The Unbalanced Reverse Wheel Method (XIV) - 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 II 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 {1,2,3,4,5,6,7,8,9,10} 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 43 → 44 → 45 → 46 → 47 → 48 → 49. (Square B1)
2. Add the right diagonal in reverse order from bottom left corner to the right upper corner choosing the pairs {4,5,6} to give Square B2.
3. This is followed by addition of the central column pairs {7,8,9) to give Square B3.
4. Then by addition of the central column pairs {1,2,3) in reverse order to give Square B4.

B1 43 44 45 46 47 48 49

⇒

B2 43 39 44 38 45 37 46 6 47 5 48 4 49

⇒

B3 43 7 39 44 8 38 45 9 37 46 6 34 47 5 35 48 4 36 49

⇒

B4 43 7 39 44 8 38 45 9 37 42 41 40 46 3 2 1 6 34 47 5 35 48 4 36 49

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

Parity Table

ROW or COLUMN	SUM	Δ 175	PAIR OF NUMBERS	PARITY (odd or even)
1	89	86	43+43	O + O
2	90	85	42+43	E + O
3	91	84	42+42	E + E
5	87	88	44+44	E + E
6	88	87	43+44	O + E
7	89	86	43+43	O + O

5. Do a summation of each column, row and diagonal on B4. The magic sum S appears to be 175 or 322 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 43 7 39 89 86 44 8 38 90 85 45 9 37 91 84 42 41 40 46 3 2 1 175 0 6 34 47 87 88 5 35 48 88 87 4 36 49 89 86 89 90 91 175 87 88 89 322 86 85 84 0 88 87 86

⇒

B5 175 43 10 12 7 31 33 39 175 44 8 38 90 45 9 37 91 42 41 40 46 3 2 1 175 6 34 47 87 5 35 48 88 4 32 30 36 13 11 49 175 89 132 133 175 131 132 89 322

⇒

B6 175 43 10 12 7 31 33 39 175 14 44 8 38 28 132 16 45 9 37 26 133 42 41 40 46 3 2 1 175 27 6 34 47 17 131 29 5 35 48 15 132 4 32 30 36 13 11 49 175 175 132 133 175 131 132 175 322

⇒

11. Fill in the rest of the cells (Square B7).
12. To convert square B7 to B8 3n² = 147 is added to the diagonal 7, 25, 22, 1, 27, 5 and 30. All the sums have been converted to the magic sum 322, and S = ½(n³ + 43n).
13. To convert square B7 to B9 n² = 49 is subtracted from the diagonal 34. 37, 40, 43, 44, 48 and 49. All the sums have been converted to the magic sum 126, and S = ½(n³ - 13n).
14. All modified numbers have been marked in green.

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