Knights Puzzle

Here is a challenge.On a board,place 12 knights so all the squares on the board are occupied or threatened.Good Luck.

🙂🙂😉😀

Originally posted by Alex011893
Here is a challenge.On a board,place 12 knights so all the squares on the board are occupied or threatened.Good Luck.

🙂🙂😉😀

This should do it.



skeeter

Originally posted by skeeter
This should do it.

[fen]8/5n2/1nn1nn2/2n5/5n2/2nn1nn1/2n5/8[/fen]

skeeter

What I've just noticed is that the board is also perfectly asymetrical.

skeeter

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

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

That's quite amazing. Anyone has a maths explanation for this? It can't be a coincidence...can it?

thanks to all.🙂

Originally posted by skeeter
This should do it.

[fen]8/5n2/1nn1nn2/2n5/5n2/2nn1nn1/2n5/8[/fen]

skeeter

how did you insert the chess board graphic?

that is easy..

place all 8 queens on the board that each other wont treaten or pass each other.

Originally posted by inipsianini
place all 8 queens on the board that each other wont treaten or pass each other.

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 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?

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 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.

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 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 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.

It's possible. Just did it. No PGN sorry. It was utterly long, but not hard. Just a matter of moving stuff around.