Basic key method (ultra magic)

 

How to construct as perfect as possible magic squares which are an odd multiple of 4?First we analyze how the basic key method works for the construction of the 16x16 magic square. Than we investigate the possibilities to use a basic key to construct a 12x12 magic square.

Why does the basic key of the 16x16 magic square work?

How is it possible that the basic key leads to most perfect (Franklin pan)magic squares for each order is a multiple of 8.

 

1+2+3+4+5+6+7+8+9+10+11+12+13+14+15+16 = 136.

 

 

(1st) sum = sum = 34 = 1/4 x 136

 

 

1

2

16

15

16

15

1

2

  

 

So, in the 16x16 magic square each 1/4 row/column gives 1/4 of the magic sum.

 

(2nd) sum = sum = 34 = 1/4 x 136

  

 

1

2

16

15

16

15

1

2

 

 

So, in the 16x16 magic square each 1/4 diagonal gives 1/4 of the magic sum.

 

 

(3rd) Between opposite combinations must be an odd number of cells (opposite combinations must not be arranged next to each other).

 

  

1

2

16

15

16

15

1

2

 

 

So the first grid can be reflected to construct the second grid and if you combine the first grid  with the second grid you get all the digits from 1 up to 256 in the magic square.

 

How to get the right basic key to construct a 12x12 magic square?

For magic squares which are an odd multiple of 4 no basic key exists to comply to all three above mentioned conditions. You have to make a choice.

 

 

[Option a]

 

(1st) sum = sum = 26 = 1/3 x 78

 

 

1

10

11

4

8

6

5

7

12

3

2

9

12

3

2

9

5

7

8

6

1

10

11

4

 

  

So each 1/3 row/column gives 1/3 of the magic sum (but 1/2 row/column gives not 1/2 of the magic sum).

 

 

(2nd) sum = sum = 39 = 1/2 x 78

  

 

1

10

11

4

8

6

12

3

2

9

5

7

 

 
So each 1/2 diagonal gives 1/2 of the magic sum (and each [[parallel] bended diagonal gives the magic sum = Franklin feauture). 


(3rd) Between each opposite combination is and odd number of cells, so the first grid can be reflected to construct the second grid and if you combine the first grid with the second grid you get all the digits from 1 up to 144 in the magic square. 

 

1x digit from first grid

1

10

11

4

8

6

5

7

12

3

2

9

12

3

2

9

5

7

8

6

1

10

11

4

1

10

11

4

8

6

5

7

12

3

2

9

12

3

2

9

5

7

8

6

1

10

11

4

1

10

11

4

8

6

5

7

12

3

2

9

12

3

2

9

5

7

8

6

1

10

11

4

1

10

11

4

8

6

5

7

12

3

2

9

12

3

2

9

5

7

8

6

1

10

11

4

1

10

11

4

8

6

5

7

12

3

2

9

12

3

2

9

5

7

8

6

1

10

11

4

1

10

11

4

8

6

5

7

12

3

2

9

12

3

2

9

5

7

8

6

1

10

11

4

 

 

+ 12 x (digit -/- 1) from second grid

9

4

9

4

9

4

9

4

9

4

9

4

2

11

2

11

2

11

2

11

2

11

2

11

3

10

3

10

3

10

3

10

3

10

3

10

12

1

12

1

12

1

12

1

12

1

12

1

7

6

7

6

7

6

7

6

7

6

7

6

5

8

5

8

5

8

5

8

5

8

5

8

6

7

6

7

6

7

6

7

6

7

6

7

8

5

8

5

8

5

8

5

8

5

8

5

4

9

4

9

4

9

4

9

4

9

4

9

11

2

11

2

11

2

11

2

11

2

11

2

10

3

10

3

10

3

10

3

10

3

10

3

1

12

1

12

1

12

1

12

1

12

1

12

 

 

= 12x12 magic square (option a)

97

46

107

40

104

42

101

43

108

39

98

45

24

123

14

129

17

127

20

126

13

130

23

124

25

118

35

112

32

114

29

115

36

111

26

117

144

3

134

9

137

7

140

6

133

10

143

4

73

70

83

64

80

66

77

67

84

63

74

69

60

87

50

93

53

91

56

90

49

94

59

88

61

82

71

76

68

78

65

79

72

75

62

81

96

51

86

57

89

55

92

54

85

58

95

52

37

106

47

100

44

102

41

103

48

99

38

105

132

15

122

21

125

19

128

18

121

22

131

16

109

34

119

28

116

30

113

31

120

27

110

33

12

135

2

141

5

139

8

138

1

142

11

136

 

 

[option b]

 

(1st) sum = sum = 39 = 1/2 x 78

 

 

1

11

10

2

7

8

12

4

3

9

6

5

12

2

3

11

6

5

1

9

10

4

7

8

 

 

So each 1/2 row/column gives 1/2 of the magic sum.

 

 

(2nd) The sum of the half winding rows is not 39, but the sum of the whole winding rows is 78.

 

So each diagonal gives the magic sum (but each 1/2 diagonal gives not 1/2 of the magic sum and each [parallel] bended diagonals gives not the magic sum).

 

(3rd) Between each opposite combination is and odd number of cells, so the first grid can be reflected to construct the second grid and if you combine the first grid with the second grid you get all the digits from 1 up to 144 in the magic square.

