| The Games and Puzzles Journal — Issue 20, May-August 2001 |
Vaclav Kotesovec sent an e-mail on 26 January with the following two original leaper results, having seen the note on the "Shortest Path Problem" in our new "Leapers at Large" page.
(1) On a board 17×17 a {5,12}-leaper in cell (1,1) takes the shortest route to (5,12).
(2) On a board 41×41 a {12,29}-leaper in cell (21,9) takes the shortest route to (33,4).
In chessic terminology these are known as "seriesmove target" problems (Serienzug Ziel in German) and are required to have an exactly determined solution. Since Vaclav's solutions take 67 and 287 moves these problems are a little too stiff to be set as "puzzles", so I give the solutions forthwith. Both solutions are given using purely numerical cooordinates for the cells of the board, taking a1 = (1,1). Vaclav claims that problem (2) shows the record length possible for all square boards less than or equal to 50×50 and for all leapers on such boards.
(1) V.Kotesovec. Board 17×17 with {5,12}-leaper at (1,1). Play to (5,12) in 67 moves.
Solution: 1.(6,13) 2.(11,1) 3.(16,13) 4.(4,8) 5.(16,3) 6.(11,15) 7.(6,3) 8.(1,15) 9.(13,10) 10.(1,5) 11.(6,17) 12.(11,5) 13.(16,17) 14.(4,12) 15.(16,7) 16.(4,2) 17.(9,14) 18.(14,2) 19.(2,7) 20.(14,12) 21.(2,17) 22.(7,5) 23.(12,17) 24.(17,5) 25.(5,10) 26.(17,15) 27.(12,3) 28.(7,15) 29.(2,3) 30.(14,8) 31.(2,13) 32.(7,1) 33.(12,13) 34.(17,1) 35.(5,6) 36.(17,11) 37.(5,16) 38.(10,4) 39.(15,16) 40.(3,11) 41.(15,6) 42.(3,1) 43.(8,13) 44.(13,1) 45.(1,6) 46.(13,11) 47.(1,16) 48.(6,4) 49.(11,16) 50.(16,4) 51.(4,9) 52.(16,14) 53.(11,2) 54.(6,14) 55.(1,2) 56.(13,7) 57.(1,12) 58.(13,17) 59.(8,5) 60.(3,17) 61.(15,12) 62.(3,7) 63.(15,2) 64.(10,14) 65.(5,2) 66.(17,7) 67.(5,12) exact way; 68.(17,17) not exact.
The diagram shows the moves 1 to 34. These form a symmetrical corner-to-corner path. The moves 34 to 68 form the same pattern rotated 90 degrees.
It may be noted that some cells are "passed through" very close to their centre points, yet they are not "visited" since no move starts or finishes there; this is especially notable in the case of cell (9,10), just above the centre cell, which is passed through eight times.
(2) V.Kotesovec. Board 41×41 with {12,29}-leaper at (21,9). Play to (33,4) in 287 moves.
Solution: 1.(33,38) 2.(4,26) 3.(33,14) 4.(4,2) 5.(16,31) 6.(28,2) 7.(40,31) 8.(11,19) 9.(40,7) 10.(28,36) 11.(16,7) 12.(4,36) 13.(33,24) 14.(4,12) 15.(16,41) 16.(28,12) 17.(40,41) 18.(11,29) 19.(40,17) 20.(11,5) 21.(23,34) 22.(35,5) 23.(6,17) 24.(35,29) 25.(6,41) 26.(18,12) 27.(30,41) 28.(1,29) 29.(30,17) 30.(1,5) 31.(13,34) 32.(25,5) 33.(37,34) 34.(8,22) 35.(37,10) 36.(25,39) 37.(13,10) 38.(1,39) 39.(30,27) 40.(1,15) 41.(30,3) 42.(18,32) 43.(6,3) 44.(35,15) 45.(6,27) 46.(35,39) 47.(23,10) 48.