18 Queens on the board in 48 moves
Ucalegon
<ecol...@inficad.com> writes:
> But what's the SHORTEST game that can be
> constructed where 18 Queens on the board are achieved?
This question was posted back on 30 Dec 1998 and I'm finally
getting around to offering some additional information.
History first.
John Olle wrote in a 25 Jan 1999 posting to an offshoot thread, 'I posed the
question in 1974 or 1975 at Monash University Chess Club, Melbourne, Victoria,
Australia.' He found a 49-move solution which a friend later shortened to 48.5.
The first I knew of it was as the '1. *f*-Preisraetsel' in issue #31 of the
German heterochess problem magazine *feenschach*. This was the last issue of
1975, though it might have appeared later than the cover date. Friedrich
Burchard
and Friedrich Hariuc independently came up with 48-move solutions leading to
almost
identical final positions. The game published in issue #33 (reprinted in #109,
from
which I take my information) was:
1. e4 f5 2. e5 Nf6 3. e:f6 e5 4. g4 e4 5. Ne2 e3 6. Ng3 e2 7. h4 f4
8. h5 f:g3 9. h6 g5 10. Rh4 g:h4 11. g5 g2 12. g6 Bg7 13. h:g7 g1=Q
14. f4 h3 15. f5 h2 16. b4 a5 17. b5 a4 18. b6 a3 19. Bb2 Ra7 20. b:a7
a:b2
21. a4 b5 22. a5 b4 23. a6 b3 24. c4 h1=Q 25. c5 h5 26. c6 Bb7 27. c:b7
c5
28. d4 c4 29. d5 Nc6 30. d:c6 c3 31. c7 c2 32. c8=Q c1=Q 33. b8=Q Qcc7
34. a8=Q d5 35. a7 d4 36. Nc3 d:c3 37. Qa6 c2 38. Qa8b7 c1=Q 39. a8=Q Qhd5
40. g:h8=Q+ Kd7 41. g7 b:a1=Q 42. g8=Q b2 43. f7 b1=Q 44. f8=Q h4
45. f6 h3 46. f7 h2 47. Qfa3 h1=Q 48. f8=Q e:f1=Q+.
Final position: QQ1q1QQQ/1Qqk4/Q7/3q4/8/Q7/8/qqqQKqqq.
It was suspected at the time that this was the only possible approach (WK never
moves, BK only once). However, the problem was published as a Christmas prize
problem in *Die Welt*, 1989, and four different games were found, all of them
48-movers. The most spectacular was one by Andreas Rein in which the kings go
travelling and fifteen promoted queens remain on their start squares:
1. h4 g5 2. h5 g4 3. Rh2 g3 4. h6 g:h2 5. g4 e5 6. g5 Ne7 7. g6 a5 8. g7 a4 9.
g8=Q a3
10. Qg2 Bg7 11. f4 c5 12. f5 c4 13. f6 c3 14. f:e7 f5 15. b4 Kf7 16. h:g7 h5
17. b5 h4
18. b6 h3 19. Bb2 f4 20. e4 f3 21. d4 d5 22. Nd2 Bf5 23. Kf2 c:d2 24. Kg3 Kg6
25. c4 Kg5 26. e:f5 e4 27. c5 a:b2 28. Bc4 d:c4 29. a4 f2 30. Qd1f3 Ra7 31.
b:a7 b5
32. a5 b4 33. a6 b3 34. c6 c3 35. c7 c2 36. d5 e3 37. d6 e2 38. d7 h:g1=Q
39. g:h8=Q b:a1=Q 40. a:b8=Q h2 41. f6 b2 42. f7 h1=Q 43. a7 b1=Q 44. e8=Q
Qd8f6
45. c8=Q c1=Q 46. d8=Q d1=Q 47. a8=Q e1=Q 48. f8=Q f1=Q+
Final position:QQQQQQ1Q/8/5q2/6k1/8/5QK1/6Q1/qqqqqqqq.
