Office Applications and Entertainment, Magic Squares

10.0   Magic Squares (10 x 10)

10.1   Medjig Solutions (10 x 10)

10.1.1 General

As described in section 6.8, for any integer n, a magic square C of order 2n can be constructed from any n x n medjig-square A with each row, column, and main diagonal summing to 3n, and any n x n magic square B, by application of the equations:

b_i + n² a_j with i = 1, 2, ... n² and j = 1, 2, ... 4n².

The Medjig method of constructing a Magic Square of order 10 is as follows:

Construct a 5 x 5 Medjig-Square A (ignoring the original game's limit on the number of times that a given sequence
is used)

Construct a 5 x 5 (Pan) Magic Square B (Already 28800 possibilities for Pan Magic only, refer Attachment 5.2.6)

Construct a 10 x 10 Magic Square C by applying the equations mentioned above.

The rows, columns and main diagonals of Square C sum to 2 times the corresponding sum of Magic Square B plus 25 times the corresponding sum of Medjig square A which results in s1 = 2 * 65 + 25 * 15 = 505.

As b(i) ≠ b(j) for i ≠ j with i, j = 1, 2, ... 25 it is obvious that also c(m) ≠ c(n) for n ≠ m with n, m = 1, 2, ... 100.

A numerical example is shown below:

10.1.2 Further Analysis

Medjig Squares are described by the same linear equations as applicable for Magic Squares, however with magic sum 15:

a( 1)+a( 2)+a( 3)+a( 4)+a( 5)+a( 6)+a( 7)+a( 8)+a( 9)+a( 10) = 15
a(11)+a(12)+a(13)+a(14)+a(15)+a(16)+a(17)+a(18)+a(19)+a( 20) = 15
a(21)+a(22)+a(23)+a(24)+a(25)+a(26)+a(27)+a(28)+a(29)+a( 30) = 15
a(31)+a(32)+a(33)+a(34)+a(35)+a(36)+a(37)+a(38)+a(39)+a( 40) = 15
a(41)+a(42)+a(43)+a(44)+a(45)+a(46)+a(47)+a(48)+a(49)+a( 50) = 15
a(51)+a(52)+a(53)+a(54)+a(55)+a(56)+a(57)+a(58)+a(59)+a( 60) = 15
a(61)+a(62)+a(63)+a(64)+a(65)+a(66)+a(67)+a(68)+a(69)+a( 70) = 15
a(71)+a(72)+a(73)+a(74)+a(75)+a(76)+a(77)+a(78)+a(79)+a( 80) = 15
a(81)+a(82)+a(83)+a(84)+a(85)+a(86)+a(87)+a(88)+a(89)+a( 90) = 15
a(91)+a(92)+a(93)+a(94)+a(95)+a(96)+a(97)+a(98)+a(99)+a(100) = 15

a( 1)+a(11)+a(21)+a(31)+a(41)+a(51)+a(61)+a(71)+a(81)+a( 91) = 15
a( 2)+a(12)+a(22)+a(32)+a(42)+a(52)+a(62)+a(72)+a(82)+a( 92) = 15
a( 3)+a(13)+a(23)+a(33)+a(43)+a(53)+a(63)+a(73)+a(83)+a( 93) = 15
a( 4)+a(14)+a(24)+a(34)+a(44)+a(54)+a(64)+a(74)+a(84)+a( 94) = 15
a( 5)+a(15)+a(25)+a(35)+a(45)+a(55)+a(65)+a(75)+a(85)+a( 95) = 15
a( 6)+a(16)+a(26)+a(36)+a(46)+a(56)+a(66)+a(76)+a(86)+a( 96) = 15
a( 7)+a(17)+a(27)+a(37)+a(47)+a(57)+a(67)+a(77)+a(87)+a( 97) = 15
a( 8)+a(18)+a(28)+a(38)+a(48)+a(58)+a(68)+a(78)+a(88)+a( 98) = 15
a( 9)+a(19)+a(29)+a(39)+a(49)+a(59)+a(69)+a(79)+a(89)+a( 99) = 15
a(10)+a(20)+a(30)+a(40)+a(50)+a(60)+a(70)+a(80)+a(90)+a(100) = 15

a( 1)+a(12)+a(23)+a(34)+a(45)+a(56)+a(67)+a(78)+a(89)+a(100) = 15
a(10)+a(19)+a(28)+a(37)+a(46)+a(55)+a(64)+a(73)+a(82)+a( 91) = 15

and following additional equations:

Resulting in the matrix representation:

