Get Your Prox Off #2

Monday, 13th Jun, 2016 by Krilling for Company in Fractals, Geometry, Mathematics, Rep-Tiles and tagged bubble-sort, divide-and-discard, fractal, fractals, geometry, incomplete bubble-sort, L-tromino, math, mathematics, maths, nearest vertex, Programming, programming error, proximity fractals, random move, randomness, recreational math, recreational mathematics, recreational maths, rep-tile, Rep-Tiles, serendipitous, serendipity, square, Squares, tessellation, tessellations, vertex, vertices | Leave a comment

Serendipity is the art of making happy discoveries by accident. I made a mistake writing a program to create fractals and made the happy discovery of an attractive new fractal. And also of a new version of an attractive fractal I had seen before.

As I described in Get Your Prox Off, you can create a fractal by 1) moving a point towards a randomly chosen vertex of a polygon, but 2) forbidding a move towards the nearest vertex or the second-nearest vertex or third-nearest, and so on. If the polygon is a square, the four possible basic fractals look like this (note that the first fractal is also produced by banning a move towards a vertex that was chosen in the previous move):

v = 4, ban = prox(1)
(ban move towards nearest vertex)

v = 4, ban = prox(2)
(ban move towards second-nearest vertex)

v = 4, ban = prox(3)

v = 4, ban = prox(4)

This program has to calculate what might be called the order of proximity: that is, it creates an array of distances to each vertex, then sorts the array by increasing distance. I was using a bubble-sort, but made a mistake so that the program ran through the array only once and didn’t complete the sort. If this happens, the fractals look like this (note that vertex 1 is on the right, with vertex 2, 3 and 4 clockwise from it):

v = 4, ban = prox(1), sweep = 1

v = 4, ban = prox(2), sweep = 1

v = 4, ban = prox(3), sweep = 1

(Animated version of v4, ban(prox(3)), sw=1)

v = 4, ban = prox(4), sweep = 1

Note that in the last case, where ban = prox(4), a bubble-sort needs only one sweep to identify the most distant vertex, so the fractal looks the same as it does with a complete bubble-sort.

These new fractals looked interesting, so I had the idea of adjusting the number of sweeps in the incomplete bubble-sort: one sweep or two or three and so on (with enough sweeps, the bubble-sort becomes complete, but more sweeps are needed to complete a sort as the number of vertices increases). If there are two sweeps, then ban(prox(1)) and ban(prox(2)) look like this:

v = 4, ban = prox(1), sweep = 2

v = 4, ban = prox(2), sweep = 2

But the fractals produced by sweep = 2 for ban(prox(3)) and ban(prox(4)) are identical to the fractals produced by a complete bubble sort. Now, suppose you add a central point to the polygon and treat that as an additional vertex. If the bubble-sort is incomplete, a ban(prox(1)) fractal with a central point looks like this:

v = 4+c, ban = prox(1), sw = 1

v = 4+c, ban = prox(1), sw = 2

When sweep = 3, an attractive new fractal appears:

v = 4+c, ban = prox(1), sw = 3

v = 4+c, ban = prox(1), sw = 3 (animated)

If you ban two vertices, the nearest and second-nearest, i.e. ban(prox(1), prox(2)), a complete bubble-sort produces a familiar fractal:

v = 4+c, ban = prox(1), prox(2)

And here is ban(prox(2), prox(4)), with a complete bubble-sort:

v = 4, ban = prox(2), prox(4)

If the bubble-sort is incomplete, sweep = 1 and sweep = 2 produce these fractals for ban(prox(1), prox(2)):

v = 4, ban = prox(1), prox(2), sw = 1

v = 4, ban = prox(1), prox(2), sw = 2*

*The second of those fractals is identical to v = 4, ban(prox(2), prox(3)) with a complete bubble-sort.

