19
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Here's a chess problem which plays very like a sliding-block puzzle, hence the rather unusual tag combo.

Ljubomir Ugren 2nd Prize, Mat, 1976; John Nunn, Solving in Style, no. 183

White Kd1 Rb4, Black Kb1 Qb2 Ra1c2 Ba2 Pa3a4c3c4d2

sh#19

To explain that stipulation "sh#19": This problem is a serieshelpmate in 19. In a serieshelpmate, Black starts, and plays a series of moves, with no White moves in between. Then, at the end, White plays one move, which checkmates Black. Black is not allowed to move so that his king is in check. Black is not allowed to give check on any move except the last move of the series. In this problem, Black plays a series of 19 moves before White plays the final checkmating move.

You, the solver, stipulate all the moves -- that's the help bit. It's not like an orthodox problem where you must specify what White does against any defence from Black.

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  • $\begingroup$ OK, so do normal chess rules apply? And can black play any number of moves in the series before white moves? $\endgroup$ – Sid Oct 14 '16 at 9:55
  • $\begingroup$ @Sid Aside from the restrictions on checks, normal chess rules apply. I tell you that Black needs to make 19 moves -- if you can make it work where Black plays fewer than 19, that's legal, but you'd then have cooked the problem! (Just as if a problem said "White to play and mate in 3" and you showed how White to play can force mate in 2.) $\endgroup$ – Rosie F Oct 14 '16 at 9:59
  • $\begingroup$ Another clarification, so, who wins? White or black? $\endgroup$ – Sid Oct 14 '16 at 10:01
  • $\begingroup$ You could say White wins, seeing as White's move checkmates Black. $\endgroup$ – Rosie F Oct 14 '16 at 10:02
  • 2
    $\begingroup$ This FEN string might come in useful: 8/8/8/8/pRp5/p1p5/bqrp4/rk1K4 b - - 0 1 $\endgroup$ – squeamish ossifrage Oct 14 '16 at 10:38
3
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After looking at the other answers, I think I found something they missed:

1. ... Bb3 2. ... Ra2 3. ... Qa1 4. ... Rb2 5. ... Ba2! 6. ... Rb3 7. ... Qb2 8. ... Ka1 9. ... Bb1 10. ... a2 11. ... Qa3 12. ... Kb2 13. ... a1=N 14. ... Qa2 15. ... Ka3 16. ... Rbb2 17. ... Nb3 18. ... Qa1 19. ... Ka2 20. Rxa4#

By blocking on b1 with the bishop, a tempo is saved, compared to the lines where both pawns are promoted.

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  • $\begingroup$ Correct. Well done! You've solved it. $\endgroup$ – Rosie F Oct 15 '16 at 6:40
1
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Here's what I have got so far.. I can get it in 22 moves but not in 19. Any edits would be appreciated.

1.Bb3
2. Rb4 Ra2
3. Rb4 Qa1
4. Rb4 Rab2
5. Rb4 a2
6. Rb4 a3
7. Rb4 Ba4
8. Rb4 Rb3
9. Rb4 Qb2
10. Rb4 a1=B
11. Rb4 a2
12. Rb4 Qa3
13. Rb4 Bc6
14. Rb4 Bb2
15. Rb4 Qa4
16. Rb4 Bc1
17. Rb4 a1=N
18. Rb4 Ka2
19. Rb4 Rb1
20. Rb4 Bd5
21. Rb4 Nb3
22. Rb4 Rcb2
23. Rxa4#

As @Sconibulus pointed out, It can be done in 20 moves as well.. (I had independently found this variation, 5 minutes ago)

Bb3 2. Rb4 Ra2 3. Rb4 Qa1 4. Rb4 Rab2 5. Rb4 a2 6. Rb4 a3 7. Rb4 Ba4 8. Rb4 Rb3 9. Rb4 Qb2 10. Rb4 a1=N 11. Rb4 a2 12. Rb4 Ra3 13. Rb4 Nb3 14. Rb4 Nc1 15. Rb4 a1=N 16. Rb4 Nab3 17. Rb4 Ra1 18. Rb4 Ka2 19. Rb4 Qb1 20. Rb4 Rb2 21. Rxa4#

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  • $\begingroup$ Looks good to me, apart of course from being 3 moves too long. Nice try, though. $\endgroup$ – Rosie F Oct 14 '16 at 12:57
1
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I've got a win in 21 20 19!

1.Bb3, 2.Ra2, 3.Qa1, 4.Rab2, 5.a2, 6.a3, 7.Ba4, 8.Rb3, 9.Qb2, 10.a1=N,
11.a2, ... (12.Ra3, 13.Nb3, 14.nc1, 15.a1=N, ...
(16.Nb3, 17.Ra2, 18.Ka1, 19.Qb1, 20.Rab2, 21.Ka2)
(16.Ka2, 17.Qb1, 18. Rb3, 19. Rbb2, 20.Nb3)
16.Nb3, 17.Ra1, 18.Ka2, 19.Qb1, 20.Rb2)
12.Qa3,13.Rbb2 14.Nb3, 15.Nc1, 16.a1=N, 17.Nb3, 18.Qa1, 19.Ka2, 20.Qb1
Then white plays Rxa4#

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  • $\begingroup$ That looks good to me too (as well as Sid's try), apart of course from being 2 moves too long. (20 Rab2, btw.) Nice try, though. $\endgroup$ – Rosie F Oct 14 '16 at 13:09
  • $\begingroup$ Ka2 and Qa2?? Some error? $\endgroup$ – Sid Oct 14 '16 at 16:43
  • $\begingroup$ @Sid oops, Qb1, not a2 $\endgroup$ – Sconibulus Oct 14 '16 at 16:44
  • $\begingroup$ You wrote 18 twice... $\endgroup$ – Sid Oct 14 '16 at 16:45

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