# IFS which construct this fractal and have affine transformation only [Image updated]

Is there an IFS which construct this fractal and have affine transformation only? (I think there must be a restriction, which is not an affine transformation. Can it be proved?)

• You changed the problem when you changed the image. Your first image could be constructed using an IFS with condensation. – Mark McClure May 11 '16 at 0:32
• @MarkMcClure Could you explain about the IFS with condensation? (I do not know that 'condensation' means.) And for the later one (with its interior) and its ifs (with condensation), does $\lim_{n\to\infty}H^{\circ n}(A)=S$ hold for any set $A$, where $H$ is the Hutchinson and $S$ is given fractal? I think it does not hold... – Kanu Kim May 11 '16 at 1:13
• I tried to answer the linked question. I don't know that I can improve it. In any case, neither image is strictly self-similar. You will need some generalization of the IFS procedure. – Mark McClure May 11 '16 at 1:44
• @MarkMcClure Okay, I see. Thanks for your comment. – Kanu Kim May 11 '16 at 2:31

## 1 Answer

I do not have a definition in IFS notation but the algorithm I write below should do the work if the definition of affine transformation is as the basic definition at the Wikipedia:

"...a function between affine spaces which preserves points, straight lines and planes. Also, sets of parallel lines remain parallel after an affine transformation. An affine transformation does not necessarily preserve angles between lines or distances between points, though it does preserve ratios of distances between points lying on a straight line."

Here it is:

1. Initial triangle is $T1$. Make a copy of the initial triangle, name it $T2$, rotate $T2$ $180$ degrees using as pivoting point the center of $T1$.

2. Resize $T2$ to fit into $T1$ and change the color to white (when the three vertices of $T2$ are exactly in a position that is part of the sides of $T1$).

3. Make a copy of $T2$, name it $T3$, and rotate $T3$ $180$ degrees using as rotation point the middle point of the segment of the upper side of $T3$ (once rotated you will not see it, because its sides are in the same position than the sides of the triangles $T1$ and $T2$, but it is there) and change the color to black.

4. Apply again the same algorithm from point ($1$) for $T3$, in other words, being now the $T1$ of the algorithm this last generated $T3$ (the new $T1$ will be the current $T3$).

• @KanuKim added the change of colors to the algorithm. – iadvd May 11 '16 at 0:40
• Sorry to bad picture (and late comment also), I updated the picture. I already know the algorithm construction, but I want to find its ifs. – Kanu Kim May 11 '16 at 1:00
• @KanuKim I did not know that point, ok! I will keep the answer just in case somebody is interested or can add that extra information. Initially if it is possible to define the algorithm in terms of affine transformations it should be possible to define the equivalent IFS. – iadvd May 11 '16 at 1:21
• OK, thanks for your answer. – Kanu Kim May 11 '16 at 2:30