The Magic Encyclopedia ™

The {Pantriagonal Associated} cubes of order 4
(by Aale de Winkel)

February 29 2008 we received an email from <Walter Trump> of his find of 37824 (= 8 * 4728) {pantriagonal associated} cubes order 4 which he filtered out from the results of a program for {associated} cubes of order 4 he had since March 2003.
This instigated a discussion between <Francis Gaspalou> <Walter Trump> and this author about these 4728 {pantriagonal associated} cubes order 4
There are 48 irregular and 4680 regular cubes. (cubes counted in normalized position)
(the factor 8 came from the panrelocations @[2i,2j,2k ; i,j,k=0..1] which have the same qualification)

Below is a depiction of this authors research in the matter as well as some of the results discussed

the {pantriagonal associated} cubes of order 4
The following cube I used to obtain formulae which allowed me to iterate in full
the 4728 {pantriagonal associated} cubes of order 4
{associated} is established from the get go in the third and fourth layer of this symbolic cube.
analitic numberrange (for regular numberrange change 63 to 65)
A B C D
E F G H
I J K L
M N O P
a b c d
e f g h
i j k l
m n o p
(63-p) (63-o) (63-n) (63-m)
(63-l) (63-k) (63-j) (63-i)
(63-h) (63-g) (63-f) (63-e)
(63-d) (63-c) (63-b) (63-a)
(63-P) (63-O) (63-N) (63-M)
(63-L) (63-K) (63-J) (63-I)
(63-H) (63-G) (63-F) (63-E)
(63-D) (63-C) (63-B) (63-A)
The following formulates all the monagonal sums in the above cube
all of which needs to sum up to the magic sum (126 analitic numberrange)
the third and fourth layers amount to the same as the second and first layers
<0,0,1> <0,1,0>
A+B+C+D
E+F+G+H
I+J+K+L
M+N+O+P
a+b+c+d
e+f+g+h
i+j+k+l
m+n+o+p
A+E+I+M
B+F+J+N
C+G+K+O
D+H+L+P
a+e+i+m
b+f+j+n
c+g+k+o
d+h+l=p
<1,0,0>
A+a+(63-p)+(63-P)
E+e+(63-l)+(63-L)
I+i+(63-h)+(63-H)
M+m+(63-d)+(63-D)
B+b+(63-o)+(63-O)
F+f+(63-k)+(63-K)
J+j+(63-g)+(63-G)
N+n+(63-c)+(63-C)
C+c+(63-n)+(63-N)
G+g+(63-j)+(63-J)
K+k+(63-f)+(63-F)
O+o+(63-b)+(63-B)
D+d+(63-m)+(63-M)
H+h+(63-i)+(63-I)
L+l+(63-e)+(63-E)
P+p+(63-a)+(63-A)
The four triagonal direction adds the following formulae
16 of which are trivial cancelation of the involved letters
<1,1,1>
A+f+(63-f)+(63-A)
E+j+(63-b)+(63-M)
I+n+(63-n)+(63-I)
M+b+(63-j)+(63-E)
B+g+(63-e)+(63-D)
F+k+(63-a)+(63-P)
J+o+(63-m)+(63-L)
N+c+(63-i)+(63-H)
C+h+(63-h)+(63-C)
G+l+(63-d)+(63-O)
K+p+(63-p)+(63-K)
O+d+(63-l)+(63-G)
D+e+(63-g)+(63-B)
H+i+(63-c)+(63-N)
L+m+(63-o)+(63-J)
P+a+(63-k)+(63-F)
<-1,1,1>
A+h+(63-f)+(63-C)
E+l+(63-b)+(63-O)
I+p+(63-n)+(63-K)
M+d+(63-j)+(63-G)
B+e+(63-e)+(63-B)
F+i+(63-a)+(63-N)
J+m+(63-m)+(63-J)
N+a+(63-i)+(63-F)
C+f+(63-h)+(63-A)
G+j+(63-d)+(63-M)
K+n+(63-p)+(63-I)
O+b+(63-l)+(63-E)
D+g+(63-g)+(63-D)
H+k+(63-c)+(63-P)
L+o+(63-o)+(63-L)
P+c+(63-k)+(63-H)
<-1,-1,1>
A+p+(63-f)+(63-K)
E+d+(63-b)+(63-G)
I+h+(63-n)+(63-C)
M+l+(63-j)+(63-O)
B+m+(63-e)+(63-J)
F+a+(63-a)+(63-F)
J+e+(63-m)+(63-B)
N+i+(63-i)+(63-N)
C+n+(63-h)+(63-I)
G+b+(63-d)+(63-E)
K+f+(63-p)+(63-A)
O+j+(63-l)+(63-M)
D+o+(63-g)+(63-L)
H+c+(63-c)+(63-H)
L+g+(63-o)+(63-D)
P+k+(63-k)+(63-P)
<1,-1,1>
A+n+(63-f)+(63-I)
E+b+(63-b)+(63-E)
I+f+(63-n)+(63-A)
M+j+(63-j)+(63-M)
B+o+(63-e)+(63-L)
F+c+(63-a)+(63-H)
J+g+(63-m)+(63-D)
N+k+(63-i)+(63-P)
C+p+(63-h)+(63-K)
G+d+(63-d)+(63-G)
K+h+(63-p)+(63-C)
O+l+(63-l)+(63-O)
D+m+(63-g)+(63-J)
H+a+(63-c)+(63-F)
L+e+(63-o)+(63-B)
P+i+(63-k)+(63-N)
With help of these formulae lifted from the symbolic cube and the
{pantriagonal associated} quality it can be established that the 8 octants
obey to the following fact (here formulated using the first two octants)
A+B=e+f
A+E=b+f
A+a=F+f
a+b=E+F
a+e=B+F
B+b=E+e
A-C=f-h
B-D=e-g
E-G=b-d
F-H=a-c
sum equations in
all octants
differences between
all octant pairs
Proof:
from <-1,1,1> A+h+(63-f)+(63-C)=126 and from <1,1,1> B+g+(63-e)+(63-D)=126
thus: A+h=C+f and B+g=D+e this gives:
A+B=(C+f-h)+(D+e-g)=(C+D)+(e+f)-(h+g)=(S-(A+B))+(e+f)-(S-(e+f))=2(e+f)-(A+B) => A+B=e+f
simular we get E+F=a+b and due to symmetry we also get:
A+E=b+f ; a+e=B+F ; A+a=F+f and B+b=E+e
the difference relations are given in the triagonal sums
QED


