Arnoux Patterns

On April 17th, 1887, the Frenchman Gabriel Arnoux
deposited a perfect (nasik) (i.e. pandiagonal and pantriagonal) magic cube of order 17
with the Académie des Sciences. It consists of 26 handwritten pages! As far as I
have been able to determine, this is the first normal perfect (nasik) magic cube ever
constructed! Each of the 4913 cells in Arnoux’s order 17 cube is part of 13 different lines that sum to the constant. However, there are many other patterns of 17 numbers that
also sum to the constant. Again I express my gratitude to Christian Boyer of Paris, France, for locating and obtaining a copy of this marvelous and historically significant cube, and making it available to me. Following is quoted from an email by Christian Boyer sent on July 7, 2003. About the other announced characteristics, Gabriel Arnoux says > that his cube is "hypermagic" : he means that all the > miscellaneous lines are magic, not only the 90° lines and 45° diagonals. > Choose randomly two numbers in the cube, draw the line going > through these two numbers, and the magic sum of this line will be > magic (excluding some rare exceptions). > An incredible feature, far better that what you call a "perfect > magic" cube. More perfect than perfect! > Look at the "ArnouxExplications1.JPG" image, he gives an example > with the line going through the 2 numbers 1759 and 3891. > I will describe an easiest sample with for example the line going > through the 2 numbers 4718 (high left corner of Arnoux01.JPG) and > 3477 (line 2, column 3 of Arnoux02.JPG): > 1,1 in Arnoux01 = 4718 > 2,3 in Arnoux02 = 3477 > 3,5 in Arnoux03 = 3443 > 4,7 in Arnoux04 = 3953 > 5,9 in Arnoux05 = 944 > 6,11 in Arnoux06 = 2916 > 7,13 in Arnoux07 = 1607 > 8,15 in Arnoux08 = 2780 > 9,17 in Arnoux09 = 4174 > 10,2 in Arnoux10 = 43 > 11,4 in Arnoux11 = 2304 > 12,6 in Arnoux12 = 2525 > 13,8 in Arnoux13 = 638 > 14,10 in Arnoux14 = 4565 > 15,12 in Arnoux15 = 1403 > 16,14 in Arnoux16 = 1879 > 17,16 in Arnoux17 = 400 In this example, Christian used jumps of 1 for coordinates Y and Z, jumps of 2 for the X coordinates.
Because of the very large number of possible patterns, I adopted the following strategy to make the number of tests required manageable. I did check both magic squares and magic cubes. Also, one plane each of the three orthogonal orientations of each cube tested.
With the following exceptions, all lines with steps from 2
to m2 exist in odd order magic squares and cubes. The sloping line with step m1 is always a pandiagonal or pantriagonal (depending on whether the line goes through 2 or 3 dimensions. Squares or cubes of even order can also contain Arnoux patterns if ‘step is not a factor of m. However, the tests summarized here only applied to odd order squares and cubes. I did check out two order 4 and two order 8 magic squares in the Sqr_9OtherLines.xls spreadsheet. For most tests, I checked only lines starting at each of the cells in columns 1 and 2 for lines moving down and to the right. I also checked lines starting at each of the cells in columns m and m1 for lines sloping down to the left. I was confident that if all these lines are correct, then ALL lines in the magic square or cube are correct. By the time I had started testing the order 17 Arnoux, I had established the fact that similar patterns starting on any column of a given plane of the cube contained the same m totals (but the series of totals start on a different line). Therefore if a column contains all correct sums, all occurrences of that pattern in the cube are correct. With this in mind, for the order 17 Arnoux cube and the order 13 Quadrant magic