Here is ban(prox(1), prox(5)) with a complete bubble-sort:

v = 4, ban = prox(1), prox(5)

Now try ban(prox(1), prox(5)) with an incomplete bubble-sort:

v = 4, ban = prox(1), prox(5), sw = 1

v = 4, ban = prox(1), prox(5), sw = 2

When sweep = 3, the fractal I had seen before appears:

v = 4, ban = prox(1), prox(5), sw = 3

v = 4, ban = prox(1), prox(5), sw = 3 (animated)

Where had I seen it before? While investigating this rep-tile (a shape that can be tiled with smaller versions of itself):

L-triomino rep-tile

L-triomino rep-tile (animated)

The rep-tile is technically called an L-triomino, because it looks like a capital L and is one of the two distinct shapes you can create by joining three squares at the edges. You can create fractals from an L-triomino by dividing it into four copies, discarding one of the copies, then repeating the divide-and-discard at smaller and smaller scales:

L-triomino fractal stage #1

L-triomino fractal stage #2

L-triomino fractal stage #3

L-triomino fractal stage #4

L-triomino fractal stage #5

$L-reptile_fractal_anim$

L-triomino fractal (animated)

$L-reptile_fractal_static$

L-triomino fractal (close-up)

And here’s part of the ban(prox(1), prox(5)) fractal for comparison:

So you can get to the same fractal (or versions of it), by two apparently different routes: random movement of a point inside a square or repeatedly dividing-and-discarding the sub-copies of an L-triomino. That’s serendipity!

Previously pre-posted:

• Get Your Prox Off

Performativizing the Polygonic

Tuesday, 17th May, 2016 by Krilling for Company in Fractals, Geometry, Mathematics and tagged animated gif, animated gifs, digits, fractal, fractal geometry, fractals, geometry, heptagons, hexagaon, Hexagons, math, mathematics, maths, Pentagons, performativizing the polygonic, Polygons, recreational math, recreational mathematics, recreational maths, Squares, triangle, Triangles | 4 Comments

Maths is a mountain: you can start climbing in different places and reach the same destination. There are many ways of proving the irrationality of √2 or the infinitude of the primes, for example. But you can also arrive at the same destination by accident. I’ve found that when I use different methods of creating fractals. The same fractals appear, because apparently different algorithms are actually the same underneath.

But different methods can create unique fractals too. I’ve found some new ones by using what might be called point-to-point recursion. For example, there are ten ways to select three vertices from the five vertices of a pentagon: (1, 2, 3), (1, 2, 4), (1, 2, 5), (1, 3, 4), (1, 3, 5), (1, 4, 5), (2, 3, 4), (2, 3, 5), (2, 4, 5), (3, 4, 5). Find the midpoint of the first three-point set, (1, 2, 3). Then select two vertices to go with this midpoint, creating a new three-point set, and find the midpoint again. And so on. The process looks like this, with the midpoints shown for all the three-point sets found at each stage:

vertices = 5, choose sets of 3 points, find mid-point of each

At stage 5, the fractal looks like this:

v = 5, p = 3

Note that when pixels are used again, the colour changes. That’s another interesting thing about maths: limits can sometimes produce deeper results. If these fractals were drawn at very high resolution, pixels would only be used once and the colour would never change. As it is, low resolution means that pixels are used again and again. But some are used more than others, which is why interesting colour effects appear.

If the formation of the fractal is animated, it looks like this (with close-ups of even deeper stages):

Here are some more examples:

v = 4 + central point, p = 2 (cf. Fingering the Frigit)

v = 4c, p = 2 (animated)

v = 4, p = 3

v = 5, p = 4

v = 5 + central point, p = 3

v = 5c, p = 4

v = 5c, p = 5

v = 6 + 1 point between each pair of vertices, p = 6

v = 6, p = 2

v = 6, p = 3

v = 6, p = 4

v = 6c, p = 2 (cf. Fingering the Frigit)

v = 6c, p = 3

v = 6c, p = 4

v = 7, p = 3

v = 7, p = 4

v = ,7 p = 5

v = 7c, p = 4

v = 3+1, p = 2

v = 3+1, p = 3

v = 3+1, p = 4

v = 3+2, p = 5

v = 3c+1, p = 2

v = 3c+1, p = 4

v = 3c, p = 2

v = 3c, p = 3

v = 4+1, p = 3

v = 4+1, p = 4

v = 4+1, p = 6

v = 4+1, p = 2

v = 4c+1, p = 4

v = 4c, p = 3

v = 5+1, p = 4 (and more)

v = 5, p = 2

Polymorphous Perverticity

Thursday, 28th Apr, 2016 by Krilling for Company in Fractals, Geometry, Mathematics and tagged animated gif, animated gifs, fractal, fractal geometry, fractals, geometry, Hexagons, math, mathematics, maths, Pentagons, polygon fractals, polygonal fractals, polygonic performativity, Polygons, premier purveyor, recreational math, recreational mathematics, recreational maths, Sierpiński pentagon, Sierpiński triangle, Squares, triangle, Triangles, vertex, vertices | Leave a comment

