New Squares from the Modified De La Loubère Method
Part IA
Discussion of New De La Loubère type squares
The previous page showed how to prepare Modified De La Loubère squares of order n where the main diagonal is constructed from a complementary table of
(n+2)x(n+2) or greater. This page will show how we can take these modified magic squares and build them up into larger squares that take
into account all of the (n+2)x(n+2) or greater numbers of the complementary table used. Two methods will be shown one using a
L-leap approach, the other a wheel approach.
The L-leap approach
To expand a modified De La Loubère square using the L-leap approach we take an nxn
and increase the number of rows and columns by 2 or greater. After the smaller square is incorporated into a larger the method involves placing the next number in the
omplementary table into either of four positions. These include either right of first top corner, left/right of top right corner or top of left bottom corner.
All four examples are shown below using the 3x3 example:
- Incorporate the smaller square as constructed previously into a larger square as shown below using a 3x3 inserted into a 5x5.
- Fill in the two corner cells to complete the main diagonal.
- Fill in the left cell adjacent to the 15 with the number 4, since 3 was the last number on the complementary list.
- Perform a L-leap, i.e. place one number in a cell leap across, in a z manner,
up/down on the squareto the next column or right/left to the next row and insert the next number.
- Stop at the first cell in the second column inserting a 6, leap across to the right lower corner and insert a 7, then leap to the second cell in the first column and
insert an 8.
- Continue L-leaping right and left across until you reach the first cell in the fourth row.
- Take the complement of 10, i.e. 16, and this time L-leap backwards across the square, filling in all the complementary cells up to
the left corner cell in the first column.
- L-leap down and up filling in all the cells with the right complementary numbers until the square is completed.
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|         |
1
|     |     |     |
    |     |
|     | 24 | 1 |
14 |     |
|     | 3 | 13 |
23 |     |
|     | 12 | 25 |
2 |     |
|     |     |     |
    |     |
|
   ⇒    |
2
|     |     |     |
    | 15 |
|     | 24 | 1 |
14 |     |
|     | 3 | 13 |
23 |     |
|     | 12 | 25 |
2 |     |
| 11 |     |     |
    |     |
|
   ⇒    |
3
|     |     |     |
4 | 15 |
|     | 24 | 1 |
14 |     |
|     | 3 | 13 |
23 |     |
|     | 12 | 25 |
2 |     |
| 11 |     |     |
    |     |
|
   ⇒    |
4
|     | 6 |     |
4 | 15 |
|     | 24 | 1 |
14 |     |
|     | 3 | 13 |
23 |     |
|     | 12 | 25 |
2 |     |
| 11 |     | 5 |
    |     |
|
   ⇒    |
5,6
|     | 6 |     |
4 | 15 |
| 8 | 24 | 1 |
14 |     |
|     | 3 | 13 |
23 | 9 |
| 10 | 12 | 25 |
2 |     |
| 11 |     | 5 |
    | 7 |
|
   ⇒    |
7
| 19 | 6 |     |
4 | 15 |
| 8 | 24 | 1 |
14 | 18 |
| 17 | 3 | 13 |
23 | 9 |
| 10 | 12 | 25 |
2 | 16 |
| 11 |     | 5 |
    | 7 |
|
   ⇒    |
8
| 19 | 6 | 21 |
4 | 15 |
| 8 | 24 | 1 |
14 | 18 |
| 17 | 3 | 13 |
23 | 9 |
| 10 | 12 | 25 |
2 | 16 |
| 11 | 20 | 5 |
22 | 7 |
|
| 1 | 2 |
3 | 4 |
5 | 6 |
7 | 8 |
9 | 10 |
11 | 12 |
| 13 |
| 25 | 24 |
23 | 22 |
21 | 20 |
19 | 18 |
17 | 16 |
15 | 14 |
********************************************************************************************************************************************************
The above gives one way of filling up a square by the L-leap approach. A second method fills it in an opposite manner gives a different
variant:
- Use square 2 from above.
- Fill in the first down cell adjacent to the 15 in this example with the number 4.
- Perform a L-leap, i.e. place one number in a cell leaping across the square left/right
to the next row and insert the next number.
