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
S8 competition
Moderator: Psyck
S8 competition
There is a chasm
of carbon and silicon
the software can't bridge
of carbon and silicon
the software can't bridge
Re: S8 competition
I have found a solution in 15.5 moves:
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.
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.
Re: S8 competition
Congratulations 101, you have proved that you were a worthy winner 
There is a chasm
of carbon and silicon
the software can't bridge
of carbon and silicon
the software can't bridge
Re: S8 competition
Here is the proof:
I postulate that the number is correct.
I postulate that the number is correct.