square, I tested only column 1 and m1. For the cube tests, I tested the 3dimensional lines, then I tested the Top, Back and Left orthogonal planes for correct 2dimensional lines. To demonstrate that Arnoux example patterns with all 3 coordinates stepping by an amount greater then 1 is universal, I show here two other examples. In both cases I started at an arbitrary cell and used arbitrary steps for all 3 coordinates. And in both cases, my first attempt was successful! Of course, the fact that these two example patterns are correct does not guarantee that all patterns in the cube (with these particular steps) will sum correctly. My two examples simply show that patterns with any particular combination of steps can exist in a particular cube. Order 11 Seimiya Perfect: Order 9 Hendricks Perfect X Y Z X Y Z 1  11  3 = 1279 4  7  1 = 501 3  6  6 = 1056 9  9  3 = 228 5  1  9 = 844 5  2  5 = 721 7  7  1 = 500 1  4  7 = 133 9  2  4 = 277 6  6  9 = 265 11  8  7 = 65 2  8  2 = 5 2  3  10 = 1173 7  1  4 = 461 4  9  2 = 950 3  3  6 = 602 6  4  5 = 606 8  5  8 = 369 8  10  8 = 394 Total = S = 3285 10  5  11 = 171 Total = S = 7315 X = step 2, Y = step 5, X = step 5, Y = step 7, and Z = step 3 and Z = step 2 Both of these cubes are listed elsewhere on these cube pages. X and Z step by 1? I mentioned above that for the most part I have limited my tests to X and Z steps of 1, varying only the steps for Y. However, in my cube test spreadsheets, of which the Arnoux pattern test is just one sheet, I already have all the planes in the two vertical orientations. Because the row and column positions of these planes are exactly the same as those for the horizontal planes, it was simple to test these arrays also. Without thinking the matter through carefully, I figured that, in effect, testing these different orientations was equivalent to varying the Y and Z coordinates. Now that I have virtually completed all the tests I wish to do, I have checked this out. I find that the steps of only the Z coordinate varies when I use the same formulae as I did when testing the horizontal planes. Testing the second orientation of the vertical planes produced only a variation of the same patterns the first orientation produced. To better illustrate this discussion, here is an example I have extracted from Cube_7HendricksJRM.xls on the basis of the X step = 3 (see the Cube Comparison chart). For the cube: I have counted the planes (Z) as 1 on the top, 7 on the bottom of the cube. The rows (Y) as 1 on the top to 7 on the bottom (when the plane is shown as in my spreadsheet. The columns (X) are counted from the left. Horizontal planes Vertical B2F Horz Vertical L2R Horz CR Value Z Y X Val. Z Y X Z Y X Val. Z Y X Z Y X B6 98 1 1 1 98 1 1 1 1 1 1 1 1 1 1 1 7 1 C19 319 2 4 2 298 2 4 2 4 2 2 775 2 4 2 4 6 2 D32 197 3 7 3 204 3 7 3 7 3 3 342 3 7 3 7 5 3 E38 131 4 3 4 117 4 3 4 3 4 4 117 4 3 4 3 4 4 F51 9 5 6 5 23 5 6 5 6 5 5 284 5 6 5 6 3 5 G57 286 6 2 6 279 6 2 6 2 6 6 59 6 2 6 2 2 6 H70 164 7 5 7 185 7 5 7 5 7 7 226 7 5 7 5 1 7 Total 1204 1204 1204 Conclusion: For the squares: Top Back Left C_R Value Y X Z Y X Value Y X Z Y X Value Y X Z Y X B6 98 1 1 1 1 1 98 1 1 1 1 1 1 1 1 1 7 1 C9 264 4 2 1 4 2 257 4 2 4 1 2 132 4 2 4 6 1 D12 136 7 3 1 7 3 122 7 3 7 1 3 207 7 3 7 5 1 E8 302 3 4 1 3 4 330 3 4 3 1 4 331 3 4 3 4 1 F11 181 6 5 1 6 5 153 6 5 6 1 5 70 6 5 6 3 1 G7 4 2 6 1 2 6 18 2 6 2 1 6 194 2 6 2 2 1 H10 219 5 7 1 5 7 226 5 7 5 1 7 269 5 7 5 1 1 Total 1204 1204 1204 Magic objects I tested for this feature are:
The above cube spreadsheets are the original test
spreadsheets with an additional worksheet appended at the end.
