ç Knight's Tour Notes Index .

Figured Tours - Part 3: The Onitiu Problem

by G. P. Jelliss

The problem is to construct a numbered knight tour on a square board of any size, in which the square numbers follow a path of knight moves,
and the tour as a whole is symmetric by 180 or 90 degree rotation.

Sections on this page: — Historical Background — The Rotary Onitiu Problem — The Birotary Onitiu Problem

‹ Historical Background

The problem of constructing a knight tour of the standard 8×8 chessboard, with the cells numbered in the sequence visited by the knight, in such a way that the square numbers appear in sequence along a rank was proposed by G. E. Carpenter and solved by S. Hertzsprung in Brentano's Magazine in 1881. The analogous problem of constructing a knight tour so that the square numbers, taken in sequence of magnitude, delineate a path of knight moves was proposed by T. R. Dawson in The Problemist Fairy Chess Supplement in 1932, and over the years he published there, and in its continuation The Fairy Chess Review, 100 examples with the square numbers showing all possible shapes of symmetric circuit that allow such a tour. These 'Dawsonian' tours show symmetric circuits within an asymmetric tour.

The difficult problem of constructing a symmetric knight tour with the square numbers in a knight circuit, on a square board of any size, is what I am here calling The Onitiu Problem. This problem was considered for the 8×8 board by the Romanian chess problemist Valeriu Onitiu in Fairy Chess Review 1939, where Dawson wrote concerning his remarkable solution: "VO notes that he examined 144 dispositions of the squares, all that are possible for diametral symmetry, and this is the only case leading to a tour. Moreover every move of the tour is determined, so that the tour is UNIQUE in all the millions possible."

The diagrams above show the tour in arithmetic and geometric forms. The circuit of the square numbers (1, 4, 9, 16, 25, , 49, 64) is marked in red and the rotated path followed by their 'complements' (33, 36, 41, 48, 57, 4, 17, 32) in blue. Corresponding numbers in the two sequences differ by 32. It will be seen that the squares 4 and 36 belong to both paths. This intersection of the paths severely restricts the possible configurations of the circuits. I have not fully checked all the possible configurations to confirm the uniqueness of the solution, but both Onitiu and Dawson are known to be reliable in such matters.

A Wikipedia page for Valeriu Onitiu, which was drawn to our attention by Pavlov Mircea of the magazine SAH, gives some brief biographical details including dates (b. April 8, 1872, d. 31 December 1948). Another question that Onitiu worked on concerned the maximum number of three-move knight lines in tours.

After constructing some Dawsonian tours on the 10×10 and larger boards, which make an attractive and relatively easy recreation [See Part 2], it occurred to me to try the Onitiu Problem on these larger boards. It proved to be a difficult question and one which gave some interesting results, which are the subject of this web-page.

‹ The Rotary Onitiu Problem

Here we reqiuire the tour to have 180 degree rotational symmetry. For this type of symmetry on a board of side 2m the diametrally opposite cells (related by the 180 degree rotation) will contain numbers, both odd or both even, that differ by 2(m^2) which is half the number of cells in the board. Consider first boards where m is odd, that is 2m = 2 (2k - 1) = 4k - 2. Is it possible to have two square numbers a^2 and b^2 that also occur in the complementary sequence (like the 4 and 36 in the Onitiu solution)? For this we require a^2 - b^2 = 2(m^2). This difference of two squares can be written (a - b)(a + b). Since a and b are both odd or both even these factors (a - b) and (a + b) must both be even, so their product is twice an even number. It follows that on boards of side 4k - 2, where m = 2k - 1 is odd, the two paths are distinct, having no intersections.

When the board edge is 4k then a pair of squares k^2 and (3k)^2 = 9(k^2) always have difference 8.(k^2) = 2[(2k)^2] and therefore appear in both the sequence of squares and the sequence of complements. So the paths have at least two intersections in every case of this type.

Board of side 2. Opposite numbers differ by 2. No knight tour is possible of course, but the simple wazir tour shows the same pattern.

