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10.2 Concentric, Eccentric and Inlaid Magic Squares
10.2.1 Concentric Magic Squares
In general an even Concentric Magic Square consists of 2 x 2 cells, around which borders can be constructed again and again.
A 10th order Concentric Magic Square consists of an embedded Magic Square of the 8th order with an embedded Magic Square of the 6th order with an embedded (Pan) Magic Square of the 4th order.
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With the inner 4 x 4 square pan magic, the embedded Magic Squares can be described by following linear equations:
a(82) = 404 - a(83) - a(84) - a(85) - a(86) - a(87) - a(88) - a(89)
a(73) = 303 - a(74) - a(75) - a(76) - a(77) - a(78)
a(72) = 101 - a(79)
a(64) = 202 - a(65) - a(66) - a(67)
a(63) = 101 - a(68)
a(62) = 101 - a(69)
a(56) = 202 - a(57) - a(66) - a(67)
a(55) = a(57) - a(65) + a(67)
a(54) = - a(57) + a(65) + a(66)
a(53) = 101 - a(58)
a(52) = 101 - a(59)
a(47) = 101 - a(65)
a(46) = -101 + a(65) + a(66) + a(67)
a(45) = 101 - a(67)
a(44) = 101 - a(66)
a(43) = 101 - a(48)
a(42) = 101 - a(49)
a(38) = 202 - a(48) - a(58) - a(68) + a(73) - a(78)
a(37) = 101 - a(57) + a(65) - a(67)
a(36) = 101 + a(57) - a(65) - a(66)
a(35) = 101 - a(57)
a(34) = -101 + a(57) + a(66) + a(67)
a(33) = -404 + a(48) + a(58) + a(68) + a(74) + a(75) + a(76) + a(77) + 2*a(78)
a(32) = 101 - a(39)
a(29) = 303 - a(39) - a(49) - a(59) - a(69) - a(79) + a(82) - a(89)
a(28) = -202 + a(74) + a(75) + a(76) + a(77) + a(78)
a(27) = 101 - a(77)
a(26) = 101 - a(76)
a(25) = 101 - a(75)
a(24) = 101 - a(74)
a(23) = 101 - a(78)
a(22) = 101 - a(29)
a(19) = 101 - a(82)
a(18) = 101 - a(88)
a(17) = 101 - a(87)
a(16) = 101 - a(86)
a(15) = 101 - a(85)
a(14) = 101 - a(84)
a(13) = 101 - a(83)
a(12) = 101 - a(89)
which can be completed with the equations describing the outer border, which results in following linear equations:
a(91) = 505 - a(92) - a(93) - a(94) - a(95) - a(96) - a(97) - a(98) - a(99) - a(100)
a(82) = 404 - a(83) - a(84) - a(85) - a(86) - a(87) - a(88) - a(89)
a(81) = 101 - a(90)
a(73) = 303 - a(74) - a(75) - a(76) - a(77) - a(78)
a(72) = 101 - a(79)
a(71) = 101 - a(80)
a(64) = 202 - a(65) - a(66) - a(67)
a(63) = 101 - a(68)
a(62) = 101 - a(69)
a(61) = 101 - a(70)
a(56) = 202 - a(57) - a(66) - a(67)
a(55) = a(57) - a(65) + a(67)
a(54) = - a(57) + a(65) + a(66)
a(53) = 101 - a(58)
a(52) = 101 - a(59)
a(51) = 101 - a(60)
a(47) = 101 - a(65)
a(46) = -101 + a(65) + a(66) + a(67)
a(45) = 101 - a(67)
a(44) = 101 - a(66)
a(43) = 101 - a(48)
a(42) = 101 - a(49)
a(41) = 101 - a(50)
a(38) = 202 - a(48) - a(58) - a(68) + a(73) - a(78)
