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:
6  7 
8  9 
10  11 
12  13 
14  15 
1  2 

 3 
25  24 
23  22 
21  20 
19  18 
17  16 
5  4 

While in the second, Reversal II, 21, 22, 23, 24 and 25 are displaced from their position:
1  2 
3  4 
5  6 
7  8 
9  10 
21  22 

 23 
20  19 
18  17 
16  15 
14  13 
12  11 
25  24 

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(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.
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.
 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)
 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.
 This is followed by addition of the central column pairs {7,8,9) to give Square B3.
 Then by addition of the central column pairs {1,2,3) in reverse order to give Square B4.

⇒ 
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 
 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.
 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).
 These light blue and pink numbers generate the pairs
in column 4. The last column shows the parity of these pairs.
 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.
 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.
 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 

⇒ 
 Fill in the rest of the cells (Square B7).
 To convert square B7 to B8 3n^{2} = 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(n^{3} + 43n).
 To convert square B7 to B9 n^{2} = 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(n^{3}  13n).
 All modified numbers have been marked in green.
B7
 175 
43  10 
12  7 
31  33 
39  175 
14  44 
18  8 
25  38 
28  132 
16  20 
45  9 
37  22 
26  133 
42  41 
40  46 
3  2 
1  175 
27  23 
6  34 
47  21 
17  131 
29  5 
24  35 
19  48 
15  132 
4  32 
30  36 
13  11 
49  175 
175  132  133 
175  131  132 
175  322 

⇒ 
B8
43  10 
12  154 
31  33 
39 
14  44 
18  8 
172  38 
28 
16  20 
45  9 
37  169 
26 
42  41 
40  46 
3  2 
148 
174  23 
6  34 
47  21 
17 
29  152 
24  35 
19  48 
15 
4  32 
177  36 
13  11 
49 

+ 
B9
6  10 
12  7 
31  33 
39 
14  5 
18  8 
25  38 
28 
16  20 
45  9 
12  22 
26 
42  41 
9  46 
3  2 
1 
27  23 
6  2 
47  21 
17 
29  5 
24  35 
19  1 
15 
4  32 
30  36 
13  11 
0 

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 

This ends the Reverse Wheel Method Part XIV.
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Copyright © 2009 by Eddie N Gutierrez. EMail: Fiboguti89@Yahoo.com