- Stop at the fourth cell in the fifth column inserting a 6, leap across to the left upper corner and insert a 7, then leap to the fifth cell in the fourth column
and insert an 8.
- Continue L-leaping up and down until you reach the fifth cell in the second column.
- Take the complement of 10, i.e. 16, and this time L-leap backwards(up/down) across the square, filling in all the complementary
cells up to the bottom right corner cell in the fifth column.
- L-leap left and right filling in all the cells with the right complementary numbers until the square is completed.
|         |
1
|     |     |     |
    | 15 |
|     | 24 | 1 |
14 |     |
|     | 3 | 13 |
23 |     |
|     | 12 | 25 |
2 |     |
| 11 |     |     |
    |     |
|
   ⇒    |
2
|     |     |     |
    | 15 |
|     | 24 | 1 |
14 | 4 |
|     | 3 | 13 |
23 |     |
|     | 12 | 25 |
2 |     |
| 11 |     |     |
    |     |
|
   ⇒    |
3,4
|     |     |     |
    | 15 |
|     | 24 | 1 |
14 | 4 |
| 5 | 3 | 13 |
23 |     |
|     | 12 | 25 |
2 | 6 |
| 11 |     |     |
    |     |
|
   ⇒    |
4,5
| 7 |     | 9 |
    | 15 |
|     | 24 | 1 |
14 | 4 |
| 5 | 3 | 13 |
23 |     |
|     | 12 | 25 |
2 | 6 |
| 11 | 10 |     |
8 |     |
|
   ⇒    |
6,7
| 7 | 16 | 9 |
18 | 15 |
| 22 | 24 | 1 |
14 | 4 |
| 5 | 3 | 13 |
23 | 21 |
| 20 | 12 | 25 |
2 | 6 |
| 11 | 10 | 17 |
8 | 19 |
|
********************************************************************************************************************************************************
The last two examples in the series (A and B) are shown below after repeating the algorithm with the difference being that the leaping across the square as in line 6
above is done towards the end of the construction:
|         |
A
| 19 | 4 | 21 |
6 | 15 |
| 10 | 24 | 1 |
14 | 16 |
| 17 | 3 | 13 |
23 | 9 |
| 8 | 12 | 25 |
2 | 18 |
| 11 | 22 | 5 |
20 | 7 |
|
                                  |
B
| 7 | 18 | 9 |
16 | 15 |
| 20 | 24 | 1 |
14 | 6 |
| 5 | 3 | 13 |
23 | 21 |
| 22 | 12 | 25 |
2 | 4 |
| 11 | 8 | 17 |
10 | 19 |
|
********************************************************************************************************************************************************
The wheel approach
To expand a modified De La Loubère square using the wheel approach we take an nxn and increase the number of
rows and columns by 2 or greater. Using the 3x3 example we:
- Incorporate the smaller square as constructed in new Loubère methods into a larger square as shown below using
a 3x3 inserted into a 5x5.
- Fill in the two corner cells to complete the main diagonal.
- Fill in the spoke cells as shown in method A variant 1, however, reverse the addition due to
rotation of the square by 90° degrees.
- Fill in the non-spoke cells using parity as shown previously, inserting the L numbers from the complementary table.
- Comparing this square to the 4 rotated version shows that the non-spoke numbers are opposite
to the normal mode of addition as was shown in variant 1.
