Timeline for Geometry Nodes: How to Read Last Digit of a Big Integer?
Current License: CC BY-SA 4.0
11 events
| when toggle format | what | by | license | comment | |
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| Feb 11 at 12:24 | history | bounty ended | Markus von Broady | ||
| Feb 11 at 10:36 | history | edited | StefLAncien | CC BY-SA 4.0 |
Finalize explanations.
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| Feb 10 at 10:52 | history | edited | StefLAncien | CC BY-SA 4.0 |
Add explanations.
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| Feb 8 at 23:16 | vote | accept | Markus von Broady | ||
| Feb 12 at 20:20 | |||||
| Feb 8 at 23:15 | comment | added | Markus von Broady | I guess the reason why it works so fast is that you don't do something obvious that I would try - iterating over points, but instead remove points - as I said, I haven't yet thoroughly analyzed it, but I think I understand it in principle, it's a smart solution and takes advantage of some of geonodes optimizations... | |
| Feb 8 at 23:13 | comment | added | Markus von Broady | Oh, you just spawn as many points as the integer AHAHAHAHA I was thinking about that, but figured it can't possibly work, too many points, too much memory, I figured a repeat zone should be much more lightweight - and yet your approach ended up working LOL, it's also quite fast, much faster than say the python equivalent of the loop I was trying (which took somewhere between 10 and 20 seconds for me) | |
| Feb 8 at 23:09 | comment | added | Markus von Broady |
I just started analyzing it, and it's late so I'm going to bed, but I just tested it for integers 123456780 to 123456789 AND IT WORKS!! - great job!
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| Feb 8 at 23:01 | comment | added | StefLAncien | NB: 2^29 = 536.870.912. Still a factor of 4 before the last signed int32: 2^31-1. | |
| Feb 8 at 22:57 | history | edited | StefLAncien | CC BY-SA 4.0 |
Typo.
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| Feb 8 at 22:55 | comment | added | StefLAncien | With my 10 years old, 8Gb of RAM MacBook, I succeed to reach 812^3 = 535.387.328. At 813^3 = 537.367.797, it just frizz and reboot... | |
| Feb 8 at 22:50 | history | answered | StefLAncien | CC BY-SA 4.0 |