Sliv and Let Tri

Fluvius, planus et altus, in quo et agnus ambulet et elephas natet,” wrote Pope Gregory the Great (540-604). “There’s a river, wide and deep, where a lamb may wade and an elephant swim.” He was talking about the Word of God, but you can easily apply his words to mathematics. However, in the river of mathematics, the very shallow and the very deep are often a single step apart.

Here’s a good example. Take the integer 2. How many different ways can it be represented as an sum of separate integers? Easy. First of all it can be represented as itself: 2 = 2. Next, it can be represented as 2 = 1 + 1. And that’s it. There are two partitions of 2, as mathematicians say:

2 = 2 = 1+1 (p=2)


Now try 3, 4, 5, 6:

3 = 3 = 1+2 = 1+1+1 (p=3)
4 = 4 = 1+3 = 2+2 = 1+1+2 = 1+1+1+1 (p=5)
5 = 5 = 1+4 = 2+3 = 1+1+3 = 1+2+2 = 1+1+1+2 = 1+1+1+1+1 (p=7)
6 = 6 = 1+5 = 2+4 = 3+3 = 1+1+4 = 1+2+3 = 2+2+2 = 1+1+1+3 = 1+1+2+2 = 1+1+1+1+2 = 1+1+1+1+1+1 (p=11)


So the partitions of 2, 3, 4, 5, 6 are 2, 3, 5, 7, 11. That’s interesting — the partition-counts are the prime numbers in sequence. So you might conjecture that p(7) = 13 and p(8) = 17. Alas, you’d be wrong. Here are the partitions of n = 1..10:

1 = 1 (p=1)
2 = 2 = 1+1 (p=2)
3 = 3 = 1+2 = 1+1+1 (p=3)
4 = 4 = 1+3 = 2+2 = 1+1+2 = 1+1+1+1 (p=5)
5 = 5 = 1+4 = 2+3 = 1+1+3 = 1+2+2 = 1+1+1+2 = 1+1+1+1+1 (p=7)
6 = 6 = 1+5 = 2+4 = 3+3 = 1+1+4 = 1+2+3 = 2+2+2 = 1+1+1+3 = 1+1+2+2 = 1+1+1+1+2 = 1+1+1+1+1+1 (p=11)
7 = 7 = 1+6 = 2+5 = 3+4 = 1+1+5 = 1+2+4 = 1+3+3 = 2+2+3 = 1+1+1+4 = 1+1+2+3 = 1+2+2+2 = 1+1+1+1+3 = 1+1+1+2+2 = 1+1+1+1+1+2 = 1+1+1+1+1+1+1 (p=15)
8 = 8 = 1+7 = 2+6 = 3+5 = 4+4 = 1+1+6 = 1+2+5 = 1+3+4 = 2+2+4 = 2+3+3 = 1+1+1+5 = 1+1+2+4 = 1+1+3+3 = 1+2+2+3 = 2+2+2+2 = 1+1+1+1+4 = 1+1+1+2+3 = 1+1+2+2+2 = 1+1+1+1+1+3 = 1+1+1+1+2+2 = 1+1+1+1+1+1+2 = 1+1+1+1+1+1+1+1 (p=22)
9 = 9 = 1+8 = 2+7 = 3+6 = 4+5 = 1+1+7 = 1+2+6 = 1+3+5 = 1+4+4 = 2+2+5 = 2+3+4 = 3+3+3 = 1+1+1+6 = 1+1+2+5 = 1+1+3+4 = 1+2+2+4 = 1+2+3+3 = 2+2+2+3 = 1+1+1+1+5 = 1+1+1+2+4 = 1+1+1+3+3 = 1+1+2+2+3 = 1+2+2+2+2 = 1+1+1+1+1+4 = 1+1+1+1+2+3 = 1+1+1+2+2+2 = 1+1+1+1+1+1+3 = 1+1+1+1+1+2+2 = 1+1+1+1+1+1+1+2 = 1+1+1+1+1+1+1+1+1 (p=30)
10 = 10 = 1+9 = 2+8 = 3+7 = 4+6 = 5+5 = 1+1+8 = 1+2+7 = 1+3+6 = 1+4+5 = 2+2+6 = 2+3+5 = 2+4+4 = 3+3+4 = 1+1+1+7 = 1+1+2+6 = 1+1+3+5 = 1+1+4+4 = 1+2+2+5 = 1+2+3+4 = 1+3+3+3 = 2+2+2+4 = 2+2+3+3 = 1+1+1+1+6 = 1+1+1+2+5 = 1+1+1+3+4 = 1+1+2+2+4 = 1+1+2+3+3 = 1+2+2+2+3 = 2+2+2+2+2 = 1+1+1+1+1+5 = 1+1+1+1+2+4 = 1+1+1+1+3+3 = 1+1+1+2+2+3 = 1+1+2+2+2+2 = 1+1+1+1+1+1+4 = 1+1+1+1+1+2+3 = 1+1+1+1+2+2+2 = 1+1+1+1+1+1+1+3 = 1+1+1+1+1+1+2+2 = 1+1+1+1+1+1+1+1+2 = 1+1+1+1+1+1+1+1+1+1 (p=42)


It’s very simple to understand what a partition is, but very difficult to say how many partitions, p(n), a particular number will have. Here’s a partition: 11 = 4 + 3 + 2 + 2. But what is p(11)? Is there a formula for the sequence of p(n)?

