# 12×12 Magic Knight's Tours

Sections on this page: Earliest Example and Border Method.Braid and Panel Methods.
Method of Biaxial Symmetry.Diagonally Magic Tours.

## Update March 2003:

Awani Kumar has constructed four diagonally magic 12×12 knight's tours, which are now shown in the final section here, thus resolving a long-standing problem. I show these tours in both numerical and graphical form. For full details of Mr Kumar's researches see The Games and Puzzles Journal Issue 26. I have reversed the numbering of the tours as given in Mr Kumar's article and reoriented them so as to conform to the standard Frénicle method, as used in our catalogue of 8×8 tours. (The other tours on this page will also be converted to this form when I find the time.) I have also added four examples of tours with one diagonal magic that I found some years ago, derived from the 8×8 magic tour 14a.

## éEarliest Example and Border Method

Apart from the four new fully magic tours we show the tours graphically, leaving the reader to put in the numbers and check the magic totals if desired. The numbering starts at the cell marked with a black square and ends at the cell marked with an arrowhead. The cells with black circles mark the other ends of the quarters of the tour. The positions of these eight cells (1, 36, 37, 72, 73, 108, 109, 144) usually help to indicate the structure of the tour. The tours in the first part are oriented as they were first published, except for a few changed for purposes of comparison.

The magic constant for the 12×12 board is 870. Reversing the numbering will also give a magic array, though other numerical features such as diagonal sums may thereby be altered; an entry E becomes 145−E, and a diagonal sum D becomes 1740−D. The new Kumar tours have both diagonals magic. These surpass earlier results which achieved one magic diagonal or the two diagonals together adding to twice the magic constant (thus differing from the constant by plus and minus the same amount).

The earliest known example of a magic knight's tour on a board larger than 8×8 is a tour on the 12×12 board attributed to Krishnaraj Wadiar (1790-1868) who was Rajah of Mysore. This tour is quoted in the manuscript by Harikrishna Sharma (1871), printed in Devanagari script in 1900 and reproduced in S.R.Iyer's Indian Chess (1982) with English commentary. The tour is constructed on the squares and diamonds principle, and it will be seen that the snake-like pattern includes a long braid component.

The earliest magic tour on the 12×12 board that was known to Murray was one by E. Lange in the book by Lehmann (1932); one diagonal is only 2 away from the magic constant (870). This was formed by adding a border braid to one of the 8×8 magic tours (05g).

Lange indicated three ways in which the links to be deleted in the parent tour can be arranged, and Murray (1951) indicates two ways in which each of these can be connected to the border braid. The dark lines are the inserted moves and the light lines the deleted moves. Murray counted 180 magic tours formed by this method from the 8×8 magic tours then known.

T.H.Willcocks (in a letter to me dated 17 June 1993) provided two examples of bordered tours using other connection schemes; the parent magic squares used are 12j and 27g.

## éBraid and Panel Methods

Based on Wihnyk's 16×16 example (1885) Murray gave the following three 12×12 tours which extend and modify the braid part of Beverley's tour. By design the 4×4 quads in these are all magic, i.e. they add to 290 in each row and column.

The fourth tour above is the first to have one diagonal magic, although Murray may not have realised this, since he does not mention the property in his 1951 manuscript. He constructed the tour as one of 130 tours of ‘gnomon’ type, formed analogously to the Lange bordered tours. He points out that the tour remains magic when renumbered from 37, 73 and 109. This cyclic property it inherits from its parent tour 00d.

The first published 12×12 tours in which one of the diagonals added to the magic constant, 870, were by T. H. Willcocks who published the following examples in Fairy Chess Review (December 1955), Recreational Mathematics Magazine (December 1962), and Journal of Recreational Mathematics (1968). For comparison with the Murray examples above I have inverted the 1955 diagram and reversed the 1962 so that the Beverley components are all similarly placed. The 1955 example distorts the lower edge of the Beverley half-board into a short braid. The 1968 example is almost identical to Murray's 1950 unpublished example (the 4×8 Beverley component is inverted).

