The Magic Encyclopedia ™

Concentric Bordering
(by Aale de Winkel)
note: investigative article

The Concentric hypercubes consist of consecutive numbers in each subhypercube for 3(4) to m centered at the hypercube center, ie. every such centered hypercube is an augmented magic hypercube of that order by a factor to allow the border filling to be such that the hightened order hypercube consist of consequtive numbers also. This process can be called "Concentric Bordering". Below I put forward some ideas about concentric bordering, notice that these notes are preliminairy regularisations of analisis of the cube concentric border shown below.

The Concentric Bordering
Augmentation factor
Δp,q = (pn-qn)/2
the augmentation factor of the order q subhypercube
The center of the order m hypercube is (mn+1)/2 so
Δp,q = (pn+1)/2 - (qn+1)/2 = (pn-qn)/2
Augmented sum
ΔS(H[(p-q)]q) = q(pn-qn)/2
the augmentation of the sum of the order q subhypercube
Each cell of the subhypercube is augmented by (Δp,q)
which therefore is counted q times.
Concentric border sum
S(Bp,q) = (p-q)(pn+1)/2
Since the order q subhypercube is an augmented magic hypercube the borders
1-agonal and n-agonal numbers need to sum up to the same constant.
The remaining sum the to be corrected by the border is the difference in the magic
sums of the target order p and the suborder q so:
S(Bp,q) = S(Hp)-S(H[(p-q)]q) = p(pn+1)/2 - q(qn+1)/2 - q(pn-qn)/2 = (p-q)(pn+1)/2
The BorderFlip
[0,0] <=> [m-1,m-1]
[0,m-1] <=> [m-1,0]
[i,0] <=> [i,m-1] i=1..m-2
[0,i] <=> [m-1,i] i=1..m-2
A "borderflipped" magic square is still a magic square
With the exchange of the numbers in the mentioned positions all 1-agonal
and 2-agonals remain having the same numbers.
Augmenting square
[Collapsed Latin square] *
(mn-m2)/(m-1)
(square factor: m)
A square augmenting the number from the square to the hypercube range
A Latin square can be collapsed by a digit change with duplicated digits
as long as the sum of those digits remain m(m-1)/2 the squares sum remains
multiplying this collapsed square with [m(mn+1)/2-m(m2+1)/2]/[m(m-1)/2] = (mn-m2)/(m-1)
is sufficient add onn to augment the sums as appropriate.
for the square this reduces to [m(m2+1)/2-m(m+1)/2]/[m(m-1)/2] = m
The above outlines the basic principles, the sample in the table below show that
imperfections in the augmenting hypercube can be corected in the augmented hypercube
(current expertise suggest these to be forced upon this scheme due to "natural overlap")
Due to lack of expertise the following are some inexact qualitative statements
Future upload these statements needs to be exactified.
"natural overlap" reason for the imperfection in the borders
The number range of the bordered hypercube is [(pn+1)/2-(qn-1)/2,[(pn+1)/2+(qn-1)/2]

Not clear yet how to formulate this phenomonon, the lowerbound of the bordered
[(pn+1)/2-(qn-1)/2 - 1(?)] % (pn-p2)/(m-1) ain't 0 so has overlap with [1..x (pn-p2)/(p-1)].
Forced imperfection Due to "natural overlap" in the numberanges of the border and the bordered
two imperfections are forced upon this scheme
The samples shown below suggest it might be not possible to find the above intended
regular borders, so the augmenting hypercube might neccesairily have an off summing
line (+/-1) the border symmetry suggest at least one other
(curently I don't see the "proof" of this, future upload might hold this)


