Octagon Algorithm for Border Wheel Squares (Part XII)

The Eight Node Way - (1,4,7) Squares

The previous section Part XI introduced the Octagon I and L algorithm for the construction of 5×5 thru 7×7 Wheel Border Squares (WBS). The type of Wheel Border squares investigated were the (a,d,f) type. This part will continue with 9×9 squares using the consecutive numbers (c,d,r) as the last cells in the central, diagonal and row spokes adjacent to the center cell in the square.

We have to decide what consecutive three numbers to place into the empty cells so that each row, column and diagonal of the central 3×3 square sums to 123. The triplet of numbers (22,23,24) to (34,35,36) in Table Ib fit the description since they will be shown not to be present as numbers in the other spoke cells. Thus, these triplets will have the requisite properties, since the rest of the even triplets will have at least one number in common with a number in every spoke of the wheel:

Table Ia
Even c
2,3,4
4,5,6
6,7,8
8,9,10
10,11,12
12,,13,14
14,15,16
16,17,18
18,19,20

Table Ib
Even c
20,21,22
22,23,24
24,25,26
26,27,28
28,29,30
30,31,32
32,33,34
34,35,36
36,37,38

It was mentioned in Part XI that 4n+1 odd order squares (5,9,13...) use the even c and the 4n+3 order squares (3,7,11...) use the odd c. This is due to the fact that the last numbers to be included in a triplet are even for the 4n+1 and odd for the 4n+3. What this means as shown in Table Ia is that the number one will always be unpaired when c is even but not when c is odd.

It was also shown in Part XI that when the initial numbers in the spoke are (1,4,7) that one 5th order square and three 7th order squares are formed. This section will show that the number of 9th order squares is seven and when tables similar to Ib are constructed for higher nth order squares that the number of terms follows the sequence:

Table II
5	7	9	11	13	15	17	19	...
1	3	7	13	21	31	43	57	...

The sequence is identical to that stored in the OEIS database under the Sloane number A002061 having the formula:

n² − n + 1

where n ≥ 1 in our case. We can convert this formula so that n is based on the order n as in Table II. Taking n to be the value as was shown in Part VIII we obtain the last equation based on the order n:

(n − 3)/2
[(n − 3)/2]² − (n − 3)/2 + 1
(n² − 8n + 19)/4

Moreover, the equation turned out to be more than just an equation for Δ=7, for it eventually led to the sequence of equations in Table I, column two of Part VIII, where most of the sequences in a row were checked in the OEIS database and the equations listed there converted to those based on nth order using (n − 3)/2 substituted into n. Thus, each row now consisted of its own quadratic equation for producing all the terms within that row and only that row.

The Seven 9×9 Squares

This section will only deal with numbers (after the main diagonal is filled) that run in the order 1 (top center), 4 (bottom right) and 7 (left center). A difference (Δ) of 7 is then added to each of the numbers followed by either of the three triplets in Table Ib those triplets without a common number in the spokes. But before that the wheel portion of the square is filled according to Part X, using the blue color values from Table Ib ensures that in the internal 3×3 square every row, column and diagonal adds up to the magic sum of 123. The second part of the build up of the square is addition of the non spoke cells employing the following two algorithms, using graph theory, as was done previously in Part X.

Final construction of the squares involve employing the Octagon I algorithm for the blue color cells and the L algorithm for the green color cells. All seven of the possible squares are shown below:

Border 9a(22,23,24)il
78	79	72	66	1	17	9	2	45
6	71	57	54	8	29	24	44	76
13	27	64	50	15	33	43	55	69
62	51	47	59	22	42	35	31	20
7	14	21	24	41	58	61	68	75
19	30	34	40	60	23	48	52	63
70	56	39	32	67	49	18	26	12
77	38	25	28	74	53	58	11	5
37	3	10	16	81	65	73	80	4

Border 9a(24,25,26)il
78	79	72	66	1	17	9	2	45
6	71	59	53	8	30	22	44	76
13	25	64	49	15	34	43	57	69
62	50	46	57	24	42	36	32	20
7	14	21	26	41	56	61	68	75
19	31	35	40	58	25	47	51	63
70	55	39	33	67	48	18	27	12
77	38	23	29	74	52	60	11	5
37	3	10	16	81	65	73	80	4

Border 9a(26,27,28)il
78	79	72	66	1	17	9	2	45
6	71	59	53	8	30	22	44	76
13	25	64	49	15	34	43	57	69
62	50	46	55	26	42	36	32	20
7	14	21	28	41	54	61	68	75
19	31	35	40	56	27	47	51	63
70	58	39	33	67	48	18	24	12
77	38	23	29	74	52	60	11	5
37	3	10	16	81	65	73	80	4

Border 9a(28,29,30)il
78	79	72	66	1	17	9	2	45
6	71	59	56	8	27	22	44	76
13	25	64	49	15	34	43	57	69
62	50	46	53	28	42	36	32	20
7	14	21	30	41	52	61	68	75
19	31	35	40	54	29	47	51	63
70	58	39	33	67	48	18	24	12
77	38	23	26	74	55	60	11	5
37	3	10	16	81	65	73	80	4

Border 9a(30,31,32)il
78	79	72	66	1	17	9	2	45
6	71	59	56	8	27	22	44	76
13	25	64	49	15	34	43	57	69
62	53	46	51	30	42	36	29	20
7	14	21	32	41	50	61	68	75
19	28	35	40	52	31	47	54	63
70	58	39	33	67	48	18	24	12
77	38	23	26	74	55	60	11	5
37	3	10	16	81	65	73	80	4

Border 9a(32,33,34)il
78	79	72	66	1	17	9	2	45
6	71	59	56	8	27	22	44	76
13	25	64	52	15	31	43	57	69
62	53	46	49	32	42	36	29	20
7	14	21	34	41	48	61	68	75
19	28	35	40	50	33	47	54	63
70	58	39	30	67	51	18	24	12
77	38	23	26	74	55	60	11	5
37	3	10	16	81	65	73	80	4

The last square shows what happens when we use an odd parity table for the internal 3×3 square. Two unpaired (non adjacent numbers in red) are left over to pair up at the end, producing a non magic square where columns 3 and 7 sum to 366 and 372, respectively.

Border 9a(34,35,36)il
78	79	72	66	1	17	9	2	45
6	71	59	56	8	27	22	44	76
13	25	64	52	15	31	43	57	69
62	53	49	47	34	42	33	29	20
7	14	21	36	41	46	61	68	75
19	28	32	40	48	35	50	54	63
70	58	39	30	67	51	18	24	12
77	38	23	26	74	55	60	11	5
37	3	10	16	81	65	73	80	4

Non Border 9a(33,34,35)il
78	79	72	66	1	17	9	2	45
6	71	59	56	8	27	22	44	76
13	25	64	52	15	31	43	57	69
62	53	46	60	33	42	36	29	20
7	14	21	35	41	59	61	68	75
19	28	32	40	61	34	50	54	63
70	58	39	30	67	51	18	24	12
77	38	23	26	74	55	60	11	5
37	3	10	16	81	65	73	80	4

This completes the Octagon I and L method. Go to Part XIII for initial numbers 3, 5, 7, 9, 11, and 13 in a 9×9 square. Go back to Part XI. Go back to homepage.