|         |
1
|     |     |     |
    |     |
|     | 24 | 1 |
14 |     |
|     | 3 | 13 |
23 |     |
|     | 12 | 25 |
2 |     |
|     |     |     |
    |     |
|
   ⇒    |
2
|     |     |     |
    | 15 |
|     | 24 | 1 |
14 |     |
|     | 3 | 13 |
23 |     |
|     | 12 | 25 |
2 |     |
| 11 |     |     |
    |     |
|
   ⇒    |
3
| 21 |     | 4 |
    | 15 |
|     | 24 | 1 |
14 |     |
| 6 | 3 | 13 |
23 | 20 |
|     | 12 | 25 |
2 |     |
| 11 |     | 22 |
    | 5 |
|
   ⇒    |
4
| 21 | 7 | 4 |
18 | 15 |
| 10 | 24 | 1 |
14 | 16 |
| 6 | 3 | 13 |
23 | 20 |
| 17 | 12 | 25 |
2 | 9 |
| 11 | 19 | 22 |
8 | 5 |
|
   ≡    |
4 rotated
| 11 | 17 | 6 |
10 | 21 |
| 19 | 12 | 3 |
24 | 7 |
| 22 | 25 | 13 |
1 | 4 |
| 8 | 2 | 23 |
14 | 18 |
| 5 | 9 | 20 |
16 | 15 |
|
| 1 | 2 |
3 | 4 |
5 | 6 |
7 | 8 |
9 | 10 |
11 | 12 |
| 13 |
| 25 | 24 |
23 | 22 |
21 | 20 |
19 | 18 |
17 | 16 |
15 | 14 |
********************************************************************************************************************************************************
To expand a modified De La Loubère square using a second wheel approach we take an nxn and increase the number of
rows and columns by 2 or greater. Using the 3x3 example we:
- Use square 2 above, but reverse the 5 and 21 corner positions.
- Fill in the rest of the spoke numbers in a reverse manner than was done above for example 3 square 3.
- Fill in the non-spoke cells using parity as shown previously, inserting the L numbers from the complementary table.
|         |
1
| 5 |     |     |
    | 15 |
|     | 24 | 1 |
14 |     |
|     | 3 | 13 |
23 |     |
|     | 12 | 25 |
2 |     |
| 11 |     |     |
    | 21 |
|
   ⇒    |
2
| 5 |     | 20 |
    | 15 |
|     | 24 | 1 |
14 |     |
| 22 | 3 | 13 |
23 | 4 |
|     | 12 | 25 |
2 |     |
| 11 |     | 6 |
    | 21 |
|
   ⇒    |
3
| 5 | 7 | 20 |
18 | 15 |
| 10 | 24 | 1 |
14 | 16 |
| 22 | 3 | 13 |
23 | 4 |
| 17 | 12 | 25 |
2 | 9 |
| 11 | 19 | 6 |
8 | 21 |
|
********************************************************************************************************************************************************
A property of these squares is that another magic square may be generated from the original by shifting pairs of numbers in the external boundary. Only pairs are
permitted since since if only one number and its complement are shifted the square is no longer magic. To perform this:
- Insert a 3x3 modified Loubère square into a 5x5 square (in this case the starting number is 3).
- Fill in the two corner cells to complete the main diagonal.
- L-leap in the boundary numbers.
- Remove the first two pairs (1,2) and and move all the numbers across to fill in the empty position.
- Fill in the last two positions with the (1,2) pairs.
|         |
1,2
|     |     |     |
    | 15 |
|     | 22 | 3 |
14 |     |
|     | 5 | 13 |
21 |     |
|     | 12 | 23 |
4 |     |
| 11 |     |     |
    |     |
|
   ⇒    |
3
| 19 | 6 | 24 |
1 | 15 |
| 8 | 22 | 3 |
14 | 18 |
| 17 | 5 | 13 |
21 | 9 |
| 10 | 12 | 23 |
4 | 16 |
| 11 | 20 | 2 |
25 | 7 |
|
   ⇒    |
4,5
| 17 | 8 | 19 |
6 | 15 |
| 10 | 22 | 3 |
14 | 16 |
| 25 | 5 | 13 |
21 | 1 |
| 2 | 12 | 23 |
4 | 24 |
| 11 | 18 | 7 |
20 | 9 |
|
To continue this method using reverse main diagonal
(Part IB) 3x3 squares. To go back to previous wheel and De la Loubère methods.
To return to homepage.
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APPENDIX (this section can be skipped)
The Number of Group of Squares Generated for both Loubère's Method
The Number using the >L-leap Boundary Method
The number of magic squares in each group may be determined by the use of two equations.
The equation for the number of total groups, magic and non-magic De La Loubère using the L-leap approach uses the modified equation
½(n2 + 1) - ½(ns2- 1) - 1 which after simplification
becomes ½(n2 - ns2). A group consists of the internal Loubère square along
with the external L-leap boundary, i.e., making up the entire larger square. Although the internal square remains the same, the boundary
may differ taking on a variety of values, some of which give rise to magic squares or to non-magic squares as shown in the last example.
