So I have two examples and I don't understand why one would be chosen over the other.

Example 1. A sends EPUb[r1] to B (asks B to prove knowledge of PRb) B replies with r1, EPUa[r2] (asks A to prove knowledge of PRa) A replies with r2 to prove it

Example 2. A sends r1 to B (asks B to prove knowledge of PRb) B replies with r2, EPRb[h(r1)], CertB (provides B with signature and certificate and own challenge) A replies with EPRa[h(r2)], CertA

Why would example 2 be better than example 1? Are there any security loopholes I'm not seeing in example 1 than example 2 fixes?

I'm studying for my cryptography exam tomorrow and been stumped by this example. Fast responses would greatly appreciated.

From what I can pick up on these two examples. Example 2, you provide each other with your public keys and you know the origin is trustworthy after you verify the certificates. When communicating you can then use each others public keys for encryption or you could establish a symmetric key which eliminates the man in the middle attack possibility.

In Example 1 all I can think of is that without the certificate they can't be sure the other entity is who they claim to be but then again to know each others public keys there must be some form of trust between them. Is example 2 better just because you can be MORE sure that the other entity is who they claim to be?


1 Answer 1


In real communications, you often don't reliably know each others' public keys. X.509 certificates provide a way to verify a public key (as the certificate is signed by another key, and that key can be in a certificate signed by a third key, and so on, until we reach a public key that the other person does trust). Certificates are unnecessary when both parties already know each others' public keys through some form of out-of-band communication.

  • $\begingroup$ Brilliant! Exactly what I was looking for, thanks :) The lecture notes are very vague and assume we know everything about cryptography $\endgroup$ Commented Jan 19, 2015 at 17:32

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.