1, 2, 3, 5, 7, 11, 15, 22, 30, 42, 56, 77, 101, 135, 176, 231, 297, 385, 490, 627, 792, 1002, 1255, 1575, 1958, 2436, 3010, 3718, 4565, 5604, 6842, 8349, 10143, 12310, 14883, 17977, 21637, 26015, 3118 5, 37338, 44583, 53174, 63261... (A000041 at the OEIS)

Yes, there is a formula, but it is very difficult to understand the Partition function that supplies it. So that part of the river of mathematics is very deep. But a step away the river of mathematics is very shallow. Here’s another question: If you multiply the numbers in a partition of n, what’s the largest possible product? Try using the partitions of 5:

4 = 1 * 4
6 = 2 * 3
3 = 1 * 1 * 3
4 = 1 * 2 * 2
2 = 1 * 1 * 1 * 2
1 = 1 * 1 * 1 * 1 * 1

The largest product is 6 = 2 * 3. So the answer is easy for n = 5, but I assumed that as n got bigger, the largest product got more interesting, using a subtler and subtler mix of prime factors. I was wrong. You don’t have to struggle to find a formula for what you might call the maximum multiplicity of the partitions of n:

1 = 1 (n=1)
2 = 2 (n=2)
3 = 3 (n=3)
4 = 2 * 2 (n=4)
6 = 2 * 3 (n=5)
9 = 3 * 3 (n=6)
12 = 2 * 2 * 3 (n=7)
18 = 2 * 3 * 3 (n=8)
27 = 3 * 3 * 3 (n=9)
36 = 2 * 2 * 3 * 3 (n=10)
54 = 2 * 3 * 3 * 3 (n=11)
81 = 3 * 3 * 3 * 3 (n=12)
108 = 2 * 2 * 3 * 3 * 3 (n=13)
162 = 2 * 3 * 3 * 3 * 3 162(n=14)
243 = 3 * 3 * 3 * 3 * 3 (n=15)
324 = 2 * 2 * 3 * 3 * 3 * 3 (n=16)
486 = 2 * 3 * 3 * 3 * 3 * 3 (n=17)
729 = 3 * 3 * 3 * 3 * 3 * 3 (n=18)


It’s easy to see why the greatest prime factor is always 3. If you use 5 or 7 as a factor, the product can always be beaten by splitting the 5 into 2*3 or the 7 into 2*2*3:

15 = 3 * 5 < 18 = 3 * 2*3 (n=8)
14 = 2 * 7 < 24 = 2 * 2*2*3 (n=9)
35 = 5 * 7 < 72 = 2*3 * 2*2*3 (n=12)

And if you’re using 7 → 2*2*3 as factors, you can convert them to 1*3*3, then add the 1 to another factor to make a bigger product still:

14 = 2 * 7 < 24 = 2 * 2*2*3 < 27 = 3 * 3 * 3 (n=9)
35 = 5 * 7 < 72 = 2*3 * 2*2*3 < 81 = 3 * 3 * 3 * 3 (n=12)


Post-Performative Post-Scriptum

The title of this post is, of course, a paronomasia on core Beatles album Live and Let Die (1954). But what does it mean? Well, if you think of the partitions of n as slivers of n, then you sliv n to find its partitions:

9 = 9 = 1+8 = 2+7 = 3+6 = 4+5 = 1+1+7 = 1+2+6 = 1+3+5 = 1+4+4 = 2+2+5 = 2+3+4 = 3+3+3 = 1+1+1+6 = 1+1+2+5 = 1+1+3+4 = 1+2+2+4 = 1+2+3+3 = 2+2+2+3 = 1+1+1+1+5 = 1+1+1+2+4 = 1+1+1+3+3 = 1+1+2+2+3 = 1+2+2+2+2 = 1+1+1+1+1+4 = 1+1+1+1+2+3 = 1+1+1+2+2+2 = 1+1+1+1+1+1+3 = 1+1+1+1+1+2+2 = 1+1+1+1+1+1+1+2 = 1+1+1+1+1+1+1+1+1 (p=30)

