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4

This has been specified by the standard, steps 4 and 5 of the protocol described in RFC 4253: S generates a random number y (0 < y < q) and computes f = g^y mod p. S receives e. It computes K = e^y mod p, H = hash(V_C || V_S || I_C || I_S || K_S || e || f || K) (these elements are encoded according to their types; see below), ...


3

The MAC is NOT redundant. As alluded to by PaĆ­lo Ebermann's comment, the word authentication has a different meaning in the two scenarios you mentioned. In the key exchange phase of SSH, the purpose of authentication is to ensure to both parties that they are indeed talking to the right peer (if using mutual authentication). Typically, the server ...


3

I'm assuming you mean a base 64 encoded key file, since removing the newlines from a binary file would obviously break things. The RSA standards (e.g. RFC 2459) only define a binary representation for keys. In practice, like OpenPGP keys (RFC 4880), they are often encoded in base 64 using the otherwise obsolete PEM standards (RFC 1421). The PEM printable ...


3

Other advantages of CTR are: easier to decrypt from a certain offset within the ciphertext no randomness requirements for the nonce nonce can be calculated, e.g. be a simple counter nonce can be a message identifier $E = D$: encryption is the same as decryption, which means only encryption or decryption required from the block cipher less logic ...


3

There seems to be an attack on SSH when using CBC: Plaintext Recovery Attacks Against SSH. I have just scanned the paper and they state, that this will not be possible when CTR mode is used. I don't think that en-/decryption parallelization is need or even utilized in SSH. Update: Link to CERT concerning the topic: Vulnerability Note VU#958563 SSH CBC ...


2

I don't think there is a dedicated name for this. If I had to find a word, it would probably be "stateful" or "with explicit state". What you observe is actually the usual case: The user initializes an encryption system with a key, resulting in some state of the system. Then, each time the user wishes to encrypt some data, he has to pass the current state ...


2

Your example is missing something: your two calls to the OpenSSL library are entirely entirely independent, but your calls to the mcrypt library reuse an existing handle. CBC has the property that identical plaintext blocks are exceedingly unlikely to encrypt to the same value due to chaining of the previous output into the next input. Your OpenSSL calls ...


1

In addition to yyyyyyy's answer, there is also probabilistic encryption, in which the encryption process incorporates some randomness. That allows many identical messages with the same key to be encrypted differently; it doesn't necessarily encrypt the same text differently in the same message (it might encrypt $a+a$ as $E(a)+E(a)$), but it does let you send ...



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