A New Procedure for Magic Squares (Part IB)

Consecutive Boustrophedonic 7x7 Mask-Generated Squares

A Discussion of the New Method

Magic squares such as the Loubère have a center cell which must always contain the middle number of a series of consecutive numbers, i.e. a number which is equal to one half the sum of the first and last numbers of the series, or ½(n² + 1). The properties of these regular or associated Loubère squares are:

That the sum of the horizontal rows, vertical columns and corner diagonals are equal to the magic sum S.
The sum of any two numbers that are diagonally equidistant from the center (DENS) is equal to n² + 1, i.e., or twice the number in the center cell and are complementary to each other.

In this method the numbers on the square are placed consecutively starting from the first leftmost column and entered across every other cell. Consecutive numbers are then added to the next rows boustrophedonically. At the end the square is filled in by adding numbers to the second cell on the first row and proceeding as before. The square which is not magic is modified into a form which can be converted into a magic one by the use of a mask. This mask generates numbers which are added to certain cells in the square to produce a final square composed of numbers which may not be in serial order. For example, negative numbers or numbers greater than n² may be present in the square.

In addition, it will also be shown that the sums of these squares follow the sum equation shown in the New block Loubère Method. :

S = ½(n³ ± an)

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Construction of a 7x7 Magic Square I

Method: Reading consecutive from left to right boustrophedonically - use of a mask

Construct the 7x7 Square 1 where 7 = 4n + 3 by adding numbers in a consecutive manner starting at row 1 cell, then swinging around at the end (boustrophedonically) until the final cell is reached (square 1).
On reaching this cell go to cell 6 row 1 (next to last) and continue adding consecutive numbers in reverse. This method is equivalent to adding the requisite numbers on a line and then using a (1 down, 1 left) break as shown for the 4 → 5 move.
Since the sums of all the columns or rows are not equal to 175 add or subtract the numbers in the last row to the center row and add or subtract the lst column to the center column. At this point three duplicates (in pink) have been generated (Square 3).

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

⇒

2
							91
1	28	2	27	3	26	4	91	84
29	7	30	6	31	5	32	140	35
8	35	9	34	10	33	11	140	35
36	14	37	13	38	12	39	189	-14
15	42	16	41	17	40	18	189	-14
43	21	44	20	45	19	46	238	-63
22	49	23	48	24	47	25	238	-63
154	196	161	189	168	182	175	91
21	-21	14	-14	7	-7	0

⇒

3
							63
1	28	2	111	3	26	4	175
29	7	30	41	31	5	32	175
8	35	9	69	10	33	11	175
57	-7	51	-15	45	5	39	175
15	42	16	27	17	40	18	175
43	21	44	-22	45	19	46	175
22	49	23	-15	24	47	25	175
175	175	175	175	175	175	175	63

⇒

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Generate a mask whereby the sums of the columns and rows are constructed as in the box below. This assures that when each of these values is added to the corresponding cell in square 10 (as in the de la Hire method) that all sums will be equal to a magic sum.

We start by subtracting the diagonals(63,63) from 175 to give 112 and 112, respectively and which will be used as what I call the "de la Hire constants".

Addition of the constants to 175, i.e., 175 + 112 gives 287 a magic pre-sum.

The following equations are used such that the following conditions are obeyed:

The right diagonal and left diagonals : 287 = 63 + 2(112)

The rows and columns: 287 = 175 + 112

Generate the mask using the 112 factor and adding this factor to the appropriate cells in square 3 generates square 4.

The equation for this square is 287 = ½(n³ + 33n).

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Mask A
112
					112
						112
			112
	112
				112
		112

⇒

4
							287
113	28	2	111	3	26	4	287
29	7	30	41	31	117	32	287
8	35	9	69	10	33	123	287
57	-7	51	97	45	5	39	287
15	154	16	27	17	40	18	287
43	21	44	-22	157	19	46	287
22	49	135	-15	24	47	25	287
287	287	287	287	287	287	287	287

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Construction of a 7x7 Magic Square II

Method: Reading consecutive from left to right boustrophedonically - use of a mask

Construct the 7x7 Square 1 where 7 = 4n + 3 by adding numbers in a consecutive manner starting at row 1 cell, then swinging around at the end (boustrophedonically) until the final cell is reached (square 1).
On reaching this cell go to cell 2 row 1 (second cell) and continue adding consecutive numbers in reverse. This method is equivalent to adding the requisite numbers on a line and then using a (1 down, 2 right) break as shown for the 4 → 5 move.
Since the sums of all the columns or rows are not equal to 175 add or subtract the numbers in the last row to the center row and add or subtract the lst column to the center column. At this point three duplicates (in pink) have been generated (Square 3).

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

⇒

2
							91
1	26	2	27	3	28	4	91	84
29	5	30	6	31	7	32	140	35
8	33	9	34	10	35	11	140	35
36	12	37	13	38	14	39	189	-14
15	40	16	41	17	42	18	189	-14
43	19	44	20	45	21	46	238	-63
22	47	23	48	24	49	25	238	-63
154	182	161	189	168	196	175	91
21	-7	14	-14	7	-21	0

⇒

3
							63
1	26	2	111	3	28	4	175
29	5	30	41	31	7	32	175
8	33	9	69	10	35	11	175
57	5	51	-15	45	-7	39	175
15	40	16	27	17	42	18	175
43	19	44	-43	45	21	46	175
22	47	23	-15	24	49	25	175
175	175	175	175	175	175	175	63

⇒

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Generate a mask whereby the sums of the columns and rows are constructed as in the box below. This assures that when each of these values is added to the corresponding cell in square 10 (as in the de la Hire method) that all sums will be equal to a magic sum.

We start by subtracting the diagonals(63,63) from 175 to give 112 and 112, respectively and which will be used as what I call the "de la Hire constants".

Addition of the constants to 175, i.e., 175 + 112 gives 287 a magic pre-sum.

The following equations are used such that the following conditions are obeyed:

The right diagonal and left diagonals : 287 = 63 + 2(112)

The rows and columns: 287 = 175 + 112

Generate the mask using the 112 factor and adding this factor to the appropriate cells in square 3 generates square 4.

The equation for this square is 287 = ½(n³ + 33n).

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Mask B
						112
	112
					112
			112
112
				112
		112

⇒

4
							287
1	26	2	111	3	28	116	287
29	117	30	41	31	7	32	287
8	33	9	69	10	147	11	287
57	5	51	97	45	-7	39	287
127	40	16	27	17	42	18	287
43	19	44	-43	157	21	46	287
22	47	135	-15	24	49	25	287
287	287	287	287	287	287	287	287

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This completes this section on a new Consecutive Boustrophedonic 7x7 Mask-Generated Squares.The next section deals with Consecutive Boustrophedonic 9x9 Mask-Generated Squares (Part II).To return to homepage.