Board of side 4. Opposite numbers differ by 8. No knight tour is possible. But similar results are possible with combined Rook and Knight moves. The two intersections occur at the square numbers 1 and 9.

Board of side 6. Opposite numbers differ by 18. All possible Dawsonian tours have been constructed. Examination of these shows that no solution to the Onitiu problem is possible.

Board of side 8. Opposite numbers differ by 32. This is the original Onitiu case. The square numbers 4 and 36 are complementary. Note that these are 4 times the previous case (1 and 9).

Board of side 10. Opposite numbers differ by 50. The following diagram shows a solution.

Board of side 12. Opposite numbers differ by 72. In this case the square numbers 9 and 81 are complementary. Nine (3^2) times the basic case. There is also a second pair of complementary squares namely 121 and 49. This second pair of intersections obeys the rule 11^2 - 7^2 = 2.(6^2). This is the first case in a series of pairs of the form (11s)^2 - (7s)^2 = 2.[(6s)^2] where s takes the successive valiues 1, 2, 3, ... This shows that on all boards whose sides are a multiple of 12 there are at least four intersections of the paths of squares with the path of complements (but there may be more as we show for the case 24×24 below).

Board of side 14. Opposite numbers differ by 98. This diagram shows a solution. There are no intersections in this case.

Board of side 16. Opposite numbers differ by 128. This is the fourth case after 4×4, 8×8, 12×12 where there are two intersections of the form (s.1)^2 and (s.3)^2. In this case (4.1)^2 and (4.3)^2 that is 16 and 144.

Board of side 18. Opposite numbers differ by 162. Here is a solution. I found this difficult despite the fact that there are no intersections.

The square numbers and complements
001 004 009 016 025 036 049 064 081 100 121 144 169 196 225 256 289 324
163 166 171 178 187 198 211 226 243 262 283 306 007 034 063 094 127 162
No intersections.

Board of side 20. Opposite numbers differ by 200.

The square numbers and complements
001 004 009 016 025 036 049 064 081 100 121 144 169 196 225 256 289 324 361 400
201 204 209 216 225 236 249 264 281 300 321 344 369 396 025 056 089 124 161 200
The intersection points are 25 and 225, that is (5.1)^2 and (5.3)^2.

Board of side 22. This has been solved for the Birotary case, shown in the next section. Any birotary solution also solves the rotary case.

The square numbers and complements
001 004 009 016 025 036 049 064 081 100 121 144 169 196 225 256 289 324 361 400 441 484
243 246 251 258 267 278 291 306 323 342 363 386 411 438 467 014 047 082 119 158 199 242
This is a case with no intersections.

Board of side 24.

The square numbers and complements
001 004 009 016 025 036 049 064 081 100 121 144 169 196 225 256 289 324 361 400 441 484 529 576
289 292 297 304 313 324 337 352 369 388 409 432 457 484 513 544 001 036 073 112 153 196 241 288

This is the first case with six or more intersections. On the 24×24 board the opposite numbers differ by 288. The three pairs of intersection points occur at the square numbers 1 and 289, 36 and 324, 196 and 484.For 1 to be an intersection point we must have 1 + 2.(m^2) = u^2, which means u must be an odd number (17 in this case).
A new series of solutions begins here of the form (s.1)^2 and (s.17)^2.
This occurs on boards whose sides are a multiple of 24, namely 48, 72, 96 and so on.

Board of side 26. This has been solved for the Birotary case, shown in the next section.

The square numbers and complements
001 004 009 016 025 036 049 064 081 100 121 144 169 196 225 256 289 324 361 400 441 484 529 576 625 676
339 342 347 354 363 374 387 402 419 438 459 482 507 534 563 594 627 662 023 062 103 146 191 238 287 338
This is a case with no intersections.

Board of side 28. This has not yet been solved.