The fourth diagram above shows a new way that the Beverley quads can be used, placing them corner-to-corner instead of edge-to-edge. The single link joining the two quads limits the movement possible.

Murray noted that a pair of quads in his 1941 example shown above could be replaced by alternative symmetric formations, as can be done in the case of the 8 by 8 Beverley tour. I show one example of this below.

In his account of 8 by 8 tours of the Beverley type, analysed in terms of contiguous contraparallel chains, Murray mentions that the method is applicable to larger boards, but strangely he does not give any example on the 12×12 board. Here is one I have constructed to show that it is possible. The two pairs of contraparallel chains are of course 1-36||108-73 and 37-72||144-109. The first two quads in the middle ranks are of beverley type (the beverley quartes are emphasised), the other three mid-edge quads are of the braid type, and the corner quads are all alike and arranged in biaxial symmetry to each other so that their variances are guaranteed to cancel out (they vary only by ±2 in the files).

Here are four magic tours I constructed which have one diagonal (top left to bottom right) adding to the magic constant 870. They are derived from the 8×8 magic tour 14a. In A the braid that forms a significant part of the parent tour is simply extended to cover the 12×12 board. In C, D, E the pattern in the bottom left corner (which is the same as the right-hand side of the Beverley tour 27a) is replaced by alternative symmetric patterns (as in the 8×8 tours 27e, 27d, 27c respectively). The non-magic diagonal adds to A 850, B 860, C 890, D 926 when numbering is from h9.

A B

C D

## éMethod of Biaxial Symmetry

A semi-magic tour (i.e. with ranks adding to the magic constant but not files), using a piece that can make a rook move of an odd length, can easily be constructed by axial symmetry, since this ensures that 1 is in the same rank as 144, 2 as 143, and so on, adding in pairs to 145, six pairs giving 870. All the moves can be arranged to be knight moves except the middle move 72-73 which has to be a rook move along the rank (also the closure move 144-1).

A magic tour (with both ranks and files adding to the magic constant) by a rook-moving piece can be constructed similarly by (a) combining the above symmetry with symmetry about the horizontal axis which ensures that 1 and 108, 2 and 107 ... 36 and 73, all adding to 109, are in the same file and 37 and 144, 38 and 143, ..., 72 and 109, all adding to 181, are in the same file, and (b) ensuring that three pairs of the first type and three of the second type are in each file, since 109 + 181 = 290 and 3×290 = 870. This can be done without using any further rook moves. All that is necessary is that the route 1-36 uses exactly three cells in each file. The resulting tour is a knight + rook (empress) tour. The moves 72-73 and 144-1 being rook moves. It also follows from this construction that the sum of the two diagonals will be 1740, i.e. twice the magic constant, because if k (in the range 1-36) is on a diagonal then so are 145 – k and 109 – k and k + 36 which add to 290.

A diagonal magic tour on the above scheme is unfortunately not possible. This can be proved as follows. Six of the numbers 1-36 must occur on the diagonals, let them be a,b,c,d,e,f. Then if there are 3 (say a,b,c) on one diagonal and 3 (d,e,f) on the other (i.e. 3 odd and 3 even) the sum of the diagonal will be a + (36+a) + b + (36+b) + c + (36+b) + (145–d) + (109–d) + (145–e) + (109–e) + (145–f) + (109–f) = 870 + 2[(a+b+c) – (d+e+f)]. This can never be 870 since (a+b+c) – (d+e+f) can never be zero, being a difference of odd and even numbers. Similarly if the numbers are distributed 4 on one diagonal and 2 on the other the formula becomes 652 + 2[(a+b+c+d) – (e+f)]. To give the total 870 the required difference is now 109, which cannot be the difference of two even numbers. The distribution 5 to 1 gives the formula 434 + 2[(a+b+c+d+e) – f] requiring an even difference of 218 but the difference can only be odd. The distribution 6 to 0 gives the formula 216 + 2[a+b+c+d+e+f] requiring a 'difference' of 327, but the sum of six odd or six even numbers is necessarily even. Thus if a diagonally magic 12×12 tour is to be possible it must deviate from this type of symmetry.

