Office Applications and Entertainment, Magic Cubes

4.3a Analytic Solution Bordered Magic Cubes

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

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

Plane 11 (Top)

Plane 12

Plane 13

Plane 14

Plane 15

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 5^th order Magic Cubes that Bordered Perfect Magic Cubes don’t 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 5^th 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 5^th 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.