As I’ve explained before on Overlord of the Über-Feral, the planet’s premier purveyor of polygonic performativity (probably (possibly (perspectivistically))), it works with triangles and pentagons, but not with squares. And what is “it”? A simple procedure in which you create a polygon, choose a point inside it, then repeatedly move half-way towards a vertex chosen at random, marking each new position as you go.

When the polygon has three vertices, you get a Sierpiński triangle. When it has five, you get what might be called a Sierpiński pentagon. When it has four, you get nothing. Or rather: you get everything, because the whole interior of the square gradually fills with points. But, as I’ve also explained before, there’s a simple way to change this. You can adapt the procedure so that a vertex can’t be chosen twice in a row, and so on.

When the rule is “No vertex twice in a row”, you get this fractal (colours change as a pixel is selected again):

But you can also use what might be a vertex increment, or vi, whereby you disallow vertices that are next to the previously chosen vertex, or two positions away, and so on. When the rule is “No vertex twice in a row”, the disallowed vertex is (v + 0), that is, vi = 0. If vi = 2 and the rule is disallow(v + 2), this fractal appears (when vi = 1, there’s no fractal):

v = 4, vi = 2

You can extend these rules to apply not just to the previously chosen vertex, but also to the vertex chosen before that. Here are some fractals produced by the rule disallow(v[1] + vi[1], v[2] + vi[2]), where v[1] is the vertex previously chosen and v[2] is the vertex chosen before that:

v = 4, vi[1] = 1, vi[2] = 2

v = 4, vi[1] = 2, vi[2] = 0

v = 4, vi[1] = 2, vi[2] = 1

v = 4, vi[1] = 2, vi[2] = 2

And here are some fractals produced by the rule disallow(v[1] + vi[1], v[2] + vi[2], v[3] + vi[3]):

v = 4, vi[1] = 1, vi[2] = 1, vi[3] = 0

v = 4, vi[1] = 1, vi[2] = 1, vi[3] = 2

Applying these rules to pentagons rather than squares doesn’t produce such a dramatic difference, because the original procedure – choose any vertex at random, taking no account of previous choices – produces a fractal when v = 5, as noted above, but not when v = 4. Nevertheless, here are some fractals for v > 4:

v = 5, vi = 0

v = 5, vi = 1

v = 5, vi = 2

v = 5, vi[1] = 0, vi[2] = 0

v = 5, vi[1] = 1, vi[2] = 0

v = 5, vi[1] = 2, vi[2] = 0

v = 5, vi[1] = 1, vi[2] = 1

v = 5, vi[1] = 1, vi[2] = 1, vi[3] = 1

v = 5, vi = various

v = 6, vi = 1

Fingering the Frigit

Thursday, 24th Mar, 2016 by Krilling for Company in Biology, Fractals, Geometry, Mathematics and tagged anatomy, animated gif, animated gifs, base 5, base 6, digits, fingers, fractal, fractal anatomy, fractal biology, fractal geometry, fractals, frigit, frigits, geometry, heptagons, Hexagons, human anatomy, integers, math, mathematics, maths, nonagons, numbers as fractals, octagons, Pentagons, Polygons, recreational math, recreational mathematics, recreational maths, Squares, triangle, Triangles | Leave a comment

Fingers are fractal. Where a tree has a trunk, branches and twigs, a human being has a torso, arms and fingers. And human beings move in fractal ways. We use our legs to move large distances, then reach out with our arms over smaller distances, then move our fingers over smaller distances still. We’re fractal beings, inside and out, brains and blood-vessels, fingers and toes.