A magic knight tour with two diagonals adding to 1740 can be constructed by the above method if we can find a way of converting one or both of the rook moves into a knight move.

The following example by T.H.Willcocks from Chessics (1986) follows a scheme similar to that described above. A quarter-path and the moves that deviate from quaternary symmetry are shown in the diagram alongside. The cells used in the quarter path are circled; note that it uses three cells in each rank.

I have constructed 25 tours in which the two diagonals add to 1740 and in which all have the moves shown in the 16-cell lozenge (or rolling-pin?) shaped areas marked in the following diagrams. The only moves that deviate from quaternary symmetry are within these areas. There are just two transpositions 1-3 and 73-75 which convert these tours to or from the tours with rook-move connections described above. From each of these 25 magic tours three others can be derived by inverting one or both of these lozenges (but this only makes 50 in all becaues of symmetry). This enumeration was done by hand and has not yet been independently checked. The four diagrams shown are those in which the deviation from 870 is less than 50. I used these on my design for a New Year and Millennium card 2000/2001.

I list here all 25 solutions, giving the coordinates of the cells 5 to 32 (files are lettered a to l, ranks numbered 1 to 12, but to equalise the lines I write 10, 11, 12 as x, y, z). The type (r,s) means that of the numbers 1-36 there are r on the odd diagonal and s on the even diagonal (r+s = 6). The final ± figure is the deviation of the diagonal totals from 870.

