<HOME> <<PREVIOUS] [NEXT>>
What can I do now with the transformation method?
With the transformation method you can produce a most perfect magic square by using a square, that
is a multiple of 4 (=4x4, 8x8, 12x12, 16x16, …) with sequencial digits. It is also possible to transform
each most perfect magic square backwards to its starting position. See for example:
Most perfect magic 8x8 square Starting position transformation method
|
1
|
60
|
23
|
46
|
3
|
48
|
21
|
58
|
|
|
1
|
19
|
17
|
21
|
3
|
7
|
5
|
23
|
|
32
|
37
|
10
|
51
|
30
|
49
|
12
|
39
|
|
|
33
|
51
|
49
|
53
|
35
|
39
|
37
|
55
|
|
42
|
19
|
64
|
5
|
44
|
7
|
62
|
17
|
|
|
41
|
59
|
57
|
61
|
43
|
47
|
45
|
63
|
|
55
|
14
|
33
|
28
|
53
|
26
|
35
|
16
|
|
|
34
|
52
|
50
|
54
|
36
|
40
|
38
|
56
|
|
9
|
52
|
31
|
38
|
11
|
40
|
29
|
50
|
|
|
9
|
27
|
25
|
29
|
11
|
15
|
13
|
31
|
|
63
|
6
|
41
|
20
|
61
|
18
|
43
|
8
|
|
|
2
|
20
|
18
|
22
|
4
|
8
|
6
|
24
|
|
34
|
27
|
56
|
13
|
36
|
15
|
54
|
25
|
|
|
10
|
28
|
26
|
30
|
12
|
16
|
14
|
32
|
|
24
|
45
|
2
|
59
|
22
|
57
|
4
|
47
|
|
|
42
|
60
|
58
|
62
|
44
|
48
|
46
|
64
|
Looking quickly at this starting position there seems to be no logica. Though each ‘ugly’or ‘pretty’ star-
ting position has the same features, just like the magic features of the most perfect magic square itself.
See, tot start with, the following scheme:
Scheme starting position,upper half Scheme starting position, complete
|
a1
|
a2
|
a3
|
a4
|
e1
|
e2
|
e3
|
e4
|
|
|
a1
|
a2
|
a3
|
a4
|
e1
|
e2
|
e3
|
e4
|
|
b1
|
b2
|
b3
|
b4
|
f1
|
f2
|
f3
|
f4
|
|
|
b1
|
b2
|
b3
|
b4
|
f1
|
f2
|
f3
|
f4
|
|
c1
|
c2
|
c3
|
c4
|
g1
|
g2
|
g3
|
g4
|
|
|
c1
|
c2
|
c3
|
c4
|
g1
|
g2
|
g3
|
g4
|
|
d1
|
d2
|
d3
|
d4
|
h1
|
h2
|
h3
|
h4
|
|
|
d1
|
d2
|
d3
|
d4
|
h1
|
h2
|
h3
|
h4
|
|
|
|
|
|
|
|
|
|
|
|
h4'
|
h3'
|
h2'
|
h1'
|
d4'
|
d3'
|
d2'
|
d1'
|
|
|
|
|
|
|
|
|
|
|
|
g4'
|
g3'
|
g2'
|
g1'
|
c4'
|
c3'
|
c2'
|
c1'
|
|
|
|
|
|
|
|
|
|
|
|
f4'
|
f3'
|
f2'
|
f1'
|
b4'
|
b3'
|
b2'
|
b1'
|
|
|
|
|
|
|
|
|
|
|
|
e4'
|
e3'
|
e2'
|
e1'
|
a4'
|
a3'
|
a2'
|
a1'
|
a1 + a1’ = a2 + a2’ = a3 + a3’ = a4 + a4’ = b1 + b1’ = b2 + b2’ = ... = h4 + h4’ = 65
If the first half has been produced (correctly), you get the second half automatically. It gives also the
first boundary condition: to produce the first half you must select each time one of the digits from
column I or column II (in the example, see the yellow marked digits):
I or II
|
1
|
|
64
|
|
2
|
|
63
|
|
3
|
|
62
|
|
4
|
|
61
|
|
5
|
|
60
|
|
6
|
|
59
|
|
7
|
|
58
|
|
8
|
|
57
|
|
9
|
|
56
|
|
10
|
|
55
|
|
11
|
|
54
|
|
12
|
|
53
|
|
13
|
|
52
|
|
14
|
|
51
|
|
15
|
|
50
|
|
16
|
|
49
|
|
17
|
|
48
|
|
18
|
|
47
|
|
19
|
|
46
|
|
20
|
|
45
|
|
21
|
|
44
|
|
22
|
|
43
|
|
23
|
|
42
|
|
24
|
|
41
|
|
25
|
|
40
|
|
26
|
|
39
|
|
27
|
|
38
|
|
28
|
|
37
|
|
29
|
|
36
|
|
30
|
|
35
|
|
31
|
|
34
|
|
32
|
|
33
|
Second boundary condition is:
a1 -/- a2 = e3 -/- e4 [in the example: 1 -/- 19 = -/- 18 versus 5 -/- 23 = -/- 18]
a2 -/- a3 = e2 -/- e3 [in the example: 19 -/- 17 = 2 versus 7 -/- 5 = 2]
a3 -/- a4 = e1 -/- e2 [in the example: 17 -/- 21 = -/- 4 versus 3 -/- 7 = -/- 4]
Third boundary condition is:
a1 -/- b1 = a2 -/- b2 = a3 -/- b3 = a4 -/- b4 = e1 -/- f1 = e2 -/- f2 = e3 -/- f3 = e4 -/- f4 [= 32]
b1 -/- c1 = b2 -/- c2 = b3 -/- c3 = b4 -/- c4 = f1 -/- g1 = f2 -/- g2 = f3 -/- g3 = f4 -/- g4 [= 8]
c1 -/- d1 = c2 -/- d2 = c3 -/- d3 = c4 -/- d4 = g1 -/- h1 = g2 -/- h2 = g3 -/- h3 = g4 -/- h4 [= -/- 7]
Try it (clue: don’t select the digits from column I or II to difficult)!!! Maybe an excellent programmer
can let the computer determine all (368640 x 8) solutions of the most perfect 8x8 square from the
(possible) starting positions of the transformation method.
Translation of the swap possibilities of the most perfect magic 8x8 square into swap of the starting
position:
[1a] swap row 1&3 and/or 2&4 and/or 5&7 and/or 6&8 of magic square = swap row 1&8 and/or 2&7 and/or 4&5 and/or 3&6 of starting position
[1b] swap column 1&3 and/or 2&4 and/or 5&7 and/or 6&8 of magic square = swap column 1&8 and/or 2&7 and/or 4&5 and/or 3&6 of starting position
[2a] swap row 1&2 and 3&4 and 5&6 and 7&8 of magic square = swap row 1&2 and 3&4 and 5&6 and 7&8 of starting position
(& vertical mirroring of starting position)
[2b] swap column 1&2 and 3&4 and 5&6 and 7&8 of magic square = swap column 1&2 and 3&4 and 5&6 and 7&8 of starting position
(& horizontal mirroring of starting position)
[3a] swap upper with down half of magic square = swap upper with down half of starting position
[3b] swap left with right half of magic square = swap left with right half of starting position
<HOME> <<PREVIOUS] [NEXT>> | 