The square numbers and complements
001 004 009 016 025 036 049 064 081 100 121 144 169 196 225 256 289 324 361 400 441 484 529 576 625 676 729 784
393 396 401 408 417 428 441 456 473 492 513 536 561 588 617 648 681 716 753 008 049 092 137 184 233 284 337 392
Two intersection points.

Board of side 30. This has been solved for the Birotary case, shown in the next section. This is about the maximum size of board that my drawing program can cope with.

Sequences of squares and complements:
001 004 009 016 025 036 049 064 081 100 121 144 169 196 225 256 289 324 361 400 441 484 529 576 625 676 729 784 841 900
451 454 459 466 475 486 499 514 531 550 571 594 619 646 675 706 739 774 811 850 891 034 079 126 175 226 279 334 391 450
No intersections in the 180 degree rotation.

The Rotary Onitiu Problem on Larger Boards.

On any board with even edge 180 degree rotary symmetry is possible, and it seems likely that a solution of the Onitiu problem can be found on all boards from side 8 upwards. The ratio of the number of moves used in a tour and its circuits of squares and complements, to the total number of knight moves available on the board, decreases towards 1/4 as the board size increases. So there is more room for manoeuvre on larger boards. However a complication is that the number of intersections between the circuit of square numbers and the circuit of their complements increases for certain numbers, which may make construction more difficult in these cases.

If a^2 and b^2 are numbers (like 4 and 36 in the 8×8 case) that occur in the list of squares and the list of complements, then their difference must be half the number of cells in the board (64/2 in the 8×8 case). Thus if the board side is 2.m we must have a^2 + 2.(m^2) = b^2, that is a^2 - b^2 = 2.(m^2).

List of differences of squares that are twice a square:
4: 16/2 = 2×4 = 8 = 9 - 1 (2 intersections)
8: 64/2 = 2×16 = 32 = 36 - 4 (2 intersections) [there is also = 81 - 49 but 81 > 64]
12: 144/2 = 2×36 = 72 = 81 - 9 = 121 - 49 (4 intersections)
16: 256/2 = 2×64 = 128 = 144 - 16 (2 intersections)
20: 400/2 = 2×100 = 200 = 225 - 25 (2 itersections)
24: 576/2 = 2×144 = 288 = 289 - 1 = 324 - 36 = 484 - 196 (6 intersections)
28: 784/2 = 2×196 = 392 = 441 - 49 (2 intersections)
32: 1024/2 = 2×256 = 512 = 576 - 64 (2 intersections)
36: 1296/2 = 2×324 = 648 = 729 - 81 = 1089 - 441 (4 intersections)
40: 1600/2 = 2×400 = 800 = 900 - 100 = 1089 - 289 (4 intersections)
44: 1936/2 = 2×484 = 968 = 1089 - 121 (2 intersections)
48: 2304/2 = 2×576 = 1152 = 1156 - 4 = 1294 - 144 = 1681 - 529 = 1936 - 784 (8 intersections)
52: 2704/2 = 2×676 = 1352 = 1521 - 169 (2 intersections)
56: 3136/2 = 2×784 = 1568 = 1764 - 196 (2 intersections)
60: 3600/2 = 2×900 = 1800 = 1849 - 49 = 2025 - 225 = 3025 - 1225 = 3481 - 1681 (8 intersections)
64: 4096/2 = 2×1024 = 2048 = 2304 - 256 (2 intersections)
68: 4624/2 = 2×1156 = 2312 = 2601 - 289 (2 intersections)
72: 5184/2 = 2×1296 = 2592 = 2601 -9 = 2916 - 324 = 4356 - 1764 (6 intersections)
76: 5776/2 = 2×1444 = 2888 = 3249 - 361 (2 intersections)
80: 6400/2 = 2×1600 = 3200 = 3249 - 49 = 3600 - 400 = 4356 - 1156 (6 intersections)
84: 7056/2 = 2×1764 = 3528 = 3969 - 441 = 4489 - 961 = 5929 - 2401 (6 intersections)
88: 7744/2 = 2×1936 = 3872 = 4356 - 484 (2 intersections)
92: 8464/2 = 2×2116 = 4232 = 4761 - 529 (2 intersections)
96: 9216/2 = 2×2304 = 4608 = 4624 - 16 = 5184 - 576 = 6724 - 2116 = 7744 - 3136 (8 intersections)
100: 10000/2 = 2×2500 = 5000 = 5625 - 625 (2 intersections)
These figures are given as a matter of interest. Whether anyone will ever construct tours of these sizes is doubtful. They are definitely beyond the capabilities of my drawing program. The figures have not been independently checked.

