The Magic Encyclopedia ™

The order 3 Hypercube
<Walter Trump>
(Aale de Winkel)

<Walter Trump> discussed with me a proof of features in the order 3 hypercube nH3. Below a formulation in hypercube notation which is recaptioned in the first section of the table below.
For the proofs I needed to specify means to state the amount of values, since I don't remember a mathematical notation for this I define #X=Y to denote the there are exactly Y X's

Note:in the course of this rewrite I encountered a few problems, these will be corrected asap (the proof is solid, it is just in the semantics) this note will be removed when this is corrected

The Hypercube notational effort
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
[ji] All positions in the hypercube
<ji> All vectors in the hypercube
<j-1,k0,l1> All pathfinder directions through the hypercube
[ji,k0]<j0,k1> ; #k = 1 The hypercubes 1-agonals
[j0]<j1>
or
[j(0i)]
The hypercubes main n-agonal
Note: the positional variants uses 0i value in the other directions
[j0,km-1]<j1,k-1>
or
[j(0i),km-1-(0i)]
The hypercubes (sub) n-agonals
Note: the positional variants uses 0i 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
Proof of center number equals the mean value of numbers
[j1,k0,l0] <j0,k1,l0> <j0,k0,l1> ; #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
[k0,l0], <k1,l0> <k0,l1> ; #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
[j1,k0] <j0,k1> <j0,k0> ; #j = 0..n-1 ; #k = 1
a monagonal through the hypercubes center
[j1,k0,l0] [j1,k1,l0] [j1,k2,l0]
[j1,k0,l2] [j1,k1,l2] [j1,k2,l2]
The above two lines form the monagonals of the forementioned squares
using the 0-centered number range -(mn-1) .. (mn-1) each
monagonal and n-agonal sums to 0. adding the two summed monagonals above gives
([j1,k0,l0] + [j1,k1,l0] + [j1,k2,l0]) + ([j1,k0,l2] + [j1,k1,l2] + [j1,k2,l2]) = 0 ==>
([j1,k0,l0] + [j1,k2,l2]) + ([j1,k2,l0] + [j1,k0,l2]) = -([j1,k1,l0] + [j1,k1,l2]) = -([s1,l0] + [s1,l2]) (#s = #j+1)
The #j=0 (oblique) squares have diagonals which corresponds to the
hypercubes n-agonals so also these diagonals sum up to 0
[k0] + [k1] + [k2]) = 0 and [k0,l2] + [k1,l1] + [k2,l0] = 0 ==>
([k0,l0] + [k2,l2]) + ([k2,l0] + [k0,l2]) = -2 [k1,l1]
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)
#j=0; #k=1 : [k1,l0] + [k1,l2] = -2 [k1,l1]
more general; #k=1: [j0,k1,l2] + [j2,k1,l0] = -2 [j1,k1,l1]

#s=#j+1; #k=1 : ([s1,l0] + [s1,l2]) =
([j1,k1,l0] + [j1,k1,l2]) = -([j1,k0,l0] + [j1,k2,l0]) + -([j1,k0,l2] + [j1,k2,l2]) =
-([j1,k0,l0] + [j1,k2,l2] + [j1,k0,l2] + [j1,k2,l0]) =
-([j1,t0] + [j1,t2] + [j1,k0,l2] + [j1,k2,l0]) (#t=l+1) = ...
-([k1,t0] + [k1,t2] + ... + [j0,k1,l2] + [j2,k1,l0]) (#t=l+1) = ...
(the displayed terms both sum to -2 [j1] 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) [k1,l1]

#j=n-1; #k=1 : ([j1,k0] + [j1,k2]) = (-2)(#n) [j1,l1]
The #l=0 expression forms a monagonal through [j1] and this monagonal sums
up to 0 leading to the conclusion [j1] = 0 Q.E.D.
Proof of associativeness
monagonals through [j1] : [j1,l0] + [j1,l2] = - [j1,l1] = 0 ==> [j1,l0] = - [j1,l2] ; #l=1
n-agonals through [j1] : [j0,l2] + [j2,l0] = - [j1,l1] = 0 ==> [j0,l2] = - [j2,l0]
from proof above #k=1 : [j0,k1,l2] + [j2,k1,l0] = -2 [j1,k1,l1] = 0 ==> [j0,k1,l2] = - [j2,k1,l0]
Q.E.D.