For the URL you mention: https://sci-hubtw.hkvisa.net/10.1177/00207314211015806
2,115 karma · joined January 25, 2016
:: https://spacetime.dev
I assume you don't care about missing out on updates since that's the free work you're complaining about
- two state charges of breaking and entering with intent to commit a felony
- wire fraud, computer fraud, unlawfully obtaining information from a protected computer, and recklessly damaging a protected computer
- breaking and entering with intent, grand larceny, and unauthorized access to a computer network
- federal prosecutors filed a superseding indictment adding nine more felony counts, which increased Swartz's maximum criminal exposure to 50 years of imprisonment and $1 million in fines
This is essentially the same as:
> The second vital piece of the puzzle was the principle of erasure. In 1961, the German American physicist Rolf Landauer showed that any logically irreversible computation, such as the erasing of information from a memory, would result in a minimal nonzero amount of work converted into heat dumped into the environment, and a corresponding rise in entropy.
https://www.reddit.com/r/scihub/comments/lofj0r/announcement...
Looking at you, Apple.
In the case of DH, g is some such element.
DH relies on the discrete logarithm problem. That means that if I give you g + g + .. + g, where g is added to itself n times, it will be difficult for you to find out what n is.
Both parties share the same prime p, and the same base point g. Then each of them select a random, secret point. Say Alice's secret is a and Bob's is b. Now each of them create their "public key", which is just g + g + g .. a times, and g + ... b times to create A and B, respectively. The (mod p) simply means divide the result by p and take the remainder as the answer.
Now if I am Alice, my secret is a, my public key is A. Bob sends me B = g^b (mod p). I exponentiate this with my own secret key to get shared secret = (B)^a = (g^b)^a = (g^a)^b = g^(ab) (mod p)
(mod can be deferred to the end of the calculation or done at each step, without changing the answer)
So you can see that, since the order of exponentiation doesn't matter, when Bob does the same calculation (multiplying his private b with alice's public key g^a) he gets the same shared secret.
For example, TLS 1.3 does this by (after performing ephemeral DH key exchange) signing the conversation transcript with the server's long term identity key. After this the client is sure that they are speaking to the correct server, but the server has no authenticity guarantees about the client.
Of course you could setup TLS and FDE but this doesn't guarantee that your data will be safe.