Bachet – Labosne

The algorithm of Bachet and Labosne uses with two auxiliary squares. In a first step, the two diagonals of the first auxiliary square are filled from left to right with the numbers 1 to n in ascending order. The columns in the left half are then filled with the already existing numbers and their numbers complementary to n+1. Each of the two possible numbers must occur exactly m=n/2 times.

When all columns of the left half are filled, the horizontally symmetrical cells of the right half are filled with the complementary numbers.

  • 16
    25
    34
    34
    25
    16
  • 1246
    6245
    1534
    6534
    6235
    1546
  • 124356
    624351
    153426
    653421
    623451
    154326

In the second auxiliary square, the secondary diagonal is filled with the numbers 0, n, 2n, … , (n-1) · n from top left to bottom right. In the example of order n=6, these would be the numbers 0, 6, 12, 18, 24 and 30. On the main diagonal, however, the numbers from the bottom left to the top right are entered in reversed order.

Similar to the procedure for the columns in the first auxiliary square, the lower m rows are now filled with the existing numbers and their numbers complementary to (n-1) · n so that each of the two possible numbers is represented exactly m times.

  • 00
    66
    1212
    1818
    2424
    3030
  • 00
    66
    1212
    121818181212
    624662424
    300300030

However, there is an additional rule to be observed: if in a row of the first auxiliary square two horizontally symmetrical numbers are complementary with respect to n+1 to their vertically symmetrical numbers, the same numbers must be placed in the corresponding row of the second auxiliary square and in the symmetrical row the complementary numbers.

For example, the numbers 5 and 2 are in the bottom row of the first auxiliary square and the complementary numbers 2 and 5 are in the symmetrical cells of the top row. The special rule now applies and the same number must be entered in the two cells of the bottom row. Although the two numbers 0 and 30 are principally available for this row, you must enter the number 0 in this case, otherwise the number 30 would occur too often.

  • 124356
    624351
    153426
    653421
    623451
    154326
  • 00
    66
    1212
    1818
    2424
    300030

Two other cases of this special rule occur in the two middle columns of the second row from below and in the two outer columns of the row above. Here too, the same numbers must be entered in the corresponding rows. The result using the special rule is shown in the following figures.

  • 124356
    624351
    153426
    653421
    623451
    154326
  • 00
    66
    1212
    1818
    246624
    300030
  • 124356
    624351
    153426
    653421
    623451
    154326
  • 00
    66
    1212
    12181812
    246624
    300030

For order n=6, the two numbers already entered on the diagonals mean that there is no choice, which numbers can be placed due to this special rule. With higher orders you have several options.

When the bottom half of the rows is filled, the vertically symmetrical cells are filled with the numbers complementary with respect to (n-1) · n.

  • 00
    66
    1212
    121818181212
    624662424
    300300030
  • 030030300
    246242466
    181212121818
    121818181212
    624662424
    300300030

Finally, the two auxiliary squares only have to be put together to create a magic square. Due to the special choice of the numbers in the two auxiliary squares, it is sufficient to simply add the corresponding cells and a magic square is created.

  • 124356
    624351
    153426
    653421
    623451
    154326
  • 030030300
    246242466
    181212121818
    121818181212
    624662424
    300300030
  • 132433356
    3082827117
    191715162024
    182321221413
    12269102925
    315343236

In the PDF document, this difficult construction is also shown for order n=10.