a(37) = 101 - a(57) + a(65) - a(67)
a(36) = 101 + a(57) - a(65) - a(66)
a(35) = 101 - a(57)
a(34) = -101 + a(57) + a(66) + a(67)
a(33) = -404 + a(48) + a(58) + a(68) + a(74) + a(75) + a(76) + a(77) + 2*a(78)
a(32) = 101 - a(39)
a(31) = 101 - a(40)
a(29) = 303 - a(39) - a(49) - a(59) - a(69) - a(79) + a(82) - a(89)
a(28) = -202 + a(74) + a(75) + a(76) + a(77) + a(78)
a(27) = 101 - a(77)
a(26) = 101 - a(76)
a(25) = 101 - a(75)
a(24) = 101 - a(74)
a(23) = 101 - a(78)
a(22) = 101 - a(29)
a(21) = 101 - a(30)
a(20) = 404 - a(30) - a(40) - a(50) - a(60) - a(70) - a(80) - a(90) + a(91) - a(100)
a(19) = 101 - a(82)
a(18) = 101 - a(88)
a(17) = 101 - a(87)
a(16) = 101 - a(86)
a(15) = 101 - a(85)
a(14) = 101 - a(84)
a(13) = 101 - a(83)
a(12) = 101 - a(89)
a(11) = 101 - a(20)
a(10) = 101 - a(91)
a( 9) = 101 - a(99)
a( 8) = 101 - a(98)
a( 7) = 101 - a(97)
a( 6) = 101 - a(96)
a( 5) = 101 - a(95)
a( 4) = 101 - a(94)
a( 3) = 101 - a(93)
a( 2) = 101 - a(92)
a( 1) = 101 - a(100)
which can be applied in an Excel spreadsheet (Ref. CnstrSngl10b).
Note: The Embedded Magic Square is based on the consecutive integers 19, 20, ... 82.
With a(74), a(78), the variables of the two exterior borders and the 4th order most inner Pan Magic Square constant, an optimized guessing routine (MgcSqr10b), produced 1440 Magic Squares in 242 seconds which are shown in Attachment 10.2.1.
With the 8th order embedded Magic Square constant, while varying a(92), a(93), a(94), a(30), a(40) and a(50),
the same routine produced 1728 Magic Squares within 161 seconds, which are shown in Attachment 10.2.2.
10.2.2 Bordered Magic Squares
Also other Magic Squares of the 8th order as described and constructed in
Section 8.1 thru Section 8.5,
can be used as Center Squares for 10th order Bordered Magic Squares.
The Embedded Magic Squares will have a Magic Sum s8 = 404 and might be based on the consecutive integers 19, 20, ... 82.
Attachment 10.2.6
contains - based on some of the described Magic Squares of order 8 - examples of Bordered Magic Squares
for the first occurring border.
10.2.3 Eccentric Magic Squares
An Eccentric Magic Square can be defined as a Magic Corner Square of order n, supplemented with two or more (i) rows and columns to a Magic Square of order (n + i).
A 10th order Eccentric Magic Square consists of one Magic Corner Square of the 8th order, supplemented with two rows and two columns.
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Rather than starting with the equations of the Magic Corner Square, the equations of the supplementary rows and columns can be used as a starting point for the generation of Eccentric Magic Squares.