Summary of intersections
0 intersections: side 2, 6, 10, 14, 18, 22, 26, 30, 34, ... (4.n + 2). Not in the above list.
2 intersections: side 8, 16, 20, 28, 32, 44, 52, 56, 64, 68, 76, 88, 92, 100, ... ratio 1/3 (squared).
4 intersections: side 12, 36, 40, ... ratios 1/3, 7/11, 17/33.
6 intersections: side 24, 72, 80, 84, ... ratios 1/3, 7/11, 1/17.
8 intersections: side 48, 60, 96, ... ratios

‹ The Birotary Onitiu Problem

On a board with oddly even sides knight tours are possible that have 90 degree rotational symmetry, which I term 'birotary' symmetry. One is naturally led to wonder if the Onitiu Problem can be solved under this extra condition. The following notes summarise my results on boards of increasing size.

Board of side 10. Having examined all 79 possible arrays of the central sections required to solve the Onitiu birotary problem on this board I have concluded that it is impossible, though I have not found a conclusive proof argument for this. It is possible to solve the problem under looser conditions. I have constructed several Emperor tour solutions, employing both Knight and Wazir moves.

The square numbers and those related by rotation:
001 004 009 016 025 036 049 064 081 100
026 029 034 041 050 061 074 089 006 025
051 054 059 066 075 086 099 014 031 050
076 079 084 091 100 011 024 039 056 075
Four intersection points shown in bold, 25, 50, 75, 100.

Board of side 14. 14×14 = 196 = 2×98 = 4×49.
Not solved. The best I have obtained is an Emperor tour using 10 wazir moves in each quarter.

The square numbers and those related by rotation:
001 004 009 016 025 036 049 064 081 100 121 144 169 196
050 053 058 065 074 085 098 113 130 149 170 193 022 049
099 102 107 114 123 134 147 162 179 002 023 046 071 098
148 151 156 163 172 183 196 015 032 051 072 095 120 147
Four intersection points shown in bold, 49, 147, 96, 194.

Board of side 18. 18×18 = 324 = 2×162 = 4×81.
Not solved. Best achieved are emperor tours with eight wazir moves, two in each quarter.

The square numbers and those related by rotation:
001 004 009 016 025 036 049 064 081 100 121 144 169 196 225 256 289 324
082 085 090 097 106 117 130 145 162 181 202 225 250 277 306 013 046 081
163 166 173 178 187 198 211 226 243 262 283 306 007 034 063 094 127 162
244 247 252 259 268 279 292 307 324 019 040 063 088 115 144 175 208 243
There are eight intersection points, printed bold, 63, 81, 144, 162, 225, 243, 306, 324.

Some statistics: 72 nodes and moves in linkages + 324 moves in tour = 396.
Number of links available on board = 32×17×2 = 1088.
Ratio 396/1088 = 0.36397 over a third.

Board of side 22. We have 22×22 = 484 = 2×242 = 4×121.

The square numbers and those related by rotation:
001 004 009 016 025 036 049 064 081 100 121 144 169 196 225 256 289 324 361 400 441 484
122 125 130 137 146 157 170 185 202 221 242 265 290 317 346 377 410 445 482 037 078 121
243 246 251 258 267 278 291 306 323 342 363 386 411 438 467 014 047 082 119 158 199 242
364 367 372 379 388 399 412 427 444 463 484 023 048 075 104 135 168 203 240 279 320 363
There are only the four basic intersection points (circled), 121, 242, 363, 484.

