Timeline for Minimum number of contractions needed to obtain a particular invariant set
Current License: CC BY-SA 2.5
5 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Aug 31, 2010 at 14:02 | vote | accept | Mark Bell | ||
| Aug 4, 2010 at 12:35 | history | edited | Lasse Rempe | CC BY-SA 2.5 |
added 1491 characters in body
|
| Aug 4, 2010 at 11:15 | comment | added | Lasse Rempe | I am not sure what you mean here. The gasket and carpet are two-dimensional sets, and of course you need at least two contractions to find them. The point about the Koch curve is that you can also write it as two copies of the same shape, as demonstrated in your post. As I state in my answer, the difference with the gasket and carpet are that you have additional topological things that need to be preserved: complementary regions. That's what makes these cases easier, and shows the IFS given are indeed optimal. | |
| Aug 3, 2010 at 13:39 | comment | added | Mark Bell | But intuitively the Koch curve appears to consists of 4 smaller copies of itself, so I'm not sure that it will "be fairly easy to see that ...". Although we have that if $G$ is the IS of an IFS consisting of $m$ contractions, then $\text{Dim}_H(G)$ < m$. This then gives us a lower bound on the number of contractions needed (for the Serpinski Gasket & Carpet, 3). So in the case of the Gasket we have an optimal solution, but for the carpet we may be able to do better. Can we? Can we prove that we can't? | |
| Jul 27, 2010 at 10:03 | history | answered | Lasse Rempe | CC BY-SA 2.5 |