Office Applications and Entertaiment, Magic Squares

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6.0 Magic Squares (6 x 6)

6.1 Analytic Solution, Pan Magic Squares (Don't Exist)

Magic squares of order 6 can be represented as follows:

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)

a(26)

a(27)

a(28)

a(29)

a(30)

a(31)

a(32)

a(33)

a(34)

a(35)

a(36)

For Pan Magic Squares the numbers a(i), i = 1 ... 36, in all rows, columns, main and broken diagonals should sum to the same constant, which results in following linear equations:

a( 1) + a( 2) + a( 3) + a( 4) + a( 5) + a( 6) = s1
a( 7) + a( 8) + a( 9) + a(10) + a(11) + a(12) = s1
a(13) + a(14) + a(15) + a(16) + a(17) + a(18) = s1
a(19) + a(20) + a(21) + a(22) + a(23) + a(24) = s1
a(25) + a(26) + a(27) + a(28) + a(29) + a(30) = s1
a(31) + a(32) + a(33) + a(34) + a(35) + a(36) = s1

a( 1) + a( 7) + a(13) + a(19) + a(25) + a(31) = s1
a( 2) + a( 8) + a(14) + a(20) + a(26) + a(32) = s1
a( 3) + a( 9) + a(15) + a(21) + a(27) + a(33) = s1
a( 4) + a(10) + a(16) + a(22) + a(28) + a(34) = s1
a( 5) + a(11) + a(17) + a(23) + a(29) + a(35) = s1
a( 6) + a(12) + a(18) + a(24) + a(30) + a(36) = s1

a( 1) + a( 8) + a(15) + a(22) + a(29) + a(36) = s1
a( 2) + a( 9) + a(16) + a(23) + a(30) + a(31) = s1
a( 3) + a(10) + a(17) + a(24) + a(25) + a(32) = s1
a( 4) + a(11) + a(18) + a(19) + a(26) + a(33) = s1
a( 5) + a(12) + a(13) + a(20) + a(27) + a(34) = s1
a( 6) + a( 7) + a(14) + a(21) + a(28) + a(35) = s1

a( 6) + a(11) + a(16) + a(21) + a(26) + a(31) = s1
a( 5) + a(10) + a(15) + a(20) + a(25) + a(36) = s1
a( 4) + a( 9) + a(14) + a(19) + a(30) + a(35) = s1
a( 3) + a( 8) + a(13) + a(24) + a(29) + a(34) = s1
a( 2) + a( 7) + a(18) + a(23) + a(28) + a(33) = s1
a( 1) + a(12) + a(17) + a(22) + a(27) + a(32) = s1

which can be reduced, by means of row and column manipulations, to: 

