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:
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 

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(n^{3} + n) to
the general equation as was shown in:
S = ½(n^{3} ± an)
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 n^{2} to some of the cells in
the square gives rise to a new magic square.
 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)
 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.
 This is followed by the central column from the pairs {9,8} (Square A3) in reverse order.
 Then by the central row from the pairs {13,12} in reverse order (Square A4).

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.

⇒ 
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 

 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.
 In square A6 n^{2} = 25 is subtracted from both 4 and 6.
All the sums have been converted to the magic sum 55, and
S = ½(n^{3}  3n)
 In square A7 n^{2} = 25 is added to 1, 5 and 7.
All the sums have been converted to the magic sum 80, and
S = ½(n^{3} + 3n)
A5
 55 
3  7 
9  20 
16  55 
5  2 
8  17 
23  55 
14  15 
1  12 
13  55 
22  10 
19  25 
4  80 
11  21 
18  6 
24  80 
55  55  55 
80  80  55 

⇒ 
A6
 55 
3  7 
9  20 
16  55 
5  2 
8  17 
23  55 
14  15 
1  12 
13  55 
22  10 
19  25 
21  55 
11  21 
18  19 
24  55 
55  55  55 
55  55  55 

+ 
A7
 80 
3  32 
9  20 
16  80 
30  2 
8  17 
23  80 
14  15 
26  12 
13  80 
22  10 
19  25 
4  80 
11  21 
18  6 
24  80 
80  80  80 
80  80  80 

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.
 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)
 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.
 This is followed by addition of the central column pairs {19,18,17) to give Square B3.
 Then by addition of the central column pairs {25,24,23) in reverse order to give Square B4.

⇒ 
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 
 Do a summation of each column, row and diagonal on B4. The magic sum S appears to be 154 but this may change.
 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).
 These light blue and pink numbers generate the pairs
in columns 4 and 6. The last column shows the parity of these pairs.
 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.
 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.
 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 nonassociatively but symmetrically opposite from the 16 in
square B5.
 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 

⇒ 
 Fill in the rest of the cells (Square B7).
 To convert square B7 to B8 n^{2} = 49 is subtracted from 7, 9 and 15.
All the sums have been converted to the magic sum 154, and
S = ½(n^{3}  5n).
 To convert square B7 to B9 n^{2} = 49 is subtracted from 1, 8, 10 and 16.
All the sums have been converted to the magic sum 203, and
S = ½(n^{3} + 9n).
 All modified numbers have been marked in green.
B7
 154 
4  16 
14  19 
37  35 
29  154 
12  3 
8  18 
43  30 
40  154 
10  6 
2  17 
31  46 
42  154 
26  27 
28  1 
23  24 
25  154 
41  45 
20  34 
49  5 
9  203 
39  21 
44  33 
7  48 
11  203 
22  36 
38  32 
13  15 
47  203 
154  154  154 
154  203  203 
203  154 

⇒ 
B8
4  16 
14  19 
37  35 
29 
12  3 
8  18 
43  30 
40 
10  6 
2  17 
31  46 
42 
26  27 
28  1 
23  24 
25 
41  45 
20  34 
49  5 
40 
39  21 
44  33 
42  48 
11 
22  36 
38  32 
13  34 
47 

+ 
B9
4  65 
14  19 
37  35 
29 
12  3 
57  18 
43  30 
40 
59  6 
2  17 
31  46 
42 
26  27 
28  50 
23  24 
25 
41  45 
20  34 
49  5 
9 
39  21 
44  33 
7  48 
11 
22  36 
38  32 
13  15 
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 

This ends the Reverse Wheel Method Part X. To continue to the next page Part XI.
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Copyright © 2009 by Eddie N Gutierrez. EMail: Fiboguti89@Yahoo.com