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3 yr. ago

is beehaw related to lemmy?

  • As someone who tried to self-host it like a month ago (and seemingly still hasn't got it fully working), I'll just write out the overview of what I've done and let you (and others) comment on how correct and feasible it is.

    Since my ISP doesn't allow me to get a static IP address, I rented a VPS connection and made a wireguard tunnel from the VPS to my computer. This tunnel forwards traffic at all the necessary ports between the two machines. I really wasn't familiar with all the necessary components for an entire mail server, so I chose mailcow since it packages everything into one single software (well, more like a bunch of docker containers). Another reason I went with mailcow was that I could easily find a github tutorial for how to set up mailcow with wireguard tunneling (it's a bit outdated IMO, but the changes are minor). Mailcow also gives a nice portal interface listing out all the DNS entries you need to put in place to get it working perfectly.

    In the end, I still see a few incoming emails getting dropped and reception time being an hour or so, and I'm not sure if it's a problem with my tunnel or DNS or something else. But overall, I'd say it was much easier than setting up all the individual services myself.

  • But does it make jack jork?

  • No love for Star Wars? :(

  • 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 🤓

  • Panik

    Jump
  • 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 * 10k + a_1 * 10{k-1} + ... + a_k * 1. Note that 10 mod 3 = 1, which means that 10i 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

  • I see your encourageMint and raise you

  • Selfhosted @lemmy.world

    Ok, how do I start self-hosting?

    1. Strairdrac is three crabs in a trenchcoat, now teaching others of his kind how to blend in with humans