 1. a1 b3 d4 b5 a7 c6 d8 f7 h6 f5 h4 j5 l4 k2 i1 j3 k1 l3 j4 l5 k7 i6 h8 f9 dx b9 ay cz (3-3) ± 22 2. a1 b3 c1 a2 b4 a6 b8 d7 f6 h7 f8 h9 j8 l9 k7 l5 j4 l3 k1 j3 i1 k2 i3 h5 f4 d5 c7 d9 (2-4) ± 34 3. a1 b3 d4 c6 a7 b5 d6 f5 h4 j5 l4 k2 i1 j3 k1 l3 j4 l5 k7 i8 h6 f7 h8 f9 dx b9 ay cz (3-3) ± 42 4. a1 b3 d4 b5 a7 c8 dx cz ay b9 d8 f7 h6 f5 h4 i6 j4 l3 k1 j3 i1 k2 l4 k6 l8 j7 h8 f9 (3-3) ± 46 5. a1 b3 c1 a2 b4 d5 f6 h7 f8 h9 i7 k6 l4 k2 i1 j3 k1 l3 j4 l5 j6 h5 f4 d3 c5 a6 b8 d9 (2-4) ± 58 6. a1 b3 d4 b5 a7 c8 dx f9 h8 j7 l8 k6 l4 k2 i1 j3 k1 l3 j4 i6 h4 f5 h6 f7 d8 b9 ay cz (3-3) ± 82 7. a1 b3 c1 a2 b4 a6 c7 d5 f4 h5 i3 k2 i1 j3 k1 l3 j4 l5 k7 l9 j8 h9 f8 h7 f6 d7 b8 d9 (2-4) ±126 8. a1 b3 c1 a2 b4 d3 f4 h5 j4 l3 k1 j3 i1 k2 l4 j5 l6 k8 i7 h9 f8 h7 f6 d5 b6 a8 c7 d9 (2-4) ±126 9. a1 b3 c1 a2 b4 d5 f4 h5 i3 k2 i1 j3 k1 l3 j4 l5 k7 l9 j8 h9 f8 h7 f6 d7 b8 a6 c7 d9 (2-4) ±134 10. a1 b3 a5 b7 d8 f7 h6 f5 h4 i6 k5 l7 j8 l9 ky iz jx kz lx j9 h8 f9 dx cz ay b9 c7 d9 (4-2) ±162 11. a1 b3 c1 a2 b4 d3 f4 h5 i7 k6 l8 j9 lx kz jx iz ky l9 j8 h9 f8 h7 f6 d5 c7 a6 b8 d9 (4-2) ±162 12. a1 b3 c1 a2 b4 d3 f4 h5 f6 h7 i5 k6 l8 j9 lx kz jx iz ky l9 j8 h9 f8 d7 b8 a6 c7 d9 (4-2) ±182 13. a1 b3 c1 a2 b4 d5 f6 h7 f8 h9 i7 j9 lx kz jx iz ky l9 k7 l5 j6 h5 f4 d3 c5 a6 b8 d9 (4-2) ±214 14. a1 b3 c1 a2 b4 d3 c5 a6 b8 d9 f8 h9 i7 h5 j4 l3 k1 j3 i1 k2 l4 k6 l8 j7 h6 f7 d8 f9 (1-5) ±230 15. a1 b3 c5 d3 c1 a2 b4 a6 b8 d9 f8 h9 i7 h5 j4 l3 k1 j3 i1 k2 l4 k6 l8 j7 h6 f7 d8 f9 (1-5) ±230 16. a1 b3 c5 a6 b8 d9 f8 h9 i7 h5 j4 l3 k1 j3 i1 k2 l4 k6 l8 j7 h6 f7 d8 b9 ay cz dx f9 (1-5) ±270 17. a1 b3 c5 a6 b8 d9 f8 h9 i7 h5 j4 l3 k1 j3 i1 k2 l4 k6 l8 j7 h6 f7 d8 f9 cx b9 ay cz (1-5) ±270 18. a1 b3 d4 b5 a7 c8 dx cz ay b9 d8 f7 h6 f5 h4 i6 k7 l9 ky iz jx kz lx j9 l8 j7 h8 f9 (5-1) ±286 19. a1 b3 d4 c6 a7 b5 d6 f7 h6 f5 h4 j5 l4 k6 l8 j9 lx kz jx iz ky ix h8 f9 d8 b9 ay cz (5-1) ±286 20. a1 b3 d4 b5 d6 f7 h6 f5 h4 j5 l4 k6 l8 j9 lx kz jx iz ky ix h8 f9 d8 c6 a7 b9 ay cz (5-1) ±294 21. a1 b3 d4 c6 a5 b7 d8 f7 h6 f5 h4 i6 k5 l7 j8 l9 ky iz jx kz lx j9 h8 f9 dx b9 ay cz (5-1) ±294 22. a1 b3 d4 b5 a7 c8 dx f9 h8 j7 l8 j9 lx kz jx iz ky l9 k7 i6 h4 f5 h6 f7 d8 b9 ay cz (5-1) ±362 23. a1 b3 d4 c6 d8 f9 h8 ix ky iz jx kz lx j9 l8 k6 l4 j5 h4 f5 h6 f7 d6 b5 a7 b9 ay cz (5-1) ±386 24. a1 b3 c1 a2 b4 d5 f6 h7 j6 l5 k7 l9 ky iz jx kz lx j9 h8 i6 h4 f5 d4 b5 a7 c8 dx f9 (6-0) ±450 25. a1 b3 c5 d3 f4 d5 f6 h7 j6 l5 k7 l9 ky iz jx kz lx j9 h8 i6 h4 f5 d4 b5 a7 b9 ay cz (6-0) ±450

Similar tours can be constructed on this plan with the lozenge areas moved inwards, but this results in one of the transposed numbers within the lozenges occurring on a diagonal, so that the sum of the two diagonals is no longer 1740 but is increased or decreased by 2. Here are two examples:

[In fact I now find that the second tour diagrammed here was composed earlier than 2000 since I used it (differently oriented) on a 1995 New Year's Greetings card (alongside the 1993 Beverley type tour shown in the preceding section).]