(11,39) 49.(40,27) 50.(11,15) 51.(40,3) 52.(28,32) 53.(16,3) 54.(4,32) 55.(33,20) 56.(4,8) 57.(16,37) 58.(28,8) 59.(40,37) 60.(11,25) 61.(40,13) 62.(11,1) 63.(23,30) 64.(35,1) 65.(6,13) 66.(35,25) 67.(6,37) 68.(18,8) 69.(30,37) 70.(1,25) 71.(30,13) 72.(1,1) 73.(13,30) 74.(25,1) 75.(37,30) 76.(8,18) 77.(37,6) 78.(25,35) 79.(13,6) 80.(1,35) 81.(30,23) 82.(1,11) 83.(13,40) 84.(25,11) 85.(37,40) 86.(8,28) 87.(37,16) 88.(8,4) 89.(20,33) 90.(32,4) 91.(3,16) 92.(32,28) 93.(3,40) 94.(15,11) 95.(27,40) 96.(39,11) 97.(10,23) 98.(39,35) 99.(27,6) 100.(15,35) 101.(3,6) 102.(32,18) 103.(3,30) 104.(15,1) 105.(27,30) 106.(39,1) 107.(10,13) 108.(39,25) 109.(10,37) 110.(22,8) 111.(34,37) 112.(5,25) 113.(34,13) 114.(5,1) 115.(17,30) 116.(29,1) 117.(41,30) 118.(12,18) 119.(41,6) 120.(29,35) 121.(17,6) 122.(5,35) 123.(34,23) 124.(5,11) 125.(17,40) 126.(29,11) 127.(41,40) 128.(12,28) 129.(41,16) 130.(12,4) 131.(24,33) 132.(36,4) 133.(7,16) 134.(36,28) 135.(7,40) 136.(19,11) 137.(31,40) 138.(2,28) 139.(31,16) 140.(2,4) 141.(14,33) 142.(26,4) 143.(38,33) 144.(9,21) 145.(38,9) 146.(26,38) 147.(14,9) 148.(2,38) 149.(31,26) 150.(2,14) 151.(31,2) 152.(19,31) 153.(7,2) 154.(36,14) 155.(7,26) 156.(36,38) 157.(24,9) 158.(12,38) 159.(41,26) 160.(12,14) 161.(41,2) 162.(29,31) 163.(17,2) 164.(5,31) 165.(34,19) 166.(5,7) 167.(17,36) 168.(29,7) 169.(41,36) 170.(12,24) 171.(41,12) 172.(29,41) 173.(17,12) 174.(5,41) 175.(34,29) 176.(5,17) 177.(34,5) 178.(22,34) 179.(10,5) 180.(39,17) 181.(10,29) 182.(39,41) 183.(27,12) 184.(15,41) 185.(3,12) 186.(32,24) 187.(3,36) 188.(15,7) 189.(27,36) 190.(39,7) 191.(10,19) 192.(39,31) 193.(27,2) 194.(15,31) 195.(3,2) 196.(32,14) 197.(3,26) 198.(32,38) 199.(20,9) 200.(8,38) 201.(37,26) 202.(8,14) 203.(37,2) 204.(25,31) 205.(13,2) 206.(1,31) 207.(30,19) 208.(1,7) 209.(13,36) 210.(25,7) 211.(37,36) 212.(8,24) 213.(37,12) 214.(25,41) 215.(13,12) 216.(1,41) 217.(30,29) 218.(1,17) 219.(30,5) 220.(18,34) 221.(6,5) 222.(35,17) 223.(6,29) 224.(35,41) 225.(23,12) 226.(11,41) 227.(40,29) 228.(11,17) 229.(40,5) 230.(28,34) 231.(16,5) 232.(4,34) 233.(33,22) 234.(4,10) 235.(16,39) 236.(28,10) 237.(40,39) 238.(11,27) 239.(40,15) 240.(11,3) 241.(23,32) 242.(35,3) 243.(6,15) 244.(35,27) 245.(6,39) 246.(18,10) 247.(30,39) 248.(1,27) 249.(30,15) 250.(1,3) 251.(13,32) 252.(25,3) 253.(37,32) 254.(8,20) 255.(37,8) 256.(25,37) 257.(13,8) 258.(1,37) 259.(30,25) 260.(1,13) 261.(30,1) 262.(18,30) 263.(6,1) 264.(35,13) 265.(6,25) 266.(35,37) 267.(23,8) 268.(11,37) 269.(40,25) 270.(11,13) 271.(40,1) 272.(28,30) 273.(16,1) 274.(4,30) 275.(33,18) 276.(4,6) 277.(16,35) 278.(28,6) 279.(40,35) 280.(11,23) 281.(40,11) 282.(28,40) 283.(16,11) 284.(4,40) 285.(33,28) 286.(4,16) 287.(33,4) exact way; 288.(21,33) not exact. Record for boards < 50×50 and all leapers.