The resulting equations show that there are only 4 sums between each layer pair, which is quite consistent with the 2-pan nature of the qualifier
(ie all panrelocations @[2i,2j,2k ; i,j,k=0..1]) qualify the same.

Construction of the regular cubes
The regular cubes have an alternate construction based on their binary content
summing patterns Binary lines with summing to 2
0 0 1 1
0 1 0 1
0 1 1 0
1 0 0 1
1 0 1 0
1 1 0 0
summing binary bisquares Binary bisquares with summing to 2
00550055
01454051
02533025
03522035
04151054
05055050
05505005
10540415
11444411
14141414
14144141
15045410
20350352
22333322
23322332
25305302
each line form two squares
as each 'digit' indexes the
binaryline table above and
four 'digit's thus form a
square.
The other two squuares
follow by complementing
in the associated position
16 more possibilities were found but these are 5-complement of the above
which amounts to the aspectial variants ~2 of the above
Investigating axial permutation the above 16 hold 4 invariants and 4 groups
of 3 members each which transform into one another under axial permutation
{pantriagonal associated}
cubes
The following 39 sixtuples of bisquares combine into valid cubes
01: 00 05 06 08 10 13
02: 00 05 06 08 10 14
03: 00 05 06 09 10 13
04: 00 05 06 09 10 14
05: 00 05 08 10 13 14
06: 00 05 09 10 13 14
07: 00 06 08 09 10 13
08: 00 06 08 09 10 14
09: 00 08 09 10 13 14
10: 01 04 06 07 10 13
11: 01 04 06 07 10 14
12: 01 04 06 10 11 13
13: 01 04 06 10 11 14
14: 01 04 07 10 13 14
15: 01 04 10 11 13 14
16: 01 06 07 10 11 13
17: 01 06 07 10 11 14
18: 01 07 10 11 13 14
19: 02 03 05 08 10 12
20: 02 03 05 08 10 15
21: 02 03 05 09 10 12
22: 02 03 05 09 10 15
23: 02 03 08 09 10 12
24: 02 03 08 09 10 15
25: 02 05 08 10 12 15
26: 02 05 09 10 12 15
27: 02 08 09 10 12 15
28: 03 05 08 10 12 15
29: 03 05 09 10 12 15
30: 03 08 09 10 12 15
31: 04 06 07 10 11 13
32: 04 06 07 10 11 14
33: 04 07 10 11 13 14
34: 05 06 08 09 10 13
35: 05 06 08 09 10 14
36: 05 06 08 10 13 14
37: 05 06 09 10 13 14
38: 05 08 09 10 13 14
39: 06 08 09 10 13 14
component permutation of the 6 binary component gives a factor 6! = 720
39 * 720 = 28080 so all the regular cubes are obtained in this manner
28080 = 6 * 4680 since all 6 (axial permutation) Aspects are in these