a(31) =        s1 - a(32) - a(33) - a(34) - a(35) - a(36)
a(25) =        s1 - a(26) - a(27) - a(28) - a(29) - a(30)
a(19) =        s1 - a(20) - a(21) - a(22) - a(23) - a(24)
a(17) =  2/3 * s1 - a(18) + a(20) + a(21) - a(23) - a(24) + a(26) + a(27) - a(29) - a(30) - a(35) - a(36)
a(16) =    2 * s1 + a(18) - a(20) - 2*a(21) - 2*a(22) - a(23) - a(26) - 2*a(27) - 2*a(28) - a(29) - a(34) + a(36)
a(15) =  2/3 * s1 - a(18) - a(33) - a(36)
a(14) =             a(18) - a(20) - a(21) + a(23) + a(24) - a(26) - a(27) + a(29) + a(30) - a(32) + a(36)
a(13) = -4/3 * s1 - a(18) + a(20) + 2*a(21) + 2*a(22) + a(23) + a(26) + 2*a(27) + 2*a(28) + a(29) - a(31) - a(36)
a(12) = 11/6 * s1 - a(20) - a(21) - a(22) - a(26) - 2*a(27) - a(28) - a(32) - a(33) - a(34)
a(11) = -7/6 * s1 + a(22) + a(23) + a(24) - a(26) + a(28) + a(29) + a(30) + a(34) + a(35) + a(36)
a(10) = -7/6 * s1 + a(21) + a(22) + a(23) - a(25) + a(27) + a(28) + a(29) + a(33) + a(34) + a(35)
a(9)  = -7/6 * s1 + a(20) + a(21) + a(22) + a(26) + a(27) + a(28) - a(30) + a(32) + a(33) + a(34)
a(8)  = 11/6 * s1 - a(22) - a(23) - a(24) - a(28) - 2*a(29) - a(30) - a(34) - a(35) - a(36)
a(7)  = 11/6 * s1 - a(21) - a(22) - a(23) - a(27) - 2*a(28) - a(29) - a(33) - a(34) - a(35)
a(6)  = -5/6 * s1 - a(18) + a(20) + a(21) + a(22) - a(24) + a(26) + 2*a(27) + a(28) - a(30) +a(32) +a(33) +a(34) -a(36)
a(5)  =  1/2 * s1 + a(18) + a(19) + a(24) - a(27) - a(28) - a(29) - a(34) - a(35)
a(4)  =  1/6 * s1 - a(18) + a(20) + a(21) - a(28) - a(29) - a(30) + a(31) + a(32)
a(3)  =  1/2 * s1 + a(18) - a(20) - 2*a(21) - a(22) - a(26) - 2*a(27) - a(28) + a(30) + a(31) + a(35) + 2*a(36)
a(2)  =  1/6 * s1 - a(18) + a(21) + a(22) - a(25) - a(26) - a(30) + a(34) + a(35)
a(1)  =  1/2 * s1 + a(18) + a(23) + a(24) - a(25) - a(26) - a(27) - a(31) - a(32)

Although the linear equations shown above can be solved for any magic constant s1, it is obvious that no solution can be found for the consecutive integers {1 ... 36}, as the related magic constant s1 = 111 is odd.

Solutions for non consecutive integers, resulting in other magic constants, will be discussed in Section 6.09.

Note: It can be proven that Pan Magic Squares based on consecutive integers don't exist for any order (4n + 2).

6.2 Analytic Solution, Simple Magic Squares

For solutions with consecutive integers we moderate the definition to all rows, columns and main diagonals sum to the same constant, which results in following linear equations:

a( 1) + a( 2) + a( 3) + a( 4) + a( 5) + a( 6) = s1
a( 7) + a( 8) + a( 9) + a(10) + a(11) + a(12) = s1
a(13) + a(14) + a(15) + a(16) + a(17) + a(18) = s1
a(19) + a(20) + a(21) + a(22) + a(23) + a(24) = s1
a(25) + a(26) + a(27) + a(28) + a(29) + a(30) = s1
a(31) + a(32) + a(33) + a(34) + a(35) + a(36) = s1

a( 1) + a( 7) + a(13) + a(19) + a(25) + a(31) = s1
a( 2) + a( 8) + a(14) + a(20) + a(26) + a(32) = s1
a( 3) + a( 9) + a(15) + a(21) + a(27) + a(33) = s1
a( 4) + a(10) + a(16) + a(22) + a(28) + a(34) = s1
a( 5) + a(11) + a(17) + a(23) + a(29) + a(35) = s1
a( 6) + a(12) + a(18) + a(24) + a(30) + a(36) = s1

a( 1) + a( 8) + a(15) + a(22) + a(29) + a(36) = s1

a( 6) + a(11) + a(16) + a(21) + a(26) + a(31) = s1

which can be reduced, by means of row and column manipulations, to: 