In early January Alfred Pfeiffer published yet another solution, by Theodor
Burian, from
*ROCHADE Europa*, Vol. 11/1997, which I'll quote here just to have all the
posted solutions in one place:
1.b4 a5 2.b5 a4 3.b6 a3 4.Bb2 axb2 5.a4 Ra7 6.bxa7 b5 7.h4 c5
8.h5 g5 9.h6 Bg7 10.hxg7 h5 11.g4 h4 12.a5 Rh5 13.gxh5 b4 14.f4
e5 15.f5 d5 16.f6 Ne7 17.fxe7 f5 18.h6 f4 19.e4 Bf5 20.exf5 Kf7
21.f6 f3 22.h7 g4 23.a6 g3 24.h8=Q g2 25.Kf2 h3 26.Kg3 c4 27.d4
c3 28.Nd2 cxd2 29.c4 b3 30.c5 e4 31.Bc4 dxc4 32.d5 h2 33.d6 f2
34.Qdh5+ Ke6 35.a8=Q gxh1=Q 36.e8=Q+ Qe7 37.a7 e3 38.d7 hxg1=Q+
39.Qg2 c3 40.a8=Q f1=Q 41.c6 e2 42.c7 bxa1=Q 43.cxb8=Q b2
44.d8=Q c2 45.f7 b1=Q 46.g8=Q c1=Q 47.Qgh7 d1=Q 48.f8=Q e1=Q+
Final position:QQ1QQQ1Q/4q2Q/4k3/7Q/8/6K1/6Q1/qqqqqqqq.
Since several attempts with varying approaches over more than two decades have
all reached the same result, 48 moves may be the absolute minimum.
Always carry a grapefruit, Treesong
Acag, Treesong (ucal...@aol.com)
Ucalegon
A brief addendum to the history, to get all the facts online:
I went home and dug out my *feenschach* #33 (Apr-May 1976) with
the results of the 18-queens prize puzzle. There I found that the
problem goes back a little further: it had been posed by V. Röpke in
*Skakbladet* in April 1972, though no result was published until
the March 1976 issue, that being a not-quite-optimum 48.5-move solution.
So at least three people posed the problem independently in 1972-75.
Seems that it was just time for the problem to appear.
Acag, Treesong (ucal...@aol.com)
John Olle
I have dug out my problem as stated at Monash Chess Club some time during
1972-5 (when I was a student there). This problem gives a numerical mark to
the solver.
It reads:
==================================================
Find x(max) such that x(max) is the maximum number of queens which can
simultaneously exist on the board in a legal game.
Find y(min) such that y(min) is the least number of moves (one move is
defined as a turn by White and Black so y ends in 1/2 if White makes the
last move) in which this can be achieved.
Compose a legal game of y moves so that there are x queens in the final
position.
Aim: Maximize x, then minimize y.
For any attempt where x >= 3, give yourself a mark, M out of 100 where:
M = 5(x+2)**3 (cubed)
---------
8(y-1)
======================================================
It is clear that my best solution at the time was 49 moves (Albert (now Dr)
Braunstein shaved a half move off my previous best solution of 49.5 moves).
The formula should be changed to reflect 48 moves ie replace (y-1) with
(y-2).
Perhaps this was the first (or only) chess problem where you can receive a
numerical mark out of 100 for your solution?
Thanks for research into this problem.
John Olle
--
Visit the Bendigo Chess Club home page:
http://ironbark.bendigo.latrobe.edu.au/staff/olle/chess/bcchome.html
and/or the first of my computer and chess columns for the Australian Chess
Federation:
http://www.somerset.qld.edu.au/chess/columns.htm follow link to 'Silicon
Difference'.
John Olle phone: (03) 5444 7350
La Trobe University, BENDIGO FAX: (03) 5444 7998
PO Box 199
BENDIGO, VIC, 3552, AUSTRALIA email: J.O...@bendigo.latrobe.edu.au
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