Random IFS

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For example, suppose we specify the resolution corresponding to addresses of length N = 3 and we start with the point (x0, y0) with address 1infinity.

To the specified resolution, (x0, y0) lies in the region with address 111.

If T2 is the first transformation applied, then resulting point (x1, y1) = T2(x0, y0) lies in the region with address 211.

If T3 is the next transformation applied, then resulting point (x2, y2) = T3(x1, y1) lies in the region with address 321.

If T4 is the next transformation applied, then resulting point (x3, y3) = T4(x2, y2) lies in the region with address 432.

Continuing will fill in all the 43 regions of address length 3.

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