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

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