~~skeeter~~515 + 30 days~~skeeter~~515 + 30 days- 05 Aug '05 09:16Since it's solved, here's something amazing that i found recently about the so called knight tour. If you scroll down a bit (to the colored section) you will find the "SUPER magic square"....awesome, how the numbers in all columns, rows and quadrants add up in perfection.

http://www.edcollins.com/chess/knights-tour.htm - 05 Aug '05 13:19 / 1 edit

That's quite amazing. Anyone has a maths explanation for this? It can't be a coincidence...can it?*Originally posted by crazyblue***Since it's solved, here's something amazing that i found recently about the so called knight tour. If you scroll down a bit (to the colored section) you will find the "SUPER magic square"....awesome, how the numbers in all columns, ro ...[text shortened]... in perfection.**

http://www.edcollins.com/chess/knights-tour.htm - 26 Aug '05 16:39

While this has already been done I did notice the word "all" that you placed before the 8 queens. Now the number of all possible queens would be 18 (2 from the beginning and 16 promoted). However I am uncertain if there is a legal sequence of moves leading to such position, and if there is, what is the shortest game that produces 18 queens?*Originally posted by inipsianini***place all 8 queens on the board that each other wont treaten or pass each other.** - 26 Aug '05 17:05

16 promoted queens is not possible. There are only 7 non-pawn pieces reach side to capture. A pawn has to capture something to get behind the opposing pawn.*Originally posted by ilywrin***While this has already been done I did notice the word "all" that you placed before the 8 queens. Now the number of all possible queens would be 18 (2 from the beginning and 16 promoted). However I am uncertain if there is a legal sequence of moves leading to such position, and if there is, what is the shortest game that produces 18 queens?** - 26 Aug '05 20:42

I may be counting wrong but here's my idea: both sides have 6 spare pieces to be captured (excluding the queens and kings).*Originally posted by Mephisto2***16 promoted queens is not possible. There are only 7 non-pawn pieces reach side to capture. A pawn has to capture something to get behind the opposing pawn.**

White gets 3 pawns at the g file with two captures (f and h pawn), 3 pawns at the d file with two more captures (c and e pawn) and lastly one more capture to get two pawns at b file (a-pawn); Black in similar manner get 3 on c and f files and h and a file pawns remain on their files. That's 5 captures from White and 4 on Black's part, accounting for 9 of the twelve spare pieces. Then the queens are promoted on the respective files. - 26 Aug '05 21:28

It's possible. Just did it. No PGN sorry. It was utterly long, but not hard. Just a matter of moving stuff around.*Originally posted by ilywrin***I may be counting wrong but here's my idea: both sides have 6 spare pieces to be captured (excluding the queens and kings).**

White gets 3 pawns at the g file with two captures (f and h pawn), 3 pawns at the d file with two more captures (c and e pawn) and lastly one more capture to get two pawns at b file (a-pawn); Black in similar manner get 3 on c and f f ...[text shortened]... counting for 9 of the twelve spare pieces. Then the queens are promoted on the respective files.