Some statistics: 88 nodes and moves in linkages + 484 moves in tour = 572.
Number of links available on board = 40×21×2 = 1680.
Ratio 572/1680 = 0.3404761 slightly over a third.

I also found a circular version of this tour. For diagram see the Jeepyjay Diary.
This was solved before the above square form. The central pattern of knight-move connections is the same in each.

Board of side 26. We have 26×26 = 676 = 2×338 = 4×169.

The sequence of squares and complements:
001 004 009 016 025 036 049 064 081 100 121 144 169 196 225 256 289 324 361 400 441 484 529 576 625 676 squares
170 173 178 185 194 205 218 233 250 269 290 313 338 365 394 425 458 493 530 569 610 653 022 069 118 169 rotated 90
339 342 347 354 363 374 387 402 419 438 459 482 507 534 563 594 627 662 023 062 103 146 191 238 287 338 rotated 180
508 511 516 523 532 543 556 571 588 607 628 651 676 027 056 087 120 155 192 231 272 315 360 407 456 507 rotated 270
There are four intersection points. 169, 338, 507, 676 (circled).

Some statistics: 100 nodes and moves in linkages + 676 moves in tour = 767.
Number of links available on board = 48×25×2 = 2400.
Ratio 776/2400 = 0.32333 less than a third.

Board of side 30. We have 30×30 = 900 = 2×450 = 4×225.

Sequences of squares and complements:
001 004 009 016 025 036 049 064 081 100 121 144 169 196 225 256 289 324 361 400 441 484 529 576 625 676 729 784 841 900 squares
226 229 234 241 250 261 274 289 306 325 346 369 394 421 450 481 514 549 586 625 666 709 754 801 850 001 054 109 166 225 rotated 90
451 454 459 466 475 486 499 514 531 550 571 594 619 646 675 706 739 774 811 850 891 034 079 126 175 226 279 334 391 450 rotated 180
676 679 684 691 700 711 724 739 756 775 796 819 844 871 900 031 064 099 136 175 216 259 304 351 400 451 504 559 616 675 rotated 270

Sixteen intersections. Besides the four principal nodes 225, 450, 675, 900. There are three quartets of secondary nodes 001, 226, 451, 676. (these are linked to the principal nodes) and 064, 289, 514, 739 and 175, 400, 625, 850. In each case two are squares. There is no intersection between opposite paths (rotated 180 degrees to each other).

Numerologists may like to notice that the mystic numbers 666 and 216 (6×6×6) are diametrally opposite each other on the central octagon.
This is because their difference is 450.

Board of side 34. We have 34×34 = 1156 = 2×578 = 4×289.

Sequences of squares and complements:
001 004 009 016 025 036 049 064 081 100 121 144 169 196 225 256 289 324 361 400 441 484 529 576 625 676 729 784 841 900 961 1024 1089 1156 squares
290 293 298 305 314 325 338 353 370 389 410 433 458 485 514 545 578 613 650 689 730 773 818 865 914 965 1018 1073 1130 1189 94 157 222 289 rotated 90
579 582 587 594 603 614 627 642 659 678 699 722 747 774 803 834 867 902 939 978 1019 1062 1107 1154 047 098 151 206 263 322 383 446 511 578 rotated 180
868 871 876 883 892 903 916 931 948 967 988 1011 1036 1063 1092 1123 1156 35 72 111 152 195 240 287 336 387 440 495 552 611 672 735 800 867 rotated 270
Four intersections (multiples of 289). Single linkages 35-36, 324-325, 613-614, 902-903.

Board of side 38. We have 38×38 = 1444 = 2×722 = 4×361.

Sequences of squares and complements:
001 004 009 016 025 036 049 064 081 100 121 144 169 196 225 256 289 324 361
400 441 484 529 576 625 676 729 784 841 900 961 1024 1089 1156 1225 1296 1369 1444 squares
362 365 370 377 386 397 410 425 442 461 482 505 530 557 586 617 650 685 722
761 802 845 890 937 986 1037 1090 1145 1202 1261 1322 1385 006 073 142 213 286 361 rotated 90
723 ... rotated 180
1084 ... rotated 270
4 intersections.

