The Unbalanced Reverse Wheel Method (XIII) - A Switcheroo

A Discussion of the Method

This method follows the normal wheel algorithm except that the complementary table from which the numbers are chosen is the what I call the unbalanced reverse of the normal complementary table. Using a 5x5 complementary table I will show two types of reversals. In the first, Reversal I, 1, 2, 3, 4 and 5 are displaced from their original positions and are used to generate the leftmost diagonal:

While in the second, Reversal II, 21, 22, 23, 24 and 25 are displaced from their position:

In addition the new squares formed are not magic but must be modified to convert them into magic squares. The square is filled as in the normal wheel fashion and the wheel spoke numbers are picked from a complementary table, e.g., the 5x5 above in the reverse fashion. However, for the sake of brevity we will display only the 7x7 squares.

Moreover, it must be stated here that the magic sum has been modified from the known equation S = ½(n³ + n) to the general equation as was shown in:

which takes into account these new squares. The variable a, an odd number, is equal to 1,3,5,7 or ... and may take on + or - values. For example when a = 1 the normal magic sum S is implied. When a takes on different odd values S gives the magic sum of a modified magic square. It will be shown that the addition or subtraction of n² to some of the cells in the square gives rise to a new magic square.

Type I Reversal - A 7x7 Magic Square

Since each conformation of a 7x7 wheel magic square can produce 7 wheel comformations, we'll take the first subset {8,9,10,11,12,13,14,15,16} and their complements as an example.

1. The magic square is first constructed by filling in the left diagonal with a group of numbers from the 7x7 complementary table below. For a 7x7 square the numbers in the left diagonal correspond to 1 → 2 → 3 → 4 → 5 → 6 → 7. (Square B1)
2. Add the right diagonal in reverse order from bottom left corner to the right upper corner choosing the pairs {11,12,13} to give Square B2.
3. This is followed by addition of the central column pairs {14,15,16) to give Square B3.
4. Then by addition of the central column pairs {8,9,10) in reverse order to give Square B4.

B1 1 2 3 4 5 6 7

⇒

B2 1 46 2 45 3 44 4 13 5 12 6 11 7

⇒

B3 1 14 46 2 15 45 3 16 44 4 13 41 5 12 42 6 11 43 7

⇒

B4 1 14 46 2 15 45 3 16 44 49 48 47 4 10 9 8 13 41 5 12 42 6 11 43 7

8	9	10	11	12	13	14	15	16	17	18	19	20	21	22	23	24	25	26	27	28	1	2	3
																								4
49	48	47	46	45	44	43	42	41	40	39	38	37	36	35	34	33	32	31	30	29	7	6	5

Parity Table

ROW or COLUMN	SUM	Δ 175	PAIR OF NUMBERS	PARITY (odd or even)
1	61	114	57+57	O + O
2	62	113	56+57	E + O
3	63	112	56+56	E + E
5	59	116	58+58	E + E
6	60	115	57+58	O + E
7	61	114	57+57	O + O

5. Do a summation of each column, row and diagonal on B4. The magic sum S appears to be 175 but this may change.
6. Set up a parity table as above and we see that the numbers are classified under two groups, one in light blue (rows 1,2,3) and one in pink (rows 5,6,7).
7. These light blue and pink numbers generate the pairs in column 4. The last column shows the parity of these pairs.
8. To fill up the magic square we notice that Square B4 below may be filled in a simple manner. Where the last entries (in light blue) of the columns coincide with the last entries of the rows also in light blue the cell is colored yellow. It is into these cells that the first number from the complementary pairs is placed. See Square B5.
9. The reason this is done is that as the squares get larger it gets more and more difficult to assign numbers to cells. This method removes that ambiguity and produces consistent results, since we now force the assignment. It is still possible to assign numbers to cells without using this method and arrive at different squares. However, this is possible only with the smaller squares.
10. Fill in the empty cells with pairs of numbers from the 7x7 complementary table to give B6.

B4 175 1 14 46 61 114 2 15 45 62 113 3 16 44 63 112 49 48 47 4 10 9 8 175 0 13 41 5 59 116 12 42 6 60 115 11 43 7 61 114 61 62 63 175 59 60 61 28 114 113 112 0 116 115 114

⇒

B5 175 1 17 19 14 38 40 46 175 2 15 45 118 3 16 44 119 49 48 47 4 10 9 8 175 13 41 5 117 12 42 6 118 11 39 37 43 20 18 7 175 61 118 119 175 117 118 61 28

⇒

B6 175 1 17 19 14 38 40 46 175 21 2 15 45 35 118 23 3 16 44 33 119 49 48 47 4 10 9 8 175 34 13 41 5 24 117 36 12 42 6 22 118 11 39 37 43 20 18 7 175 175 118 119 175 117 118 175 28

⇒

11. Fill in the rest of the cells (Square B7).
12. To convert square B7 to B8 n² = 49 is added to 2, 3, 4, 7, 14, 24 and 36. All the sums have been converted to the magic sum 224, and S = ½(n³ + 15n).
13. To convert square B7 to B9 3n² = 147 is subtracted from the diagonal 14, 3, 29, 8, 34, 12 and 37. All the sums have been converted to the magic sum 28, and S = ½(n³ - 41n).
14. All modified numbers have been marked in green.

This ends the Reverse Wheel Method Part XIII. To continue to Part XIV. Go back to homepage.