The diagram below, which is about at the limits of my drawing capability, shows the moves 1 to 72. The moves from there to 143 reflect this pattern in the (1,1)-(41,41) diagonal. Then move 144 takes the leaper to (9,21) on the horizontal median and 143 takes it to (38,9) which is a 90 degree clockwise rotation from the initial cell (33,38). The rest of the moves then repeat the previous pattern via the (1,41) corner (top left) which is reached on move 216. The dotted lines show moves analogous to 144-145 but via (21,9) on the vertical median.
Juha Saukkola sent this note by e-mail on 21 September 2000: I reproduce this directly from his HTML code, including the colour scheme (apart from the background colour and slight editing).
Constructing shortest Nightrider-tour with no knight-steps:
First fill in the ranks 2,4,6,8:
18 26 15 21 29 06 09 01
** ** ** ** ** ** ** **
30 05 12 02 17 27 14 24
** ** ** ** ** ** ** **
16 22 19 25 10 04 32 07
** ** ** ** ** ** ** **
11 03 31 08 13 23 20 28
** ** ** ** ** ** ** **
(from to indicates a long step of three knight moves)
Completing the rest of the ranks as well:
18 26 15 21 29 06 09 01
60 52 55 45 40 63 35 43
30 05 12 02 17 27 14 24
39 64 36 42 57 51 54 48
16 22 19 25 10 04 32 07
56 46 59 49 34 44 37 62
11 03 31 08 13 23 20 28
33 41 38 61 53 47 58 50
Only 7 steps of three knight moves are used (if the tour is open), but 8 in the closed form.
The construction of this tour wasn't as easy as it now looks!
An Unsolved Problem: In my reply I raised the problem of whether it is possible to construct a diagonally magic 8×8 nightrider tour (but not with the ban on single knight moves). Of course an ordinary knight tour is also a nightrider tour but the nightrider's extra powers should make a tour with the diagonals magic easier to construct.
Readers may be aware of my occasional forays into the mathematical aspects of various forms of esoteric symbolism, for instance: I Ching Hexagrams (in Chessics 13 p.12) and Runes (in G&PJ vol.1 pp.144, 162), not to mention Magic Squares. I now turn my attention to the Cabalistic "Tree of Life" (A) which is a network diagram usually shown as ten nodes (termed "sephiroth" = spheres) joined by 22 paths, which are identified with the letters of the Hebrew alphabet (and also with the "major arcana" cards of the Tarot deck)
Some versions of the diagram (e.g. wisdoms door) also show an eleventh node between the top and central nodes on the middle "pillar". The term "cabala" (which has numerous alternative transliterations) apparently means "given" i.e. handed down by tradition. What is puzzling to me is why this particular network should be thought to be a source of mystic knowledge rather than any of many others that could be designed. Just because some ancient authority said so seems an inadequate reason.
The nearest I have come to 'explaining' it is as a garbled version of the "doubled cube" diagram (B) shown above which shows the relationships between the various divisors of sixty. Note that the three directions of the sides of the cubes indicate multiplication by the primes 2, 3 and 5, while the three horizontal lines indicate multiplication by 15 (= 3x5). Other lines (diagonals of the faces of the cubes) can be inserted indicating multiplications by 6 (= 2x3) and by 10 (= 2x5), and (diagonals of the cubes) by 30 (= 2x3x5). Such a diagram may have been of assistance in making calculations in base sixty, which was the system of numeration used by the Babylonians (surviving in our system of time and angle units), from whom the cabalistic tradition is believed to have derived.
