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4.3a Analytic Solution Bordered Magic Cubes

Another 5th order Magic Cube published by Walter Trump in 2003 is the s-Magic Bordered - or Concentric Magic Cube.

A 5th order Bordered Magic Cube consists of one Embedded Magic Cube of the 3th order with one border around it:

Magic Cube (5 x 5 x 5)

Plane 11 (Top)

a101 a102 a103 a104 a105
a106 a107 a108 a109 a110
a111 a112 a113 a114 a115
a116 a117 a118 a119 a120
a121 a122 a123 a124 a125

Plane 12

a76 a77 a78 a79 a80
a81 a82 a83 a84 a85
a86 a87 a88 a89 a90
a91 a92 a93 a94 a95
a96 a97 a98 a99 a100

Plane 13

a51 a52 a53 a54 a55
a56 a57 a58 a59 a60
a61 a62 a63 a64 a65
a66 a67 a68 a69 a70
a71 a72 a73 a74 a75

Plane 14

a26 a27 a28 a29 a30
a31 a32 a33 a34 a35
a36 a37 a38 a39 a40
a41 a42 a43 a44 a45
a46 a47 a48 a49 a50

Plane 15

a1 a2 a3 a4 a5
a6 a7 a8 a9 a10
a11 a12 a13 a14 a15
a16 a17 a18 a19 a20
a21 a22 a23 a24 a25


The Embedded Simple Magic Cube can be described by following linear equations (Section 2.2):

Rows:

Columns:

Pillars:

Space Diagonals:

a82 + a83 + a84 = 189
a87 + a88 + a89 = 189
a92 + a93 + a94 = 189
a57 + a58 + a59 = 189
a62 + a63 + a64 = 189
a67 + a68 + a69 = 189
a32 + a33 + a34 = 189
a37 + a38 + a39 = 189
a42 + a43 + a44 = 189

a82 + a87 + a92 = 189
a83 + a88 + a93 = 189
a84 + a89 + a94 = 189
a57 + a62 + a67 = 189
a58 + a63 + a68 = 189
a59 + a64 + a69 = 189
a32 + a37 + a42 = 189
a33 + a38 + a43 = 189
a34 + a39 + a44 = 189

a82 + a57 + a32 = 189
a87 + a62 + a37 = 189
a92 + a67 + a42 = 189
a83 + a58 + a33 = 189
a88 + a63 + a38 = 189
a93 + a68 + a43 = 189
a84 + a59 + a34 = 189
a89 + a64 + a39 = 189
a94 + a69 + a44 = 189

a42 + a63 + a84 = 189
a44 + a63 + a82 = 189
a34 + a63 + a92 = 189
a32 + a63 + a94 = 189

As mentioned in Section 4.1 it can be proven for 5th order Magic Cubes that Bordered Perfect Magic Cubes donít exist.

However if for a Magic Cube of order 5, the equations of the magic border squares are combined with the required symmetry conditions:

a(91) + a( 95) = 126
a(86) + a( 90) = 126
a(81) + a( 85) = 126
a(80) + a( 96) = 126
a(79) + a( 99) = 126
a(78) + a( 98) = 126
a(77) + a( 97) = 126
a(76) + a(100) = 126
a(66) + a( 70) = 126
a(61) + a( 65) = 126
a(56) + a( 60) = 126
a(55) + a( 71) = 126

a(54) + a(74) = 126
a(53) + a(73) = 126
a(52) + a(72) = 126
a(51) + a(75) = 126
a(41) + a(45) = 126
a(36) + a(40) = 126
a(31) + a(35) = 126
a(30) + a(46) = 126
a(29) + a(49) = 126
a(28) + a(48) = 126
a(27) + a(47) = 126
a(26) + a(50) = 126

a(25) + a(101) = 126
a(24) + a(104) = 126
a(23) + a(103) = 126
a(22) + a(102) = 126
a(21) + a(105) = 126
a(20) + a(116) = 126
a(19) + a(119) = 126
a(18) + a(118) = 126
a(17) + a(117) = 126
a(16) + a(120) = 126
a(15) + a(111) = 126
a(14) + a(114) = 126

a(13) + a(113) = 126
a(12) + a(112) = 126
a(11) + a(115) = 126
a(10) + a(106) = 126
a( 9) + a(109) = 126
a( 8) + a(108) = 126
a( 7) + a(107) = 126
a( 6) + a(110) = 126
a( 5) + a(121) = 126
a( 4) + a(124) = 126
a( 3) + a(123) = 126
a( 2) + a(122) = 126
a( 1) + a(125) = 126

this will result, after deduction, in following set of linear equations, describing the border of a Bordered Magic Cube of the 5th order:

