goddard_guryon

joined 1 year ago
[–] [email protected] 6 points 1 month ago

Since you're seeking for answers within physics buzzwords, you're missing a lot of nuance which is causing you to come up with nonsensical theories. For one, the wavefunction doesn't exist 'in' the particle, it is the particle. A wavefunction collapsing is what causes a particle to show up in a specific location (going by the most prominent interpretation of the wavefunction); as long as the wavefunction is non-zero at more than one location, the particle exists in all those locations.

Equipped with this knowledge, phrases like "a wavefunction collapsing in such a way that it absorbs light" or "change the composition of atoms" make no sense, so I'd suggest you to rethink your assumptions, and, if possible, not look at quantum mechanics as a magic black box that can explain everything a mind can conjure up.

[–] [email protected] 1 points 8 months ago

Indeed, an integer is divisible by 3 if and only if the sum of its digits is divisible by 3.

For proof, take the polynomial representation of an integer n = a_0 * 10^k + a_1 * 10^{k-1} + ... + a_k * 1. Note that 10 mod 3 = 1, which means that 10^i mod 3 = (10 mod 3)^i = 1. This makes all powers of 10 = 1 and you're left with n = a_0 + a_1 + ... + a_k. Thus, n is divisible by 3 iff a_0 + a_1 + ... + a_k is. Also note that iff answers your question then; all multiples of 3 have to, by definition, have digits whose sum is a multiple of 3

[–] [email protected] 1 points 8 months ago (1 children)

I see your encourageMint and raise you a stone

[–] [email protected] 7 points 9 months ago

Now that's one channel that'll always deserve more viewers than it has

[–] [email protected] 18 points 10 months ago* (last edited 10 months ago) (1 children)

A flair-like implementation would be nice for certain communities. From past experience on reddit, I can see how it could be beneficial to filter a community's posts through tags, say, to check latest announcements or new support questions. I'd personally prefer community-specific tags as opposed to global post tags ~~(which is what I inferred from this post's content, I haven't read the RFC yet though)~~ edit: the RFC talks about both instance-based tags and community-based tags, which is even better

[–] [email protected] 8 points 10 months ago

Everything else aside, you do see what you're asking, right? This is basically "hey guys, I know I'm being actively tracked by security agencies (but I'm not a threat, trust me). How do I get rid of these trackers? Does anybody have any tips and tricks I can follow to get security agencies off my back?". Again, without getting into any of the context, what do you expect to get here? What kind of person would reply with "sure I gotchu, here's a list of tools I regularly use to keep cops distracted from whatever it is I do on the internet"?

[–] [email protected] 2 points 10 months ago

I've been using it for a couple weeks but haven't used RCS, so I can't say specifically about that. Overall though, it's still a work in progress and is not as polished but it gets the job done (more or less). If you're really concerned about privacy using their closed source app, you can just host your own bridges in your Matrix server (the app is the only proprietary part of Beeper, the protocol is just Matrix). The app doesn't support logging in from another Matrix account, so you'll have to stick with Element (I think Element derivatives would work too) when using your own bridges. But that's probably a better option given that their own app lacks a few features.

[–] [email protected] 23 points 10 months ago* (last edited 10 months ago)

That's some high IQ usage of a meme. Lemme see if I'm getting this right:

  • the total area of the image ( = RHS of the equation) is 1
  • you divide the image into 4 parts so that the area of 1 part is is 1/4 ( = 1/2^(2*1)). You take the first three quarters and leave the fourth quarter for recursion (I'll call it x1). That gives you 3(1/4) + x1 = 1
  • now you take x1 and do the same with it. This time, the area of each sub-quarter is 1/16 ( = 1/2^(2*2)). Three such sub-quarters and a leftover x2 gives you 3(1/16) + x2 = x1. Put this back into the first equation to get 3(1/4 + 1/16) + x2 = 1.
  • repeat until infinity; each time the area of the resulting tile is 1/4 of the previous tile (which is the 2n in the exponent part)

Edit: imma remove all markdown since it doesn't seem to work, at least on liftoff. Enjoy the lisp-like mess

[–] [email protected] 11 points 10 months ago* (last edited 10 months ago) (3 children)

DISINGENUOUS COUNTER-ARGUMENT WITH SLIGHT PERSONAL ATTACK

EDIT: TYPO

[–] [email protected] 3 points 10 months ago

Basically this. I just read it as "exhibition + sale". Also, yo is this not common knowledge? I've seen this thrown around everywhere (I mean the phrase) and thought the whole world uses this phrasing (o_O )

[–] [email protected] 55 points 10 months ago (7 children)

More like "-20459 will be my year"

[–] [email protected] 9 points 10 months ago (1 children)

Oh dangit, it's simpler than I thought. So the only data being sent is...just whatever is sent in your average GET request.

 

I only saw the app once while scrolling around on f-droid; tried it but it seemed too empty to be useful. The only place I've since seen it even being mentioned is this post, but that's also not specifically about the app. I'm genuinely curious about what exactly the point is of GNU/Jami. Is it just a p2p version of calling and messaging?

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