Case study
The following case studies generate conditions for the main part of the border
however I think it also applies to latin content of the Augmenting hypercube
The Concentric Bordered square
The above construction is general and so also applicable for the square
This section provide some notes for this situation
Magic line pair
[a,..,b] [a,..,m+1-b]
The set of permutation [a,..,b] [a,..,m+1-b] in the range [1..m] sum automatically
to the magic sum m(m+1)/2 and are the filling of the first row resp first column
the pair (b,m+1-b) thus automatically fulfill the subdiagonal restriction, further
content of the last row and column follow from the 1-agonal and main-diagonal condition
The Concentric Bordered cube
The following section is a case study of the situation for the cube to provide understanding
of further conditions which might be genarisable to the highter dimensions
Magic Square triplet
{Sf,Sl,St}
The triplet of squares to the front, left and top share the number at position [0,0,0]
further the line [i,0,0] is common to the front and top square, while the line
[0,i,0] is shared by the front and left facial squares of the target cube and
[0,0,i] is shared by the left and top facial squares of the target cube
aside from the above, note that the back square is a complemented borderflipped copy
of the front square which connects the leftside square line [0,i,m-1] with the front
squares line [m-1,i,0] and the top square line [i,0,m-1] with the front line [i,m-1,0]
further reflection also connect the left squares bottom line [0,m-1,i] with the top
squares line [m-1,0,i] in a simular manner.
thus summarising in the following conditions:
Sl[0,i,0] = Sf[0,i,0], Sl[0,0,i] = St[0,0,i], St[i,0,0] = Sf[i,0,0]; i = 0..m-1
Sl[0,i,m-1] = m2+1 - Sf[m-1,i,0], St[i,0,m-1] = m2+1 - Sf[i,m-1,0]; i = 2..m-2
Sl[0,0,m-1] = m2+1 - Sf[m-1,m-1,0], Sl[0,m-1,m-1] = m2+1 - Sf[m-1,0,0]
St[0,0,m-1] = m2+1 - Sf[m-1,m-1,0], St[m-1,0,m-1] = m2+1 - Sf[0,m-1,0]
Sl[0,m-1,i] = m2+1 - St[m-1,0,i]; i = 2..m-2

<Walter Trump> found the first concentric border for order 5 concentric cubes. His investigation can be found at his site: magic-squares

Concentric Square Border
01 02 18 21 23
19 ............ 07
20 ............ 06
22 ............ 04
03 24 08 05 25
div 5 border
0 0 3 4 4
3 ....... 1
3 ....... 1
4 ....... 0
0 4 1 0 4
mod 5 border
0 1 2 0 2
3 ....... 1
4 ....... 0
1 ....... 3
2 3 2 4 4
Note in this sample the imperfection of the
first and last row (noticable in the div 5 square)

simularly the 2 imperfections in the Trumps border
in the front and back facial squares
Trumps Concentric Border: {S(B5,3) = (5-3)(53+1)/2 = 126}
033 022 113 042 105
002 089 106 019 099
085 082 004 119 025
098 090 006 112 009
097 032 086 023 077
095 080 111 011 018
091 ................. 035
016 ................. 110
005 ................. 121
108 046 015 115 031
100 017 048 034 116
114 ................. 012
083 ................. 043
008 ................. 118
010 109 078 092 026
038 102 003 125 047
081 ................. 045
030 ................. 096
087 ................. 039
079 024 123 001 088
049 094 040 103 029
027 037 020 107 124
101 044 122 007 041
117 036 120 014 028
021 104 013 084 093
mod 25 Border:
07 21 12 16 04
01 13 05 18 23
09 06 03 18 24
22 14 05 11 08
21 06 10 22 01
19 04 10 10 17
15 ............ 09
15 ............ 09
04 ............ 20
07 20 14 14 05
24 16 22 08 15
13 ............ 11
07 ............ 17
07 ............ 17
09 08 02 16 00
12 01 02 24 21
05 ............ 19
04 ............ 20
11 ............ 13
03 23 22 00 12
23 18 14 02 03
01 11 19 06 23
00 18 21 06 15
16 10 19 13 02
20 03 12 08 17
div 25 Border:
1 0 4 1 4
0 3 4 0 3
3 3 0 4 0
3 3 0 4 0
3 1 3 0 3
3 3 4 0 0
3 ....... 1
0 ....... 4
0 ....... 4
4 1 0 4 1
3 0 1 1 4
4 ....... 0
3 ....... 1
0 ....... 4
0 4 3 3 1
1 4 0 4 1
3 ....... 1
1 ....... 3
3 ....... 1
3 0 4 0 3
1 3 1 4 1
1 1 0 4 4
4 1 4 0 1
4 1 4 0 1
0 4 0 3 3

Notice that Trumps concentric border defies the outlined theory in that the sums of the mod 25 border are playing catchup with the off sums the div 25 border is presenting