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Post by Admin on May 31, 2014 3:02:40 GMT
each number in the bracket is an entry
a group of entries is called a line
a group of lines is called a plane
[2,3 (1) 3,4]HC the (1) means that [2,3]HC and [3,4]HC are on different lines
[2,3 (1) 3,4]HC simplifies to [2,3,2,3,...,2,3]HC where there are [3,4]HC repetitions of [2,3]HC
EXAMPLE
[2,3 (1) 3,4]HC
[3,4]HC = 1.312*10184 = 13.12 sexginatillion
[2,3 (1) 13.12 sexgintaillion]HC = [2,3,2,3,...,2,3]HC with 13.12 sexagintillion repetitions of [2,3]HC = Big number
In the case where there are multiple lines
[2,3 (1) 3,4 (1) 4,5]HC simplifies to [2,3 (1) 3,4,3,4,...,3,4]HC where there are [4,5]HC repetitions of [3,4]HC
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Post by Admin on May 31, 2014 3:47:31 GMT
If you have a bracket that is separated by a plane
[2,3 (2) 3,4]HC this will simplify to [2,3 (1) 2,3 (1)...(1) 3,4]HC with [3,4]HC repetitions of "2,3 (1)"
this holds constant for entries separated by higher dimensional objects
What happens when you have entries split by several different dimensional objects?
[2,3 (1) 3,4 (2) 4,5 (1) 5,6]HC
This is a 3 dimensional bracket because there are two planes in 3-space. One plane being [2,3 (1) 3,4]HC and the other being [4,5 (1) 5,6]HC
To solve a 3-D bracket identify the last 2-D structure and solve it
[2,3 (1) 3,4 (2) 4,5 (1) 5,6]HC
[4,5 (1) 5,6]HC = [4,5,4,5,...,4,5]HC with [5,6]HC repetitions of [4,5]HC = A
now we have
[2,3 (1) 3,4 (2) A]HC
which simplifies to [2,3 (1) 3,4 (1) 2,3 (1) 3,4 (1)...(1) 2,3 (1) 2,3]HC where there are A repetitions of [2,3 (1) 3,4]HC
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Post by Admin on Jun 4, 2014 17:52:02 GMT
Say you have the case where lines are not only on different planes, but are in different 3-spaces
[1,3 (1) 2,4 (3) 3,5 (1) 4,6]HC
Since this is a 4-D bracket we must start with locating the last 3-D structure and solve it
[1,3 (1) 2,4 (3) 3,5 (1) 4,6]HC
[3,5 (1) 4,6]HC
now we have a nice and simple 2-D bracket which can be solved
[4,6]HC=4.657*10778
[3,5,3,5,...,3,5]HC (with the "3,5" repeated 4.657778 times)=A
So now we have
[1,3 (1) 2,4 (3) A]HC
If the lines were separated by 2-space instead of 3-space
[1,3 (1) 2,4 (2) A]HC
we could just repeat [1,3 (1) 2,4]HC A amount of times
But since the two lines are separated by something higher than 2-space, what you do is drop the separating dimension by 1 (in this case the 3) and use that when you repeat.
EXAMPLE
[1,2 (1) 3,4 (2) 5,6 (1) 7,8 (4) 9,10 (1) 11,12 (2) 13,14 (1) 15,16]HC
Here we have 2 planes each with two lines, and another 2 planes with 2 lines each on completely different 4-D structures
Since this is a 5-D bracket, we must find the last 4-D structure and solve it.
[1,2 (1) 3,4 (2) 5,6 (1) 7,8 (4) 9,10 (1) 11,12 (2) 13,14 (1) 15,16]HC
[9,10 (1) 11,12 (2) 13,14 (1) 15,16]HC
Since this is a 3-D bracket we have to find the last 2-D structure and solve it
[13,14 (1) 15,16]HC
[13,14 (1) A]HC
[13,14,13,14,...,13,14]HC where there are A repetitions of "13,14"=B
So now we have
[9,10 (1) 11,12 (2) B]HC
[9,10 (1) 11,12 (1) 9,10 (1) 11,12 (1)...(1) 9,10 (1) 11,12]HC where there are B repetitions of "9,10 (1) 11,12"=C
Now we have
[1,2 (1) 3,4 (2) 5,6 (1) 7,8 (4) C]HC
[1,2 (1) 3,4 (2) 5,6 (1) 7,8 (3) 1,2 (1) 3,4 (2) 5,6 (1) 7,8 (3)...(3) 1,2 (1) 3,4 (2) 5,6 (1) 7,8]HC repeated C times
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Post by Admin on Jun 5, 2014 15:09:12 GMT
To solve an n-dimensional bracket follow these steps - Locate the last (n-1)-dimensional structure
- Solve it to receive some number "a"
- Repeat the preceding (n-1)-dimensional structure "a" times with an (n-2) dimension separator, you now have an n-dimensional bracket with one less (n-1)-dimensional structure
- Repeat until bracket is solved
Example[1,4 (1) 3,5 (3) 2,4 (1) 6,2 (2) 3,1] HCSince this is a 4-D bracket, we need to locate the last 3-D structure [1,4 (1) 3,5 (3) 2,4 (1) 6,2 (2) 3,1] HCSolve this 3-D bracket (see above for how to) [2,4 (1) 6,2 (2) 3,1] HC= A[1,4 (1) 3,5 (3) A] HCNow we have to repeat [1,4 (1) 3,5] HC A amount of times with a (2) separator (4-2=2) [1,4 (1) 3,5 (2) 1,4 (1) 3,5 (2)...(2) 1,4 (1) 3,5] HC with A repetitions |
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