.h3 30.hxg2 31.Rxh2 32.Rxh1 33.Bd6 34.Bh2 35.Ra6 36.Rg6 37.Rg3 38.Bh3 39.f5 40.f4 41.f3 42.g5 43.g4 Qxd4 =
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| (16) Posted by Michel Caillaud [Monday, Nov 22, 2021 18:14] |
Amazing. I had forgotten the problem by Ernst Eygenloop when I proposed another sh=43 in 2004 :
MC France Echecs IV 2004
sh=43 (presented with color reversed)
1.e6 2.Sf6 3.Se4 4.Sxd2 5.Bd6 6.Bxh2 7.Bf4 8.Bh6 9.Qf6 10.Qg6 11.c5 12.c4 13.c3 14.cxb2 15.b5 16.b4 17.b3 18.bxc2 19.cxb1=B 20.Bb7 21.Bf3 22.Bh5 23.Ke7 24.Kf6 25.Kf5 26.Kg4 27.a5 28.a4 29.a3 30.Ra5 31.Rg5 32.Rc8 33.Rcc5 34.Rcf5 35.Rf6 36.Bf5 37.b1=R 38.Rb5 39.Re5 40.d5 41.Sc6 42.Sd4 43.Se4 Qxd4=
I assumed that the final position was unique, but it is now (pre)-cooked...
I also proposed in the same place a ser-!=51
MC France Echecs IV 2004
ser-!=51
1.a4 2.a5 3.a6 4.Ra3 5.Rf3 6.Rxf7 7.Sc3 8.Sd5 9.Sxe7 10.Sxg8 11.b4 12.b5 13.b6 14.bxc7 15.cxb8=R 16.Ba3 17.Bxf8 18.c4 19.c5 20.c6 21.cxb7 22.bxc8=S 23.Sce7 24.Rb6 25.Re6 26.d4 27.d5 28.d6 29.Qd2 30.Qh6 31.e4 32.e5 33.h4 34.h5 35.Rh3 36.Rg3 37.Rgxg7 38.Kd2 39.Kd3 40.Ke4 41.Kf5 42.Kf6 43.Bd3 44.Bg6 45.f4 46.f5 47.Sh3 48.Sg5 49.Sxh7 50.g4 51.g5 !=
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| (17) Posted by James Malcom [Tuesday, Nov 23, 2021 01:07] |
Try doing Hauke's challenges, but with no promotions allowed.
Arno's idea can be done in three more moves: 1. d4 2. a4 3. Ra3 4. Rc3 5. Rxc7 6. Rxc8 7. Rxb8 8. Rxa8 9. Rxa7 10. Rxb7 11. Rxd7 12. Bg5 13. Bxe7 14. Bxd8 15. Rxf7 16. Rxg7 17. Rxh7 18. Rxh8 19. Qd3 20. Qf5 21. Qd5 22. Rxg8
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| (18) Posted by Arno Tungler [Tuesday, Nov 23, 2021 08:14] |
Without promotion takes also only 19 moves:
 (= 16+4 )
End position for a Ser-=19 with for example 1.d4 2.Bf4 3.Bxc7 4.d5 5.Bxd8 6.Qd4 7.Qxa7 8.Qxa8 9.Qxb7 10.Qxc8 11.Qxb8 12.Qg3 13.Qxg7 14.Qxh7 15.d6 16.Qxh8 17.Bxe7 18.Bf6 19.Qxg8=
So the question remains - are 18 moves possible?
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| (19) Posted by Hauke Reddmann [Tuesday, Nov 23, 2021 09:48] |
Just for fun, for a maximally pacifist version (letting the whitefield
pawns and the two bishops live) I get 23 moves.
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| (20) Posted by Joost de Heer [Thursday, Nov 25, 2021 12:14] |
Isn't there something that can be done with Jacobi by stipulating a ser-dia17 with no black pieces on a8,b8,d8,g8,h8 followed by a =1? (any of the black pawns could theoretically still be present and blocked by a white piece on the 6th row)
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