Board of side 42. We have 42×42 = 1764 = 2×882 = 4×441.

Sequences of squares and complements:
0001 0004 0009 0016 0025 0036 0049 0064 0081 0100 0121 0144 0169 0196 0225 0256 0289 0324 0361 0400 0441
0484 0529 0576 0625 0676 0729 0784 0841 0900 0961 1024 1089 1156 1225 1296 1369 1444 1521 1600 1681 1764 squares
0442 0445 0450 0457 0466 0477 0490 0505 0522 0541 0562 0585 0610 0637 0666 0697 0730 0765 0802 0841 0882
0925 0970 1017 1066 1117 1170 1225 1282 1341 1402 1465 1530 1597 1666 1737 0046 0121 0198 0277 0358 0441 rotated 90
0883 ... rotated 180
1324 ... rotated 270
16 intersections.

Board of side 46. We have 46×46 = 2116 = 2×1058 = 4×529.

Sequences of squares and complements:
0001 0004 0009 0016 0025 0036 0049 0064 0081 0100 0121 0144 0169 0196 0225 0256 0289 0324 0361 0400 0441 0484 0529
0576 0625 0676 0729 0784 0841 0900 0961 1024 1089 1156 1225 1296 1369 1444 1521 1600 1681 1764 1849 1936 2025 2116 squares
0530 0533 0538 0545 0554 0565 0578 0593 0610 0629 0650 0673 0698 0725 0754 0785 0818 0853 0890 0929 0970 1013 1058
1105 1154 1205 1258 1313 1370 1429 1490 1553 1618 1685 1754 1825 1898 1973 2050 0013 0094 0177 0262 0349 0438 0529 rotated 90
1059 ... rotated 180
1588 ... rotated 270
4 intersections.

Board of side 50. We have 50×50 = 2500 = 2×1250 = 4×625.

Sequences of squares and complements:
0001 0004 0009 0016 0025 0036 0049 0064 0081 0100 0121 0144 0169 0196 0225 0256 0289 0324 0361 0400 0441 0484 0529 0576 0625
0676 0729 0784 0841 0900 0961 1024 1089 1156 1225 1296 1369 1444 1521 1600 1681 1764 1849 1936 2025 2116 2209 2304 2401 2500 squares
0626 0629 0634 0641 0650 0661 0674 0689 0706 0725 0746 0769 0794 0821 0850 0881 0914 0949 0986 1025 1066 1109 1154 1201 1250
1301 1354 1409 1466 1525 1586 1649 1714 1781 1850 1921 1994 2069 2146 2225 2306 2389 2474 0061 0150 0241 0334 0429 0526 0625 rotated 90
1251 ... rotated 180
1876 ... rotated 270
4 intersections.

Birotary Onitiu Problem on Larger Boards

The side of the board must be n = 4.m + 2 to allow birotary symmetry. This is 2.(2.m + 1), twice an odd number. We have shown earlier that on this board the circuits of squares and 180 degree complements have no intersection.

The number of cells in the board is the square of 4.m + 2. This is 4.[(2m + 1)^2]. A rotation of 90 degrees thus adds the quarter number of cells (2.m + 1)^2 to each number in the sequence of square numbers (unil the sum exceeds n^2 in which case n^2 is subtracted). For an intersection we therefore require a^2 + (2.m + 1)^2 = b*2 for square numbers a and b. This is a Pythagorean relationship. So to find the intersections we need to find the Pythagorean triples that include (2.m + 1)^2.

If the line of squares intersects the 90 degree rotated line, then each of the four lines will similarly intersect the next 90 degree rotated line, so the number of intersections is always a multiple of 4.

Summary of intersections
4 intersections: side 10, 14, 22, 26, 34, 38, 46, 50,
8 intersections: side 18,
12 intersections: no cases found so far
16 intersections: side 30, 42,

‹Back to Top