And when you find the greatest product among those partitions, you let 3 or “tri” work its multiplicative magic. So you “Sliv and Let Tri”:

8 = 1 * 8
14 = 2 * 7
18 = 3 * 6
20 = 4 * 5
7 = 1 * 1 * 7
12 = 1 * 2 * 6
15 = 1 * 3 * 5
16 = 1 * 4 * 4
20 = 2 * 2 * 5
24 = 2 * 3 * 4
27 = 3 * 3 * 3 ←
6 = 1 * 1 * 1 * 6
10 = 1 * 1 * 2 * 5
12 = 1 * 1 * 3 * 4
16 = 1 * 2 * 2 * 4
12 = 1 * 2 * 3 * 3
24 = 2 * 2 * 2 * 3
5 = 1 * 1 * 1 * 1 * 5
8 = 1 * 1 * 1 * 2 * 4
9 = 1 * 1 * 1 * 3 * 3
12 = 1 * 1 * 2 * 2 * 3
16 = 1 * 2 * 2 * 2 * 2
4 = 1 * 1 * 1 * 1 * 1 * 4
6 = 1 * 1 * 1 * 1 * 2 * 3
8 = 1 * 1 * 1 * 2 * 2 * 2
3 = 1 * 1 * 1 * 1 * 1 * 1 * 3
4 = 1 * 1 * 1 * 1 * 1 * 2 * 2
2 = 1 * 1 * 1 * 1 * 1 * 1 * 1 * 2
1 = 1 * 1 * 1 * 1 * 1 * 1 * 1 * 1 * 1

Sampled (Underfoot)

Some interesting statistics from the American sociologist Elizabeth Wrigley-Field:

Here are three puzzles.

• American fertility fluctuated dramatically in the decades surrounding the Second World War. Parents created the smallest families during the Great Depression, and the largest families during the postwar Baby Boom. Yet children born during the Great Depression came from larger families than those born during the Baby Boom. How can this be?

• About half of the prisoners released in any given year in the United States will end up back in prison within five years. Yet the proportion of prisoners ever released who will ever end up back in prison, over their whole lifetime, is just one third. How can this be?

• People whose cancers are caught early by random screening often live longer than those whose cancers are detected later, after they are symptomatic. Yet those same random screenings might not save any lives. How can this be?

And here is a twist: these are all the same puzzle.

• Answers here: Length-Biased Sampling by Elizabeth Wrigley-Field


Proxi-Performative Post-Scriptum

The title of this post is, of course, a radical reference to core Led Zeppelin track “Trampled Underfoot” (1975).

Dime Time

Everyone knows the shapes for one and two dimensions, far fewer know the shapes for three and four dimensions, let alone five, six and seven. And what shapes are those? The shapes that answer this question:

• How many equidistant points are possible in 1d, 2d, 3d, 4d…?

In one dimension it’s obvious that the answer is 2. In other words, you can get only two equidistant points, (a,b), on a straight line. Point a must be as far from point b as point b is from point a. You can’t add a third point, c, such that (a,b,c) are equidistant. Not on a straight line in 1d. But suppose you bend the line into a circle, so that you’re working in two dimensions. It’s easy to place three equidistant points, (a,b,c), on a circle.

equidistant points on a circle

Three equidistant points around a circle forming the vertices of an equilateral triangle


And it’s also easy to see that the three points will form the vertices of an equilateral triangle. Now try adding a fourth point, d. If you place it in the center of the triangle, it will be equidistant from (a,b,c), but it will be nearer to (a,b,c) than they are to each other. So you can have only 2 equidistant points in 1d and 3 equidistant points in 2d.

But what are the co-ordinates of the equidistant points in 1d and 2d? Suppose (a,b) in 1d are given the co-ordinates (0) and (1), so that a is 1 unit distant from b. When you move to 2d and add point c, the co-ordinates for (a,b) become (0,0) and (1,0). They’re still 1 unit distant from each other. But what are the co-ordinates for c? Start by placing c exactly midway between a and b, so that it has the co-ordinates (0.5,0) and is 0.5 units distant from both a and b. Now, if you move c in the first dimension, it will become nearer either to a or b: (0.49,0) or (0.51,0) or (0.48,0) or (0.52,0)…

But if you move c in the second dimension, it will always be equidistant from a and b, because (a,b) stay in the first dimension, as it were, and c moves equally away from both into the second dimension. So where in 2d will c be 1 unit distant from both a and b just as a and b are 1 unit distant from each other in 1d? You can see the answer here:

equilateral_triangle heightHeight of an equilateral triangle


The co-ordinates for c are (0.5,√3/2) or (0.5,0.8660254…), because the second co-ordinate satisfies the Pythagorean equation 1^2 = 0.5^2 + (√3/2)^2 = 0.25 + 0.75. That is, to find the second co-ordinate of c for 2d, you find the answer to √(1 – 0.5^2) = √(1-0.25) = √0.75 = √(3/4) = √3/2 = 0.8660254….

