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Author Topic: 2 bug reports  (Read 1043 times)
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plynch27
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« on: December 25, 2014, 07:58:38 AM »

Hey, Kalle, just a couple things:

1. The "Resume zoom sequence" command changes the maximum iterations field to 1000 no matter what the configuration before it begins resuming.
2. The general zoom sequence render routine fails to continue behaving robotically when it gets to magnifications under 1e10

I'm not sure how to usefully elaborate on the first one, so please, ask any questions you may have.

About the second complaint, though, basically, when it finishes calculating the first frame that has a magnification of less than 1e10 ─ i.e. the first frame whose magnification has a single-digit exponent ─ it will, under some conditions, read "R: 5000% and, after that, under all conditions, get caught in an infinite loop before ever drawing to the screen. Sometimes, calling the zoom sequence render routine again without cancelling the render will cause the program to draw the last image to the screen before getting caught in another loop.

Hopefully, that helps you pinpoint the problem. If you're having a hard time duplicating the problem, let me know and I'll give my system info.
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Kalles Fraktaler
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« Reply #1 on: December 25, 2014, 10:26:38 AM »

Thanks. I will have a look when I get to a computer in a week or so.
But I have resumed many zooms and with a high iteration. Maybe you meant that you wanted it to use a lower value than 1000?
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plynch27
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« Reply #2 on: December 25, 2014, 12:11:47 PM »

Alright, I just had another light bulb in my head turn on after I read your reply. Basically, I found out that the routine was interrupted while it was in the middle of writing the last frame's .kfb file, so I needed to delete last labelled .kfb file and the recovery.kfb file. I'm hoping the same kind of thing was happening when I got to the frames with zooms less than 1e10, but I can't be sure till it gets there again. Yes, I realize how confusing that last sentence probably was, but it's a long story and, trust me, you don't care.
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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
plynch27
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Posts: 71


« Reply #3 on: December 25, 2014, 07:06:05 PM »

Okay, yeah, them problem with the low exponents is still there.
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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
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