But fingers are fractal are in another way. A digit – digitus in Latin – is literally a finger, because we once counted on our fingers. And digits behave like fractals. If you look at numbers, you’ll see that they contain patterns that echo each other and, in a sense, recur on smaller and smaller scales. The simplest pattern in base 10 is (0, 1, 2, 3, 4, 5, 6, 7, 8, 9). It occurs again and again at almost very point of a number, like a ten-hour clock that starts at zero-hour:

0, 1, 2, 3, 4, 5, 6, 7, 8, 9…
10, 11, 12, 13, 14, 15, 16, 17, 18, 19…
200… 210… 220… 230… 240… 250… 260… 270… 280… 290…

These fractal patterns become visible if you turn numbers into images. Suppose you set up a square with four fixed points on its corners and a fixed point at its centre. Let the five points correspond to the digits (1, 2, 3, 4, 5) of numbers in base 6 (not using 0, to simplify matters):

1, 2, 3, 4, 5, 11, 12, 13, 14, 15, 21, 22, 23, 24, 25, 31, 32, 33, 34, 35, 41, 42, 43, 44, 45, 51, 52, 53, 54, 55, 61, 62, 63, 64, 65… 2431, 2432, 2433, 2434, 2435, 2441, 2442, 2443, 2444, 2445, 2451, 2452…

Move between the five points of the square by stepping through the individual digits of the numbers in the sequence. For example, if the number is 2451, the first set of successive digits is (2, 4), so you move to a point half-way between point 2 and point 4. Next come the successive digits (4, 5), so you move to a point half-way between point 4 and point 5. Then come (5, 1), so you move to a point half-way between point 5 and point 1.

When you’ve exhausted the digits (or frigits) of a number, mark the final point you moved to (changing the colour of the pixel if the point has been occupied before). If you follow this procedure using a five-point square, you will create a fractal something like this:
$fractal4_1single$

$fractal4_1$
A pentagon without a central point using numbers in a zero-less base 7 looks like this:
$fractal5_0single$

$fractal5_0$
A pentagon with a central point looks like this:
$fractal5_1single$

$fractal5_1$
Hexagons using a zero-less base 8 look like this:
$fractal6_1single$

$fractal6_1$

$fractal6_0single$

$fractal6_0$
But the images above are just the beginning. If you use a fixed base while varying the polygon and so on, you can create images like these (here is the program I used):
$fractal4$

$fractal5$

$fractal6789$

Get Your Prox Off

Thursday, 8th Oct, 2015 by Krilling for Company in Fractals, Geometry, Mathematics and tagged fractal, fractals, geometry, math, mathematics, maths, pentagonal fractal, pentagonal fractals, Pentagons, recreational math, recreational mathematics, recreational maths, Squares, Triangles | Leave a comment

Create a triangle. Find a point somewhere inside it. Choose a corner at random and move halfway towards it. Mark the new point. Repeat the procedure: choose, move, mark. Repeat again and again. In time, a fractal will appear:

However, if you try the same thing with a square – choose a corner at random, move halfway towards it, mark the new point, repeat – no fractal appears. Instead, the points fill the interior of the square:

But what happens if you impose restrictions on the randomly chosen corner (or chorner)? Suppose you can’t choose the same corner twice in a row. If this rule is applied to the square, this fractal appears:

Now apply the no-corner-twice-in-a-row rule to a square that contains a central chorner. This fractal appears:

And if the rule is that you can choose a corner twice in a row but not thrice? This fractal appears:

Here is the rule is that a corner can’t be chosen if it was chosen two moves ago:

But what if the restriction is based not on how often or when a corner is chosen, but on its proximity, i.e. how near it is to the marked point? If the nearest corner can’t be chosen, the result is the same as the no-corner-twice-in-a-row rule:

But if the second-nearest corner can’t be chosen, this fractal appears:

This is the fractal when the third-nearest corner can’t be chosen:

And this is the fractal when the fourth-nearest, or most distant, corner can’t be chosen:

Here are the same restrictions applied to a pentagon:

Nearest corner forbidden

Second-nearest corner forbidden

Third corner forbidden

Fourth corner forbidden

Fifth corner forbidden

Fifth corner forbidden (animated)