 

1x digit from first grid

1

11

10

2

7

8

12

4

3

9

6

5

12

2

3

11

6

5

1

9

10

4

7

8

1

11

10

2

7

8

12

4

3

9

6

5

12

2

3

11

6

5

1

9

10

4

7

8

1

11

10

2

7

8

12

4

3

9

6

5

12

2

3

11

6

5

1

9

10

4

7

8

1

11

10

2

7

8

12

4

3

9

6

5

12

2

3

11

6

5

1

9

10

4

7

8

1

11

10

2

7

8

12

4

3

9

6

5

12

2

3

11

6

5

1

9

10

4

7

8

1

11

10

2

7

8

12

4

3

9

6

5

12

2

3

11

6

5

1

9

10

4

7

8

 

 

+ 12x (digit -/- 1) from second grid

5

8

5

8

5

8

5

8

5

8

5

8

6

7

6

7

6

7

6

7

6

7

6

7

9

4

9

4

9

4

9

4

9

4

9

4

3

10

3

10

3

10

3

10

3

10

3

10

4

9

4

9

4

9

4

9

4

9

4

9

12

1

12

1

12

1

12

1

12

1

12

1

8

5

8

5

8

5

8

5

8

5

8

5

7

6

7

6

7

6

7

6

7

6

7

6

2

11

2

11

2

11

2

11

2

11

2

11

10

3

10

3

10

3

10

3

10

3

10

3

11

2

11

2

11

2

11

2

11

2

11

2

1

12

1

12

1

12

1

12

1

12

1

12

 

 

= 12x12 magic square (option b)

49

95

58

86

55

92

60

88

51

93

54

89

72

74

63

83

66

77

61

81

70

76

67

80

97

47

106

38

103

44

108

40

99

45

102

41

36

110

27

119

30

113

25

117

34

112

31

116

37

107

46

98

43

104

48

100

39

105

42

101

144

2

135

11

138

5

133

9

142

4

139

8

85

59

94

50

91

56

96

52

87

57

90

53

84

62

75

71

78

65

73

69

82

64

79

68

13

131

22

122

19

128

24

124

15

129

18

125

120

26

111

35

114

29

109

33

118

28

115

32

121

23

130

14

127

20

132

16

123

21

126

17

12

134

3

143

6

137

1

141

10

136

7

140

 

 

Ultra panmagic 12x12 square

It is also possible to construct an ultra panmagic 12x12 square in which each 4x4 sub-square has the following symmetric structure:

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Two digits of the same colour sums to (the lowest + highest digit of the magic 12x12 square = 1 + 144 =) 145.

 

 

1x digit

 

 

 

 

 

 

 

 

 

12

2

2

12

5

6

6

5

10

4

4

10

1

11

11

1

8

7

7

8

3

9

9

3

12

2

2

12

5

6

6

5

10

4

4

10

1

11

11

1

8

7

7

8

3

9

9

3

12

2

2

12

5

6

6

5

10

4

4

10

1

11

11

1

8

7

7

8

3

9

9

3

12

2

2

12

5

6

6

5

10

4

4

10

1

11

11

1

8

7

7

8

3

9

9

3

12

2

2

12

5

6

6

5

10

4

4

10

1

11

11

1

8

7

7

8

3

9

9

3

12

2

2

12

5

6

6

5

10

4

4

10

1

11

11

1

8

7

7

8

3

9

9

3

 

 

+ 12 x (digit -/- 1)

 

 

 

 

 

 

 

10

3

10

3

10

3

10

3

10

3

10

3

4

9

4

9

4

9

4

9

4

9

4

9

4

9

4

9

4

9

4

9

4

9

4

9

10

3

10

3

10

3

10

3

10

3

10

3

5

8

5

8

5

8

5

8

5

8

5

8

6

7

6

7

6

7

6

7

6

7

6

7

6

7

6

7

6

7

6

7

6

7

6

7

5

8

5

8

5

8

5

8

5

8

5

8

12

1

12

1

12

1

12

1

12

1

12

1

2

11

2

11

2

11

2

11

2

11

2

11

2

11

2

11

2

11

2

11

2

11

2

11

12

1

12

1

12

1

12

1

12

1

12

1

 


= ultra panmagic 12x12 square

 

 

 

 

120

26

110

36

113

30

114

29

118

28

112

34

37

107

47

97

44

103

43

104

39

105

45

99

48

98

38

108

41

102

42

101

46

100

40

106

109

35

119

25

116

31

115

32

111

33

117

27

60

86

50

96

53

90

54

89

58

88

52

94

61

83

71

73

68

79

67

80

63

81

69

75

72

74

62

84

65

78

66

77

70

76

64

82

49

95

59

85

56

91

55

92

51

93

57

87

144

2

134

12

137

6

138

5

142

4

136

10

13

131

23

121

20

127

19

128

15

129

21

123

24

122

14

132

17

126

18

125

22

124

16

130

133

11

143

1

140

7

139

8

135

9

141

3

 

 

Notify that each 1/2 row/column gives 1/2 of the magic sum, each 1/3 diagonal gives 1/3 of the magic sum, the 12x12 magic square is panmagic, 2x2 compact and symmetric within each 4x4 sub-square.

 

Try the following basic key:

 

 

12

12

2

2

10

10

4

4

5

5

6

6

1

1

11

11

3

3

9

9

8

8

7

7

 


You get an ultra panmagic 12x12 square in which each 4x4 sub-square has the following structure: 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

To construct an ultra panmagic square, which is symmetric within each 6x6 sub-square (and 2x2 compact), use the following basic key: 

 

12

2

10

10

2

12

4

6

5

5

6

4

1

11

3

3

11

1

9

7

8

8

7

9

 

 

To construct an ultra panmagic square, which is symmetric (and 2x2 compact) and each 1/2 row/column gives 1/2 of the magic sum, use the following basic key: 

 

12

2

10

4

5

6

6

5

4

10

2

12

1

11

3

9

8

7

7

8

9

3

11

1

 

 

Have fun!!! 

 

Download
12x12, Basic key method (1).xls
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