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| (1) Posted by James Malcom [Thursday, Nov 14, 2019 03:02]; edited by James Malcom [19-11-14] |
The Staircase Of Discovery
Many of you know of an old double check staircase mechanism first shown in 1916 by A.C. White-WBg1, WBh1, Rg2, & BKf3. I’ve seen a few problems that use it to it’s full extent, including Morse’s 2004 problem that has 13 consective double checks in the main line.
But I wondered. had it been done in a selfmate yet? I only found one such problem.
P1298310
Filip S. Bondarenko
2189 Sachova skladba 12/1989
s#9
 (= 11+8 )
To me, it was obvious that the doublle check count could be increased. Thus I set out to use what I like to call “The Staircase Of Discovery” to it’s full extent.
s#13
 (= 9+11 )
The solution should be obvious, and it makes a total to 12 consective double checks in a selfmate. I wouldn’t be suprised if my s#13 was cooked.
Could 13 possibly be done at all in a selfmate, as Morse did in his directmate? Have more double checks been done in a s# that are non-consective?
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| (2) Posted by Olaf Jenkner [Thursday, Nov 14, 2019 06:46] |
It's cooked. It's a selfmate in 9 moves.
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| (3) Posted by James Malcom [Thursday, Nov 14, 2019 13:11]; edited by James Malcom [19-11-14] |
I’m afraid that I can’t see it.
But I did find a cook in 12.
What if we just toss in a bundle of Black pieces to try the ensure the double checking?
s#13
 (= 9+13 )
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| (4) Posted by Harry Fougiaxis [Thursday, Nov 14, 2019 14:46] |
Cooked in 11, as well: 1.c8=B+ 2.Rc7+ 3.Rc6+ 4.Rd6+ 5.Rd5+ 6.Re5+ 7.Re4+ 8.Rf4+ 9.Rf3+ 10.Rf2+ 11.Ke2+ Qxd1#
Or in 12 moves: 10.Rg3+ 11.Rg2+ 12.Kd2+ Qxd1#
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| (5) Posted by James Malcom [Thursday, Nov 14, 2019 16:18] |
Well at least the staircase still works, and there is a count of 10 consecutive double checks.
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| (6) Posted by Olaf Jenkner [Thursday, Nov 14, 2019 18:35] |
The first problem is cooked in 9 moves:
1. c8=B+ Kc6 2. Rc7+ Kxd6 3. Rd7+ Ke6 4. Bd5+ Kf5 5. Qd3+ Kg4 6. Bf3+ Kh3 7. Bh5+ Kg2 8. Qf3+ Kg1 9. Ke2+ Qxd1#
2. Bd7+ Kd5 3. Qc4+ Ke5 4. Se4+ Kxe4 5. Ra7+ Sc6 6. Qe6+ Kf3 7. Qf5+ Kg2 8. Qf2+ Kh1 9. Kd2+ Qxd1#
5. ... Sb7 6. Qe6+ Kf3 7. Qf5+ Kg2 8. Qf2+ Kh1 9. Kd2+ Qxd1#
The second problem is cooked in 10 moves:
1. c8=B+ Kc6 2. Rc7+ Kxd6 3. Rc6+ Kd5 4. Rd6+ Ke5 5. Rd5+ Ke4 6. Rxg5+ Sd5 7. Bd3+ Kf3 8. Rf5+ Kg2 9. Be4+ Kg1 10. Kd2+ Qxd1#
6. ... Sc6 7. Bd3+ Kf3 8. Rf5+ Kg2 9. Rf2+ Kg1 10. Ke2+ Qxd1#
9. ... Kh1 10. Kd2+ Qxd1#
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| (7) Posted by Miodrag Mladenović [Thursday, Nov 14, 2019 20:35]; edited by Miodrag Mladenović [19-11-14] |
What about this one:
 (= 10+9 )
S#14
There are some capture moves but still there are 13 double checks.
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| (8) Posted by Olaf Jenkner [Thursday, Nov 14, 2019 20:55] |
Cooked in 13 moves:
1. Sd6+ Kd7 2. Qb5+ Kxd6 3. c8=D+ Kd5 4. Ra7+ Qc6 5. Qd7+ Ke4 6. Qe6+ Kf3 7. Qf7+ Ke4 8. Qb1+ d3 9. Qf4+ Kd5 10. Qe5+ Kc4 11. Ra4+ Qxa4 12. Qe6+ Kd4 13. Qa1+ Qxa1#
2. ... Ke6 3. c8=D+ Ke5 4. Sf7+ Kd5 5. Qb3+ Ke4 6. Qxg4+ Qf4 7. Sd6+ Ke5 8. Rxe7+ Kf6 9. Rf7+ Ke5 10. Sb7+ Ke4 11. Qxg2+ Qf3 12. d3+ Ke3 13. Qe2+ Qxe2#
7. ... exd6 8. Qge6+ Qe5 9. Rf7+ d5 10. Qg4+ Qf4 11. Qxg2+ Qf3 12. d3+ Ke3 13. Qe2+ Qxe2#
5. ... c4 6. Qbxc4+ Ke4 7. Qxg4+ Qf4 8. Qc6+ Kd3 9. Se5+ Qxe5+ 10. Qe2+ Qxe2#
4. ... Ke4 5. Qe2+ Be3 6. d3+ Kd5 7. Qa2+ c4 8. Qxg2+ Qf3 9. Qxc4+ Ke4 10. Ra7+ Kd3 11. Qd2+ Bxd2#
5. ... Kd5 6. Qd7+ Qd6 7. Qxg2+ Kc4 8. Qa4+ Kd3 9. Se5+ Qxe5+ 10. Qe2+ Qxe2#
3. ... Kd5 4. Qc4+ Ke5 5. Sf7+ Ke4 6. Qe2+ Be3 7. d3+ Kd5 8. Qa2+ c4 9. Qxg2+ Qf3 10. Qxc4+ Ke4 11. Ra7+ Kd3 12. Qd2+ Bxd2#
6. ... Kd5 7. Qd7+ Qd6 8. Qxg2+ Kc4 9. Qa4+ Kd3 10. Se5+ Qxe5+ 11. Qe2+ Qxe2#
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| (9) Posted by Miodrag Mladenović [Thursday, Nov 14, 2019 21:00] |
How about this position:
 (= 10+9 )
The black Knights are defending c8Q+ move. Hopefully this is correct.
