The Reverse Wheel Method (X) - 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 reverse of the normal complementary table. Reversal produces:

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. 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.

1. The magic square is first constructed by filling in the left diagonal with the group of numbers, order (top left corner to the right lower corner) from the end numbers in the 5x5 complementary table above. For a 5x5 square the numbers in the left diagonal correspond to 3 → 2 → 1 → 25 → 24. (Square A1)
2. Add the right diagonal in reverse order from bottom left corner to the right upper corner choosing only from the pair {11,10} (as in the little square below) to give Square A2.

11	→	10
		↓
16	←	17

3. This is followed by the central column from the pairs {9,8} (Square A3) in reverse order.
4. Then by the central row from the pairs {13,12} in reverse order (Square A4).
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.

A1 3 2 1 25 24

⇒

A2 3 16 2 17 1 10 25 11 24

⇒

A3 3 9 16 2 8 17 1 10 19 25 11 18 24

⇒

A4 3 9 16 2 8 17 14 15 1 12 13 10 19 25 11 18 24

⇒

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

6. Fill in the spoke portions of the square with the four pairs that are left over (Square A5). At this point we have two sums 55 and 80. Two methods may be used to arrive at two different squares.
7. In square A6 n² = 25 is subtracted from both 4 and 6. All the sums have been converted to the magic sum 55, and S = ½(n³ - 3n)
8. In square A7 n² = 25 is added to 1, 5 and 7. All the sums have been converted to the magic sum 80, and S = ½(n³ + 3n)

One example of a 7x7 magic square

Since each conformation of a 7x7 wheel magic square can produce 7 wheel comformations, we'll take the first subset {25,24,23,22,21,20,19,18,17} 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 4 → 5 → 2 → 1 → 49 → 48 → 47. (Square B1)


2. Add the right diagonal in reverse order from bottom left corner to the right upper corner choosing the pairs {22,21,20} to give Square B2.


3. This is followed by addition of the central column pairs {19,18,17) to give Square B3.


4. Then by addition of the central column pairs {25,24,23) in reverse order to give Square B4.



B1 4 3 2 1 49 48 47

⇒

B2 4 29 3 30 2 31 1 20 49 21 48 22 47

⇒

B3 4 19 29 3 18 30 2 17 31 1 20 34 49 21 33 48 22 32 47

⇒

B4 4 19 29 3 18 30 2 17 31 26 27 28 1 23 24 25 20 34 49 21 33 48 22 32 47

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

Parity Table

ROW or COLUMN	SUM	Δ 154	PAIR OF NUMBERS	Δ 203	PAIR OF NUMBERS	PARITY (odd or even)
1	52	102	51+51	52	-	O + O
2	51	103	51+52	51	-	O + E
3	50	104	52+52	50	-	E + E
5	103	51	-	100	50+50	E + E
6	102	52	-	101	50+51	E + O
7	101	53	-	102	51+51	O + O

5. Do a summation of each column, row and diagonal on B4. The magic sum S appears to be 154 but this may change.


6. Set up a parity table as above and we see that the numbers are classified under two groups (if one of the S= 203 which we surmise in step 10), 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 columns 4 and 6. The last column shows the parity of these pairs.


8. To fill up the magic square we notice that Square B6 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. As we begin to fill up the square B4 to get B5 we see that two sums have been generate, i.e., 154 and 203. The last penultimate column and rows give the running total, while the last columns/rows the amounts needed to give 154 or 203. The little square below shows how the numbers are chosen. For example 16 is added to the first row at a yellow cell and its complement 35 is added at a white cell non-associatively but symmetrically opposite from the 16 in square B5.



16		15
↓	↗	↓
35		36

11. Fill in the empty cells with pairs of numbers from the 7x7 complementary table to give B6.



B4 154 4 19 29 52 102 3 18 30 51 103 2 17 31 50 104 26 27 28 1 23 24 25 154 0 20 34 49 103 100 21 33 48 102 101 22 32 47 101 102 52 51 50 154 103 102 101 154 102 103 104 0 100 101 102

⇒

B5 154 4 16 14 19 37 35 29 154 3 18 30 51 2 17 31 50 26 27 28 1 23 24 25 154 20 34 49 103 21 33 48 102 22 36 38 32 13 15 47 203 52 103 102 154 153 152 101 154

⇒

B6 154 4 16 14 19 37 35 29 154 12 3 18 30 40 103 10 2 17 31 42 102 26 27 28 1 23 24 25 154 41 20 34 49 9 153 39 21 33 48 11 152 22 36 38 32 13 15 47 203 154 103 102 154 153 152 203 154

⇒

12. Fill in the rest of the cells (Square B7).
13. To convert square B7 to B8 n² = 49 is subtracted from 7, 9 and 15. All the sums have been converted to the magic sum 154, and S = ½(n³ - 5n).
14. To convert square B7 to B9 n² = 49 is subtracted from 1, 8, 10 and 16. All the sums have been converted to the magic sum 203, and S = ½(n³ + 9n).
15. All modified numbers have been marked in green.

This ends the Reverse Wheel Method Part X. To continue to the next page Part XI. Go back to homepage.