Snow IFS

Snowflakes have approximate six-fold symmetry: rotating a snowflake 60 deg about its center leaves it more-or-less unchanged.

Two pieces of the decomposition are easy to see:

the middle is a copy of the snowflake shrunk by a factor of r = s = 0.6, and

the outer half of the right branch (at 3 o'clock, for those who remember the faces of analogue watches) is a copy shrunk by r = 0.4 and s = 0.2 in the y-direction (note this piece is flatter than the whole snowflake), and translated in the x-direction (e = 0.6).

What about the other five branches?

Note their orientation is not the same as the whole snowflske, so unlike the Queen Anne's Lace example, translations will not suffice.

We must use rotations and translations, or is there an easier way?

Rotating the entire snowflake 60 deg about its center will move the 3 o'clock branch to 1 o'clock.

Rotating again will move the 1 o'clock branch to 11 o'clock, and so on.

The third rule below will fill in the rest of the snowflake.

More carefully, with these rules the Random Algorithm generates the snowflake.

The Deterministic Algorithm gives a different result, at least if the starting picture is an extended shape such as a square.

(Make a guess, then try it.)

The Random Algorithm still will generate the snowflake, so long as on average the transformations applied are contractions.

Return to Natural Fractal Inverse Problems.