New Method for Odd Magic Squares
Interchangeable Odd Magic Squares
On page 226 of the book entitled "The Neuroscience of You" Copyright © 2022 by Cantel Prat PhD provides a picture puzzle to the reader in which the goal is to select the best answer from a set of four options. The completed picture consists of a matrix of nine 1-3 vertical or horizontal markings which has the appearance of a 3×3 non-magic square; non-magic because the square consists of the numbers 1-3 repeated three times. For example, the simplest known 3×3 non-magic square A with sequential numbers 1,2,3 can be constructed using the StairCase (or Loubère) method on a 3×3 grid:
We can numerically depict the stick version. i.e., the square matrix of vertical and horizontal markings by mixing (sort of like hybridizing) A with B, a square having identical numbers in each cell of the same right diagonals. The original Cantel Prat design was modified (including placement of markings) so that we could focus just on one type of marking, the "sticks".
Stick Version
This stick version may be represented in tuple format as in square C by combining numbers from the same cells of square A and square B into a tuple (v,h) where v and h are the number of vertical and horizontal stick markings, respectively.
C
(2,3) | (1,1) | (3,2) |
(3,1) | (2,2) | (1,3) |
(1,2) | (3,3) | (2,1) |
|
➙ |
D1
23 | 11 | 32 |
31 | 22 | 13 |
12 | 33 | 21 |
|
➙ |
D2
32 | 11 | 23 |
13 | 22 | 31 |
21 | 33 | 12 |
|
Square C has the additional property that when the comma and parentheses from each of the tuples are removed, a magic square D1 is produced where the magic sums of all the vertical, horizontal and left diagonal cells equal 66. In addition, if the numbers in D1 are reversed the square D2 is formed where columns 1 and 3 have been interchanged so that D1 = D2. However, this is not the property of just the 3×3 squares but is a property that can be applied to all the odd squares using not only the Staircase but also the Knight (as in chess) method as well.
Larger Odd Squares
Both the Staircase and Knight methods can be used to produce 5×5 squares where the knight move is chosen to be one cell to the right (r) or left (l)
and two cells to the top (t) or down (d). The squares are constructed with a square similar to B consisting of numbers ranging from 1-5 with each number in the appropriate diagonal. To form E1, for example, we start with the 5×5 analog of B and start at the leftmost corner bottom cell (in yellow) and fill in the diagonal with tuples starting at (1,3) and terminating at (5,3) (Square E1).
Beginning at (1,3) four (1r,2d) knight moves are used to enter all tuples starting with 1. After the last 1 is entered, the next move is blocked by the initial (1,3), and thus a break move back to the diagonal at position (2,3) is required in order to fill in all tuples starting with 2 (Square E2). Note that Square E2 is at a point where only a partial number of 2's have been added. E below depicts the completed square.
B5
4 | 5 | 1 |
2 | 3 |
5 | 1 | 2 |
3 | 4 |
1 | 2 | 3 |
4 | 5 |
2 | 3 | 4 |
5 | 1 |
3 | 4 | 5 |
1 | 2 |
|
➙ |
E1 (r,d)
4 | 5 | 1 |
2 | (5,3) |
5 | 1 | 2 |
(4,3) | 4 |
1 | 2 | (3,3) |
4 | 5 |
2 | (2,3) | 4 |
5 | 1 |
(1,3) | 4 | 5 |
1 | 2 |
|
➙ |
E2 (r,d)
4 | 5 | (2,1) |
(1,2) | (5,3) |
5 | (1,1) | 2 |
(4,3) | 4 |
1 | 2 | (3,3) |
(2,4) | (1,5) |
2 | (2,3) | (1,4) |
5 | 1 |
(1,3) | 4 | 5 |
1 | 2 |
|
➙ |
... |
Similarly the Staircase method can be used on B5, starting at the top central cell labeled 1 (in yellow) and the numbers beginning with 1 are added sequentially on the same diagonal. If the next position on the diagonal is blocked and a break is required, viz., a move down one cell to the next empty cell where the number count is restarted at 1. Note that diagonal containing 2's is only partially filled. For the completed square, see F below.
B5
4 | 5 | 1 |
2 | 3 |
5 | 1 | 2 |
3 | 4 |
1 | 2 | 3 |
4 | 5 |
2 | 3 | 4 |
5 | 1 |
3 | 4 | 5 |
1 | 2 |
|
➙ |
F Staircase
4 | 5 | (1,1) |
2 | 3 |
5 | (5,1) | (2,2) |
3 | 4 |
(4,1) | (1,2) | 3 |
4 | 5 |
2 | 3 | 4 |
5 | (3,1) |
3 | 4 | 5 |
(2,1) | 2 |
|
➙ |
... |
The four squares that would be formed consists of two sets of identical squares one of which (square F) is equivalent to the Staircase method. As can be seen only one configuration of sticks is present for each cell.