But you can’t add a fourth point, d, in 2d such that (a,b,c,d) are equidistant. So let’s move to 3d for the points (a,b,c,d). Begin with point d in the center of the triangle formed by (a,b,c), where it will have the co-ordinates (0.5,√3/6,0) = (0.5,0.28867513…,0) and will be equidistant from (a,b,c). But d will be nearer to (a,b,c) than they are to each other. However, if you move d in the third dimension, it will be moving equally away from (a,b,c). So where in 3d will d be 1 unit from (a,b,c)? By analogy with 2d, the third co-ordinate for d will satisfy the generalized Pythagorean equation √(1 – 0.5^2 – (√3/6)^2). And √6/3 = √(1 – 0.5^2 – (√3/6)^2) = 0.81649658… So point d will have the co-ordinates (0.5,√3/6,√6/3) = (0.5, 0.288675135…, 0.816496581…).

And the four points (a,b,c,d) will be the vertices of a three-dimensional shape called the tetrahedron:

Rotating tetrahedron

Rotating tetrahedron


But you can’t add a fifth point, e, in 3d such that (a,b,c,d,e) are equidistant. So let’s move to 4d, the fourth dimension, for the points (a,b,c,d,e). Begin with point e in the center of the tetrahedron formed by (a,b,c,d), where it will have the co-ordinates (0.5,√3/6,√6/12,0) = (0.5,0.28867513…, 0.2041241…, 0) and will be equidistant from (a,b,c,d). But e will be nearer to (a,b,c,d) than they are to each other. However, if you move e in the fourth dimension, it will be moving equally away from (a,b,c,e). So where in 4d will e be 1 unit from (a,b,c,d)? By analogy with 2d and 3d, the co-ordinate for 4d will satisfy the equation √(1 – 0.5^2 – (√3/6)^2 – (√6/12)^2). And √10/4 = √(1 – 0.5^2 – (√3/6)^2 – (√6/12)^2) = 0.79056941… So point e will have the co-ordinates (0.5,√3/6,√6/3,√10/4) = (0.5, 0.288675135…, 0.816496581…, 0.79056941…).

And the five points (a,b,c,d,e) will be the vertices of a four-dimensional shape called variously the hyperpyramid, the 5-cell, the pentachoron, the 4-simplex, the pentatope, the pentahedroid and the tetrahedral pyramid. It’s impossible for 3d creatures like human beings (at present) to visualize the hyperpyramid, but we can see its 3d shadow, as it were. And here is the 3d shadow of a rotating hyperpyramid:

Rotating hyperpyramid or 5-cell

Rotating hyperpyramid


N.B. Wikipedia reveals the mathematically beautiful fact that the “simplest set of coordinates [for a hyperpyramid] is: (2,0,0,0), (0,2,0,0), (0,0,2,0), (0,0,0,2), (φ,φ,φ,φ), with edge length 2√2, where φ is the golden ratio.”

So that’s the hyperpyramid, with 5 points in 4d. But you can’t add a sixth point, f, in 4d such that (a,b,c,d,e,f) are equidistant. You have to move to 5d. And it should be clear by now that in any dimension nd, the maximum possible number of equidistant points, p, in that dimension will be p = n+1. And here are the co-ordinates for p in dimensions 1 to 10 (the co-ordinates are given in full for 1d to 4d, then for 5d to 10d only the co-ordinates of the additional point are given):

d1: (0), (1)
d2: (0,0), (1,0), (0.5,0.866025404)
d3: (0,0,0), (1,0,0), (0.5,0.866025404,0), (0.5,0.288675135,0.816496581)
d4: 0.5, 0.288675135, 0.204124145, 0.790569415
d5: 0.5, 0.288675135, 0.204124145, 0.158113883, 0.774596669
d6: 0.5, 0.288675135, 0.204124145, 0.158113883, 0.129099445, 0.763762616
d7: 0.5, 0.288675135, 0.204124145, 0.158113883, 0.129099445, 0.109108945, 0.755928946
d8: 0.5, 0.288675135, 0.204124145, 0.158113883, 0.129099445, 0.109108945, 0.0944911183, 0.75
d9: 0.5, 0.288675135, 0.204124145, 0.158113883, 0.129099445, 0.109108945, 0.0944911183, 0.0833333333, 0.745355992
d10: 0.5, 0.288675135, 0.204124145, 0.158113883, 0.129099445, 0.109108945, 0.0944911183, 0.0833333333, 0.0745355992, 0.741619849

In each dimension d, the final co-ordinate, cd+1, of the additional point satisfies the generalized Pythagorean equation cd+1 = √(1 – c1^2 – c2^2 – … cd^2).


