The Reverse Wheel Method (XI)  A Switcheroo
A Discussion of the Method
This method follows the modified 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 modified 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 → 25 → 1 → 2 → 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, i.e,
11 → 17 → 10 → 16 to give Square A2).
 This is followed by the central column from the pairs {9,19} (Square A3) in reverse order.
 Then by the central row from the pairs {13,15} 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 
 25 
 10 

 
1  

 17 
 2 

11  
 
24 

⇒ 
A3
3  
9  
16 
 25 
19  10 

 
1  

 17 
8  2 

11  
18  
24 

 ⇒ 
A4
3  
9  
16 
 25 
19  10 

14  12 
1  15 
13 
 17 
8  2 

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 

Parity Table
ROW or COLUMN  SUM  Δ 55  Sum 1 pair  Δ 80  Sum 1 pair  PARITY (odd or even) 
1  28  27  27  52    O 
2  54  1    26  26  E 
4  27  28  28  53    E 
5  53  2    27  27  O 
 Do a summation of each column, row and diagonal on B4. The magic sum S appears to be 55 or 80 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 1,4) and one pink (rows 2,5).
 These light blue and pink numbers generate the sum of one pair
in columns 4 and 6. The last column shows the parity of these pairs.
 To fill up the magic square we notice that Square A4 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 A5.
 The reason we follow this procedure 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 and is shown below in Square A6'.
 As we begin to fill up the square A4 to get A5 we see that two definite sums have been generate, i.e., 55 and 80.
The little square below shows how the numbers are chosen.
For example 7 is added to the first row at a yellow cell and its complement 20 is added at a white cell nonassociatively but symmetrically opposite from the 7 in
square A5.
 Fill in the empty cells with pairs of numbers from the 5x5 complementary table to give A5.
 To convert all sums to 55 subtract n^{2} = 25 from both 4 and 5 (Square A6) and
S = ½(n^{3}  3n).
 To convert all sums to 80 add n^{2} = 25 to 5, 1 and 7 (Square A7) and
S = ½(n^{3} + 7n).
A4
 55  
3  
9  
16  28  27 
 25 
19  10 
 54  26 
14  12 
1  15 
13  55  0 
 17 
8  2 
 27  28 
11  
18  
24  53  27 
28  54  55 
27  53  55  
27  26  0  28 
27   

⇒ 
A5
 55 
3  20 
9  7 
16  55 
22  25 
19  10 
4  80 
14  12 
1  15 
13  55 
5  17 
8  2 
23  55 
11  6 
18  21 
24  80 
55  80  55 
55  80  55 

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

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

A Different 5x5 square by the Nonsimple Route
 The next step shows how placing the initial numbers on two other yellow cells results in a different square A6' where four different sums
(52,53,55,80,82) are obtained. This shows that the intersection of two partial summations, one light blue
and the other pink, makes this approach more difficult.
 Fill in the spoke portions of the square with the four pairs that are left over (Square A5'). At this point we have five sums 53, 55, 80 and 82.
Two methods may be used to arrive at two different squares.
 Convert all 82 sums to 80 by subtracting 2 from 25 and adding 2 to 2 (Square A6').
 To convert all sums to 55 subtract n^{2} = 25 from both 21 and 23 (Square A7').
 Since 4 is duplicated add n^{2} = 25 to 4, 5, 6 and 7 (Square A8').
 All the sums have been converted to the magic sum 80, and
S = ½(n^{3} + 7n).
A4
 55  
3  
9  
16  28  27 
 25 
19  10 
 54  26 
14  12 
1  15 
13  55  0 
 17 
8  2 
 27  28 
11  
18  
24  53  27 
28  54  55 
27  53  55  
27  26  0  28 
27   

⇒ 
A5
 55 
3  7 
9  20 
16  55 
5  25 
19  10 
23  82 
14  12 
1  15 
13  55 
22  17 
8  2 
4  53 
11  21 
18  6 
24  80 
55  82  55 
53  80  55 

⇒ 
A6'
 55 
3  7 
9  20 
16  55 
5  23 
19  10 
23  80 
14  12 
1  15 
13  55 
22  17 
8  4 
4  55 
11  21 
18  6 
24  80 
55  80  55 
55  80  55 

⇒ 
A7'
 55 
3  7 
9  20 
16  55 
5  23 
19  10 
2  55 
14  12 
1  15 
13  55 
22  17 
8  4 
4  55 
11  4 
18  6 
24  55 
55  55  55 
55  55  55 

⇒ 
A8'
 80 
3  32 
9  20 
16  80 
30  23 
19  10 
2  80 
14  12 
26  15 
13  80 
22  17 
8  4 
29  80 
11  4 
18  31 
24  80 
80  80  80 
80  80  80 

 Otherwise we take square A6' and add n^{2} = 25 to 1, 20 and 22 in order to convert all sums to 80
(Square A9').
 In order to remove the duplicate 4 add n^{2} = 25 to 4, 9, 11, 12 and 24 (Square A10').
 All sums are now 105, with no duplicates and S = ½(n^{3} + 17n).
A6'
 55 
3  7 
9  20 
16  55 
5  23 
19  10 
23  80 
14  12 
1  15 
13  55 
22  17 
8  4 
4  55 
11  21 
18  6 
24  80 
55  80  55 
55  80  55 

⇒ 
A9'
 80 
3  7 
9  45 
16  80 
5  23 
19  10 
23  80 
14  12 
26  15 
13  80 
47  17 
8  4 
4  80 
11  21 
18  6 
24  80 
80  80  80 
80  80  80 

⇒ 
A10'
 105 
3  7 
34  45 
16  105 
5  23 
19  10 
48  105 
14  27 
26  15 
13  105 
47  17 
8  29 
4  105 
36  21 
18  6 
24  105 
105  105  105 
105  105  105 

This ends the Reverse Wheel Method Part XI. To continue to 7x7 square Part XII.
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Copyright © 2008 by Eddie N Gutierrez. EMail: Fiboguti89@Yahoo.com