Office Applications and Entertainment, Magic Squares

Vorige Pagina Volgende Pagina Index About the Author

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:

bi + n2 aj 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.

B (5 x 5)

b1

b2

b3

b4

b5

b6

b7

b8

b9

b10

b11

b12

b13

b14

b15

b16

b17

b18

b19

b20

b21

b22

b23

b24

b25

Medjig Square A (5 x 5)

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

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

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

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

Magic Square C (10 x 10)

b1+25*a1

b1+25*a2

b2+25*a3

b2+25*a4

b3+25*a5

b3+25*a6

b4+25*a7

b4+25*a8

b5+25*a9

b5+25*a10

b1+25*a11

b1+25*a12

b2+25*a13

b2+25*a14

b3+25*a15

b3+25*a16

b4+25*a17

b4+25*a18

b5+25*a19

b5+25*a20

b6+25*a21

b6+25*a22

b7+25*a23

b7+25*a24

b8+25*a25

b8+25*a26

b9+25*a27

b9+25*a28

b10+25*a29

b10+25*a30

b6+25*a31

b6+25*a32

b7+25*a33

b7+25*a34

b8+25*a35

b8+25*a36

b9+25*a37

b9+25*a38

b10+25*a39

b10+25*a40

b11+25*a41

b11+25*a42

b12+25*a43

b12+25*a44

b13+25*a45

b13+25*a46

b14+25*a47

b14+25*a48

b15+25*a49

b15+25*a50

b11+25*a51

b11+25*a52

b12+25*a53

b12+25*a54

b13+25*a55

b13+25*a56

b14+25*a57

b14+25*a58

b15+25*a59

b15+25*a60

b16+25*a61

b16+25*a62

b17+25*a63

b17+25*a64

b18+25*a65

b18+25*a66

b19+25*a67

b19+25*a68

b20+25*a69

b20+25*a70

b16+25*a71

b16+25*a72

b17+25*a73

b17+25*a74

b18+25*a75

b18+25*a76

b19+25*a77

b19+25*a78

b20+25*a79

b20+25*a80

b21+25*a81

b21+25*a82

b22+25*a83

b22+25*a84

b23+25*a85

b23+25*a86

b24+25*a87

b24+25*a88

b25+25*a89

b25+25*a90

b21+25*a91

b21+25*a92

b22+25*a93

b22+25*a94

b23+25*a95

b23+25*a96

b24+25*a97

b24+25*a98

b25+25*a99

b25+25*a100

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:

B (5 x 5)

12

6

5

24

18

4

23

17

11

10

16

15

9

3

22

8

2

21

20

14

25

19

13

7

1

Medjig Square A (5 x 5)

0

2

0

1

0

2

3

2

2

3

1

3

3

2

3

1

0

1

1

0

0

3

1

3

3

0

0

1

1

3

2

1

0

2

2

1

3

2

0

2

3

0

0

1

2

3

0

3

3

0

1

2

3

2

1

0

2

1

2

1

1

2

3

0

1

3

2

1

2

0

3

0

2

1

0

2

0

3

3

1

3

0

2

0

3

2

2

0

0

3

1

2

1

3

0

1

3

1

1

2

Magic Square C (10 x 10)

12

62

6

31

5

55

99

74

68

93

37

87

81

56

80

30

24

49

43

18

4

79

48

98

92

17

11

36

35

85

54

29

23

73

67

42

86

61

10

60

91

16

15

40

59

84

3

78

97

22

41

66

90

65

34

9

53

28

72

47

33

58

77

2

46

96

70

45

64

14

83

8

52

27

21

71

20

95

89

39

100

25

69

19

88

63

57

7

1

76

50

75

44

94

13

38

82

32

26

51

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:

a(1)+a( 2)+a(11)+a(12) = 6
a(3)+a( 4)+a(13)+a(14) = 6
a(5)+a( 6)+a(15)+a(16) = 6
a(7)+a( 8)+a(17)+a(18) = 6
a(9)+a(10)+a(19)+a(20) = 6

a(21)+a(22)+a(31)+a(32) = 6
a(23)+a(24)+a(33)+a(34) = 6
a(25)+a(26)+a(35)+a(36) = 6
a(27)+a(28)+a(37)+a(38) = 6
a(29)+a(30)+a(39)+a(40) = 6

