The Hypercube notational effort | ||
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The notations below were designed by the author to depict positions and vectors in the hypercube. The front subscript runs through the dimensions from 0 to n-1, the normal letters run through the order 0 to m-1. With the understanding that same letters between the n-Points [..] and n-Vectors <..> run through the same range simultaneously, all kinds of lines can be defined. also unequal lettering denote unequal numbers, some entries might be left out. no ordering is infered by the factual placement of front subscripted values Note: the total amount sum of the presubscripts is always the dimension n |
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[_{j}i] | All positions in the hypercube | |
<_{j}i> | All vectors in the hypercube | |
<_{j}-1,_{k}0,_{l}1> | All pathfinder directions through the hypercube | |
[_{j}i,_{k}0]<_{j}0,_{k}1> ; #k = 1 | The hypercubes 1-agonals | |
[_{j}0]<_{j}1> or [_{j}(_{0}i)] |
The hypercubes main n-agonal | |
Note: the positional variants uses _{0}i value in the other directions | ||
[_{j}0,_{k}m-1]<_{j}1,_{k}-1> or [_{j}(_{0}i),_{k}m-1-(_{0}i)] |
The hypercubes (sub) n-agonals | |
Note: the positional variants uses _{0}i value in the other directions also #k=0 equals the main n-agonal, #j=0 reverses the main n-agonal, so effectively every n-agonal is mentioned here in both directions |
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Proof of center number equals the mean value of numbers | ||
[_{j}1,_{k}0,_{l}0] <_{j}0,_{k}1,_{l}0> <_{j}0,_{k}0,_{l}1> ; #j = 0..n-1 ; #k = 1 ; #j + #l = n-1 | ||
The above plane defines an order 3 square within the hypercube spanned by a monagonal and an (#l)-agonal when #j = 0 this general formula reduces to [_{k}0,_{l}0], <_{k}1,_{l}0> <_{k}0,_{l}1> ; #k = 1 ; #l = n-1 which is an oblique plane of the hypercube, with increasing j subsequent squares take the midpoint of the previous squares monagonal as a corner point until #j = n-1 when the general formula reduces to [_{j}1,_{k}0] <_{j}0,_{k}1> <_{j}0,_{k}0> ; #j = 0..n-1 ; #k = 1 a monagonal through the hypercubes center |
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[_{j}1,_{k}0,_{l}0]
[_{j}1,_{k}1,_{l}0]
[_{j}1,_{k}2,_{l}0] [_{j}1,_{k}0,_{l}2] [_{j}1,_{k}1,_{l}2] [_{j}1,_{k}2,_{l}2] |
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The above two lines form the monagonals of the forementioned squares using the 0-centered number range -(m^{n}-1) .. (m^{n}-1) each monagonal and n-agonal sums to 0. adding the two summed monagonals above gives |
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([_{j}1,_{k}0,_{l}0] +
[_{j}1,_{k}1,_{l}0] +
[_{j}1,_{k}2,_{l}0]) +
([_{j}1,_{k}0,_{l}2] +
[_{j}1,_{k}1,_{l}2] +
[_{j}1,_{k}2,_{l}2]) = 0 ==> ([_{j}1,_{k}0,_{l}0] + [_{j}1,_{k}2,_{l}2]) + ([_{j}1,_{k}2,_{l}0] + [_{j}1,_{k}0,_{l}2]) = -([_{j}1,_{k}1,_{l}0] + [_{j}1,_{k}1,_{l}2]) = -([_{s}1,_{l}0] + [_{s}1,_{l}2]) (#s = #j+1) |
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The #j=0 (oblique) squares have diagonals which corresponds to the hypercubes n-agonals so also these diagonals sum up to 0 |
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[_{k}0] + [_{k}1] + [_{k}2]) = 0 and
[_{k}0,_{l}2] + [_{k}1,_{l}1] +
[_{k}2,_{l}0] = 0 ==> ([_{k}0,_{l}0] + [_{k}2,_{l}2]) + ([_{k}2,_{l}0] + [_{k}0,_{l}2]) = -2 [_{k}1,_{l}1] |
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Combining these two #j=0 equations gives the start of an iterative expression note the second line is the first due to symmetry reason, argument put on aspectial variant this form also comes in the more general iterative steps somewhere (I'll try to make this clearer in future upload) |
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#j=0; #k=1 :
[_{k}1,_{l}0] + [_{k}1,_{l}2] =
-2 [_{k}1,_{l}1] more general; #k=1: [_{j}0,_{k}1,_{l}2] + [_{j}2,_{k}1,_{l}0] = -2 [_{j}1,_{k}1,_{l}1] #s=#j+1; #k=1 : ([_{s}1,_{l}0] + [_{s}1,_{l}2]) = ([_{j}1,_{k}1,_{l}0] + [_{j}1,_{k}1,_{l}2]) = -([_{j}1,_{k}0,_{l}0] + [_{j}1,_{k}2,_{l}0]) + -([_{j}1,_{k}0,_{l}2] + [_{j}1,_{k}2,_{l}2]) = -([_{j}1,_{k}0,_{l}0] + [_{j}1,_{k}2,_{l}2] + [_{j}1,_{k}0,_{l}2] + [_{j}1,_{k}2,_{l}0]) = -([_{j}1,_{t}0] + [_{j}1,_{t}2] + [_{j}1,_{k}0,_{l}2] + [_{j}1,_{k}2,_{l}0]) (#t=l+1) = ... -([_{k}1,_{t}0] + [_{k}1,_{t}2] + ... + [_{j}0,_{k}1,_{l}2] + [_{j}2,_{k}1,_{l}0]) (#t=l+1) = ... (the displayed terms both sum to -2 [_{j}1] terms dipicted by ... also sum to this value but also have factors depending on #j with increasing #j thus a factor -2 emerges uptil:) (-2)^{(#s)} [_{k}1,_{l}1] #j=n-1; #k=1 : ([_{j}1,_{k}0] + [_{j}1,_{k}2]) = (-2)^{(#n)} [_{j}1,_{l}1] |
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The #l=0 expression forms a monagonal through [_{j}1] and this monagonal sums up to 0 leading to the conclusion [_{j}1] = 0 Q.E.D. |
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Proof of associativeness | ||
monagonals through [_{j}1] :
[_{j}1,_{l}0] + [_{j}1,_{l}2] =
- [_{j}1,_{l}1] = 0 ==>
[_{j}1,_{l}0] = - [_{j}1,_{l}2] ; #l=1 n-agonals through [_{j}1] : [_{j}0,_{l}2] + [_{j}2,_{l}0] = - [_{j}1,_{l}1] = 0 ==> [_{j}0,_{l}2] = - [_{j}2,_{l}0] from proof above #k=1 : [_{j}0,_{k}1,_{l}2] + [_{j}2,_{k}1,_{l}0] = -2 [_{j}1,_{k}1,_{l}1] = 0 ==> [_{j}0,_{k}1,_{l}2] = - [_{j}2,_{k}1,_{l}0] Q.E.D. |