E (r,d) or (l,u)
(4,4) | (3,5) | (2,1) |
(1,2) | (5,3) |
(2,5) | (1,1) | (5,2) |
(4,3) | (3,4) |
(5,1) | (4,2) | (3,3) |
(2,4) | (1,5) |
(3,2) | (2,3) | (1,4) |
(5,5) | (4,1) |
(1,3) | (5,4) | (4,5) |
(3,1) | (2,2) |
|
|
F ((r,u) or (l,d) or Staircase
(2,4) | (4,5) | (1,1) |
(3,2) | (5,3) |
(3,5) | (5,1) | (2,2) |
(4,3) | (1,4) |
(4,1) | (1,2) | (3,3) |
(5,4) | (2,5) |
(5,2) | (2,3) | (4,4) |
(1,5) | (3,1) |
(1,3) | (3,4) | (5,5) |
(2,1) | (4,2) |
|
If we remove the commas and parentheses from the tuples the magic squares EM1 and FM1 with a magic sum of 165 are obtained.
EM1
44 | 35 | 21 |
12 | 53 |
25 | 11 | 52 |
43 | 34 |
51 | 42 | 33 |
24 | 15 |
32 | 23 | 14 |
55 | 41 |
13 | 54 | 45 |
31 | 22 |
|
|
FM1
24 | 45 | 11 |
32 | 53 |
35 | 51 | 22 |
43 | 14 |
41 | 12 | 33 |
54 | 25 |
52 | 23 | 44 |
15 | 31 |
13 | 34 | 55 |
21 | 42 |
|
Reversing all the numbers in EM1 and FM1 produces squares EM2 and FM2 in which both squares are still magic having the magic sum of 165. This would have the effect of reversing the tuples in E and F and generating a different configuration of sticks, where an h becomes v and a v becomes h.
EM2
44 | 53 | 12 |
21 | 35 |
52 | 11 | 25 |
34 | 43 |
15 | 24 | 33 |
42 | 51 |
23 | 32 | 41 |
55 | 14 |
31 | 45 | 54 |
13 | 22 |
|
|
FM2
42 | 54 | 11 |
23 | 35 |
53 | 15 | 22 |
34 | 41 |
14 | 21 | 33 |
45 | 52 |
25 | 32 | 44 |
51 | 13 |
31 | 43 | 55 |
12 | 24 |
|
Note that EM1 and EM2 can also be classified as Staircase methods if we know beforehand where to place the initial number 11.
In fact, the Staircase method uses variable Knight break moves depending on the position of the initial number 11, covered in Part I of General Methods and Rules of Staircase Squares, where the initial number is 1. For both EM2 and EM2 the break is calculated as (3,3), i.e. 3 cells right and 3 cells down.
Applying the above methods and employing the same Knight moves to the 7×7 squares we can skip the tuple intermediate and go directly to the magic squares GM and HM, having magic sums of 308. Note that given the squares below we can reform the tuples by adding commas between the centers of each number. Note that if interest is solely on the magic squares the tuple forming step may be omitted since its used mainly for the generation of the stick markings.
GM1 (r,d) or (l,u)
35 | 66 | 27 |
51 | 12 | 43 | 74 |
56 | 17 | 41 |
72 | 33 | 64 | 25 |
77 | 31 | 62 |
23 | 54 | 15 | 46 |
21 | 52 | 13 | 44 |
75 | 36 | 67 |
42 | 73 | 34 |
65 | 26 | 57 | 11 |
63 | 24 | 55 |
16 | 47 | 71 | 32 |
14 | 45 | 76 |
37 | 61 | 22 | 53 |
|
|
HM1 ((r,u) or (l,d) or Staircase
25 | 46 | 67 |
11 | 32 | 53 | 74 |
36 | 57 | 71 |
22 | 43 | 64 | 15 |
47 | 61 | 12 |
33 | 54 | 75 | 26 |
51 | 72 | 23 | 44 |
65 | 16 | 37 |
62 | 13 | 34 |
55 | 76 | 27 | 41 |
73 | 24 | 45 |
66 | 17 | 31 | 52 |
14 | 35 | 56 | 77 |
21 | 42 | 63 |
|
And finally reversing all the numbers in GM1 and HM1 the result is GM2 and HM2 in which both squares are still magic having the magic sum of 308.
GM2 (r,d) or (l,u)
53 | 66 | 72 |
15 | 21 | 34 | 47 |
65 | 71 | 14 |
27 | 33 | 46 | 52 |
77 | 13 | 26 |
32 | 45 | 51 | 64 |
12 | 25 | 31 | 44 |
57 | 63 | 76 |
24 | 37 | 43 |
56 | 62 | 75 | 11 |
36 | 42 | 55 |
61 | 74 | 17 | 23 |
41 | 54 | 67 |
73 | 16 | 22 | 35 |
|
|
HM2 ((r,u) or (l,d) or Staircase
52 | 64 | 76 |
11 | 23 | 35 | 47 |
63 | 75 | 17 |
22 | 34 | 46 | 51 |
74 | 16 | 21 |
33 | 45 | 57 | 62 |
15 | 27 | 32 | 44 |
56 | 61 | 73 |
26 | 31 | 43 |
55 | 67 | 72 | 14 |
37 | 42 | 54 |
66 | 71 | 13 | 25 |
41 | 53 | 65 | 77 |
12 | 24 | 36 |
|
Note that again as in Part I, GM1 and GM2 can also be classified as a Staircase method with variable Knight break moves of (6,2); 6 cells right followed by 2 cells down.
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Copyright © 2023 by Eddie N Gutierrez. E-Mail: enaguti1949@gmail.com