a(41)+a(42)+a(51)+a(52) = 6
a(43)+a(44)+a(53)+a(54) = 6
a(45)+a(46)+a(55)+a(56) = 6
a(47)+a(48)+a(57)+a(58) = 6
a(49)+a(50)+a(59)+a(60) = 6

a(61)+a(62)+a(71)+a(72) = 6
a(63)+a(64)+a(73)+a(74) = 6
a(65)+a(66)+a(75)+a(76) = 6
a(67)+a(68)+a(77)+a(78) = 6
a(69)+a(70)+a(79)+a(80) = 6

a(81)+a(82)+a(91)+a( 92) = 6
a(83)+a(84)+a(93)+a( 94) = 6
a(85)+a(86)+a(95)+a( 96) = 6
a(87)+a(88)+a(97)+a( 98) = 6
a(89)+a(90)+a(99)+a(100) = 6



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 5th 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 10th 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:

B (5 x 5)

25 5 8 24 3
4 16 11 12 22
6 9 13 17 20
7 14 15 10 19
23 21 18 2 1

Medjig Square A (5 x 5)

1 2 0 1 0 2 2 3 1 3
0 3 3 2 3 1 1 0 2 0
0 3 2 1 0 3 3 0 1 2
2 1 0 3 2 1 2 1 0 3
3 0 1 0 0 3 2 3 3 0
1 2 3 2 2 1 1 0 2 1
1 2 3 2 2 0 0 2 2 1
3 0 0 1 3 1 1 3 3 0
3 0 2 0 3 2 0 2 0 3
1 2 1 3 0 1 3 1 1 2

Magic Square C (10 x 10)

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

The Concentric Magic Squares resulting from this Medjig Square and 128 of the 23040 possible 5th 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:

B (5 x 5)

5 19 3 20 18
7 21 23 6 8
4 22 16 11 12
25 1 9 13 17
24 2 14 15 10

Medjig Square A (5 x 5)

0 2 0 1 0 3 1 3 2 3
1 3 3 2 2 1 2 0 1 0
3 0 1 3 3 0 0 1 3 1
1 2 0 2 1 2 3 2 0 2
2 3 0 1 2 1 0 3 3 0
1 0 3 2 0 3 2 1 2 1
1 2 3 0 1 0 0 3 2 3
3 0 2 1 3 2 2 1 1 0
0 1 2 3 3 2 2 0 0 2
3 2 1 0 0 1 3 1 1 3

Magic Square C (10 x 10)

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

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

10.1.5 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:

B (5 x 5)

25 10 3 5 22
4 23 7 15 16
12 11 14 19 9
18 8 20 2 17
6 13 21 24 1

Medjig Square A (5 x 5)

3 2 0 2 0 2 0 2 1 3
0 1 1 3 1 3 1 3 2 0
0 2 2 3 0 1 2 3 0 2
1 3 1 0 3 2 1 0 1 3
0 1 1 0 3 2 1 2 2 3
3 2 3 2 0 1 3 0 1 0
1 0 3 1 1 0 2 3 3 1
3 2 0 2 3 2 0 1 2 0
3 2 1 0 1 0 2 1 2 3
1 0 3 2 3 2 3 0 1 0

Magic Square C (10 x 10)

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

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.

10.1.6 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:

B (5 x 5)

7 19 1 13 25
15 23 10 6 11
2 8 14 21 20
24 3 18 16 4
17 12 22 9 5

Medjig Square A (5 x 5)

1 2 0 2 2 3 0 1 1 3
0 3 1 3 0 1 2 3 0 2
0 2 1 2 2 3 0 2 0 3
1 3 0 3 0 1 1 3 2 1
2 1 2 0 1 0 3 1 3 2
3 0 3 1 3 2 2 0 1 0
1 0 1 0 2 3 0 3 3 2
3 2 3 2 1 0 2 1 1 0
1 0 1 0 3 2 2 1 3 2
3 2 3 2 1 0 3 0 1 0

Magic Square C (10 x 10)

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

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.

10.1.7 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).


Vorige Pagina Volgende Pagina Index About the Author