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| (10) Posted by Olaf Jenkner [Thursday, Nov 14, 2019 21:15] |
It might be correct:
1. Sd6+ Kd7 2. c8=B+ Kc6 3. Rc7+ Kxd6 4. Rc6+ Kd5 5. Rd6+ Ke5 6. Rd5+ Ke4 7. Re5+ Kf4 8. Re4+ Kf3 9. Rf4+ Kg3 10. Rxg4+ Kh3 11. Rg3+ Kh2 12. Rxg2+ Kh1 13. Rxg1+ Kxg1 14. Qf1+ Qxf1#
Nice mate position:
 (= 7+6 )
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| (11) Posted by Miodrag Mladenović [Thursday, Nov 14, 2019 22:42]; edited by Miodrag Mladenović [19-11-14] |
If correct this one has 14 consecutive double checks:
 (= 11+9 )
S#15
1.e8Q++ Kxe8 2.Sd6++ Kxd7 3.b8B++ Kc6 4.Rc7++ Kxd6 5.Rc6++ Kd5
6.Rd6++ Ke5 7.Rd5++ Ke4 8.Re5++ Kf4 9.Rf5++ Kg4 10.Rf4++ Kg3
11.Rg4++ Kh3 12.Rg3++ Kh2 13.Rg2++ Kh1 14.Rxg1++ Kxg1 15.Qg3+ Qxg3#
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| (12) Posted by James Malcom [Thursday, Nov 14, 2019 22:58]; edited by James Malcom [19-11-14] |
Very nice! Much more than I thought possible!
EDIT: As to your second diagram, the better economy the better it is!
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| (13) Posted by Miodrag Mladenović [Thursday, Nov 14, 2019 22:59]; edited by Miodrag Mladenović [19-11-14] |
Here is improved position:
 (= 10+10 )
S#15
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| (14) Posted by Miodrag Mladenović [Thursday, Nov 14, 2019 23:01]; edited by Miodrag Mladenović [19-11-14] |
This last position does not work. The square e7 is not guarded after 5.Rc6++.
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| (15) Posted by James Malcom [Thursday, Nov 14, 2019 23:02] |
That is unfortunate.
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| (16) Posted by Olaf Jenkner [Thursday, Nov 14, 2019 23:16] |
The other position is cooked as well:
1. e8=Q+ Kxe8 2. Sd6+ Kxd7 3. c8=B+ Kc6 4. Rb1+ Sb7 5. Qxb7+ Kc5 6. Bb6+ Kxd6 7. Qc6+ Ke5 8. Bc7+ Rd6 9. Bxd6+ Kd4 10. Qb6+ Kc3 11. Rc1+ Qxc1#
5. ...Kxd6 6. Bb4+ Ke5 7. Qb8+ Rd6 8. Bxd6+ Kd4 9. Qb6+ Kc3 10. Rc1+ Qxc1#
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| (17) Posted by Miodrag Mladenović [Thursday, Nov 14, 2019 23:16]; edited by Miodrag Mladenović [19-11-14] |
The position from post #11 is incorrect. Perhaps this may be correct:
Perhaps this works:
 (= 11+9 )
S#15
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| (18) Posted by Olaf Jenkner [Thursday, Nov 14, 2019 23:20] |
Cooks are getting shorter:
1. e8=Q+ Kxe8 2. Sd6+ Kxd7 3. c8=B+ Kc6 4. Rc7+ Kxd6 5. Rc6+ Kd5 6. Qd6+ Ke4 7. Rxc5+ Rd5 8. Rxc4+ Qd4 9. Bxd5+ Ke3 10. Bd2+ Qxd2#
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| (19) Posted by Miodrag Mladenović [Thursday, Nov 14, 2019 23:30] |
I need wB to guard square d8. But it looks like it creates a lots of issues. Perhaps this may be correct:
 (= 11+9 )
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| (20) Posted by Olaf Jenkner [Thursday, Nov 14, 2019 23:34] |
No flaws found.
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