To determine whether the group is magic or non-magic we use the following scheme:
- Determine the number of group of squares using ½(n2 - ns2).
- Determine the halfway mark of the group of squares by dividing the equation by 2.
These two values will be used in the construction of the group of squares set, where each group in the set may have a value of 1 if the group is magic and
or 0 if the group is non-magic.
The halfway point ¼(n2 - ns2) and the next
group after this
¼(n2 - ns2) + 1 are the inversion points and both take on the value of 0.
Numbers to the left of the halfway point vary from 1, 3 ...¼(n2 - ns2) - 1,
i.e., odd groups are magic. Those to the right of the halfway + 1 point vary from
¼(n2 - ns2) + 2 ...
½(n2 - ns2), i.e., even groups are magic. For example, for a 7x7
Loubère modified square expanded into a 9x9 square the set is as follows:
| 1  | 2  | 3  |
4  | 5  | 6  |
7  | 8  | 9  |
10 | 11 | 12 |
13 | 14 |
15 | 16 |
| 1 | 0 | 1 |
0 | 1 | 10 | 1 |
0 | 0 |
1 | 0 | 1 |
0 | 1 | 0 | 1 |
|
All of the expanded squares , except for 5x5 expanded into a 7x7, follow the above rules. Because the 5x5 contains the pair 20/30 on the external boundary an extra two groups
while not magic per se become so on the exchange of these two numbers so that the modified groups are:
| 1  | 2  | 3  |
4  | 5  | 6  |
7  | 8  | 9  |
10 | 11 | 12 |
| 1 | 0 | 1 |
0 | 1 |
0 | 1 |
1 | 1 | 1 |
0 | 1 |
|
The Number using the Wheel Boundary Method
The total number of groups of magic De La Loubère squares using the
wheel approach to finish off the square (the third example), however, uses the equation
ns(n - 1), to account for the
larger number of possibilities of generating these squares. Again ns and
n correspond to the smaller and the larger square, respectively. Since this last equation gives only the total number of groups of squares, the following tables
show the distribution of magic squares for the entire group, i.e ns = 3, 5, 7, ... only that half of the table is shown
(to conserve space and show how the numbers are related to the rest of the entries). To obtain the rest of the entries the rest of the table is expanded using symmetry.
See example following the tables.
Number of Loubère Square Groups with the wheel
outer structure
| magic square | 1 | 2 |
3 | 4 | 5 | 6 |
7 | 8 | 9 | 10 |
| ns      n | n0 = 3 |
| 3        5 | n0 |
n0-3 |
n0-1 | n0-2 | - |
- | - | - | - | - |
| 5        7 | n0+2 |
n0-3 |
n0+1 | n0-2 |
n0 |
n0-1 | - | - | - | - |
| 7        9 | n0+4 |
n0-3 |
n0+3 | n0-2 |
n0+2 |
n0-1 | n0+1 | n0 |
- | - |
| 9       11 | ns+6 |
n0-3 |
n0+5 | n0-2 |
n0+4 |
n0-1 | n0+3 | n0 |
n0+2 | n0+1 |
********************************************************************************************************************************************************
|
Number of Loubère Square Groups with the wheel outer structure
(calculated from the above table)
| magic square | 1 | 2 |
3 | 4 | 5 | 6 |
7 | 8 | 9 | 10 |
| ns      n |     |
| 3        5 |   3   |   0   |
  2   |   1   |
  -   |
  -   |   -   |   -   |
  -   |   -   |
| 5        7 | 5 | 0 |
4 | 1 | 3 |
2 | - | - | - | - |
| 7        9 | 7 | 0 |
6 | 1 | 5 |
2 | 4 | 3 |
- | - |
| 9       11 | 9 | 0 |
8 | 1 | 7 |
2 | 6 | 3 |
5 | 4 |
For example the entry for line 1 is 3 0 2 1 1 2 0 3 after taking account of symmetry, which includes the numbers of the above table plus the reverse of these numbers.
The sum of these numbers is 12 identical to the answer obtained from the equation
ns(n - 1). This total is doubled since the two corner positions on the external boundary may be inverted as shown in example 4.
Copyright © 2008 by Eddie N Gutierrez. E-Mail: Fiboguti89@Yahoo.com
|