Readers’ advisory: I am not a mathematician and the discussion above cannot be trusted to be free of errors, whether major or minor.

Think Inc

This is a T-square fractal:

T-square fractal


Or you could say it’s a T-square fractal with the scaffolding taken away, because there’s nothing to show how it was made. And how is a T-square fractal made? There are many ways. One of the simplest is to set a point jumping 1/2 of the way towards one or another of the four vertices of a square. If the point is banned from jumping towards the vertex two places clockwise (or counter-clockwise) of the vertex, v[i=1..4], it’s just jumped towards, you get a T-square fractal by recording each spot where the point lands.

You also get a T-square if the point is banned from jumping towards the vertex most distant from the vertex, v[i], it’s just jumped towards. The most distant vertex will always be the diagonally opposite vertex, or the vertex, v[i+2], two places clockwise of v[i]. So those two bans are functionally equivalent.

But what if you don’t talk about bans at all? You can also create a T-square fractal by giving the point three choices of increment, [0,1,3], after it jumps towards v[i]. That is, it can jump towards v[i+0], v[i+1] or v[i+3] (where 3+2 = 5 → 5-4 = 1; 3+3 = 6 → 2; 4+1 = 5 → 1; 4+2 = 6 → 2; 4+3 = 7 → 3). Vertex v[i+0] is the same vertex, v[i+1] is the vertex one place clockwise of v[i], and v[i+3] is the vertex two places clockwise of v[i].

So this method is functionally equivalent to the other two bans. But it’s easier to calculate, because you can take the current vertex, v[i], and immediately calculate-and-use the next vertex, without having to check whether the next vertex is forbidden. In other words, if you want speed, you just have to Think Inc!

Speed becomes important when you add a new jumping-target to each side of the square. Now the point has 8 possible targets to jump towards. If you impose several bans on the next jump, e.g the point can’t jump towards v[i+2], v[i+3], v[i+5], v[i+6] and v[i+7], you will have to check for five forbidden targets. But using the increment-set [0,1,4] you don’t have to check for anything. You just inc-and-go:

inc = 0, 1, 4


Here are more fractals created with the speedy inc-and-go method:

inc = 0, 2, 3


inc = 0, 2, 5


inc = 0, 3, 4


inc = 0, 3, 5


inc = 1, 4, 7


inc = 2, 4, 7


inc = 0, 1, 4, 7


inc = 0, 3, 4, 5


inc = 0, 3, 4, 7


inc = 0, 4, 5, 7


inc = 1, 2, 6, 7


With more incs, there are more possible paths for the jumping point and the fractals become more “solid”:

inc = 0, 1, 2, 4, 5


inc = 0, 1, 2, 6, 7


inc = 0, 1, 3, 5, 7


Now try applying inc-and-go to a pentagon:

inc = 0, 1, 2

(open in new window if blurred)


inc = 0, 2, 3


And add a jumping-target to each side of the pentagon:

inc = 0, 2, 5


inc = 0, 3, 6


inc = 0, 3, 7


inc = 1, 5, 9


inc = 2, 5, 8


inc = 5, 6, 9


And add two jumping-targets to each side of the pentagon:

inc = 0, 1, 7


inc = 0, 2, 12


inc = 0, 3, 11


inc = 0, 3, 12


inc = 0, 4, 11


inc = 0, 5, 9


inc = 0, 5, 10


inc = 2, 7, 13


inc = 2, 11, 13


inc = 3, 11, 13


After the pentagon comes the hexagon:

inc = 0, 1, 2


inc = 0, 1, 5


inc = 0, 3, 4


inc = 0, 3, 5


inc = 1, 3, 5


inc = 2, 3, 4


Add a jumping-target to each side of the hexagon:

inc = 0, 2, 5


inc = 0, 2, 9


inc = 0, 6, 11


inc = 0, 3, 6


inc = 0, 3, 8


inc = 0, 3, 9


inc = 0, 4, 7


inc = 0, 4, 8


inc = 0, 5, 6


inc = 0, 5, 8


inc = 1, 5, 9


inc = 1, 6, 10


inc = 1, 6, 11


inc = 2, 6, 8


inc = 2, 6, 10


inc = 3, 5, 7


inc = 3, 6, 9


inc = 6, 7, 11


Boole(b)an #3

In the posts “Boole(b)an #1″ and “Boole(b)an #2” I looked at fractals created by certain kinds of ban on a point jumping (quasi-)randomly towards the four vertices, v=1..4, of a square. For example, suppose the program has a vertex-history of 2, that is, it remembers two jumps into the past, the previous jump and the pre-previous jump. There are sixteen possible combinations of pre-previous and previous jumps: [1,1], [1,2], 1,3] … [4,2], [4,3], [4,4].