The three order 5 cubes permitted a direct comparison between a cube with pantriagonals correct in all 4 directions, in 1 direction only, and in NO directions. Because these 3 magic cubes do not appear on any of my other cube pages, I present the listings here. Cube_5_Aale1.xls (Aale de Winkel) Plane 1  Top II III 1 50 69 88 107 110 4 48 67 86 89 108 2 46 70 125 19 38 57 76 79 123 17 36 60 58 77 121 20 39 94 113 7 26 75 73 92 111 10 29 27 71 95 114 8 63 82 101 25 44 42 61 85 104 23 21 45 64 83 102 32 51 100 119 13 11 35 54 98 117 120 14 33 52 96 IV V  Bottom 68 87 106 5 49 47 66 90 109 3 37 56 80 124 18 16 40 59 78 122 6 30 74 93 112 115 9 28 72 91 105 24 43 62 81 84 103 22 41 65 99 118 12 31 55 53 97 116 15 34 Cube_5_Boyer2.xls Plane 1  Top II III 54 31 13 115 97 10 117 99 51 33 96 53 30 12 119 15 122 79 56 38 76 58 35 17 124 37 19 121 78 55 81 63 40 22 104 42 24 101 83 60 103 80 62 44 21 47 4 106 88 65 108 85 67 49 1 69 46 3 105 87 113 90 72 29 6 74 26 8 110 92 5 112 94 71 28 IV V  Bottom 32 14 116 98 50 118 95 52 34 11 123 75 57 39 16 59 36 18 120 77 64 41 23 100 82 20 102 84 61 43 0 107 89 66 48 86 68 45 2 109 91 73 25 7 114 27 9 111 93 70 Cube_5_Trump2. Plane 1  Top II III 34 6 115 35 120 11 122 92 47 38 78 108 96 23 5 85 123 26 45 31 110 54 49 55 42 15 56 76 57 106 8 91 20 94 97 81 74 73 72 10 80 63 62 61 44 95 37 25 112 41 22 58 64 59 107 18 67 48 68 109 88 53 124 24 21 86 2 32 77 113 119 16 28 101 46 IV V  Bottom 84 3 7 105 111 103 71 0 100 36 17 65 60 66 102 83 12 99 87 29 114 52 51 50 43 27 30 104 33 116 82 69 75 70 14 93 79 98 1 39 13 121 117 19 40 4 118 9 89 90 As an afterthought, I decided to check out order 16 cubes because order 16 is 42, and also to investigate Arnoux patterns in the even orders. As with the preceding tests, I tested X and Z steps of 1,
Y steps of from 2 to m1 and for X and Y steps of 1, Z steps of from 2 to
m1 Here I will compare results of 4 different types of order
16 magic cubes. The number of magic squares shown in the table are the orthogonal squares included in the cube .
The following notes and table summarize the findings from my tests. For magic cubes Patterns starting in column m1 will always be equivalent to a pantriagonal This feature is NOT limited to the Arnoux perfect order 17
cube! It seems to be dependent to some degree on correct
pantriagonals Notice also the Hendricks lowly order 7 pantriagonal cube has ALL solutions correct for steps 3 and 4! (This cube, by definition, has all triagonals correct in ALL 4 directions!) Frost’s order 7 pandiagonal cube also has all correct solutions for steps 3 and 4. This feature is NOT limited to magic hypercubes whose order is a prime number. However, only steps that are relatively prime to the order can possibly work. So, for example, for order 9, steps of 3 and 6 will not work. That is because it is impossible to cycle through a series of 9 numbers. I tested two other cubes with all three coordinates using steps other then 1! For magic squares Patterns with a step for X of 1 and Y of 1 or m1 will always be equivalent to a pandiagonal This feature is very common in magic squares and orthogonal arrays of magic cubes Hendricks order 7 pantriagonal cube (no magic squares) – ALL solutions for steps 2, 3, 4, and 5 Arnoux order 17 perfect cube (orthogonal planes)  ALL solutions are correct for steps 2, and 3 (and step 4 for one , step 5 for two orientations). Pandiagonal magic order 13 square (my quadrant magic square) – All solutions for steps 2, 5, 6. 7, 8, 11, and 12. For this square, I also tested patterns with X confined to a step of 1 and Y with steps 2, 3, 4, and 5. They show exactly the same results as the normal steps This feature is NOT dependent on pandiagonals because I tested it on two order 5 pandiagonal magic squares, and none of the steps had ALL solutions correct (except step 4 which is m1)
