Some Miscellaneous Results

6 September 2006

Dear webmaster I am a math and physics teacher from Germany. I managed to find a new record for the "longest-noncrossing-leaper-tour" on a 16*16 board by means of a computer algorithm (and partly by using the results already known). My work concerns the "183-step-jump" Robin Merson already conjectured to be possible on a 16*16 board. I attach a *.gif file.

Alexander Fischer, Eschborn, Germany

11 September 2006

Here is another improvement concerning the longest noncrossing leaper tour on a 14*14 chess board, it's of 135 steps length. By board extension some other records can possibly be improved.

Alexander Fischer

Laszlo originally submitted this item to the Brights forum under the name "Ansem". I thought it was of interest to readers of the Games and Puzzles Journal and he has given permission for it to appear here.

In this subforum I feel encouraged to share my mathematic discovery. It's very little, but I still would be sorry to get it lost, so I share it on every (a very few) forum which seems good for the aim. The interesting thing about this discovery how for at least 2000 years mathematicians passed by it unnoticed, believing that the subject is closed and nothing is left to discover. It's like finding a pyramid in the center of Cairo. I had made the discovery when playing with LEGO bricks, but I have to start the explanation with the already known facts.

If a, b, c are the sides of a right-angled triangle, then a^2 + b^2 = c^2 (here a^2 means a square, etc). If a, b, c are all integer numbers, we call them a Pythagorean triplet. The only known method to find Pythagorean triplets (there is more than one method, but they are only variants of the original), is the following:

Choose two positive integer numbers: a bigger p and a less q. Then these three numbers:

• a: p^2 - q^2
• b: 2.p.q
• c: p^2 + q^2

It's easy to validate:

• a^2 + b^2 = (p^2 - q^2)^2 + (2.p.q)^2 = p^4 + q^4 - 2.p^2.q^2 + 4.p^2.q^2 = p^4 + q^4 + 2.p^2.q^2
• c^2 = (p^2 + q^2)^2 = p^4 + q^4 + 2.p^2.q^2

But how had they found the method? Most mathematican-historians (those who haven't met me yet) think the road was something like this [in a later post Laszlo cited: Bartel Leendert van der Waerden "Science Awakening" (1954)]:

• a^2 + b^2 = c^2
• b^2 = c^2 - a^2
• b^2 = (c+a).(c-a)
• 1 = (c/b + a/b).(c/b - a/b)
• If a, b, c are integers, (c/b + a/b) and (c/b - a/b) are rational numbers, which give 1 if multiplied.
• So there are p, q numbers, such that
• c/b + a/b = p/q
• c/b - a/b = q/p

They didn't use these forms and symbols, but this pair of equations (when the same pair of numbers is added and subtracted) was very familiar to them. So what if we regard c/b and a/b as unknown variables, and solve this system of equations for them.

• a/b = (p/q - q/p)/2 = (p^2 - q^2)/(2.p.q)
• c/b = (p/q + q/p)/2 = (p^2 + q^2)/(2.p.q)

These equations are easily satisfiable with the first-mentioned a, b, c triplet. For example, if p=2, q=1, a/b must be 3/4, c/b must be 5/4, so let be a=3, c=5, b=4.

Why do they think this was the method of the Babylonians? Because they don't know any other method. (All other publicly known methods are just variants of this.) I can't refute their statement, but I had found a much more simple and Babylonian-friendly method:

Let's imagine a Babylonian mathematician thinking above a squared grid, for example, a floor covered with squares of stone. We can use a squared paper to figure things out. The grid consists of vertical and horizontal lines, and the matching points are called nodes, if I'm correct.

Choose and mark a node. Go 2 units to the right, and 1 unit up. Mark the node there. Go 2 up, 1 left. Mark the node. Go 2 left, 1 down. Mark. Go 2 down, 1 right. You have arrived to the first marked node. The 4 marked nodes make up the angles of a tilted square. You can cover the whole paper with such tilted squares, but for this simplest example we will need only two.

So let's start again with an empty paper. Mark a node. 2 right, 1 up, mark. 2 right, 1 up, mark. 2 up, 1 left, mark. 2 left, 1 down, mark. 2 left, 1 down, mark. These nodes make up two squares, which make up one rectangle. Now, look what we have done. We started with the diagonal of a 2X1 rectangle. From four such line, we made a square. From such squares, we made a bigger 2X1 rectangle. One of the diagonals of this big rectangle lies on a horizontal line of the original grid. So we know it is 5 units long. The other diagonal is tilted in the original grid, but we know it must be also 5 units long. We can make it the hypotenuse of a right-angled triangle, which has a horizontal leg with 3 unit length, and a vertical leg with 4 units length [shown bold in the diagram].

