New answers tagged

5

OpenPGP is a hybrid cryptosystem. The actual message is encrypted applying a symmetric cipher like AES with a random session key. This session key again is encrypted using a public/private key cryptography algorithm like RSA. This is mostly because symmetric encryption is much faster than public/private key cryptography, especially for large messages. As the ...


3

Exchanging full RSA keys can be very inconvenient, as they consist of very long numbers. In the OpenPGP ASCII-armored version, my public key (without any user IDs and certifications!) already is very long stream: mQQNBFDaK/sBIADm2gjnw7aPoNIoCy7gj85btwZU+zGkvtGonznlLrXELdU6zR3u VHNCn9vAl6OoU32r+suFvGdX+7MjiPbGKwJFOvICpAVh6bV55+hdqJbS02cpPmJH ...


1

The padding used for RSA is not the PKCS #5/#7 padding (as you seem to suggest in your own answer), but the Wikipedia entry seems to refer to PKCS #1 v1.5 (RFC2313)) which uses a padding 00 || BT || PS || 00 || D where for RSA encryption we start with a 0x00-byte (to guarantee that the resulting number is below the modulus), then use BT (Block Type) equal ...


0

I naively forgot that with a "random" amount of padding added the receiver would not know how many bytes to remove. That is why PKCS #5/#7 add the number of bytes as the padding value. For example, if we need to add 4 bytes then the last part of the message would look like: [... 04 04 04 04]. So the answer is PKCS #7 is the only possible way to add padding ...


1

TL;DR: This attack extends the standard chosen ciphertext attacks on RSA and ElGamal to the hybrid encryption setting, but requires some huge IFs which no longer even have a slight possibility of being fulfilled. To understand this attack, one first needs to understand how data is encrypted for PGP according to this snippet. Let's assume you have a ...


5

RFC 4880, OpenPGP (superseded RFC 2440 which was up to date in 2002) contains a chapter on security considerations, which also discusses the decryption oracle attack Jallad et al described: In late summer 2002, Jallad, Katz, and Schneier published an interesting attack on the OpenPGP protocol and some of its implementations [JKS02]. In ...


3

This is actually a rather interesting question, whiches solution is obvious to all cryptographers but I guess nobody cared yet to write it down. After all, our computers who are generating secret keys (not just GPG / RSA) are deterministic machines. These deterministic machines implement well-defined routines to generate keys of well-defined format which ...



Top 50 recent answers are included