The Magic Encyclopedia ™

The Generating Hypercube
(by Aale de Winkel)

Some people have use for squares they call "Generating squares" which seem to be defined on composite orders and can be obtained by basic multiplications of the composing orders "normal" and "transposed normal" squares. Also rectangles can be used to obtain GS's
This article generalizes the idea into all dimension into "Generating Hypercubes"
NOTE: currently these samples in upload, will augment when time permits. drawn order 6 conclusion is preliminairy at best.

The Generating Hypercube
The Normal Hyperbeam is a hyperbeam with all numbers in sequential order
N[ji] = l=0j (k=0l-1mk) li
generating hypercube

square samples
Given a composite order m = k=0nmk the normal
hyperbeam is given by the above mentioned remark, aplying the basic multiplication
formula on aspectial variants of these hypercubes gives genrerating hypercubes.
order 2 Normal square
N2
1 2
3 4
order 4 Normal square
N4
01 02 03 04
05 06 07 08
09 10 11 12
13 14 15 16
N2N2
01 02 05 06
03 04 07 08
09 10 13 14
11 12 15 16
N2N2t
01 03 05 07
02 04 06 08
09 11 13 15
10 12 14 16
order 2 by 3 Normal rectangle
2N3
1 2
3 4
5 6
order 3 by 2 Normal rectangle
3N2
1 2 3
4 5 6
N6
01 02 03 04 05 06
07 08 09 10 11 12
13 14 15 16 17 18
19 20 21 22 23 24
25 26 27 28 29 30
31 32 33 34 35 36
Next to N6 5 other order 6 GS's
can be simply derived
this excercise show the use of rectangles
in this situation, the general hypercube
situation is just a bit more complex
note the curious identity of the 6th
N2 N3 = (N2t N3t)t
01 02 03 10 11 12
04 05 06 13 14 15
07 08 09 16 17 18
19 20 21 28 29 30
22 23 24 31 32 33
25 26 27 34 35 36
N2 N3t = (N2t N3)t
01 04 07 10 13 16
02 05 08 11 14 17
03 06 09 12 15 18
19 22 25 28 31 34
20 23 26 29 32 35
21 24 27 30 33 36
2N3 3N2
01 02 03 07 08 09
04 05 06 10 11 12
13 14 15 19 20 21
16 17 18 22 23 24
25 26 27 31 32 33
28 29 30 34 35 36
2N3 2N3t
01 03 05 07 09 11
02 04 06 08 10 12
13 15 17 19 21 23
14 16 18 20 22 24
25 27 29 31 33 35
26 28 30 32 34 36
3N2 2N3
01 02 07 08 13 14
03 04 09 10 15 16
05 06 11 12 17 18
19 20 25 26 31 32
21 22 27 28 33 34
23 24 29 30 35 36
3N2 3N2t = N2 N3t
01 04 07 10 13 16
02 05 08 11 14 17
03 06 09 12 15 18
19 22 25 28 31 34
20 23 26 29 32 35
21 24 27 30 33 36
generating hypercube

cube sample
3N2
01 02
03 04
05 06
07 08
3N2 3N2
01 02 09 10
03 04 11 12
17 18 25 26
19 20 27 28
05 06 13 14
07 08 15 16
21 22 29 30
23 24 31 32
33 34 41 42
35 36 43 44
49 50 57 58
51 52 59 60
37 38 45 46
39 40 47 48
53 54 61 62
55 56 63 64


<Gil Lamb> constructs these squares by using horizontal and vertical numberic lines representing the high order component vertically and the low order component vertically
currently a bit foggy how to present this view systematically (omission I might resolve in future upload)
Also <George Chen> this author knows makes frequaent use of these kind of objects

As a samle of my own use for them the folllowing samples of the pan n-agonal transform for doubly even orders generating most-perfect hypecube
NOTE: a most-perfect hypercube is only pan n-agonal and need not be perfect.

The Pan n-agonal transform
General (order 4k) transform on the generating hypercube forming most-perfect hypercubes
1st step: mirror all 1-agonals upper halves in 3/4 line
2nd step: swapp alternate elements with element m/2 apart onsame 1-agonal
further steps: repeat 2nd step for all further 1-agonal directions
pan diagonal
transform

square samples
complete derivaton of order 4 most-perfect square families
N4
01 02 03 04
05 06 07 08
09 10 11 12
13 14 15 16
high 1-agonal mirror
01 02 04 03
05 06 08 07
13 14 16 15
09 10 12 11
half alternate swap
01 03 04 02
08 06 05 07
13 15 16 14
12 10 09 11
Pan(N4)
01 15 04 14
12 06 09 07
13 03 16 02
08 10 05 11
N2N2
01 02 05 06
03 04 07 08
09 10 13 14
11 12 15 16
high 1-agonal mirror
01 02 06 05
03 04 08 07
11 12 16 15
09 10 14 13
half alternate swap
01 05 06 02
08 04 03 07
11 15 16 12
14 10 09 13
Pan(N2N2)
01 15 06 12
14 04 09 07
11 05 16 02
08 10 03 13
N2N2t
01 03 05 07
02 04 06 08
09 11 13 15
10 12 14 16
high 1-agonal mirror
01 03 07 05
02 04 08 06
10 12 16 14
09 11 15 13
half alternate swap
01 05 07 03
08 04 02 06
10 14 16 12
15 11 09 13
Pan(N2N2t)
01 14 07 12
15 04 09 06
10 05 16 03
08 11 02 13
pan triagonal
transform

cube sample
complete derivaton of order 4 most-perfect cube
high 1-agonal mirror 3N2 3N2
01 02 10 09
03 04 12 11
19 20 28 27
17 18 26 25
05 06 14 13
07 08 16 15
23 24 32 31
21 22 30 29
37 38 46 45
39 40 48 47
55 56 64 63
53 54 62 61
33 34 42 41
35 36 44 43
51 52 60 59
49 50 58 57
alternate swap of m/2 aparts (done 3 times)
01 09 10 02
12 04 03 11
19 27 28 20
26 18 17 25
14 06 05 13
07 15 16 08
32 24 23 31
21 29 30 22
37 45 46 38
48 40 39 47
55 63 64 56
62 54 53 61
42 34 33 41
35 43 44 36
60 52 51 59
49 57 58 50
01 27 10 20
26 04 17 11
19 09 28 02
12 18 03 25
32 06 23 13
07 29 16 22
14 24 05 31
21 15 30 08
37 63 46 56
62 40 53 47
55 45 64 38
48 54 39 61
60 34 51 41
35 57 44 50
42 52 33 59
49 43 58 36
Pan(3N2 3N2)
01 63 10 56
62 04 53 11
19 45 28 38
48 18 39 25
60 06 51 13
07 57 16 50
42 24 33 31
21 43 30 36
37 27 46 20
26 40 17 47
55 09 64 02
12 54 03 61
32 34 23 41
35 29 44 22
14 52 05 59
49 15 58 08
Note: This cube is a most-perfect cube thus merely pantriagonal
each order 2 sub(hyper)cube has a total sum of 260 (= 22 (43+1) )
each triagonally cells 2 apart sum 65 (= 43+1 )
magic sum 130 (= 2 (43+1) ) of course