Let’s suppose the program bans 4 of those 16 combinations by reference to the current possible jump. For example, it might ban [0,0]; [0,1]; [0,3]; [2,0]. To see what that means, let’s suppose the program has to decide at some point whether or not to jump towards v=3. It will check whether the combination of pre-previous and previous jumps was [3+0,3+0] = [3,3] or [3+0,3+1] = [3,4] or [3+0,3+3] = [3,2] or [3+2,3+0] = [1,3] (when the sum > 4, v = sum-4). If the previous combination is one of those pairs, it bans the jump towards v=3 and chooses another vertex; otherwise, it jumps towards v=3 and updates the vertex-history. This is the fractal that appears when those bans are used:

ban = [0,0]; [0,1]; [0,3]; [2,0]


And here are more fractals using a vertex-history of 2 and a ban on 4 of 16 possible combinations of pre-previous and previous jump:

ban = [0,0]; [0,1]; [0,3]; [2,2]


ban = [0,0]; [0,2]; [1,0]; [3,0]


ban = [0,0]; [0,2]; [1,1]; [3,3]


ban = [0,0]; [0,2]; [1,3]; [3,1]


ban = [0,0]; [1,0]; [2,2]; [3,0]


ban = [0,0]; [1,1]; [1,2]; [3,2]

Continue reading “Boole(b)an #3”…


Elsewhere other-engageable

Boole(b)an #1
Boole(b)an #2

Root Routes

Suppose a point traces all possible routes jumping half-way towards the three vertices of an equilateral triangle. A special kind of shape appears — a fractal called the Sierpiński triangle that contains copies of itself at smaller and smaller scales:

Sierpiński triangle, jump = 1/2


And what if the point jumps 2/3rds of the way towards the vertices as it traces all possible routes? You get this dull fractal:

Triangle, jump = 2/3


But if you add targets midway along each side of the triangle, you get this fractal with the 2/3rds jump:

Triangle, jump = 2/3, side-targets


Now try the 1/2-jump triangle with a target at the center of the triangle:

Triangle, jump = 1/2, central target


And the 2/3-jump triangle with side-targets and a central target:

Triangle, jump = 2/3, side-targets, central target


But why stop at simple jumps like 1/2 and 2/3? Let’s take the distance to the target, td, and use the function 1-(sqrt(td/7r)), where sqrt() is the square-root and 7r is 7 times the radius of the circumscribing circle:

Triangle, jump = 1-(sqrt(td/7r))


Here’s the same jump with a central target:

Triangle, jump = 1-(sqrt(td/7r)), central target


Now let’s try squares with various kinds of jump. A square with a 1/2-jump fills evenly with points:

Square, jump = 1/2 (animated)


The 2/3-jump does better with a central target:

Square, jump = 2/3, central target


Or with side-targets:

Square, jump = 2/3, side-targets


Now try some more complicated jumps:

Square, jump = 1-sqrt(td/7r)


Square, jump = 1-sqrt(td/15r), side-targets


And what if you ban the point from jumping twice or more towards the same target? You get this fractal:

Square, jump = 1-sqrt(td/6r), ban = prev+0


Now try a ban on jumping towards the target two places clockwise of the previous target:

Square, jump = 1-sqrt(td/6r), ban = prev+2


And the two-place ban with a central target:

Square, jump = 1-sqrt(td/6r), ban = prev+2, central target


And so on:

Square, jump = 1-sqrt(td/6.93r), ban = prev+2, central target


Square, jump = 1-sqrt(td/7r), ban = prev+2, central target


These fractals take account of the previous jump and the pre-previous jump:

Square, jump = 1-sqrt(td/4r), ban = prev+2,2, central target


Square, jump = 1-sqrt(td/5r), ban = prev+2,2, central target


Square, jump = 1-sqrt(td/6r), ban = prev+2,2, central target


Elsewhere other-accessible

Boole(b)an #2 — fractals created in similar ways

This Charming Dis-Arming

One of the charms of living in an old town or city is finding new routes to familiar places. It’s also one of the charms of maths. Suppose a three-armed star sprouts three half-sized arms from the end of each of its three arms. And then sprouts three quarter-sized arms from the end of each of its nine new arms. And so on. This is what happens:

Three-armed star


3-Star sprouts more arms


Sprouting 3-Star #3


Sprouting 3-Star #4


Sprouting 3-Star #5


Sprouting 3-Star #6


Sprouting 3-Star #7


Sprouting 3-Star #8


Sprouting 3-Star #9


Sprouting 3-Star #10


Sprouting 3-Star #11 — the Sierpiński triangle


Sprouting 3-star (animated)


