Hot answers tagged

6

Yes, the basic idea of hardcoding a public key is secure. It is sometimes recommended as an alternative to the complexity TLS and PKI bring – otherwise it can be easy to skip a crucial step and end up with little or no security. However, the "encrypt a secret for server" scheme has some weaknesses compared to TLS. The clearest is lack of forward secrecy ...


3

It is secure against private key exposure but not against replay attacks by Eve. A three-way protocol avoids this, and doesn't need to use timestamps. The description below is from Delfs and Knebl's book Introduction to Cryptography. Each user, say Alice, has a key pair $(e_A, d_A)$ for encryption and another key pair $(s_A, v_A)$ for digital ...


2

If Eve knows $C, \space N, \space K_{CA}(N)$ then she is able to impersonate $C$ just by following the protocol you described above. If she just knows $K_{CA}(N)$ for a single $N$ then the effectiveness of the attack depends on the set wherein $N$ is picked. But as you are supposing that the communication channel is not secure, an attacker could spy ...


2

First of all, you should make a more formal definition of the protocol. Security cannot be assessed without a proper definition. Second you don't specify an key sizes. RSA-512 is such a low key size that it may be considered broken. On the other hand, you may run into performance issues if you choose a higher key size (Elliptic Curve crypto would make more ...


1

Short answer: No, it is not vulnerable to man-in-the-middle attacks, assuming that Alice and Bob each have the right signature verification key of the other party. Yet, the man-in-the-middle attack could have taken place at the moment of exchanging the signature verification key. So if $sig_{X}$ is party X's signature key, the attack on the exchange itself ...


1

Here is how it can be Vulnerable. Alice: $x$ Bob $y$ Eve $z$ Alice$\rightarrow$ $g^x$ $\rightarrow$ Eve->$g^z$->Bob Bob$\rightarrow$ $g^y$$\rightarrow$Eve$\rightarrow$ $g^z$$\rightarrow$Alice What Alice thinks key is $g^{(xz)}$ what Bob thinks the key is $g^{(zy)}$ Eve can compute both of these values $(g^x)^z$ and $(g^y)^z$ This is why we need ...


1

In the Smart Card industry, it is often used a technique called diversification: individual Smart Cards get a serial number $S$, and a device-unique secret key $K_S$ computed from a Master Key $K$, and $S$, using a (typically: non-entropy-stretching) Key Derivation Function. For example, $K_S=\operatorname{AES-ENC}(K,S)$ where the first parameter of ...



Only top voted, non community-wiki answers of a minimum length are eligible