And a pentagon with a central chorner:

Now try excluding more than one corner. Here are pentagons excluding the n-nearest and n+1-nearest corners (for example, the nearest and second-nearest corners; the second-nearest and third-nearest; and so on):

But what if the moving point is set equal to the n-nearest corner before it moves again? If the corner is the second-nearest and the shape is a triangle with a central chorner, this is the fractal that appears:

Animated version

And here is the same rule applied to various n-nearest corners in a pentagon:

Boldly Breaking the Boundaries

Thursday, 21st May, 2015 by Krilling for Company in Fractals, Geometry, Mathematics, Rep-Tiles and tagged block fractals, divide-and-discard, fractal, fractal geometry, fractals, fractals based on squares, geometry, hexagonal fractal, math, mathematics, maths, much in little, over-fractals, recreational math, recreational mathematics, recreational maths, square, Squares | Leave a comment

In “M.I.P. Trip”, I looked at fractals like this, in which a square is divided repeatedly into a pattern of smaller squares:

As you can see, the sub-squares appear within the bounds of the original square. But what if some of the sub-squares appear beyond the bounds of the original square? Then a new family of fractals is born, the over-fractals:

Magistra Rules the Waves

Monday, 30th Mar, 2015 by Krilling for Company in Base Behavior, Binary numbers, Digit-sums, Integer Sequences, Magistra Mundi, Mathematics and tagged base 10, base 13, base 16, base 17, base 2, base 25, base 33, base 49, base 9, digit sum, digit sums, integer patterns, Integer Sequences, integer spirals, integers, math, mathematics, maths, number patterns, primes, randomness, recreational math, recreational mathematics, recreational maths, regularity, spiral of integers, Square Numbers, Squares, Ulam Spirals | Leave a comment

One of my favourite integer sequences has the simple formula n(i) = n(i-1) + digitsum(n(i-1)). If it’s seeded with 1, its first few terms go like this:

n(1) = 1
n(2) = n(1) + digitsum(n(1)) = 1 + digitsum(1) = 2
n(3) = 2 + digitsum(2) = 4
n(4) = 4 + digitsum(4) = 8
n(5) = 8 + digitsum(8) = 16
n(6) = 16 + digitsum(16) = 16 + 1+6 = 16 + 7 = 23
n(7) = 23 + digitsum(23) = 23 + 2+3 = 23 + 5 = 28
n(8) = 28 + digitsum(28) = 28 + 2+8 = 28 + 10 = 38

As a sequence, it looks like this:

1, 2, 4, 8, 16, 23, 28, 38, 49, 62, 70, 77, 91, 101, 103, 107, 115, 122, 127, 137, 148, 161, 169, 185, 199, 218, 229, 242, 250, 257, 271, 281, 292, 305, 313, 320, 325, 335, 346, 359, 376, 392, 406, 416, 427, 440, 448, 464, 478, 497, 517, 530, 538, 554, 568, 587, 607, 620, 628, 644, 658, 677, 697, 719, 736, 752, 766, 785, 805, 818, 835, 851, 865, 884, 904, 917, 934, 950, 964, 983, 1003…

Given a number at random, is there a quick way to say whether it appears in the sequence seeded with 1? Not that I know, with one exception. If the number is divisible by 3, it doesn’t appear, at least in base 10. In base 2, that rule doesn’t apply:

n(1) = 1
n(2) = 1 + digitsum(1) = 10 = 1 + 1 = 2
n(3) = 10 + digitsum(10) = 10 + 1 = 11 = 2 + 1 = 3
n(4) = 11 + digitsum(11) = 11 + 1+1 = 101 = 3 + 2 = 5
n(5) = 101 + digitsum(101) = 101 + 1+0+1 = 111 = 5 + 2 = 7
n(6) = 111 + digitsum(111) = 111 + 11 = 1010 = 7 + 3 = 10
n(7) = 1010 + digitsum(1010) = 1010 + 10 = 1100 = 10 + 2 = 12
n(8) = 1100 + digitsum(1100) = 1100 + 10 = 1110 = 12 + 2 = 14

