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, rows and quadrants add up in perfection.
http://www.edcollins.com/chess/knights-tour.htm
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Originally posted by crazyblueThat's quite amazing. Anyone has a maths explanation for this? It can't be a coincidence...can it?
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
Originally posted by inipsianiniWhile 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?
place all 8 queens on the board that each other wont treaten or pass each other.
Originally posted by ilywrin16 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.
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 Mephisto2I may be counting wrong but here's my idea: both sides have 6 spare pieces to be captured (excluding the queens and kings).
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.
Originally posted by ilywrinIt's possible. Just did it. No PGN sorry. It was utterly long, but not hard. Just a matter of moving stuff around.
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.