New answers tagged key-exchange
2
Here's the problem with this scheme: suppose $A$ is the honest client, and $B$ is a dishonest server (who doesn't know $s$). Then, $A$ tries to log in, he selects $a$ and transmits
$$A = g^{as} \bmod p$$
Then, $B$ just picks a random value $b$, and transmits
$$B = g^b \bmod p$$
Then, $A$ will compute a 'shared secret' $$S = B^a \bmod p$$ (which would be ...
1
(in DH) Bob used a small secret key that can be brute-forced.
If the private key is $k$-bit, a meet in the middle attack allows private key recovery with cost $O(2^{k/2})$ group operations (here, multiplication in the modulo $p$ group $\Bbb Z_p^*$).
In its simplest form: we know Bob's public key $y=g^x\bmod p$ and want the $k$-bit $x$. We choose $a$ with $\...
0
The sender must know and trust something about the intended receiver. Otherwise, the sender can't know who s/he shares a secret with.
Also, the intended receiver must know and trust something about the intended sender; otherwise, the receiver will share a secret but can't know with who.
The standard solution is that the above known and trusted "somethings" ...
0
It could have three consequences.
1) If you are very unlucky, and you pick a "zero" ($a$ such that $a=0\mod p-1 $), it will break your system : (but this will happen with a negligible probability, and it could be detected) An external observer will easily guess the shared secret
2) You lose in efficiency
3) Your integer had to be chosen upper-bounded (you ...
Top 50 recent answers are included
