cultural reviewer and dabbler in stylistic premonitions

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Joined 4 years ago
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Cake day: January 17th, 2022

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  • Thanks. Sorry to see my assumption was correct; that does indeed sound a lot like when they were called OSSO two decades ago.

    Notably absent from the list of things they might open source soon is their current “Lipstick” UI, the graphical shell itself.

    All of the stuff they plan to open source are things I didn’t even figure out were still closed from my 5-10 minutes of research before writing my previous comments. It is difficult to estimate the number (do you know how?) of other small closed components which they can dribble out over the next years to maintain users’ false hope that they will one day have an actually-open-source operating system.

    we’ll see though

    my advice is: don’t hold your breath.

    Sorry if this sounds bitter, but it’s because I am - I naively believed that OSSO might actually ship a free OS one day (to be fair they didn’t say they would either, but they helped us believe that they might… in effect saying “we’ll see” for years while releasing bits here and there) and it was frustrating to realize that it was never a real possibility.






  • Arthur Besse@lemmy.mlMtoLinux@lemmy.mlAnd so it begins
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    1 day ago

    1 reason it’s wrong to me: https://nosystemd.org/

    Under “Notable bugs and security issues” there is a big list of issues which were all (afaict) fixed many years ago.

    There have been reasonable philosophical objections to systemd, some of which are still relevant, and as that site shows there are still many distros without it, but for the vast majority of desktop users who want something that JustWorks… using a mainstream distro with systemd is the way to go.

    This blog post from pmOS covers some of the pain of trying to use KDE or GNOME without it.




  • I can’t really imagine a benefit to --autoremove except for keeping old packages a bit longer before removing them.

    Eg, if you run apt --update --autoremove upgrade -y once a day you’ll keep your prior-to-currently-running-version kernel packages a day longer than if you ran autoremove immediately after each upgrade.

    To make things more confusing: the new-ish apt full-upgrade command seems to remove most of what apt autoremove wants to… but not quite everything. 🤷













  • If a payment processor implemented this (or some other anonymous payment protocol), and customers paid them on their website instead of on the website of the company selling the phone number, yeah, it could make sense.

    But that is not what is happening here: I clicked through on phreeli’s website and they’re loading Stripe js on their own site for credit cards and evidently using their own self-hosted thing for accepting a hilariously large number of cryptocurrencies (though all of the handful of common ones i tried yielded various errors rather than a payment address).


  • So like, it’s a situation where the “lock” has 2 keys, one that locks it and one that unlocks it

    Precisely :) This is called asymmetric encryption, see https://en.wikipedia.org/wiki/Public-key_cryptography to learn more, or read on for a simple example.

    I thought if you encrypt something with a key, you could basically “do it backwards” to get the original information

    That is how it works in symmetric encryption.

    In many real-world applications, a combination of the two is used: asymmetric encryption is used to encrypt - or to agree upon - a symmetric key which is used for encrypting the actual data.

    Here is a simplified version of the Diffie–Hellman key exchange (which is an asymmetric encryption system which can be used to agree on a symmetric key while communicating over a non-confidential communication medium) using small numbers to help you wrap your head around the relationship between public and private keys. The only math you need to do to be able to reproduce this example on paper is exponentiation (which is just repeated multiplication).

    Here is the setup:

    1. There is a base number which everyone uses (its part of the protocol), we’ll call it g and say it’s 2
    2. Alice picks a secret key a which we’ll say is 3. Alice’s public key A is ga (23, or 2*2*2) which is 8
    3. Bob picks a secret key b which we’ll say is 4. Bob’s public key B is gb (24, or 2*2*2*2) which is 16
    4. Alice and Bob publish their public keys.

    Now, using the other’s public key and their own private key, both Alice and Bob can arrive at a shared secret by using the fact that Ba is equal to Ab (because (ga)b is equal to g(ab), which due to multiplication being commutative is also equal to g(ba)).

    So:

    1. Alice raises Bob’s public key to the power of her private key (163, or 16*16*16) and gets 4096
    2. Bob raises Alices’s public key to the power of his private key (84, or 8*8*8*8) and gets 4096

    The result, which the two parties arrived at via different calculations, is the “shared secret” which can be used as a symmetric key to encrypt messages using some symmetric encryption system.

    You can try this with other values for g, a, and b and confirm that Alice and Bob will always arrive at the same shared secret result.

    shia labeouf magic gif

    Going from the above example to actually-useful cryptography requires a bit of less-simple math, but in summary:

    To break this system and learn the shared secret, an adversary would want to learn the private key for one of the parties. To do this, they can simply undo the exponentiation: find the logarithm. With these small numbers, this is not difficult at all: knowing the base (2) and Alice’s public key (8) it is easy to compute the base-2 log of 8 and learn that a is 3.

    The difficulty of computing the logarithm is the difficulty of breaking this system.

    It turns out you can do arithmetic in a cyclic group (a concept which actually everyone has encountered from the way that we keep time - you’re performing mod 12 when you add 2 hours to 11pm and get 1am). A logarithm in a cyclic group is called a discrete logarithm, and finding it is a computationally hard problem. This means that (when using sufficiently large numbers for the keys and size of the cyclic group) this system can actually be secure. (However, it will break if/when someone builds a big enough quantum computer to run this algorithm…)