New answers tagged

1

Key exchange algorithms attempt to protect against eavesdropping. You have to assume what you send over the wire (C1, C2, and C3) are intercepted. That's a problem with the method because C2 is simply C1 xor Bob's random bytes and C3 is simply the key xor Bob's random bytes. An attacker with C1, C2, and C3 could take C1 xor C2 to get Bob's random bytes, ...


32

I've simplified the Alice random bytes to ARB and Bob random bytes to BRB. Then the protocol follows as; Alice knows $key$ and $ARB$ and sends $$C_1 = key \oplus ARB$$ Bob knows $C_1$ and $BRB$ and sends $$C_2 = C_1 \oplus BRB = key \oplus ARB \oplus BRB$$ Alice calculates $C_2 \oplus key \oplus ARB = key \oplus key \oplus ARB \oplus BRB = BRB$ Alice knows $...


2

You're wrong. This is a difference between TLS 1.2 and 1.3. In 1.2 SignatureAndHashAlgorithm identifies only the algorithm (not curve) and hash. In 1.3 SignatureScheme does identify the curve for ECDSA, and the certificate OID for RSA-PSS. See the next to last para on page 44: [1.3] ECDSA signature schemes align with TLS 1.2's ECDSA hash/signature pairs. ...


0

Maarten Bodewes claims that TLS provides no way for an application to use TLS to extract keying material; however, this is not the case. TLS does provide the key material exporter interface, which allows the two sides to derive secret keys (based on the TLS session) without exchanging any data records. This exported keys won't be the same as the traffic ...


1

1 - Handshake protocol: uses public-key cryptography (ECDH, DH) to establish a shared secret between the client and server. The handshake protocol is also used to establish a cipher suite, some configuration parameters and (usually) to perform entity authentication, amongst others. So, can I simply complete the handshake protocol and then drop the ...


Top 50 recent answers are included