1, 2, 3, 5, 7, 10, 12, 14, 17, 19, 22, 25, 28, 31, 36, 38, 41, 44, 47, 52, 55, 60, 64, 65, 67, 70, 73, 76, 79, 84, 87, 92, 96, 98, 101, 105, 109, 114, 118, 123, 129, 131, 134, 137, 140, 143, 148, 151, 156, 160, 162, 165, 169, 173, 178, 182, 187, 193, 196, 199, 204, 208, 211, 216, 220, 225, 229, 234, 239, 246, 252, 258, 260, 262, 265, 268, 271, 276, 279, 284, 288, 290, 293, 297, 301, 306, 310, 315, 321, 324, 327, 332, 336, 339, 344, 348, 353, 357, 362, 367, 374…

What patterns are there in these sequences? It’s easier to check when they’re represented graphically, so I converted them into patterns à la the Ulam spiral, where n is represented as a dot on a spiral of integers. This is the spiral for base 10:

Base 10

And these are the spirals for bases 2 and 3:

Base 2

Base 3

These sequences look fairly random to me: there are no obvious patterns in the jumps from n(i) to n(i+1), i.e. in the values for digitsum(n(i)). Now try the spirals for bases 9 and 33:

Base 9

Base 33

Patterns have appeared: there is some regularity in the jumps. You can see these regularities more clearly if you represent digitsum(n(i)) as a graph, with n(i) on the x axis and digitsum(n(i)) on the y axis. If the graph starts with n(i) = 1 on the lower left and proceeds left-right, left-right up the screen, it looks like this in base 10:

Base 10 (click to enlarge)

Here are bases 2 and 3:

Base 2

Base 3

The jumps seem fairly random. Now try bases 9, 13, 16, 17, 25, 33 and 49:

Base 9

Base 13

Base 16

Base 17

Base 25

Base 33

Base 49

In some bases, the formula n(i) = n(i-1) + digitsum(n(i-1)) generates mild randomness. In others, it generates strong regularity, like waves rolling ashore under a steady wind. I don’t understand why, but regularity seems to occur in bases that are one more than a power of 2 and also in some bases that are primes or squares.

Elsewhere other-posted:

• Mathematica Magistra Mundi

M.i.P. Trip

Thursday, 8th Jan, 2015 by Krilling for Company in Fractals, Geometry, Mathematics and tagged animated gif, animated gifs, divide-and-discard, fractal, fractals, fractals based on squares, geometry, hexagon, hexagonal fractal, math, mathematics, maths, much in little, multum in parvo, recreational math, recreational mathematics, recreational maths, square, Squares | Leave a comment

The Latin phrase multum in parvo means “much in little”. It’s a good way of describing the construction of fractals, where the application of very simple rules can produce great complexity and beauty. For example, what could be simpler than dividing a square into smaller squares and discarding some of the smaller squares?

Yet repeated applications of divide-and-discard can produce complexity out of even a 2×2 square. Divide a square into four squares, discard one of the squares, then repeat with the smaller squares, like this:

Increase the sides of the square by a little and you increase the number of fractals by a lot. A 3×3 square yields these fractals:

And the 4×4 and 5×5 fractals yield more:

N-route

Tuesday, 30th Sep, 2014 by Krilling for Company in Binary numbers, Digit-sums, Geometry, Mathematics and tagged Arithmetic, binary, binary numbers, digit sums, digits, integer, integer spiral, integers, math, mathematics, maths, number spiral, recreational math, recreational mathematics, recreational maths, rectangles, Squares, ternary, ternary numbers | Leave a comment

In maths, one thing leads to another. I wondered whether, in a spiral of integers, any number was equal to the digit-sum of the numbers on the route traced by moving to the origin first horizontally, then vertically. To illustrate the procedure, here is a 9×9 integer spiral containing 81 numbers:

| 65 | 64 | 63 | 62 | 61 | 60 | 59 | 58 | 57 |
| 66 | 37 | 36 | 35 | 34 | 33 | 32 | 31 | 56 |
| 67 | 38 | 17 | 16 | 15 | 14 | 13 | 30 | 55 |
| 68 | 39 | 18 | 05 | 04 | 03 | 12 | 29 | 54 |
| 69 | 40 | 19 | 06 | 01 | 02 | 11 | 28 | 53 |
| 70 | 41 | 20 | 07 | 08 | 09 | 10 | 27 | 52 |
| 71 | 42 | 21 | 22 | 23 | 24 | 25 | 26 | 51 |
| 72 | 43 | 44 | 45 | 46 | 47 | 48 | 49 | 50 |
| 73 | 74 | 75 | 76 | 77 | 78 | 79 | 80 | 81 |

