Yup it seems crazy to me how deep insights one needs to have to be able to, say, connect the dots between compression and machine learning. And now it looks to me like he has done a lot of the foundational work in these fields. Super cool stuff

Too lazy to check, but is this the Rivest from the RSA algorithm?

If only it worked; the mentos would start disintegrating by your saliva and the coke would already form bubbles as soon as it enters your mouth thanks to the rough surface inside 🤓

That won’t work tho, you need to make it sys.maxsize//2 to coerce the output into int form

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

For Meet, I’d also suggest Jitsi. For Drive, I’d recommend giving cryptpad.fr a shot; that seems to be the closest to Drive’s file editing capabilities.

I see your encourageMint and raise you

Commenting so you see your post one more time

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

Not exactly what you wanted, but njalla is a privacy-focussed domain registrar that basically buys domains on your behalf under their own name and gives you all the access to it that you need

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

That’s fair too. I was just trying to get some use out of my old device, but I think it’ll be better if I simply use to run some CLI tools via SSH that I don’t want running on my main device :P

Going for a RasPi might be a bigger hassle for me due to the market, but I suppose I can try the rest on my laptop as easily. Thanks!

Ah now it makes a lot more sense. I’ll have to stick with things like overlay network simply because my ISP is super unreliable (for example, I’m out of wifi right now because of mildly incovenient weather lmao)

Haha I think it’s best if I stop running towards just getting my own server up and actually learn this stuff instead, regardless of how long it takes. I’ll try to follow through on this, thanks again for all the help :D

That’s super cool, I’ll check it out. Thanks a lot!

Oh I’ve never heard about openwrt, but it sounds interesting. I’ll check it out, thanks!

Understood. Guess I’ll figure out setting up all the services before I get to figuring out putting them beyond my local network. Thanks!

I’ve been putting off learning docker for a while now, guess it’s finally time to dive right into it. Thanks for the info about freshrss, I’d been looking for ways to get more into cybersecurity and this might be it

Welp, port-forwarding seems to be the major issue then, since I’m soon going to shift to an institutional wifi where I may (not) have access to the router. But you’re right, I should try getting familiar with what I have first lol. Thanks!