As mentioned using the resulting set of equations and adding restrictions to obtain only normalized positioned cubes 4728 cubes where obtained.
<Francis Gaspalou> established 6 sets in the regular cubes.
The 48 irregular cubes where first mentioned by <Walter Trump>.

the {pantriagonal associated} cube groups of order 4
<Francis Gaspalou> arrived at a full formulae
cube based on the 8 parameters A,B,C,E,F,G,I and a:
A = 0,1 ; S = 126+4A

D = S-A-B-C
H = S-E-F-G
J = -B-C+E+G+I
K = S-E-G-I
L = B+C-I
M = S-A-E-I
N = S+C-E-F-G-I
O = -C+E+I
P = A+E+F+G+I-S
b = E+F-a
c = S-E-2F-G+a
d = F+G-a
e = B+F-a
f = A-F+a
g = S-A-B-C+F-a
h = C-F+a
i = S+C-E-2F-G-I+a
j = S-A-E+F-I-a
k = A+E+G+I+a-S
l = -C+E+F+I-a
m = -B-C+E+F+G+I-a
n = -F+I+a
o = B+C+F-I-a
p = S-E-F-G-I+a
Equations reformed to symbolic cube above
The cells of the other 2 squares folow
from the associated condition as mentioned
in that cube.
(note: '63'(A=0) => '65'(A=1))
derivations: given A,B,C,E,F,G,I and a
A+B+C+D=S => D=S-A-B-C; E+F+G+H=S => H=S-E-F-G; A+E+i+m=S => M=S-A-E-I;
E+F=a+b => b=E+F-a; B+F=a+e => e=B+F-a; F+f=A+a => f=A+a-F;
F-H=a-c => c=a-F+H=a-F+(S-E-F-G)=S-E-2F-G+a
a+b+c+d=S => d=S-a-b-c=S-a-(E+F-a)-(S-E-2F-G+a)=F+G-a
g-e=D-B => g=D-B+e=(S-A-B-C)-B+(B+F-a)=S-A-B-C+F-a
e+f+g+h=S =>h=S-e-f-g=S-(B+F-a)-(A+a-F)-(S-A-B-C+F-a)=C-F+a
I+i=H+h => i=H+h-I=(S-E-F-G)+(C-F+a)-I=S+C-E-2F-G-I+a
M+m=D+d => m=D+d-M=(S-A-B-C)+(F+G-a)-(S-A-E-I)=-B-C+E+F+G+I-a
J-B=m-e => J=m-e+B=(-B-C+E+F+G+I-a)-(B+F-a)+B=-B-C+E+G+I
B+F+J+N=S => N=S-B-F-J=S-B-F-(-B-C+E+G+I)=S+C-E-F-G-I
M+N=i+j => j=M+N-i=(S-A-E-I)+(S+C-E-F-G-I)-(S+C-E-2F-G-I+a)=S-A-E+F-I-a
I+i=N+n => n=I+i-N=I+(S+C-E-2F-G-I+a)-(S+C-E-F-G-I)=-F+I+a

POSTULATING: K = S-E-G-I
K+k=P+p=A+a => k=A+a-K=A+a-(S-E-G-I)=A+E+G+I+a-S
C+G+K+O=S => O=S-C-G-K=S-C-G-(S-E-G-I)=-C+E+I
I+J+K+L=S => L=S-I-J-K=S-I-(-B-C+E+G+I)-(S-E-G-I)=B+C-I
M+N+O+P=S => P=S-M-N-O=S-(S-A-E-I)-(S+C-E-F-G-I)-(-C+E+I)=A+E+F+G+I-S
P+p=A+a => p=A+a-P=A+a-(A+E+F+G+I-S)=S-E-F-G-I+a
O+o=B+b => o=B+b-O=B+(E+F-a)-(-C+E+I)=B+C+F-I-a
L+l=E+e => l=E+e-L=E+(B+F-a)-(B+C-I)=-C+E+F+I-a
<Francis Gaspalou> splitted the regular in 6 sets
given by the following equations, these types cover the entire regular cubes
Sets A B and C intersect with sets D E and F
Set A: A + C - E - 2F - G + 2a = 0
Set B: A + C + E + G = S
Set C: A - C + E + G + 2I = S
Set D: -B - C + 2E + F + G + 2I = S
Set E: B + C + F + G = S
Set F: B + C - F - G = 0
Alternate set definition
The "supplementary transformation" (below) establishes
A <=> F ; B <=> E ; C <=> D
Set A: A + K + h + n = S
Set B: A - K = n - h
Set C: A - K = h - n
Set D: A + M = D + P
Set E: A + D = M + P
Set F: A + D + M + P = S
All 6 sets hold 5760 cubes, while the 9 crossovers hold 720 cubes each
6 * 5760 - 9 * 720 = 28080
<Francis Gaspalou>'s "Geometric Transformations"
The following transformation formulates a bijection (group of order 2)
if {A,B,C,E,F,G,I,a} is a solution then
{A,B+C-I,I,-C+E+I,F,G,C,a} is also a solution

