|
Furan
|
 |
« Reply #15 on: March 23, 2013, 11:33:16 AM » |
|
I was wondering, some kind of DE method could be used to find points of the M-set at log(log(zoom))~5;6
It should take larger steps to get deeper inside the boundary so as not to end at the very edge.
|
|
|
|
|
Logged
|
|
|
|
makc
Strange Attractor
Posts: 272
|
|
« Reply #16 on: March 25, 2013, 10:17:42 PM » |
|
why not just do what you do manually? i.e. calculate some NxN grid, pick the cell with highest iteration count, calculate NxN grid with 1/N scale around it, repeat.
|
|
|
|
|
Logged
|
|
|
|
|
plynch27
|
|
« Reply #17 on: March 25, 2013, 10:25:11 PM » |
|
Don't forget to have the algorithm look at the pixel's surroundings. If you just look for the highest iteration count, all you're gonna find in the end is a bunch of visual noise.
|
|
|
|
|
Logged
|
If you'd like to leave me a text message, my 11-digit phone number can be found in π starting at digit 224,801,520,878
((π1045,111,908,392) mod 10)πi + 1 ≈ 0
|
|
|
makc
Strange Attractor
Posts: 272
|
|
« Reply #18 on: March 26, 2013, 02:18:53 AM » |
|
there's also an option to find roots of (((z^2+z)^2+z)^2+z)^2+z and so on. there is a method to find all roots of such an equation, but obviously when you go deeper, the number of roots is prohibitive.
|
|
|
|
|
Logged
|
|
|
|
|
Dinkydau
|
|
« Reply #19 on: March 26, 2013, 03:47:49 PM » |
|
why not just do what you do manually? i.e. calculate some NxN grid, pick the cell with highest iteration count, calculate NxN grid with 1/N scale around it, repeat.
This doesn't always work. As soon as you find a julia or a dense infinite spiral you will get stuck there. The center of a shape in a practical n×n grid doesn't always have the most iterations. |
|
|
|
|
Logged
|
|
|
|
makc
Strange Attractor
Posts: 272
|
|
« Reply #20 on: March 27, 2013, 01:45:14 AM » |
|
If you just look for the highest iteration count, all you're gonna find in the end is a bunch of visual noise. as long as he simply wants to get to the boundary, visual quality doesn't matter  |
|
|
|
|
Logged
|
|
|
|
|
Dinkydau
|
|
« Reply #21 on: March 28, 2013, 01:11:31 AM » |
|
Then (-2 + 0i) is still easier.
|
|
|
|
|
Logged
|
|
|
|
|
plynch27
|
|
« Reply #22 on: March 28, 2013, 02:10:52 AM » |
|
lol Took the words right out of my mouth. ;-)
|
|
|
|
|
Logged
|
If you'd like to leave me a text message, my 11-digit phone number can be found in π starting at digit 224,801,520,878
((π1045,111,908,392) mod 10)πi + 1 ≈ 0
|
|
|
|
grobblewobble
|
|
« Reply #23 on: May 09, 2013, 11:09:02 PM » |
|
I feel that looking at depth only is a very unfair way to compare zooms. Some areas are much harder to zoom in on then others. The very end of the main Mandelbrot antenna, for example, is extremely easy, while the endpoints of the valleys are super hard because the number of iterations you need explodes.
Then there is also the question of quality - the resolution matters a lot. Technically, if you could call a single pixel an "image", then you can produce an "image" of unlimited depth very easily.
|
|
|
|
|
Logged
|
|
|
|
|
plynch27
|
|
« Reply #24 on: May 09, 2013, 11:56:26 PM » |
|
I feel that looking at depth only is a very unfair way to compare zooms.
Well, yes, of course. That's why when most people are discussing fractal renders, they divulge several dimensions of resolution: 1. zoom depth, 2. the minimum or average iteration count of the frame or keyframe -- this piece coupled with the zoom depth provides some information on the image's or zoom movie's structure, 3. whether or not it's anti-aliased and if so, the resolution of the oversampling and 4. the final output resolution if it's an image, or the keyframe and output resolution if it's a movie -- typically also followed by the values used for the video's encoding parameters. The thing is: the topic of this specific thread only discusses one of those parameters: the zoom depth which is what makes your remark confusing, no offense. |
|
|
|
|
Logged
|
If you'd like to leave me a text message, my 11-digit phone number can be found in π starting at digit 224,801,520,878
((π1045,111,908,392) mod 10)πi + 1 ≈ 0
|
|
|
|
grobblewobble
|
|
« Reply #25 on: May 10, 2013, 12:53:48 AM » |
|
Well, the reason for my remark is that without taking those parameters into account, a discussion of deepest zoom is not really possible, it becomes meaningless. There are areas of which you can predict what they look like on any scale, no matter how small. A trivial example is the middle of the Mandelbrot, which is monotone black. A less trivial example is (-2,0); once you are zoomed in deep enough, the pattern just keeps repeating itself. Someone could construct a theoretical image without calculating any iterations and say ''this is a zoom factor of googleplex".
|
|
|
|
|
Logged
|
|
|
|
|
plynch27
|
|
« Reply #26 on: May 10, 2013, 01:27:04 AM » |
|
Ahh, yes, I agree with you, there. I've seen a number of zooms to depth of say 2^900x, that are centered so close to the value -2 that, when I'm done watching them, I'm left thinking, "Okay, cool, but that doesn't really count." I also feel that a good ultra-deep zoom is one that's going to take a number of months to render. I've got a movie at 2^1135.7x that took almost 3 months and that was even after sacrificing resolution.
|
|
|
|
|
Logged
|
If you'd like to leave me a text message, my 11-digit phone number can be found in π starting at digit 224,801,520,878
((π1045,111,908,392) mod 10)πi + 1 ≈ 0
|
|
|
|
grobblewobble
|
|
« Reply #27 on: May 10, 2013, 09:27:50 AM » |
|
The most impressive zoom I have seen was done by a certain "Phaumann". It zooms into a non-trivial area by a factor of 10^999. Amazingly, it was done in Mathematica, so it appears he wrote his own code. The resolution is low, but still reasonable. Here is a link:
<![CDATA[<a href="http://www.youtube.com/v/1sSm53Q9Jws&rel=1&fs=1&hd=1" target="_blank">http://www.youtube.com/v/1sSm53Q9Jws&rel=1&fs=1&hd=1</a>]]> |
|
|
|
|
Logged
|
|
|
|
|
November 30, 2006, 12:38:08 PM
IMAGE ART 2
Fractal Art
Deepest - e10000
Movies Showcase (Rate My Movie)
« 1 2 3 »
Dilber MC Theme by HarzeM