The supplementary rows and columns can be described by following linear equations:
a( 1) + a( 2) + a( 3) + a( 4) + a( 5) + a( 6) + a( 7) + a( 8) + a( 9) + a(10) = 505
a(11) + a(12) + a(13) + a(14) + a(15) + a(16) + a(17) + a(18) + a(19) + a(20) = 505
a( 1) + a(11) + a(21) + a(31) + a(41) + a(51) + a(61) + a(71) + a(81) + a(91) = 505
a( 2) + a(12) + a(22) + a(32) + a(42) + a(52) + a(62) + a(72) + a(82) + a(92) = 505
a(10) + a(19) + a(28) + a(37) + a(46) + a(55) + a(64) + a(73) + a(82) + a(91) = 505
a( 1) + a(12) = 101
a(21) + a(22) = 101
a(31) + a(32) = 101
a(41) + a(42) = 101
a(51) + a(52) = 101
a(61) + a(62) = 101
a(71) + a(72) = 101
a(81) + a(82) = 101
a(91) + a(92) = 101
a( 3) + a(13) = 101
a( 4) + a(14) = 101
a( 5) + a(15) = 101
a( 6) + a(16) = 101
a( 7) + a(17) = 101
a( 8) + a(18) = 101
a( 9) + a(19) = 101
a(10) + a(20) = 101
Which can be reduced, by means of row and column manipulations, to:
a(91) = 101 - a(92)
a(81) = 101 - a(82)
a(71) = 101 - a(72)
a(61) = 101 - a(62)
a(51) = 101 - a(52)
a(41) = 101 - a(42)
a(31) = 101 - a(32)
a(21) = 101 - a(22)
a(19) = 404 + a(20) - a(28) - a(37) - a(46) - a(55) - a(64) - a(73) - a(82) - a(91)
a(12) = 50.5 + 0.5*(- a(13) - a(14) - a(15) - a(16) - a(17) - a(18) - a(19) - a(20) +
+ a(21) + a(31) + a(41) + a(51) + a(61) + a(71) + a(81) + a(91))
a(11) = 505 - a(12) - a(13) - a(14) - a(15) - a(16) - a(17) - a(18) - a(19) - a(20)
a(10) = 101 - a(20)
a( 9) = 101 - a(19)
a( 8) = 101 - a(18)
a( 7) = 101 - a(17)
a( 6) = 101 - a(16)
a( 5) = 101 - a(15)
a( 4) = 101 - a(14)
a( 3) = 101 - a(13)
a( 2) = 101 - a(11)
a( 1) = 101 - a(12)
The linear equations, deducted above, can be applied in an Excel spreadsheet. Following typical cases have been considered:
- The 8th order Magic Corner Square is an Eccentric Magic Square itself (ref. CnstrSngl10c);
- The 8th order Magic Corner Square is a Pan Magic Square (ref. CnstrSngl10d).
Note: The Magic Corner Square is based on the consecutive integers 19, 20, ... 82.
In both cases it is obvious that the number of Eccentric Magic Squares is determined by the sum s2 of the values of the key variables
a(28), a(37), a(46), a(55), a(64) and a(73).
An optimized guessing routine (MgcSqr10c) produced,
with the Eccentric Magic Corner Square of the 8th order as shown in the first Spreadsheet Solution above constant, while varying the variables a(13), a(14), a(15), a(22), a(32) and a(42),
1800 Eccentric Magic Squares within 150 seconds, which are shown in Attachment 10.2.3.
The same routine produced, based on óne Pan Magic Corner Square of the eighth order constant and while varying the same variables
504 Eccentric Magic Squares within 52 seconds, which are shown in Attachment 10.2.4.
10.2.4 Order 3 Square Inlays (4 ea)
The 10th order Inlaid Magic Square shown below,
contains four each 3th order Simple Magic Square Inlays with Magic Sums
s(1) = 111, s(2) = 201, s(3) = 102 and s(4) = 192.
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9 |
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53 |
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50 |
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7 |
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84 |
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The relation between the Magic Sums s(1), s(2), s(3) and s(4) is:
s(1) + s(2) + s(3) + s(4) = 6 * s10 / 5
With s10 = 505 the Magic Sum of the 10th order Inlaid Magic Square.
The 3th order Simple Magic Square Inlays can be constructed by means of
suitable selected Latin Squares
based on the two 3th order magic series
{2, 3, 4} and
{5, 6, 7}
as illustrated below:
A
| 5 |
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| 7 |
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| 6 |
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| 4 |
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| 2 |
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| 3 |
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B = R(A)
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| 4 |
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| 2 |
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| 2 |
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M = A + 10 * B + 1
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66 |
78 |
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67 |
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| 27 |
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The remainder of the 10th order Inlaid Magic Square ('Window') can be completed based on the defining equations
as incorporated in procedure Inlaid103.
The square shown above corresponds with
84 * (3!)4 = 5.308.416 solutions,
which can be obtained by selecting other aspects of the four inlays and variation of the window.