The final stage is a famous fractal called the Sierpiński triangle — the sprouting 3-star is a new route to a familiar place. But what happens when you trying sprouting a four-armed star in the same way? This does:

Four-armed star #1


Sprouting 4-Star #2


Sprouting 4-Star #3


Sprouting 4-Star #4


Sprouting 4-Star #5


Sprouting 4-Star #6


Sprouting 4-Star #7


Sprouting 4-Star #8


Sprouting 4-Star #9


Sprouting 4-Star #10


Sprouting 4-star (animated)


There’s no obvious fractal with a sprouting 4-star. Not unless you dis-arm the 4-star in some way. For example, you can ban any new arm sprouting in the same direction as the previous arm:

Dis-armed 4-star (+0) #1


Dis-armed 4-Star (+0) #2


Dis-armed 4-Star (+0) #3


Dis-armed 4-Star (+0) #4


Dis-armed 4-Star (+0) #5


Dis-armed 4-Star (+0) #6


Dis-armed 4-Star (+0) #7


Dis-armed 4-Star (+0) #8


Dis-armed 4-Star (+0) #9


Dis-armed 4-Star (+0) #10


Dis-armed 4-star (+0) (animated)


Once again, it’s a new route to a familiar place (for keyly committed core components of the Overlord-of-the-Über-Feral community, anyway). Now try banning an arm sprouting one place clockwise of the previous arm:

Dis-armed 4-Star (+1) #1


Dis-armed 4-Star (+1) #2


Dis-armed 4-Star (+1) #3


Dis-armed 4-Star (+1) #4


Dis-armed 4-Star (+1) #5


Dis-armed 4-Star (+1) #6


Dis-armed 4-Star (+1) #7


Dis-armed 4-Star (+1) #8


Dis-armed 4-Star (+1) #9


Dis-armed 4-Star (+1) #10


Dis-armed 4-Star (+1) (animated)


Again it’s a new route to a familiar place. Now trying banning an arm sprouting two places clockwise (or anti-clockwise) of the previous arm:

Dis-armed 4-Star (+2) #1


Dis-armed 4-Star (+2) #2


Dis-armed 4-Star (+2) #3


Dis-armed 4-Star (+2) #4


Dis-armed 4-Star (+2) #5


Dis-armed 4-Star (+2) #6


Dis-armed 4-Star (+2) #7


Dis-armed 4-Star (+2) #8


Dis-armed 4-Star (+2) #9


Dis-armed 4-Star (+2) #10


Dis-armed 4-Star (+2) (animated)


Once again it’s a new route to a familiar place. And what happens if you ban an arm sprouting three places clockwise (or one place anti-clockwise) of the previous arm? You get a mirror image of the Dis-armed 4-Star (+1):

Dis-armed 4-Star (+3)


Here’s the Dis-armed 4-Star (+1) for comparison:

Dis-armed 4-Star (+1)


Elsewhere other-accessible

Boole(b)an #2 — other routes to the fractals seen above

Boole(b)an #2

In “Boole(b)an”, I looked at some of the things that happen when you impose bans of different kinds on a point jumping half-way towards a randomly chosen vertex of a square. If the point can’t jump towards the same vertex twice (or more) in a row, you get the fractal below (or rather, you get a messier version of the fractal below, because I’ve used an algorithm that finds all possible routes to create the fractals in this post):

ban = v(i) + 0


If the point can’t jump towards the vertex one place clockwise of the point it has just jumped towards, you get this fractal:

ban = v(i) + 1


If the point can’t jump towards the vertex two places clockwise (or anti-clockwise) of the point it has just jumped towards, you get this fractal:

ban = v(i) + 2


Finally, you get a mirror-image of the one-place-clockwise fractal when the ban is on jumping towards the vertex three places clockwise (or one place anti-clockwise) of the previous vertex:

ban = v(i) + 3


Now let’s introduce the concept of “vertex-history”. The four fractals above use a vertex-history of 1, vh = 1, because they look one step into the past, at the previously chosen vertex. Because there are four vertices, there are four possible previous vertices. But when vh = 2, you’re taking account of both the previous vertex, v(i), and what you might call the pre-previous vertex, v(i-1). There are sixteen possible combinations of previous vertex and pre-previous vertex (16 = 4 x 4).