This double cube can also be viewed from other perspectives. If the 6 and 12 are occluded behind the 5 and 10 we get the pattern (C) which has been proposed by J.S.Graham as a possible origin of the diagram [t was on his web site "numerometria" which is now lost] chapter 8, page headed Qabbala I). Another view would occlude the 12 behind the 5, leaving five nodes on the middle pillar (D). The middle node in (D) could also be taken to represent the opposite corners 1 and 60. Finally (E) shows a non-occluded version. If the upper circle is taken as 1 the lower is 60, but in this case the horizontal lines connecting the left and right columns would represent a fractional multiplication by 5/3 or 3/5.
This note derives from a diagram on the page headed Qabbala II in J.S.Graham's [lost] web site "numerometria" chapter 8, mentioned in the previous item. He shows a trapezoidal network consisting of 22 nodes joined by 48 links which are the sides of 27 equilateral triangular regions, This network has just two odd nodes, the two upper corners, and therefore by Euler's well known theorem can be drawn 'unicursally' beginning at one odd node and ending at the other. In other words a tour of the 48 edges is possible. Graham shows a 37-move 'rook' tour along these edges, which fulfills the extra conditions that: (a) the path never crosses itself and (b) parallel moves are always taken in the same sense; e.g. beginning the tour at the top right, all sloping moves are down to the right or up to the right, while all horizontal moves are to the left.
The minimum number of rook moves needed for such a tour seems to be 32. I show three examples of this, one using all the horizontal lines as single moves, one using all the 'down' diagonals, and one in which the four external lines are intact. In each case it can be seen that there are just two nodes that do not have a straight pass through them. Eliminating these reduces the diagram to 31 or 30 moves, but also has the effect of splitting off one or two 'short circuits' from the main path, so that the diagram becomes a pseudotour. Many other 32-move tours are possible. Finally I give a 33-move example with 11 moves in each direction. In this the three 'unpassed' nodes are at the vertices of an equlilateral triangle.
This note is also inspired by a comment in J.S.Graham's Qabbala II page on the [lost] website "numerometria" chapter 8, referred to above. He states that the numbers 27 and 37 are related by 1/27 = 0.037~ and 1/37 = 0.027~, where the tilde (~) is short for 'recurring'. This is a property that at first sight seems remarkable. He follows it with the statement that 27 + 37 = 64, which however is a red herring. The equation that implies the reciprocal relationship between the recurring decimals is that 27×37 = 999.
The general principle behind this kind of 'amicable' relationship between two numbers X and Y such that 1/X = 0.Y~ and 1/Y = 0.X~, where X and Y may contain leading zeroes, is that 1/X = 0.Y~ = Y/99...9 and so 99...9 = X×Y where the number of 9s is the number of digits in Y.
A trivial case of this is the relationship 1/3 = 0.333333... = 0.3~ where X = Y = 3 and the denominator is a single 9. With two digits we have 99 = 3×33 = 9×11 and we find 1/3 = 0.33~ with 1/33 = 0.03~, and 1/9 = 0.11~ with 1/11 = 0.09~. Evidently the cases 3×33...3 and 9×11...1 occur as pairs of factors in 99...9 for any number of digits. Of more interest are cases where the 'repunit' factor 11...1 is not a prime. The first case of this is 111 = 3×37 which gives the result cited by Graham. With four digits we have 9999 = 3×3×11×101 which gives the reciprocating pairs 1/99 = 0.0101~ with 1/101 = 0.0099~ among others.
Repunits and their factorisation or primality have been much studied. A table of factors of repunits was available at: "http://www.swox.com/gmp/repunit.html" (the author's name was not given as far as I could determine) but the site no longer exists. This revealed that 11111 = 41×271 which generates the reciprocating fractions: 1/369 = 0.00271~ with 1/271 = 0.00369~ among others. The first repunit other than 11 which is prime is the one with 19 digits, and most of the intermediate repunits have multiple factors, so quite a lot of reciprocating fractions can be found.
The 6-digit repunit = 111×1001 = (3×37)×(7×11×13) = (7×37)×(3×11×13) = 259×429 so we have 999999 = 777×1287 and 1/777 =0.001287~ with 1/1287 = 0.000777~ among others.
The 7-digit repunit = 239×4649, so 9999999 = 717×13947 and 1/717 = 0.0013947~ with 1/13947 = 0.0000717~ among others.
I leave higher cases for the reader to investigate. The next repunits with just two factors are the 11-digit = 21649×513239 and the 17-digit = 2071723×5363222357.