a121 =  315 - a122 - a123 - a124 - a125
a116 =  315 - a117 - a118 - a119 - a120
a111 =  315 - a112 - a113 - a114 - a115
a109 =        a110 - a113 + a115 - a117 + a120 - a121 + a125
a107 = (315 - a108 - a109 - a110 + a111 - a113 + a116 - a119 + a121 - a125)/2
a106 =  315 - a107 - a108 - a109 - a110
a105 =  315 - a109 - a113 - a117 - a121
a104 =  315 - a109 - a114 - a119 - a124
a103 =  315 - a108 - a113 - a118 - a123
a102 =  315 - a107 - a112 - a117 - a122
a101 =  315 - a102 - a103 - a104 - a105
a96  =  315 - a97  - a98  - a99 - a100
a85  =  189 - a90  - a95  + a96 - a100
a73  = -315 + a74  - a97  + a99 + a101 + a110 - a113 + a114 + a115 - a117 + a119 + a120 - 2 * a121 + 2 * a124 + a125
a72  =  630 + 2 * a73 - a74 - a112 + 2 * a113 - a114 - 4 * a122 - 2 * a123 - 4 * a124
a71  =  315 - a72 - a73 - a74 - a75
a70  =(8316-3*a74-2*a75-3*a90-6*a95-3*a98-6*a99-6*a100-2*a108-8*a110-2*a112+2*a113-5*a114-10*a115- 5*a118- 6*a119 +
                                                                          -14*a120-8*a122-10*a123-14*a124-18*a125)/3
a65  = -378 - a66 + a90 + 2 * a95 - a96 + a100 + a101 + a110 + a115 - a116 + 2 * a120 + a125
a60  =  126 + a65 + a90  +2 * a95 - a96 + a100 - a107 - a108 - 3*a110 + a113 - a115 + a117 - a120
a50  =  189 - a75 - a100 +a101 - a125
a49  =  819 - a74 - a99 - a110 + a113 - a114 - a115 + a117 - a119 - a120 - a122 - a123 - 3*a124 - 2*a125
a48  =  504 - a73 - a98 - a108 - a113 - a118 - 2 * a123
a47  =  504 - a73 - a99 - a110 - a115 - a120 - 2 * a125
a46  =  315 - a47 - a48 - a49 - a50
a45  = -945 + a56 + 2*a90 + 3*a95 + 2*a97 + 2*a98 + 2*a99 + 4*a100 - a111 + a115 - a116 + a120 - 2*a121 + 2*a125
a40  =  630 - a41 - a49 - a73 + a85 - a90 - a97 - a110 + a111 - 2 * a115 - a120 - a121 - a125
a35  =  126 - a36 + a90 - a95 - a111 + a115 + a121 - a125

a(91) = 126 - a( 95)
a(86) = 126 - a( 90)
a(81) = 126 - a( 85)
a(80) = 126 - a( 96)
a(79) = 126 - a( 99)
a(78) = 126 - a( 98)
a(77) = 126 - a( 97)
a(76) = 126 - a(100)
a(66) = 126 - a( 70)
a(61) = 126 - a( 65)
a(56) = 126 - a( 60)
a(55) = 126 - a( 71)

a(54) = 126 - a(74)
a(53) = 126 - a(73)
a(52) = 126 - a(72)
a(51) = 126 - a(75)
a(41) = 126 - a(45)
a(36) = 126 - a(40)
a(31) = 126 - a(35)
a(30) = 126 - a(46)
a(29) = 126 - a(49)
a(28) = 126 - a(48)
a(27) = 126 - a(47)
a(26) = 126 - a(50)

a(25) = 126 - a(101)
a(24) = 126 - a(104)
a(23) = 126 - a(103)
a(22) = 126 - a(102)
a(21) = 126 - a(105)
a(20) = 126 - a(116)
a(19) = 126 - a(119)
a(18) = 126 - a(118)
a(17) = 126 - a(117)
a(16) = 126 - a(120)
a(15) = 126 - a(111)
a(14) = 126 - a(114)

a(13) = 126 - a(113)
a(12) = 126 - a(112)
a(11) = 126 - a(115)
a(10) = 126 - a(106)
a( 9) = 126 - a(109)
a( 8) = 126 - a(108)
a( 7) = 126 - a(107)
a( 6) = 126 - a(110)
a( 5) = 126 - a(121)
a( 4) = 126 - a(124)
a( 3) = 126 - a(123)
a( 2) = 126 - a(122)
a( 1) = 126 - a(125)

The linear equations shown above, can be incorporated in a guessing routine, which might be used to find other suitable borders.

However the total number of independent variables (22) is to large to find more results within a reasonable time.

With the border variables constant, a simplified guessing routine (MgcCube5b1) produced 192 Bordered Magic Cubes within 40 seconds, which are shown in Attachment 4.3.1 and Attachment 4.3.2.

The collection of 48 elements, based on a Bordered Magic Cube with the Inner Cube variables constant, is shown in Attachment 4.3.3 and Attachment 4.3.4.

Based on these collections 192 * 48 = 9216 Concentric Magic Cubes of the 5th order can be constructed.