Attachment 10.2.7
shows a few more order 10 Inlaid Magic Squares for miscellaneous order 3 Inlays, which can be constructed based on order 8 Inlaid Magic Squares as discussed in
Section 8.8.6.
10.2.5 Order 4 Pan Magic Square Inlays (4 ea)
The 10th order Inlaid Magic Square shown below,
contains four each 4th order Pan Magic Square Inlays with Magic Sums
s(1) = 220, s(2) = 180, s(3) = 224 and s(4) = 184.
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80 |
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3 |
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7 |
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4 |
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The relation between the Magic Sums s(1), s(2), s(3) and s(4) is:
s(1) + s(2) + s(3) + s(4) = 8 * s10 / 5
With s10 = 505 the Magic Sum of the 10th order Inlaid Magic Square.
The 4th order Pan Magic Square Inlays can be constructed by means of
suitable selected Latin Diagonal Squares
based on the two 4th order magic series
{1, 3, 5, 7} and
{2, 4, 6, 8}
as illustrated below:
A
| 3 |
1 |
7 |
5 |
3 |
1 |
7 |
5 |
| 5 |
7 |
1 |
3 |
5 |
7 |
1 |
3 |
| 1 |
3 |
5 |
7 |
1 |
3 |
5 |
7 |
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3 |
1 |
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5 |
3 |
1 |
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6 |
4 |
2 |
8 |
6 |
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2 |
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6 |
8 |
2 |
4 |
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8 |
2 |
4 |
6 |
8 |
| 8 |
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4 |
2 |
8 |
6 |
4 |
2 |
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B = R(A)
| 8 |
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6 |
4 |
7 |
1 |
5 |
3 |
| 6 |
4 |
8 |
2 |
5 |
3 |
7 |
1 |
| 4 |
6 |
2 |
8 |
3 |
5 |
1 |
7 |
| 2 |
8 |
4 |
6 |
1 |
7 |
3 |
5 |
| 8 |
2 |
6 |
4 |
7 |
1 |
5 |
3 |
| 6 |
4 |
8 |
2 |
5 |
3 |
7 |
1 |
| 4 |
6 |
2 |
8 |
3 |
5 |
1 |
7 |
| 2 |
8 |
4 |
6 |
1 |
7 |
3 |
5 |
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M = A + 10 * B + 1
| 84 |
22 |
68 |
46 |
74 |
12 |
58 |
36 |
| 66 |
48 |
82 |
24 |
56 |
38 |
72 |
14 |
| 42 |
64 |
26 |
88 |
32 |
54 |
16 |
78 |
| 28 |
86 |
44 |
62 |
18 |
76 |
34 |
52 |
| 85 |
23 |
69 |
47 |
75 |
13 |
59 |
37 |
| 67 |
49 |
83 |
25 |
57 |
39 |
73 |
15 |
| 43 |
65 |
27 |
89 |
33 |
55 |
17 |
79 |
| 29 |
87 |
45 |
63 |
19 |
77 |
35 |
53 |
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The remainder of the 10th order Inlaid Magic Square ('Border') can be completed based on the defining equations
as incorporated in procedure Inlaid104.
The square shown above corresponds with
3844 * (4!)4 = 7,214 1015 solutions,
which can be obtained by selecting other aspects of the four inlays and variation of the border.
10.2.6 Order 5, 6 and 7 Single Magic Square Inlays
The construction of Order 10 Simple Magic Squares with Order 5, 6 or 7 Single Magic Square Inlays will be discussed in
Sections 10.2.12, 13 and 14.
10.2.7 Spreadsheet Solutions
The linear equations deducted in previous sections, have been applied in following Excel Spread Sheets:
-
CnstrSngl10b, Magic Squares of order 10, Concentric Squares
-
CnstrSngl10c, Magic Squares of order 10, Eccentric Corner Square
-
CnstrSngl10d, Magic Squares of order 10, Pan Magic Corner Square
Only the red figures have to be “guessed” to construct one of the applicable Magic Squares of the 10th order
(wrong solutions are obvious).
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