or:
B => B+C-I ; C => I
E => -C+E+I ; I => C
This transformation can be applied on
all three planes connected with the 0-position
which generates a dividing factor of 6
4728 / 6 = 788

(the reformulation shows the symmetry
of the bijection in the xy-plane.)
a further "supplementary transformation" gives yet another factor 2
if {A,B,C,E,F,G,I,a} is a solution then
{A,B,126-A-B-C+F-a,E,F,126-E-2F-G+a,
126-A-E+F-I-a,a} is also a solution.

or:
C => 126-A-B-C+F-a
G => 126-E-2F-G+a
I => 126-A-E+F-I-a
This transformation effectively performs
a reflection ~6 on the 2nd and 7th octants
a reflection ~5 on the 3rd and 6th octants and
a reflection ~3 on the 4th and 5th octants
which shows the bijective nature of this
transformation, and thus the factor 2
thus <Francis Gaspalou> reduced the 4728 cubes
to 394 elementairy cubes (390 regular and 4 irregular)
these are pictures of the two transformations above
the first one is shown on one plane, but applies on all

the reflections of the octants in the second transformation
are shown on the octants ribs, and form a continuous line of
arrows around the cube, the reflections are:
octants #0 and #7: none or ~0
octants #1 and #6: y and z-axes: ~(21+22) = ~6
octants #2 and #5: x and z-axes: ~(20+22) = ~5
octants #3 and #4: x and y-axes: ~(20+21) = ~3
(probable notation C~{0,6,5,3,3,5,6,0})


<Walter Trump>'s contribution to this was much larger then what is shown below, but more or less amounts to what is discussed above.

the {pantriagonal associated} basic cubes of order 4
<Walter Trump>. arrived at the following
4 basic cubes for the regular cubes
TrumpBase #1: 3TB04
180 "elementary cubes"
00 07 57 62
27 45 34 20
53 50 12 11
46 24 23 33
47 25 22 32
05 02 60 59
26 44 35 21
48 55 09 14
49 54 08 15
42 28 19 37
04 03 61 58
31 41 38 16
30 40 39 17
52 51 13 10
43 29 18 36
01 06 56 63
TrumpBase #2: 3TB14
90 "elementary cubes"
00 07 57 62
26 29 35 36
45 42 20 19
55 48 14 09
31 24 38 33
05 02 60 59
50 53 11 12
40 47 17 22
41 46 16 23
51 52 10 13
04 03 61 58
30 25 39 32
54 49 15 08
44 43 21 18
27 28 34 37
01 06 56 63
TrumpBase #3: 3TB24
60 "elementary cubes"
00 07 57 62
26 41 23 36
45 30 32 19
55 48 14 09
43 24 38 21
05 02 60 59
50 53 11 12
28 47 17 34
29 46 16 35
51 52 10 13
04 03 61 58
42 25 39 20
54 49 15 08
44 31 33 18
27 40 22 37
01 06 56 63
TrumpBase #4: 3TB34
60 "elementary cubes"
00 15 49 62
54 57 07 08
25 22 40 39
47 32 30 17
59 52 10 05
13 02 60 51
34 45 19 28
20 27 37 42
21 26 36 43
35 44 18 29
12 03 61 50
58 53 11 04
46 33 31 16
24 23 41 38
55 56 06 09
01 14 48 63
Due to the geometric transformations of <Francis Gaspalou>
The 720 component permutations of these cubes amount to the mentioned elementary cubes
180 + 90 + 60 + 60 = 390