Note: The Inner Cubes are based on the consecutive integers 50, 51 ... 76 = (1, 2 ... 27) + 49.

4.3b Alternative Solution Bordered Magic Cubes

Alternatively Bordered Magic Cubes can be constructed based on Complementary Anti Symmetric Magic Squares of order 5.

Examples of such squares, which can be used as top squares for Bordered Magic Cubes, are shown in Attachment 4.3.5.

The relation between opposite surface squares (symmetry) can be represented as follows:

c1 c2 c3 c4 c5
c6 c7 c8 c9 c10
c11 c12 c13 c14 c15
c16 c17 c18 c19 c20
c21 c22 c23 c24 c25
Pr5 - c25 Pr5 - c22 Pr5 - c23 Pr5 - c24 Pr5 - c21
Pr5 - c10 Pr5 - c7 Pr5 - c8 Pr5 - c9 Pr5 - c6
Pr5 - c15 Pr5 - c12 Pr5 - c13 Pr5 - c14 Pr5 - c11
Pr5 - c20 Pr5 - c17 Pr5 - c18 Pr5 - c19 Pr5 - c16
Pr5 - c5 Pr5 - c2 Pr5 - c3 Pr5 - c4 Pr5 - c1

with Pr5 = 2 * s1 / 5 the pair sum for the corresponding Magic Sum s1.

With c(i) the cube variables and the substitution:

a(1) a(2) a(3) a(4) a(5)
a(6) a(7) a(8) a(9) a(10)
a(11) a(12) a(13) a(14) a(15)
a(16) a(17) a(18) a(19) a(20)
a(21) a(22) a(23) a(24) a(25)
=
c(1) c(2) c(3) c(4) c(5)
c(26) c(27) c(28) c(29) c(30)
c(51) c(52) c(53) c(54) c(55)
c(76) c(77) c(78) c(79) c(80)
c(101) c(102) c(103) c(104) c(105)

the defining equations of the Magic Back Square can be written as:

a( 6) = s1 - a( 7) - a( 8) - a( 9) - a(10)
a( 7) = s1 - a(13) - a(19) - a( 1) - a(25)
a( 8) = s1 - a(13) - a(18) - a( 3) - a(23)
a( 9) = s1 - a(14) - a(19) - a( 4) - a(24)
a(10) = s1 - a(15) - a(20) - a( 5) - a(25)
a(11) = s1 - a( 6) - a(16) - a( 1) - a(21)
a(12) = s1 - a( 7) - a(17) - a( 2) - a(22)
a(13) =      a(14) - a(17) + a(19) + a( 4) - a(5) - a(21) + a(24)
a(16) = s1 - a(17) - a(18) - a(19) - a(20)

with a(i) independent for i = 14, 15††and i = 17 ... 20;
and††a(i) defined†††††for i = 1 ... 5 and i = 21 ... 25.

Based on a comparable substitution:

a(1) a(2) a(3) a(4) a(5)
a(6) a(7) a(8) a(9) a(10)
a(11) a(12) a(13) a(14) a(15)
a(16) a(17) a(18) a(19) a(20)
a(21) a(22) a(23) a(24) a(25)
=
c(1) c(6) c(11) c(16) c(21)
c(26) c(31) c(36) c(41) c(46)
c(51) c(56) c(61) c(66) c(71)
c(76) c(81) c(86) c(91) c(96)
c(101) c(106) c(111) c(116) c(121)

the defining equations of the Left Magic Square can be written as:

a( 7) = s1 - a(13) -  a(19) - a( 1) - a(25)
a( 8) = s1 - a(13) -  a(18) - a( 3) - a(23)
a( 9) = s1 - a(14) -  a(19) - a( 4) - a(24)
a(12) = s1 - a(13) -  a(14) - a(11) - a(15)
a(13) =      a(14) -  a(17) + a(19) + a(4) -   a( 5) -   a(21) +  a(24)
a(14) =  (-3*a(18) -6*a(19) - a(11) - a(15)- 3*a(16) - 3*a(20) +  a( 1) + 
                                           + 3*a( 2) + 2*a( 3) +4*a( 5) + 4*a(21) + 3*a(22) + 2*a(23) + a(25))/3
a(17) = s1 - a(18) - a(19) - a(16) - a(20)

with a(i) independent for i = 18 and 19;
and††a(i) defined†††††for i = 1 ... 6, 10, 11, 15, 16 and 20 ... 25.

Based on the equations listed above, a guessing routine can be written to generate Bordered Magic Cubes of order 5 within a reasonable time (MgcCube5i).

Attachment 4.3.6 shows, for the Anti Symmetric Magic Squares enclosed in Attachment 4.3.5, the first occurring Bordered Magic Cube with 6 Magic Surface Planes.

Each cube shown corresponds with 9216 (= 4 * 48 * 48) Bordered Magic Cubes.


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