Now, suppose the jump-ban is imposed when one of two conditions is met: the vertex is 1) one place clockwise of the previous vertex and the same as the pre-previous vertex; 2) three places clockwise of the previous chosen vertex and the same as the pre-previous vertex. So the boolean test is (condition(1) AND condition(2)) OR (condition(3) AND condition(4)). When you apply the test, you get this fractal:

ban = v(i,i-1) + [0,1] or v(i,i-1) + [0,3]


The fractal looks more complex, but I think it’s a blend of some combination of the four classic fractals shown at the beginning of this post. Here are more multiple-ban fractals using vh = 2 and bani = 2:

ban = [0,1] or [1,1]


ban = [0,2] or [2,0]


ban = [0,2] or [2,2]


ban = [1,0] or [3,0]


ban = [1,1] or [3,3]


ban = [1,2] or [2,2]


ban = [1,2] or [3,2]


ban = [1,3] or [2,0]


ban = [1,3] or [3,1]


ban = [2,0] or [2,2]


ban = [2,1] or [2,3]


For the fractals below, vh = 2 and bani = 3 (i.e., bans are imposed when one of three possible conditions is met). Again, I think the fractals are blends of some combination of the four classic ban-fractals shown at the beginning of this post:

ban = [0,0] or [1,2] or [3,2]


ban = [0,0] or [1,3] or [3,1]


ban = [0,0] or [2,1] or [2,3]


ban = [0,1] or [0,2] or [0,3]


ban = [0,1] or [0,3] or [1,1]


ban = [0,1] or [0,3] or [2,0]


ban = [0,1] or [0,3] or [2,2]


ban = [0,1] or [1,1] or [1,2]


ban = [0,1] or [1,1] or [3,0]


ban = [0,1] or [1,2] or [3,2]


ban = [0,2] or [1,0] or [3,0]


ban = [0,2] or [1,1] or [3,3]


ban = [0,2] or [1,2] or [2,2]


ban = [0,2] or [1,2] or [3,1]


ban = [0,2] or [1,2] or [3,2]


ban = [0,2] or [1,3] or [2,0]


ban = [0,2] or [1,3] or [3,1]


ban = [0,2] or [2,0] or [2,2]


ban = [0,2] or [2,1] or [2,3]


ban = [0,2] or [2,2] or [3,2]


ban = [0,3] or [1,0] or [2,0]


ban = [1,0] or [1,2] or [3,0]


ban = [1,0] or [2,2] or [3,0]


ban = [1,1] or [2,0] or [3,3]


ban = [1,1] or [2,1] or [3,1]


ban = [1,1] or [2,2] or [3,3]


ban = [1,1] or [2,3] or [3,3]


ban = [1,2] or [2,0] or [3,1]


ban = [1,2] or [2,0] or [3,2]


ban = [1,2] or [2,1] or [2,3]


ban = [1,2] or [2,3] or [3,2]


ban = [1,2] or [3,2] or [3,3]


ban = [1,3] or [2,0] or [2,2]


ban = [1,3] or [2,0] or [3,0]


ban = [1,3] or [2,0] or [3,1]


ban = [1,3] or [2,2] or [3,1]


ban = [2,0] or [3,1] or [3,2]


ban = [2,1] or [2,3] or [3,2]


Previously pre-posted

Boole(b)an — an early look at ban-fractals

Jonglietzsche


Post-Performative Post-Scriptum

“Jonglietzsche” is a portmanteau of German Jongleur / jonglieren, “juggler, juggling”, and the surname of core counter-cultural philosopher Friedrich Nietzsche (1844-1900). Jongleur is pronounced something like “zhawngloer”, as in French.

Boole(b)an

Suppose you allow a point to jump at random half-way towards one of the four vertices of a square. But not entirely at random — you ban the point from jumping towards the same vertex twice (or more) in a row. You get this pattern:

ban on v(i) + 0


It’s a fractal, that is, a shape that contains smaller and smaller copies of itself. Next you ban the point from jumping towards the vertex one place clockwise of the vertex it last jumped towards (i.e., it can jump towards, say, vertex 2 as many times as it likes, but it can’t jump towards vertex 2+1 = 3, and so on). You get this fractal:

ban on v(i) + 1


Now ban it from jumping towards the vertex two places clockwise of the vertex it last jumped towards (i.e., it can’t jump towards the diagonally opposite vertex). You get this fractal:

ban on v(i) + 2


And if you ban the point from jumping towards the vertex three places clockwise of the last vertex, you get a mirror-image of the v(i)+1 fractal (see above):

ban on v(i) + 3


The fractals above have a memory one vertex into the past: the previous vertex. Let’s try some fractals with a memory two vertices into the past: the previous vertex and the pre-previous vertex (and even the pre-pre-previous vertex).

But this time, let’s suppose that sometimes the point can’t jump if the previous or pre-previous isn’t equal to v(i) + n. So sometimes the jump is banned when the test is true, sometimes when the test is false — you might call it a boolean ban or boole(b)an. Using boole(b)ans, you can get this set of fractals:
















With these fractals, the boolean test sends the point back to the center of the square:











Posteriously post-posted

Boole(b)an #2 — a later look at ban-fractals