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Here is a description of how to color pictures of the Mandelbrot set, more accurately the complement of the Mandelbrot set. Suppose we have a rectangular array of points. Say the array is $m$ by $n$. Suppose also we have a number of color names. Now suppose we assign the color name $j$ to a point in the array if the $j$-th iteration exceeds $2$. If the iterates do not exceed $2$ we color the point black. This process will yield a picture. By careful positioning the array of points can we get any picture we want? In particular can we get a digital representation of the Mona Lisa.

I do not know how to begin to prove or disprove this. My guess is that we can probably get any pictures.

Edit

A different way to color the array would be to use color $c$ if the first iterate to exceed $2$ is iterate $i_{1}$, $i_{2}$, $\cdots$, $i_{j_{c}}$. The iterates for different colors should be distinct. If someone wishes to use infinites lists for the number of iterates that are assigned to a color that would also be acceptable.

With this change the problem reduces to finding an $m$ by $n$ array where each point in the array has a different number of iterates before the value exceeds $2$.

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It appears you are envisioning a two-dimensional analog of the idea that eventually one can find any arbitrary sequence of digits in an infinite sequence with a uniform random distribution. The question, however, does not seem clearly formed. Most importantly, if the escape values in your array are unique, how can they generate the Mona Lisa whose array must contain identical values? If, as you see it, the Mona Lisa's array does not contain identical values, then this is akin to claiming that, for example, a 1000 x 1000 array of the numbers 1 to 10,000 is equivalent to the Mona Lisa. – Harlan Aug 7 '13 at 6:25
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I assume that initially we have an $m$ by $n$ array of color names. This is how we choose to represent the Mona Lisa. Some points in this array will likely be assigned identical color names. The idea is to find an $m$ by $n$ array of points in the complement of the Mandelbrot set and a method of assigning color names to escape values(thanks for the terminology) so that the $m$ by $n$ array, when colored by this assignment, looks like the Mona Lisa. To be specific, points with the same escape value will have the same color. – Jay Aug 7 '13 at 14:16
up vote 3 down vote accepted

I think yes

Consider the sequence of "westernmost" islands increasing in period:

westward ho

Here are some examples, you can see them increasing in hairyness / spinyness.

Period 20:

period 20

Period 30:

period 30

Period 40:

period 40

Period 50:

period 50

Zooming in near to a sufficiently hairy high-period island, you can get very nearly parallel spines:

period 50 zoom annotated

There is a natural grid with a 4:1 aspect ratio between successive escape time level sets in neighbouring spine gaps. (Determined by measuring images: I have no proof of this seeming fact. But it's not a strictly necessary detail.)

If you choose the angle of the spines carefully, you can create an NxN grid with square cells that satisfies "distinct level set at each pixel sampling point in an (m,n) image array".

schematic grid

There is a slight fuzziness due to the imperfect parallelism and the quantized angles available for any given period island, but increasing the period far enough and that won't matter - eventually it gets good enough for the finite width of the level sets to take care of it.

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Your answer is an idea for a proof. I think that something like that will work. Unfortunately thinking something will work is not the same as proving it works. The hard part is describing the judicious positioning. – Jay Feb 5 at 15:07
    
I edited my answer, using a slightly different approach to make judicious positioning easier. Thanks for the push towards more rigour. – Claude Feb 5 at 20:03
    
Thank you. That seems like it will work. There will be a certain amount of tediousness to fill in the details. – Jay Feb 5 at 20:27

You are coloring the interval between the lemniscates http://mathworld.wolfram.com/MandelbrotSetLemniscate.html

When zooming into the outer parts of the mandelbot set (where z[k] tends to eventually diverge) , you will render the picture into the boundaries of a convex polygon (likely a square or a rectangle). By looking carefully at the different points where |z[j]| < 2 and z[j+1] > 2, you will notice that the minimal values for j lies on the border of the rendering surface.

This mean that mandelbrot rendering is no more than drawing the contour lines on the map of an area without lake or holes. It always looks like that http://en.wikipedia.org/wiki/Contour_line#/media/File:Contour2D.svg

This requires the model to look as pretty as an elevation map. Lisa Gherardini might not appreciate the comparison (nor my humour).

As a annoying side effect, this must prevent Leonardo to arbitrarily use color transitions, the artist must always follow the transitions color[j+min], color[j+min+1] color[j+min+2]...color[j+max], in that very order, everywhere in the paint. Isolated areas within the face (like the eyes) must follow the same color transitions. Look at the bottom of her eyes pupils, some colors are "missing" (blame Mandelbrot, not Leonardo).

The other approach could be to start numbering the colors from the masterwork, and verify (or enforce) the numbers to meet the topography constraints of the connected contours. THAT alone is a challenging problem.

Maybe, a realistic request would be to locate the topographic map of any location and scale you wish in the world.

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The point (ahem) is that by point-sampling the grid you can break much of the continuity that's necessary for this topological argument; in particular, all of the transitions that you describe as necessary for an isolated area could 'bleed out between the samples' and have the contouring be essentially invisible. – Steven Stadnicki Apr 16 '15 at 15:51
    
If you look at the first picture at this link: artyfactory.com/art_appreciation/portraits/chuck_close.html you will see something like the type of picture I mean. The difference is that Chuck Close has doo-dads in the squares. I would use solid colors. – Jay Apr 19 '15 at 22:41

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