         →   →
     A * a = s

which can be reduced, by means of row and column manipulations, to the minimum number of linear equations:

a(91) = 15 - a(92) - a(93) - a(94) - a(95) - a(96) - a(97) - a(98) - a(99) - a(100)
a(89) =  6 - a(90) - a(99) - a(100)
a(87) =  6 - a(88) - a(97) - a(98)
a(85) =  6 - a(86) - a(95) - a(96)
a(83) =  6 - a(84) - a(93) - a(94)
a(81) =  6 - a(82) - a(91) - a(92)
a(71) = 15 - a(72) - a(73) - a(74) - a(75) - a(76) - a(77) - a(78) - a(79) - a(80)
a(69) =  6 - a(70) - a(79) - a(80)
a(67) =  6 - a(68) - a(77) - a(78)
a(65) =  6 - a(66) - a(75) - a(76)
a(63) =  6 - a(64) - a(73) - a(74)
a(61) =  6 - a(62) - a(71) - a(72)
a(51) = 15 - a(52) - a(53) - a(54) - a(55) - a(56) - a(57) - a(58) - a(59) - a(60)
a(49) =  6 - a(50) - a(59) - a(60)
a(47) =  6 - a(48) - a(57) - a(58)
a(45) =  6 - a(46) - a(55) - a(56)
a(43) =  6 - a(44) - a(53) - a(54)
a(41) =  6 - a(42) - a(51) - a(52)
a(31) = 15 - a(32) - a(33) - a(34) - a(35) - a(36) - a(37) - a(38) - a(39) - a(40)
a(29) =  6 - a(30) - a(39) - a(40)
a(27) =  6 - a(28) - a(37) - a(38)
a(25) =  6 - a(26) - a(35) - a(36)
a(23) =  6 - a(24) - a(33) - a(34)
a(21) =  6 - a(22) - a(31) - a(32)
a(19) = 18 + a(20) - a(28) + a(30) - a(37) + a(40) - a(46) - a(49) - a(55) - a(59) - a(64) - a(69) - a(73) - a(79) +
           - a(82) - a(89) - a(91) - a(99)
a(12) = 0.5 * (33  - a(13) - a(14) - a(15) - a(16) - a(17) - a(18) - a(19) - a(20) - a(22) + a(24) - a(32) + a(33) +
                   - a(42) - a(45) - a(52) - a(56) - a(62) - a(67) - a(72) - a(78) - a(82) - a(89) - a(92) - a(100))
a(11) =-18 + a(12) + a(22) - a(24) + a(32) - a(33) + a(42) + a(45) + a(52) + a(56) + a(62) + a(67) + a(72) + a(78) +
           + a(82) + a(89) + a(92) + a(100)
a(10) = 15 - a(20) - a(30) - a(40) - a(50) - a(60) - a(70) - a(80) - a(90) - a(100)
a( 9) = -9 - a(20) + a(28) + a(37) + a(46) + a(55) + a(64) + a(73) + a(82) + a(91)
a( 8) = 15 - a(18) - a(28) - a(38) - a(48) - a(58) - a(68) - a(78) - a(88) - a(98)
a( 7) =  6 - a( 8) - a(17) - a(18)
a( 6) = 15 - a(16) - a(26) - a(36) - a(46) - a(56) - a(66) - a(76) - a(86) - a(96)
a( 5) =  6 - a( 6) - a(15) - a(16)
a( 4) = 15 - a(14) - a(24) - a(34) - a(44) - a(54) - a(64) - a(74) - a(84) - a(94)
a( 3) =  9 - a(13) + a(24) + a(34) - a(43) - a(53) - a(63) - a(73) - a(83) - a(93)
a( 2) = 0.5 * (30  - a(11) - a(12) - a(19) + a(20) - a(22) - a(24) - a(28) + a(30) - a(32) - a(33) - a(37) + a(40) +
                   - a(42) + a(45) - a(46) - a(49) - a(52) - a(55) + a(56) - a(59) - a(62) - a(64) + a(67) - a(69) +
                   - a(72) - a(73)+ a(78) - a(79) - 2 * a(82) - a(91) - a(92) - a(99) + a(100))
a( 1) = -6 + a( 2) + a(22) + a(24) + a(32) + a(33) + a(42) - a(45) + a(52) - a(56) + a(62) - a(67) + a(72) - a(78) +
                   + a(82) - a(89) + a(92) - a(100)

The linear equations shown above are ready to be solved, for the magic constant 15.

The solutions can be obtained by guessing a(i) for

     i = 13 ... 18, 20, 22, 24, 26, 28, 30, 32 ... 40, 42, 44, 46, 48, 50, 52 ... 60
                        62, 64, 66, 68, 70, 72 ... 80, 82, 84, 86, 88, 90, 92 ... 100

and filling out these guesses in the abovementioned equations.

To obtain the integers 0, 1, 2 and 3 also following relations should be applied:

     0 =< a(i) =< 3        for i = 1 ... 12, 19, 21, 23, 25, 27, 29, 31, 41, 43, 45, 47, 49, 51
                                                 61, 63, 65, 67, 69, 71, 81, 83, 85, 87, 89, 91
     Int(a(i)) = a(i)      for i = 2 and 12

which can be incorporated in a guessing routine, which can be used to generate a defined number of Medjig Squares within a reasonable time.

