Brute forcing the key would hardly be an issue: 128-bit keys (assuming they have been properly generated) are in a space which is way too large to be successfully explored by brute force; and 256-bit keys (the kind you put in AES-256) are even more larger. Whether AES is "faster" than HMAC or not does not make such brute force more feasible: even if each key ...
Definitions / Introduction
We define (this is solely for our example):
$enc()$ and $dec()$ as the encryption and decryption function using CBC mode, with a constant key. any block cipher will do
$\oplus$ is the XOR operation.
$n$ the amount of plain text blocks.
the length of a block in bytes is $16$ (i.e. 128 bit).
$m_1$ through $m_n$ the plain text ...
Clearly, if you had been using AES-256-CBC for confidentiality and AES-256-CBC-MAC for authentication, it would not be secure to use the same key for both confidentiality and authentication. Hence, using the same key for confidentiality and authentication cannot generally be secure; you need additional premises to arrive at that conclusion. In your case it ...
How does the new attack work at top level?
They used BEAST-like Man in the Browser attack by using Cache-like attacks to perform a downgrade attack against any TLS connection to a vulnerable server. With this, they showed the feasibility of using Cache-like attacks.
Even over the years, numerous mitigation techniques are ...
In the padding oracle attack you have an oracle that only tells you whether a particular chosen ciphertext decrypts to a correctly padded plaintext. That oracle is used to build a last word oracle, which used iteratively can reveal a whole message.
The reason it works in CBC mode is that we can make predictable, arbitrary changes to the plaintext of the ...
Am I not understanding padding correctly?
You're not understanding the padding attack correctly.
Yes, only the last block is (typically) padded in CBC mode, however that doesn't mean that we can only attack the last block; what that means is that we can only use the last block.
In CBC-mode, the decryption of the block $i$ is computed as $P_i = D_k( C_i ) \...
With pure asymmetric encryption there is no way to ensure integrity and authenticity, since anyone who knows your public key can encrypt any message for you. For that you would need either a symmetric key to use for a MAC (in which case you could use it/derivatives for symmetric encryption too) or a signature from the sender. And in the latter case the ...
It's a question of attack model. Vaudenay places himself in a model where the attacker is "outside", and trying to work out the password that a human user enters in a Web site (or another similar password-protected protocol). Both client and server systems are honest and truthfully run the TLS protocol; the attacker can inspect and alter packets, but that's ...
Does PKCS#5 allow you to pad 16 bytes with 0x10 value?
No, but that's because PKCS#5 only specifies padding for DES, and that has an 8 byte block size. Even PKCS#5 itself refers to PKCS#7 when it comes to padding AES. More information about the differences here.
Now for the real answer. PKCS#7 padding, which is identical to PKCS#5 but allows any block size,...
Well, when TLS encrypts a stream of traffic, it breaks up the data into records. Each record contains up to 16k of data, and includes its own HMAC. Hence, there is no HMAC over the entire downloaded file; instead, there are a series of HMACs; every single byte of data is covered by one of those HMACs.
…are any other modes of operation vulnerable to padding oracle attacks?
According to Vaudenay, it’s purely restricted to CBC. (Two years later, this was shown to be somewhat incorrecr.)
A padding oracle attack is also known as “Vaudenay attack” because it was originally published by Serge Vaudenay in 2002 and introduced at EUROCRYPT 2002, is an attack ...
You have to differentiate between it being theoretically OK and practically. Without fully checking, I can say that such an approach may be possible to fully prove. However, practically, it is almost impossible to implement it. Specifically, you have to be able to make it impossible to differentiate between an error due to the symmetric decryption or the ...
I implemented this padding oracle attack some time ago in Python and remember this part being a bit confusing to wrap my head around, my code is as follows:
newM = 
for (a, b) in M:
# util.ceiling rounds arg1 / arg2 to the next highest integer
rlow = util.ceiling((a*s - 3*B + 1), n)
rhigh = (b*s - 2*B) / n
for r in range(rlow, rhigh + 1):
Their attack does not recover the private key. Instead, it gives the attacker a way to decrypt an arbitrary ciphertext of the attacker's choosing. (This is not the same thing.)
If the attacker has a ciphertext $c$, the attacker can query the hardware device tens of thousands of times and then based upon the responses, deduce what the decryption of $c$ is. ...
