segunda-feira, 13 de junho de 2016
Xadrez - 64 Torres
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ResponderExcluirEste comentário foi removido pelo autor.
ExcluirHint: in the beginning of the process, a Rook in one of the 4 corners cannot be removed, as it attacks 2 other Rooks. And it will continue to attack 2 other Rooks, unless you remove all the other Rooks from the same row, or from the same column. But that would include removing another "corner Rook".
ResponderExcluirWhat can we conclude from this?
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ExcluirThe maximum possible number of rook removals is fifty-nine (59).
ResponderExcluirHere's the procedure for accomplishing this:
First, you choose a row and column that aren't edges, then you remove, one by one, starting with the edges, all of the rooks other than the corner rooks that are not in that row or that column.
Now, one by one, starting the outermost rooks, remove the rooks in that row and column, leaving the one located where that row and column intersect.
Now you've got only five rooks left and you can't remove any more.
Well done, Jake. We can remove 59 Rooks at most. Five Rooks must stay on the board: 4 in the corners, and 1 somewhere else.
ExcluirAs I wrote above, in the beginning of the process, a Rook in one of the 4 corners cannot be removed, as it attacks 2 other Rooks. And it will continue to attack 2 other Rooks, unless we remove all the other Rooks from the same row, or from the same column. But that would include removing another "corner Rook".
ResponderExcluirFrom that, we conclude that the 4 Rooks in the corners of the chessboard will never be removed, as each one of them will always attack 2 other Rooks, no matter how many Rooks we remove. For a corner rook to attack only one Rook, another corner Rook would have to be removed, which is impossible in the first place.
Next, I note that we cannot leave behind only the four Rooks in the corners, because the last Rook we remove from the chessboard would either attack no Rook (if it is on one of the 36 central squares) or two corner Rooks (if it is on one of the 24 border squares). Either way, it is an even number of Rooks.
Now, I just need to find a step-by-step solution in which 59 Rooks are removed. There are many possible ways to do this. I did it this way (the numbers denote the order in which the Rooks were removed):
** 01 02 03 04 05 06 **
13 50 51 52 53 54 55 07
14 15 16 17 18 19 56 08
20 21 22 23 24 25 57 09
26 27 28 29 30 31 58 10
32 33 34 35 36 37 59 11
38 39 40 41 42 43 !! 12
** 44 45 46 47 48 49 **
The "!!" is the Rook that remains, together with the corner Rooks, marked as "**". :)
Hi, Gents.
ResponderExcluirIt reminded me a book I had 30 and so years ago called "Ajedrez y las Matematicas - las relaciones matematicas en el ajedrez", too complex for my understanding, I must confess.
Fluzão 2x0 Mulambos Paulistas
Mauro,
ExcluirI would love to take a look at this book. Chess and math are two passions of my life.
Fluzão 1 x 0, it was enough. :)
Seconded.
Excluir