OpenChess

Martin held an 'S8' competition, to draw the the characters S and 8 beginning from the starting position and ending with the Kings at the ends of 8, on f2 and f6.

The competition ended with a win by 101, in about 18 moves.

This is the shortest solution I have seen so far, submitted after Martin closed the competition:

1.h4 e5 2.Nc3 Qxh4 3.a4 Qxf2+ 4.Kxf2 a5 5.Rxh7 g5 6.Rxf7 Ra6 7.Nh3 Rxh3 8.d3 Rxd3 9.Bf4 Rxd1 10.Rxd7 Rxa1 11.Rxc7 Nc6 12.e3 Bb4 13.Rxc8+ Ke7 14.Rxg8 b6 15.Bc4 Ra2 16.g3 Kf6 17.Rxg5
Solution by user - Thanar

I have found a solution in 15.5 moves:
1. a4!? c6 2. Ra2 Qb6!? 3. f4 Qxg1 4. e3 Qxh2 5. g3 Qxd2+ 6. Bxd2 a5 7. Rxh7 e5 8.
Bxa5 Rxa5 9. Rxg7 Bb4+ 10. Nc3 b6 11. Rxf7 Kxf7! 12. Qxd7+ Kf6 13. Qxc8 Na6 14.
Qxg8 Rxg8 15. Kf2 Rg5 16. Bc4

I have written a WordPad document which together with a mess of variations (not sufficiently annotated for proof standard, though) show that this is indeed the minimum number of moves needed. I will publish the analysis when it actually comes close to a rigorous proof, but until then, you can take my word for 15.5 moves being the minimum.

See http://rybkaforum.net/cgi-bin/rybkaforu ... #pid557167 for more details on the problem.

Congratulations 101, you have proved that you were a worthy winner

Here is the proof:

I postulate that the number is correct.