## éDiagonally Magic Tours

These four tours are the first diagonally magic knight tours on the 12×12 board to be constructed. They are the work of Awani Kumar of Lucknow, India, and have just been published in The Games and Puzzles Journal Issue 26, to which reference should be made for more details of the work leading to their discovery. I have chosen to number and orient the tours in accordance with the Frénicle method as used for our catalogue of 8×8 tours, and they are arranged in numerical sequence according to the same method (note that the first difference of B from A (and of D from C) is the number 55 in place of 51 at (7, 7) i.e. the top left corner of the bottom right quarter of the board, while C differs from A (and D from B) only in the 4×4 area in the middle of the top four ranks.

A
 1 70 107 112 5 66 103 116 9 64 99 118 108 111 2 69 104 115 6 65 100 117 10 63 71 106 109 4 113 102 67 8 61 98 119 12 110 3 72 105 68 7 114 101 120 11 62 97 73 142 35 40 77 140 17 60 93 122 13 58 36 39 74 141 32 41 92 121 16 59 96 123 143 34 37 76 139 78 51 18 129 94 57 14 38 75 144 33 42 31 130 91 52 15 124 95 81 136 29 44 79 138 19 50 89 128 21 56 28 45 80 137 30 43 90 131 20 53 88 125 135 82 47 26 133 84 49 24 127 86 55 22 46 27 134 83 48 25 132 85 54 23 126 87
B
 1 70 107 112 5 66 103 116 9 64 99 118 108 111 2 69 104 115 6 65 100 117 10 63 71 106 109 4 113 102 67 8 61 98 119 12 110 3 72 105 68 7 114 101 120 11 62 97 73 142 35 40 77 140 17 60 93 122 13 58 36 39 74 141 32 41 92 121 16 59 96 123 143 34 37 76 139 78 55 18 125 94 57 14 38 75 144 33 42 31 126 91 56 15 124 95 81 136 29 44 79 138 19 54 89 128 21 52 28 45 80 137 30 43 90 127 20 53 88 129 135 82 47 26 133 84 49 24 131 86 51 22 46 27 134 83 48 25 132 85 50 23 130 87
C
 1 70 107 112 5 68 101 116 9 64 99 118 108 111 2 69 104 113 8 65 100 117 10 63 71 106 109 4 67 6 115 102 61 98 119 12 110 3 72 105 114 103 66 7 120 11 62 97 73 142 35 40 77 140 17 60 93 122 13 58 36 39 74 141 32 41 92 121 16 59 96 123 143 34 37 76 139 78 51 18 129 94 57 14 38 75 144 33 42 31 130 91 52 15 124 95 81 136 29 44 79 138 19 50 89 128 21 56 28 45 80 137 30 43 90 131 20 53 88 125 135 82 47 26 133 84 49 24 127 86 55 22 46 27 134 83 48 25 132 85 54 23 126 87
D
 1 70 107 112 5 68 101 116 9 64 99 118 108 111 2 69 104 113 8 65 100 117 10 63 71 106 109 4 67 6 115 102 61 98 119 12 110 3 72 105 114 103 66 7 120 11 62 97 73 142 35 40 77 140 17 60 93 122 13 58 36 39 74 141 32 41 92 121 16 59 96 123 143 34 37 76 139 78 55 18 125 94 57 14 38 75 144 33 42 31 126 91 56 15 124 95 81 136 29 44 79 138 19 54 89 128 21 52 28 45 80 137 30 43 90 127 20 53 88 129 135 82 47 26 133 84 49 24 131 86 51 22 46 27 134 83 48 25 132 85 50 23 130 87

A question still unanswered is: can diagonal magic be achieved in a 12×12 closed tour?

Back to: Top