Example 2102 coloring a tree 97 210 coloring ifs by

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Unformatted text preview: ost obvious loop is the square with vertices (0, 0), (1/2, 0), (1/2, 1/2), and (9, 1/2). The simplest way to break that loop is to incorporate a reflection across the y -axis into TB . The top left image of Fig. 2.61 shows this idea works. The attractor is simply-connected and connected, hence a dendrite. The bottom left image of Fig. 2.61 is the attractor of the three function IFS {TA , TB , TC }, also a dendrite. So in this case, we see that a three function dendrite gives rise to a four function dendrite. The next problem asks if this always is the case. Figure 2.61: Top. Left to right, examples of solutions for Prxs 2.9.1, 2.9.2, and 2.9.3. Bottom: the attractors of the corresponding IFS {TA , TB , TC } PrxsSol 2.9.2 And the answer is No. Nothing involving fractals could be that simple. Recalling S denotes the filled-in unit square, the loops in this four function IFS were destroyed because TB (S ) does not contain the point (3/4, 1/2), the lower right corner of TD (S ). Suppose we keep the reflection across the y -axis i...
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This document was uploaded on 02/14/2014 for the course MATH 290B at Yale.

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