# Strachey

Ralph Strachey's method divides a square of order n=4k+2 into four quadrants with odd order m=n/2, each of which is assigned a specific base number.

 A C D B
 0 2 m2 3 m2 m2

The quadrant A is filled with numbers 1 to m2. Strachey chose the method of de la Loubère's, when introducing his method, but you can also choose any other method for odd order magic squares. The three other quadrants are then filled with the same magic square, where the base number belonging to the quadrant is added to each number.

This special arrangement ensures that the columns already have the same sum. However, this does not yet apply to the rows and the diagonals. To achieve this goal, a few more cells have to be replaced. It is

n=2 · m=2 · (2 · k + 1)=4k + 2

where k specifies the number of numbers per row in quadrant A, which must be exchanged with the corresponding vertically symmetrical numbers of quadrant D. The numbers are selected one after the other from the left margin of the rows. As the only exception, the first numbers in the center row of quadrants A and C remain unaffected and the k numbers from the second position are exchanged.

In quadrants B and C, on the other hand, all numbers in the k − 1 columns seen from the right mergin are exchanged with the vertically symmetrical number from the lower quadrant. Regarding order 6 is k − 1=0, so that these exchanges are only used from order n=10.

Since only numbers in one column are swapped, the column sums do not change. But the rows and diagonal sums are now also magic, since the sum of the basic numbers in the rows and diagonals is now always m2 after the swaps. Added are the numbers from the square of de la Loubère, so that the resulting square is magic.

•  8 1 6 26 19 24 3 5 7 21 23 25 4 9 2 22 27 20 35 28 33 17 10 15 30 32 34 12 14 16 31 36 29 13 18 11
•  35 1 6 26 19 24 3 32 7 21 23 25 31 9 2 22 27 20 8 28 33 17 10 15 30 5 34 12 14 16 4 36 29 13 18 11

In the example shown for n=6 there were no exchanges on the right margin. So, another example for order n=10 shall be given. This time, the method of Moschopoulos is chosen as the base square for quadrant A.

This is followed by swaps at the left and right margins to balance the base numbers in rows and diagonals. To do this, k numbers on the left margin and k − 1 numbers on the right margin must be exchanged vertically symmetrically

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

and the magic square is created.