The connection is: if you choose a diagonal of a pXq rectangle in the original grid, and make a square from four such lines, and make a "secondary grid" of such squares, and in this "secondary grid" also mark a pXq rectangle, the diagonals of this rectangle will produce the phenomenon. One of them will lie on a horizontal or vertical line of the original grid, the other will be tilted, so (having the same - integer - length) being able to make a Pythagorean triangle. The measures of this triangle will be identical to the ones we can obtain from the firstly mentioned formula, started with the p,q numbers.

If you really followed me by drawing on a paper, you might agree with me that a Babylonian more easily could get the result this way (which does not make sure that it really happened so).

I found this of interest, although of course it is nothing to do with Brights. Ansem may like to look at some more geometry of this type on my Knight's Tour Notes site [links updated here]:

• (Theory of moves)
• (Knight-move geometry)

The construction given by Ansem can be simplified a little, and presented in terms of "leapers" (otherwise known as generalised chess knights).

Take O as origin (0,0). A as point (p,0), B as point (p,q), so AB = q. And let the length of OB = h. By using the principle of similar triangles we will prove that h^2 = p^2 + q^2, the Pythagorean theorem.

Let the (p,q) leaper, starting from O, make p moves, reaching point C = (p^2, pq) at distance OC = ph. Now let the leaper, starting from C make q moves at right angles to OC, reaching D if it goes to the left, and reaching D' if it goes to the right. Then the triangles OAB and OCD are similar right angled triangles. OC = ph = h.OA and CD = qh = h.AB. Thus it follows that OD = h.OB = h.h = h^2.

The point D' lies on the line OA extended, that is on the x-axis. Its coordinates are D' = (p^2 + q^2, 0), the x-coordinate being equal to p horizontal steps of length p plus q horizontal steps of length q.

Finally, the two triangles OCD and OCD' are congruent, OD and OD' being the equal sides of an isosceles triangle, with OC its perpendicular bisector. So OD = OD', that is h^2 = p^2 + q^2. Thus we have proved the Pythagorean theorem (though only for the case when p and q are integers).

The coordinates of the point D are (p^2 - q^2, 2.p.q). Thus we also have the result that the sum of the squares of these coordinates equals p^2 + q^2. This is the method for generating Pythagorean triples that Ansem started with.

I enjoyed this, anyway!

Since I noticed that I can put drawings here, I present another possible way for the ancient Babylonian to discover the triplets. It's only a variant of the ones already mentioned by you and me, and maybe not the most probable scenario, but I like it the most.

You can cut from paper a number of r,s legged right-angled triangles; the ancient Babylonian mathematician (well, he might be a priest at the same time) could make them from a not yet dried clay tablet. He wondered that from these little triangles he might be able to construct a bigger triangle which has not only its legs but also its hypotenuse as a whole number. After some trying, he got this:

(in this case, from four 1,2 legged right triangles) The two kites put together this way look promising: it's a figure with all whole number long sides and some of them tilted. But how could he make it more smooth? He noticed, that where the two kites are put together, one section is r (here: 1) the other is s (here: 2) long. So, if he multiplies one kite by s (here: 2) and the other by r (here: 1, so it remains the same) the two kites give a better form:

from which he just must cut a piece to get a right triangle with all whole sides. And in a further post:

Here is another possibility to discover the (s^2 – r^2) : 2.r.s : (s^2 + r^2) method for creating Pythagorean triplets geometrically. It's a little bit un-Babylonian, more Greekish, but still a possible candidate as the first such method. Somebody just drew an ABC rectangular triangle, where the right angle was at point C, and AC was longer than BC.

(S)he just stated that the shorter leg BC be 1 unit long, and the longer AC leg x long. Then drew another right angled triangle from points A, B, and D, where D lies in a straight line with A and C, and the right angle is at B.

What then is the length of CD? The ABC and BDC triangles are similar because they have identical angles, so CD:BC=BC:AC so CD is 1/x long. Let's draw a circle around the ABD triangle, and sign the center of the triangle as point E.

