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=0∑j (k=0∏l-1mk) li
generating hypercube square samples	Given a composite order m = k=0∏nmk 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