• metiulekm@sh.itjust.works
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    8 months ago

    I’m betting there’s probably something that generates the key from a vastly smaller player input, i.e what gameobjects you interacted with, in what order, or what did you press/place somwhere. But that also means that the entropy is probably in the bruteforcable range, and once you find the function that decrypts the secrets, it should be pretty easy to find the function that generates the key, and the inputs it takes.

    When handling passwords, it is standard practice to use an intentionally costly (in CPU, memory, or both) algorithm to derive the encryption key from the password. Maybe the dev can reuse this? The resulting delay could easily be masked with some animation.

    • Mikina@programming.dev
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      8 months ago

      Most of the costly algortihms I know of are still reasonably fast, i.e you can try thousands of checks per second, so it would definitely help, but you would then still have to design a puzzle that has a large number od possible answers or combinations - which I still think really limits what you can create with it.

      He could design a check/hash that would take a lot longer than common algs like bcrypt, but then theres a risk of someone reverse engineering it and simplyfiing it, or even finding a vulnerabilty that makes guessing the key easier. Because its suprisingly really difficult to make a hash that is matematically ok and doesnt have side-effects. Especially since crypto is dealling with some obscure advanced math, and some of the vulnerabilities in existing algorithms are pretty mind-blowing - especially since the more math you stack up, the more chances are there of you unknowingly using some kind of obscure math laws that can be used to simlplyfi or predict the results of your algorithm.

      For a really bad and simple example (that kind of illustrates the point) from the top of my head, if i was just multipliing the input by 2 to get the key, and i did it 1000 times, it would mean that 1) the attacker could make it faster by multiplying it just once by 2^1000, and 2) the result would always be even, so now he knows he only has to bruteforce half of the keys, since it cannot be any odd key.