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Topic: Two Attacking Queens (Read 2215 times) |
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ThudnBlunder
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Two Attacking Queens
« on: Oct 8th, 2003, 4:48am » |
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When two queens are randomly placed on an nxn chessboard the probability that they attack each other is 3/11.
What is n?
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BNC
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Re: Two Attacking Queens
« Reply #1 on: Oct 8th, 2003, 5:42am » |
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Either I missed something, or it should be in "easy"...
::
A queen placed randomly on an nxn chessboard controles 3n-3 squares (I don't consider the square on which it stand "controlled")
The second queen may be placed on any of n^2-1 locations.
Solving 3(n-1)/(n^2-1)=3/11 gives n=1/10, so
n=10.
::
What do you say?
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ThudnBlunder
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Re: Two Attacking Queens
« Reply #2 on: Oct 8th, 2003, 6:10am » |
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Quote:| A queen placed randomly on an nxn chessboard controles 3n-3 squares |
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That is true only of queens placed randomly on the edge of the board.
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towr
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Re: Two Attacking Queens
« Reply #3 on: Oct 8th, 2003, 7:24am » |
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It's pretty easy, I think.. the answer is ::11::
Just a matter of finding the right formula and then find the n for which you get the given probability..
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ThudnBlunder
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Re: Two Attacking Queens
« Reply #4 on: Oct 8th, 2003, 9:21am » |
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Quote:| Just a matter of finding the right formula |
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And how do you easily do that? I don't see the point in posting bald answers.
(I suppose the fact that the method is obvious makes it Easy.)
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| « Last Edit: Oct 8th, 2003, 9:57am by ThudnBlunder » |
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towr
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Re: Two Attacking Queens
« Reply #5 on: Oct 8th, 2003, 10:55am » |
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on Oct 8th, 2003, 9:21am, THUDandBLUNDER wrote:And how do you easily do that? I don't see the point in posting bald answers.
(I suppose the fact that the method is obvious makes it Easy.)
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The way I allways start is first looking at small problems, n=2, n=3, n=4. Once you see the regularity it's (in this case) easy enough to generalize it, and a math-program can help simplify it (though that's not even necessary)
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| « Last Edit: Oct 8th, 2003, 10:55am by towr » |
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This may be taking it the long way around, but here goes.
First figure the total number of squares attacked by a queen placed on every square of the board. That works out to
(n^2)*3(n-1)+ 2(n-2)^2+2(n-4)^2+2(n-6)^2... until n-? reaches 1 or 0. (Can that equation be simplified?)
Divide that by n^2 to get the average attacks.
Divide that by n^2 -1 to get the probability.
Then plugging different numbers in for n leads you to the correct solution.
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James Fingas
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Re: Two Attacking Queens
« Reply #7 on: Oct 8th, 2003, 1:00pm » |
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It's definitely possible to come up with a simpler formula. If you compute the answer for n=1 to 10 or so, you can find a pattern fairly easily. Hint: it doesn't grow exponentially.
I factored out the 2n-2 squares that are covered horizontally, then the n-1 squares that are always covered diagonally, then assigned a number to each square on the board specifying how many squares more than those 3n-3 are covered. The sum of these values has a fairly simple formula.
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