This is the same as counting sets of 3 numbers (±a, ±b, ±c) where a, b, c have no prime factor among all of them, such as (5, 6, 10). Any such point would then be the first lattice point visible from (0,0,0), hiding all other multiples (±an, ±bn, ±cn). Unfortunately, even for the simpler N x N case, there's no easy formula for finding this number. To get an idea of the difficulty involved, let's try to count the set of numbers (a,b) where a ≤ b, and have no prime factors in common. We have:
(1,0)
(1,1)
(1,2)
(1,3), (2,3)
(1,4), (3,4)
(1,5), (2,5), (3,5), (4,5)
(1,6), (5,6)
etc.
Now, the curious thing is, as you count them up, you get the series
1,2,3,5,7,11,13,....
Looks familiar? Alas! Not so fast. The NEXT term is 19! Followed by 23, 29, 33, 43, etc. This is called the Farey series. And this one has no simple formula, but you might want to check out the link below for more information on this series. This would be a good place to start in order generalize this problem to the N x N x N case.
2007-04-04 04:09:53
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answer #1
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answered by Scythian1950 7
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The problem transforms into following in a 3d space :
Let the given point be (x,y,z)
The lattice is described by the set of all points
(a,b,c) where
a can take integer values from -N/2 to + N/2
and similar for b and C
Total number of points in lattice is (N+1)^3
Now we have to find the number of distinct straight lines passing through (a,b,c) containing at least one of the lattic points
A computer simulation can be done to arrive at this number.
2007-04-03 07:47:06
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answer #2
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answered by Nishit V 3
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