For a key recovery attack, you'd basically need to break AES itself. There are no known practical key recovery attacks on AES (and if there were, it would not be considered safe to use), so your pretty much only hope would be to find some kind of side-channel attack on the AES implementation, or on the overall crypto framework it is part of.
Here's a real-world example: A captcha service behaves as follows. First, it generates a distorted image of some known text $t$. It sends then image to a web browser, and waits for a response. It compares the response to $t$ to decide whether the captcha is successful.
In a stateful solution, the captcha service itself would have to store $t$ while the web ...
These should both be valid paddings:
... ... ... 68 02 02
... ... ... 68 02 01
But if this ambiguity comes up, we can distinguish between the two by modifying the second-to-last byte in some way. After flipping, say, the highest four bits:
... ... ... 68 f2 02
... ... ... 68 f2 01
Only the second one is a valid padding.
I am afraid you didn't understand the algorithm of the padding oracle. I suggest studying the algorithm given at Vaudenay's original article on this. He is very concise in what he writes in the article, but it is there.
It is deterministic, doesn't assume anything about plain bytes and works like this:
The last block of ciphertext always has to contain ...
Padding oracle attacks are not the padding scheme's fault! They are not fixed by changing the padding! They are fixed by not leaking partial information about the result of decryption, or (preferably) by just using CCA encryption.
Any time you provide this to an adversary:
* Accept arbitrary c
* Decrypt m = Dec(c)
* Reveal formatcheck(m)
... if it's only ...
So, how do you reveal the first byte of a block, since you’re out of padding length?
We are are not out of padding length. However,
PKCS#5 (rfc 2898) is defined for 8-byte blocks i.e. for 64 block size like DES, and
PKCS#7 (rfc2315) is defined up to 256-byte blocks that covers AES, too.
Therefore, for 16-byte blocks, let assume we are using PKCS#7.
We have a Padding Oracle if there is a different response from the server gives us an indication of the correctness of the pad (say if this needs proving). We can establish this by playing a game where we send badly padded cipher-text and random strings to the server, finally submitting some at random and seeing if we can get a non-negligible Advantage is ...
Why does this prevent the attack? Why doesn't the attacker just infer that the connection failed because of the bad padding? Why else could the connection fail?
Well, the connection may fail because the host decrypted a valid pre-master secret, and it wasn't the pre-master secret that we expect.
That is, when the attacker injects his encrypted message, one ...
Bleichenbacher's attack relies on being able to determine whether the padding was correct or not. The patch tries to ensure that the following two (previously distinguishable) cases look identical to an attacker:
the padding was correct, but the attacker has no knowledge of the transmitted pre-master secret — hence he can't use the resulting symmetric keys ...
Your suggestion is essentially what ISO 10126 does, since there's no way to verify the random bytes that make up the rest of the padding. You could do the same with e.g. PKCS #7 padding, as you suggest.
However, this would leave a covert channel. If those other padding bytes are not verified, they can be chosen by the sender and even modified by an attacker ...
First of all, a more usual padding scheme would add 5 times the same byte 0x05 (in your example) so the check not just removes 5 bytes, but also checks that the 4 bytes before it have the same value. But let's assume your scheme (which is underspecified: what to put in the bytes before? Zeroes, or random values?) for now.
What if you cannot remove that ...
Quoting the answer here:
Padding Oracle attacks are mainly a problem in cases, where e.g. an encrypted message is modified and send to a target. These attacks try to measure the difference when decrypting and validating the message.
The steps are:
decrypting the message
checking the padding > error if wrong
checking or processing the data > error if ...
Actually, we don't care about the amount of padding the original message had; we care whether the modified plaintext (that is, the result of the decryption of the modified ciphertext) has good padding or not.
The padding will be valid if the last block had one of these patterns:
XX XX XX XX XX XX XX 01
XX XX XX XX XX XX 02 02
XX XX XX XX XX 03 ...
First of all, this attack assumes that we have a decryption Oracle; that is, an Oracle that we can submit any ciphertext to (except for, apparently, the challenge ciphertext), and it will tell us that the padding is bad, or it will give us the decrypted message (minus the padding).
In the attack, we're given the challenge ciphertext $(Pad + M)^e$, and we ...
You won't find the answer in the RFC; instead, here's what your SSL implementation is doing: when you send a single byte of data, it actually generates two records.
The first record is empty; so it consists of the 5 byte record header, and the encryption of the 20 byte HMAC tag (which, after encryption padding, is 32 bytes long); this is a total of 37 bytes....