NON-CONSECUTIVE MAGIC SQUARES WHEEL METHOD (Part II)

A Discussion of the non-consecutive Magic Square Wheel Method

Construction of these magic squares requires is an approach which differs from the traditional step methods such as the Loubère and Méziriac. The magic square is constructed as follows using a complimentary table as a guide. The cells in the same color are associated with one another; for example, none are pair-associated consecutively as in the first variation but are separated by at least one different colored pair. Because of the way these pairs are grouped the 5x5 square produced will not be initially magic but must be converted into one by a series of steps.

1. The left diagonal is filled with the group of numbers ½ (n²-n+2) to ½(n²+n) in consecutive order (top left corner to the right lower corner) from the numbers listed in the complementary table described above, for example using n = 5. For a 5x5 square the numbers in the left diagonal correspond to 11 → 12 → 13 → 14 → 15 (Square A1).
2. Add the right diagonal in reverse order from bottom left corner to the right upper corner choosing only from the pair {3,5,21,23} (as in the little square below) to give Square A2.
   
   

   3	→	5
   		↓
   23	←	21

   
   
   
3. This is followed by the central column from the pairs {2,4,22,24} (Square A3) in regular order.
4. Then by the central row from the pairs {1,6,20,25} in reverse order (Square A4). We now have a partial square with not all sums equal to 65.
5. The result of these operations figuratively speaking resembles the hub and spokes of a wheel where the cells in color correspond to the spoke and hub of the wheel, with the non-consecutive pairs in different colors.

********************************************************************************************************************************************************

A1 11                     12                     13                     14                     15

⇒

A2 11             23     12     21             13             5     14     3             15

⇒

A3 11     2     23     12 4 21             13             5 22 14     3     24     15

⇒

A4 65 11     2     23 36     12 4 21     37 25 20 13 6 1 65     5 22 14     41 3     24     15 42 39 37 65 41 39 65



********************************************************************************************************************************************************

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




6. Fill in the spoke portions of the square with the four pairs that are left over (Square A5 and A6). This is done as follows:
   
   Use the small squares (from the complementary table) as an aid in in how to place the adjacent pair of numbers (in white), which may be placed in either a clockwise or anticlockwise manner as shown below:

7		8		7		8
↓	↗	↓		↑	↙	↑
19		18		19		18



********************************************************************************************************************************************************
7. At this point all sums are not equal to 65 so modifications are carried out to generate a magic square. Convert 7 and 19, respectively, to 10 and 16. This converts all sums in the last grey row to 65 (Square A7). Note that at this point there are several routes which may be taken but this one happens to be the simplest.



A5 65 11 7 2 19 23 62     12 4 21     37 25 20 13 6 1 65     5 22 14     41 3 18 24 8 15 68 39 62 65 68 39 65

⇒

A6 65 11 7 2 19 23 62 9 12 4 21 16 62 25 20 13 6 1 65 17 5 22 14 10 68 3 18 24 8 15 68 65 62 65 68 65 65

⇒

A7 65 11 10 2 16 23 62 9 12 4 21 16 62 25 20 13 6 1 65 17 5 22 14 10 68 3 18 24 8 15 68 65 65 65 65 65 65

⇒



*******************************************************************************************************************************************************
8. Convert 10 and 18 in column 2 to 13 and 15, respectively, (Square A8).


9. Convert 16 and 10 in column 5 to 19 and 7, respectively, (Square A9). At this point all sums are 65.


10. Since duplicates are still present (13 and 15) (n² = 25 is added to the numbers {7,9,13,15,16} to give the magic square A10 with the magic sum = 90 and S = ½(n³ + 11n).
.

This completes Part II of the non-consecutive Magic Square Wheel method. Part III contains a 7x7 first variation.
To go back to homepage.