Subject:
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Combining 2x4 bricks, again
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Newsgroups:
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lugnet.general
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Date:
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Fri, 22 Nov 2002 07:40:30 GMT
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Highlighted:
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Hi everybody,
There was a topic of the number of combinations of 2x4 bricks.
http://news.lugnet.com/general/?n=29016&t=i&v=a
This is an answer to that.
*Problem*
Compute the number of all combinations of n bricks with 2x4 size and
same color. Count rotation images as the same solid, and mirror images
as different solids.
In fact, LEGO company computed it under the following:
*Condition*
The solids are n-story buildings. For example (n=3),
_[]_
__[]
_[]_ is the case, which is a view from the side.
But
_[][]_
__[]__ is not.
Lack of information of this condition was the origin of
misunderstandings and difficulty of the computation. But under
*Condition*, the computation is easy. The following *Answer*
coincides with LEGO's result:
*Answer*
# = (2^(n-1) + 46^(n-1))/2, where ^ denotes the power.
For n = 2, 3, 6, # = 24, 1060, 102981500, respectively.
*Proof*
Fix the brick on the 1st floor.
Counting rotation images as DIFFERENT solids,
the number of combinations of n bricks is 46^(n-1),
since there are 46 choices to put an brick on another (fixed) brick.
There are 2^(n-1) combinations such that the solid is the same as its
rotation image, since there are 2 choices to put an brick with keeping
symmetry: the same place as the downstairs brick or the place rotated
by 90 degrees.
So the number of non-symmetric combinations is
46^(n-1) - 2^(n-1) regarding rotation images as DIFFERENT solids,
and (46^(n-1) - 2^(n-1))/2 regarding rotation images as the SAME.
Summing up the symmetric and the non-symmetric, we have the answer. []
There remains the problem for the case where the solid is not
necessarily an n-story building. I only have a result 1560 for n=3
using a computer. I think it is computable until n=5 or 6.
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