Category Archives: Fractals

two circles

Lately I have played around with inversion at circles trying to find some new kind of fractals. Thus I found a simple mapping that gives interesting designs. They are not fractal, instead overlapping circles appear. Inversion at a single circle … Continue reading

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improved code for fractals in high resolution

I was not satisfied with the earlier code for generating high-resolution fractals and I improved on it to make experimentation more rapid. Now the program first generates only a low-resolution image for the computer screen.Then the code stops the “draw()” … Continue reading

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fractals in high resolution – the code

// needs the class OutputBuffer and the improved Vector class OutputBuffer outputBuffer, activeOutputBuffer; int n,iteMax; Vector c; float rLimitSq; void setup() {   size(600, 600);   noLoop();   int magnification=10;   outputBuffer=new OutputBuffer(magnification);   outputBuffer.setUnitLength(230);   outputBuffer.setOffset(-0.05,0);   n=6;   … Continue reading

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fractals in high resolution

Fractal images are a good reason to draw off-screen in high-resolution, as discussed in an earlier post. Looking at the low-resolution image of “self-similar fractals …” we need some imagination to see that it is really self-similar. Too much details … Continue reading

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high resolution images with off-screen drawing

In an earlier post I have shown how to make smooth images at any scale using the pdf-renderer. But you can do this only with graphics objects such as line, point, shape, ellipse and so on. It won’t work if … Continue reading

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Coloring the Julia set

The Julia set of a function f(z) in the complex plane has all points z that remain finite upon iterations of the function. In the last posts I have used expanding functions to get fractal images from iteration, as discussed … Continue reading

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Rainbow colors

We can define a continuous number x of iterations needed to reach the critical radius R. Note that if the n-th iteration of f(z) equals R then x=n, and if the (n-1)th iteration equals R then x=n-1. For values in-between … Continue reading

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