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Topic: Mutilated Go Board (Read 314 times) |
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ThudnBlunder
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OK, a mutilated mini-go board.
The four corners of the 11x11 board are removed. The object is to place four stones on the corners of a square. The square formed may have edges that are parallel to those of the board or may be tilted (see below). How many possible squares can be thus formed?
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towr
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Re: Mutilated Go Board
« Reply #1 on: Jan 4th, 2004, 7:52am » |
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unless I'm mistaken the answer is ::1173::
For an m-size board (m [ge] 2) with the corners missing it would be ::(m^4 - m^2 - 48 m + 84)/12::
Unless, of course, I'm mistaken, in which case it wouldn't be.. 
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ThudnBlunder
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Re: Mutilated Go Board
« Reply #2 on: Jan 4th, 2004, 9:33am » |
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Quote:| Unless, of course, I'm mistaken |
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...or your source is mistaken. 
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| « Last Edit: Jan 5th, 2004, 6:10am by ThudnBlunder » |
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towr
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Re: Mutilated Go Board
« Reply #3 on: Jan 4th, 2004, 9:52am » |
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I didn't use any sources this time..
I just examind the problem, and generalized for other cases, and tried to deduce a formula.
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TimMann
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Re: Mutilated Go Board
« Reply #4 on: Jan 4th, 2004, 2:05pm » |
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Quibble: The diagram is drawn improperly if you want this to be a Go board. Go is played on the intersections of the lines, not inside the squares. (Also, the stones are black and white, not black and red.) Drawing the diagram properly wouldn't change the problem any, though. You'd make it one row and one column smaller to keep it 11x11 when counting intersections instead of squares.
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| « Last Edit: Jan 4th, 2004, 2:08pm by TimMann » |
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