Oh absolutely, I don’t blame KDE or arch repos lol. I did see that it was a KDE update but somehow didn’t clock the version number. I had it in my head that KDE6 was much farther off.
Oh absolutely, I don’t blame KDE or arch repos lol. I did see that it was a KDE update but somehow didn’t clock the version number. I had it in my head that KDE6 was much farther off.
Didn’t realize this was happening and yay -Syu went brrr and it broke my shit. Probably doesn’t help that I’m running nvidia with linux (endeavouros). Wayland doesn’t work at all (black screen on login with only mouse ptr, wrong resolution), while Xorg is now much less smooth e.g. on the switching desktop animations. Moving windows around and in-window graphics are fine. Some graphical config stuff changed too; I’m still taking inventory.
I’m also currently playing with nvidia vs nvidia-dkms with different kernels to see if that solves anything.
EDIT: Looks like that my configuration was failing to set nvidia_drm modeset=1
correctly due to my unfamiliarity with dracut. Manually adding nvidia_drm.modeset=1
to my kernel cmdline makes Wayland work (and quite well at that), though Xorg is still laggy.
My condolences, buddy.
You may be interested to know that these kinds of paper adhesives are usually intentionally designed so that the substrate (paper) tears before the adhesive does. This is meant to ensure robust packing and to give proof that the package has not been tampered with. Couple this with ever thinner and shittier substrates, and, well…
Lol OW was the first thing I thought of.
Me: hey look a protan filter
OW: okay, red is now pink, and everything else is washed out :)
Me: okay tritan it is
OW: lol have fun on your acid trip
So many solver solutions that day, either Z3 or Gauss-Jordan lol. I got a little obsessed about doing it without solvers or (god forbid) manually solving the system and eventually found a relatively simple way to find the intersection with just lines and planes:
It’s a suboptimal solution in that it uses 4 hailstones instead of the theoretical minimum of 3, but was a lot easier to wrap my head around. Incidentally, it is not too hard to adapt the above algorithm to not need C (i.e., to use only 3 hailstones) by using line intersections. Such a solution is not much more complicated than what I gave and still has a simple geometric interpretation, but I’ll leave that as an exercise for the reader :)