As a result we can see a rectangular triangle (CBE) of which all three sides we can calculate. The leg BC is 1 long as we chose. The hypotenuse BE is a radius in the circle. The diameter is AD, which is AC=x plus CD=1/x, so the radius is (x + 1/x)/2 long. And the CE leg is the DE radius = (x + 1/x)/2 minus CD=1/x that is (x - 1/x)/2. So CBE is a (x - 1/x)/2 : 1: (x + 1/x)/2 rectangular triangle.

We can chose x to be a rational number let's say s/r. So, if we multiply our triagle with 2.r.s we can get from the (s/r-r/s)/2 : 1: (s/r+r/s)/2 triangle the (s^2 – r^2) : 2.r.s : (s^2 + r^2) triangle.

concerning van de Waerden's treatment of the problem Laszlo also wrote:

[Perhaps] van der Waerden thought that his (just one) proof for the method presented by him is so strong that even mentioning the possibility of the geometric discovery is needless. If this was the case I would say this proof is very weak. He just refers to the 11. line of the Plimpton 322 tablet. It contains either the 45,60,75 or the 3/4,1,5/4 triplet. We can't know because they used a 60 based ("sexagezimal") system with no "sexagezimal point" marked. I still agree the 3/4,1,5/4 interpretation is slightly more probable. (If indeed we must interpret it, because for the Babylonians only the proportions were important here.) But as much as this triplet (that is, that one of the numbers is 1) is in accordance with van der Waerden's supposed Babylonian method, so is my latest presented geometrical method, and God knows with how many more, not yet persented ones. Not to mention the possibility that the concrete numbers of the triplet have nothing to do with the circumstances of the discovery, but are the result of a later, unknown viewpoint.

In a later email Laszlo writes:

On the net I have found this [a PDF file, requiring Adobe Reader] - unfortunately no longer at the address cited: http://www.math.siu.edu/kocik/pracki/44Cliffpdf.pdf (Clifford algebra)

The author reduces the finding of Pythagorean triplets to finding a complex number with whole numbers as real part, imaginary part, and norm, that is, if we take the complex number as vector, whole numbers as the two components and the length of the vector. Since the norm of a complex product is the product of the norms of the factors, a square of a complex number will have a whole norm. (The latter parts of the document - unfortunately - are too complex for my knowledge.)

But this complex-number-approach is very closely related to mine, where from a diagonal of a rectangle (a complex number-vector) I create a "secondary grid", and using this grid as another coordinate-system, I draw another vector. Actually by this I multiply the two complex number-vectors, even if I never heard about complex numbers. Since I used the same vector in the secondary grid, I made a square of the original complex number-vector.

Similarly, when you take a 2,1 leaper knight and take a 2,1 journey two ways, you multiply the 2+i complex number with its complex conjugate in one case and with itself in the other case. The norms of the products are equal.

Laszlo had a website: http://www.zorpia.com/klacika - unfortunately no longer at that address.

4 and 14 August 2006

This is the tour sent by Mr Singh, together with a geometrical diagram:

1 126 191 196 5 122 187 200 9 120 183 202 11 118 181 204 192 195 2 125 188 199 6 121 184 201 10 119 182 203 12 117 127 190 193 4 197 186 123 8 211 18 111 176 113 180 205 14 194 3 128 189 124 7 198 185 110 175 210 17 208 13 116 179 129 254 63 68 171 252 21 70 173 212 19 112 177 114 15 206 64 67 130 253 22 69 172 251 20 109 174 209 16 207 178 115 255 132 65 62 249 170 71 24 213 168 107 26 215 166 105 28 66 61 256 131 72 23 250 169 108 25 214 167 106 27 216 165 133 244 47 76 147 248 45 88 149 232 43 92 151 228 29 104 60 77 146 245 46 73 148 233 44 89 150 229 42 93 164 217 243 134 75 48 247 144 87 50 231 154 91 40 227 152 103 30 78 59 246 145 74 49 254 143 90 39 230 153 94 41 218 163 135 242 55 82 139 238 51 86 155 224 35 96 159 226 31 102 58 79 138 239 54 83 142 235 38 97 158 225 34 95 162 219 241 136 81 56 237 140 85 52 223 156 99 36 221 160 101 32 80 57 240 137 84 53 236 141 98 37 222 157 100 33 220 161