Office Applications and Entertainment, Magic Squares

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 10^th order Concentric Magic Square consists of an embedded Magic Square of the 8^th order with an embedded Magic Square of the 6^th order with an embedded (Pan) Magic Square of the 4^th order.

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

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 4^th 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 8^th 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 8^th order as described and constructed in Section 8.1 thru Section 8.5, can be used as Center Squares for 10^th 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 10^th order Eccentric Magic Square consists of one Magic Corner Square of the 8^th order, supplemented with two rows and two columns.

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

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 8^th order Magic Corner Square is an Eccentric Magic Square itself (ref. CnstrSngl10c);
- The 8^th 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 8^th 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 10^th order Inlaid Magic Square shown below, contains four each 3^th order Simple Magic Square Inlays with Magic Sums s(1) = 111, s(2) = 201, s(3) = 102 and s(4) = 192.

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

111 201 102 192

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 10^th order Inlaid Magic Square.

The 3^th order Simple Magic Square Inlays can be constructed by means of suitable selected Latin Squares based on the two 3^th order magic series {2, 3, 4} and {5, 6, 7} as illustrated below:

A 5 7 6 5 7 6 7 6 5 7 6 5 6 5 7 6 5 7 4 2 3 4 2 3 2 3 4 2 3 4 3 4 2 3 4 2

B = R(A) 3 2 4 6 7 5 4 3 2 5 6 7 2 4 3 7 5 6 3 2 4 6 7 5 4 3 2 5 6 7 2 4 3 7 5 6

M = A + 10 * B + 1 36 28 47 66 78 57 48 37 26 58 67 76 27 46 38 77 56 68 35 23 44 65 73 54 43 34 25 53 64 75 24 45 33 74 55 63

The remainder of the 10^th 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 10^th order Inlaid Magic Square shown below, contains four each 4^th order Pan Magic Square Inlays with Magic Sums s(1) = 220, s(2) = 180, s(3) = 224 and s(4) = 184.

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

220 180 224 184

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 10^th order Inlaid Magic Square.

The 4^th order Pan Magic Square Inlays can be constructed by means of suitable selected Latin Diagonal Squares based on the two 4^th order magic series {1, 3, 5, 7} and {2, 4, 6, 8} as illustrated below:

The remainder of the 10^th 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 10^th order (wrong solutions are obvious).