With 18 of the 25 Medjig pieces constant, an optimized guessing routine (MgcSqr10a), produced 2112 Medjig Squares within 201 seconds, which are shown in Attachment 10.1.1.

The resulting Magic Squares, based on the 5^th order Magic Square shown in the numerical solution above are shown in Attachment 10.1.2.

It should be noted that, although much faster, not all Magic Squares of the 10^th order can be found by means of the Medjig Solution.

10.1.3 Concentric Magic Squares

The Medjig method of constructing a Concentric Magic Square of order 10 is as follows:

Construct a  5 x  5 Concentric Magic Medjig-Square A;
Construct a  5 x  5 Concentric Magic Square B (ref. Attachment 5.5.2);
Construct a 10 x 10 Concentric Magic Square C by applying the equations mentioned in Section 10.1.1 above.

A numerical example is shown below:

The Concentric Magic Squares resulting from the Medjig Square A shown above and 128 of the 23040 possible 5^th order Concentric Magic Squares, are shown in Attachment 10.1.3.

10.1.4 Eccentric Magic Squares

The Medjig method of constructing an Eccentric Magic Square of order 10 is as follows:

Construct a  5 x  5 Eccentric Magic Medjig-Square A;
Construct a  5 x  5 Eccentric Magic Square B (ref. Attachment 5.5.3);
Construct a 10 x 10 Eccentric Magic Square C by applying the equations mentioned in Section 10.1.1 above.

A numerical example is shown below:

The Eccentric Magic Squares resulting from the Medjig Square A shown above and 128 of the 3072 possible 5^th order Eccentric Magic Squares, are shown in Attachment 10.1.4.

10.1.5 Almost Associated Magic Squares

The Medjig method of constructing an Almost Associated Magic Square of order 10 is as follows:

Construct a  5 x  5 Almost Associated Medjig-Square A composed of:
   An Associated Medjig Border
   An Almost Associated Medjig Center Square   (ref. Section 6.6.4);
Construct a  5 x  5 Associated Magic Square B (ref. Attachment 5.4.2);
Construct a 10 x 10 Almost Associated Magic Square C by applying the equations mentioned in Section 10.1.1 above.

A numerical example is shown below:

Attachment 10.2.5 shows a few examples of Almost Associated Magic Squares resulting from the Associated Magic Square B shown above and miscellaneous Almost Associated Medjig Squares.

The Almost Associated Medjig Squares are based on the border of Square A above and the 120 Almost Associated Medjig Center Squares shown in Attachment 6.8.5.

It should be noted that although the Almost Associated Medjig Square A is bordered, the resulting Square C will be only Almost Associated as the 5 x 5 Square B is Associated.

10.1.6 Magic Squares with Bimagic Main Diagonals

It can be proven that the Medjig method is not suitable for the construction of Bimagic Squares of order 10.

However Magic Squares of order 10 with Bimagic Main Diagonals can be constructed as follows:

• Select order 5 Magic Lines suitable for the construction of order 10 Bimagic Lines (ref. Attachment 10.1.5);
• Generate order 5 Magic Squares B using these lines as diagonals (ref. Attachment 10.1.6);
• Generate the order 10 Magic Squares C with an appropriate guessing routine, which checks the Bimagic Diagonals while generating A and calculating C (ref. MgcSqr10a2).

A numerical example is shown below:

Attachment 10.1.7 shows a few 10 x 10 Magic Squares with Bimagic Diagonals, which could be constructed based on the 5 x 5 Magic Squares described above.

10.1.7 Magic Squares with Bimagic Center Lines

Eight of the order 5 Magic Lines shown in Attachment 10.1.5 might return two adjacent order 10 Bimagic Lines.

Order 5 Magic Squares with these Magic Lines as center lines will return order 10 Magic Squares with Bimagic Center Lines.

The Medjig method of constructing order 10 Magic Squares with Bimagic Center Lines is as follows:

• Generate, based on preselected center lines as described above, order 5 Magic Squares B (ref. MgcSqr5k);
• Generate the order 10 Magic Squares C with an appropriate guessing routine, which checks the Bimagic Center Lines while generating A and calculating C (ref. MgcSqr10a3).

A numerical example is shown below:

Attachment 10.1.9 shows a few 10 x 10 Magic Squares with Bimagic Center Lines, which could be constructed based on the 5 x 5 Magic Squares described above.

10.1.8 Spreadsheet Solution

The linear equations shown in section 10.1.2 above can be applied in an Excel spreadsheet (Ref. CnstrSngl10a).

The red figures have to be “guessed” to construct a 5 x 5 Medjig Square (wrong solutions are obvious).