Take the number 21, which is three places across and up from the bottom left corner of the spiral. The route to the origin contains the numbers 21, 22, 23, 8 and 1, because first you move right two places, then up two places. And 21 is what I call a route number, because 21 = 3 + 4 + 5 + 8 + 1 = digitsum(21) + digitsum(22) + digitsum(23) + digitsum(8) + digitsum(1). Beside the trivial case of 1, there are two more route numbers in the spiral:

58 = 13 + 14 + 6 + 7 + 7 + 6 + 4 + 1 = digitsum(58) + digitsum(59) + digitsum(60) + digitsum(61) + digitsum(34) + digitsum(15) + digitsum(4) + digitsum(1).

74 = 11 + 12 + 13 + 14 + 10 + 5 + 8 + 1 = digitsum(74) + digitsum(75) + digitsum(76) + digitsum(77) + digitsum(46) + digitsum(23) + digitsum(8) + digitsum(1).

Then I wondered about other possible routes to the origin. Think of the origin as one corner of a rectangle and the number being tested as the diagonal corner. Suppose that you always move away from the starting corner, that is, you always move up or right (or up and left, and so on, depending on where the corners lie). In a x by y rectangle, how many routes are there between the diagonal corners under those conditions?

It’s an interesting question, but first I’ve looked at the simpler case of an n by n square. You can encode each route as a binary number, with 0 representing a vertical move and 1 representing a horizontal move. The problem then becomes equivalent to finding the number of distinct ways you can arrange equal numbers of 1s and 0s. If you use this method, you’ll discover that there are two routes across the 2×2 square, corresponding to the binary numbers 01 and 10:

Across the 3×3 square, there are six routes, corresponding to the binary numbers 0011, 0101, 0110, 1001, 1010 and 1100:

Across the 4×4 square, there are twenty routes:

(Please open in new window if it fails to animate)

Across the 5×5 square, there are 70 routes:

(Please open in new window etc)

Across the 6×6 and 7×7 squares, there are 252 and 924 routes:

After that, the routes quickly increase in number. This is the list for n = 1 to 14:

1, 2, 6, 20, 70, 252, 924, 3432, 12870, 48620, 184756, 705432, 2704156, 10400600… (see A000984 at the Online Encyclopedia of Integer Sequences)

After that you can vary the conditions. What if you can move not just vertically and horizontally, but diagonally, i.e. vertically and horizontally at the same time? Now you can encode the route with a ternary number, or number in base 3, with 0 representing a vertical move, 1 a horizontal move and 2 a diagonal move. As before, there is one route across a 1×1 square, but there are three across a 2×2, corresponding to the ternary numbers 01, 2 and 10:

There are 13 routes across a 3×3 square, corresponding to the ternary numbers 0011, 201, 021, 22, 0101, 210, 1001, 120, 012, 102, 0110, 1010, 1100:

And what about cubes, hypercubes and higher?

Fractal Fourmulas

Monday, 25th Nov, 2013 by Krilling for Company in Fractals, Geometry, Mathematics and tagged animated fractal, animated fractals, animated gif, animated gifs, fractal, fractal method, fractal methods, fractals, geometric, geometry, math, mathematics, maths, recreational math, recreational mathematics, recreational maths, right triangle, right triangles, square, Squares, triangle, Triangles | Leave a comment

A square can be divided into four right triangles. A right triangle can be divided into a square and two more right triangles. These simple rules, applied again and again, can be used to create fractals, or shapes that echo themselves on smaller and smaller scales.

Overlord In Terms of Core Issues Around Maximal Engagement with Key Notions of the Über-Feral

მეფე (2) • მსოფლიოს (3) • მათემატიკა (5) •

Tag Archives: Squares