a(31) = s1 - a(32) - a(33) - a(34) - a(35) - a(36)
a(25) = s1 - a(26) - a(27) - a(28) - a(29) - a(30)
a(19) = s1 - a(20) - a(21) - a(22) - a(23) - a(24)
a(13) = s1 - a(14) - a(15) - a(16) - a(17) - a(18)
a(11) =    + a(12) - a(16) + a(18) - a(21) + a(24) - a(26) + a(30) - a(31) + a(36)
a( 8) =   {- a( 9) - a(10) - 2*a(11) - a(14) - 2*a(15) - 2*a(16) - a(17) + 2*a(19) + a(20) + a(23) + 2*a(24) + 2*a(25)
           + a(27) + a(28) + 2*a(30)}/2
a( 7) =      a( 8) - a(13) + a(15) - a(19) + a(22) - a(25) + a(29) - a(31) + a(36)
a( 6) = s1 - a(11) - a(16) - a(21) - a(26) - a(31)
a( 5) = s1 - a(11) - a(17) - a(23) - a(29) - a(35)
a( 4) = s1 - a(10) - a(16) - a(22) - a(28) - a(34)
a( 3) = s1 - a( 9) - a(15) - a(21) - a(27) - a(33)
a( 2) = s1 - a( 8) - a(14) - a(20) - a(26) - a(32)
a( 1) = s1 - a( 8) - a(15) - a(22) - a(29) - a(36)

The linear equations shown above are ready to be solved, for the magic constant 111.

However the solutions can only be obtained by guessing a(9), a(10), a(12), a(14) .. a(18), a(20) .. a(24), a(26) .. a(30) and a(32) .. a(36) and filling out these guesses in the abovementioned equations.

For distinct integers also following relations should be applied:

0 < a(i) =< 36        for i = 1, 2, ... 8, 11, 13, 19, 25
Int(a(i)) = a(i)      for i = 8
a(i) ≠ a(j)           for i ≠ j

which can be incorporated in a guessing routine, which can be used to generate - if not all - at least collections of 6th order squares with distinct integers within a reasonable time (MgcSqr6b).

The number of simple magic squares of order 6 is gigantic and estimated to be 1,77 1019, which is likely because the squares are determined by 23 independent variables (red).

6 32 3 34 35 1
7 11 27 28 8 30
19 14 16 15 23 24
18 20 22 21 17 13
25 29 10 9 26 12
36 5 33 4 2 31

With the highlighted variables constant, above mentioned guessing routine (MgcSqr6b), produced 856 Simple Magic Squares of the 6th order within 56 minutes, which are shown in Attachment 6.5.1.

Note:

An alternative routine - in which after the bottom row, first the main diagonals and afterward the remaining rows and columns are calculated - appeared to be much faster (MgcSqr6a).

With row a(31) - a(36), diagonal a(1) - a(36) and column a(5) - (35) constant, subject routine generated 16808 Simple Magic Squares of the 6th order within 17,75 minutes, of which the first 856 are shown in Attachment 6.6.1.

6.3 Transformations

For simple magic squares, related squares can be found by means of rotation and/or reflection (ref. Attachment 6.5.2).

Further, each 6th order simple magic square corresponds with 24 transformations as described below:

  • Any line n can be interchanged with line (7 - n). As any combination of these 3 simple permutations can be applied, the resulting number of transformations is 23 = 8, which are shown in Attachment 6.5.3.
    It should be noted that for each square the 180o rotated aspect is included in this collection.

  • Any permutation can be applied to the lines 1, 2, 3 provided that the same permutation is applied to the lines 6, 5, 4. The possible number of transformations is 3! = 6, which are shown in Attachment 6.5.4.

  • The resulting number of transformations, excluding the 180o rotated aspects, is 8/2 * 6 = 24, which are shown in Attachment 6.5.5

Based on this set of transformations and the eight squares which can be found by means of rotation and/or reflection (ref. Attachment 6.5.2) any 6th order simple magic square corresponds with a Class of 8 * 24 = 192 magic squares.

Magic squares with symmetrical main diagonals, as described in next section and highlighted in red in Attachment 6.5.1, will allow for another Class definition.

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