Take the 2-minute tour ×
Cryptography Stack Exchange is a question and answer site for software developers, mathematicians and others interested in cryptography. It's 100% free, no registration required.

I'm preparing myself to exam, but I have a lot of troubles with rigorous proofs.

Let $\Pi=(Gen,Enc,Dec)$ be an efficient secret-key encryption scheme that is not fixed-length. That is, for any $n$ and any $k \leftarrow Gen(1^n)$ the encryption algorithm $Enc_k(\cdot)$ can encrypt arbitrary length messages $m \in \lbrace 0, 1 \rbrace^*$.

Prove that $\Pi$ cannot satisfy definition of being one-time computationally-secret when the adversary $\mathcal{A}$ in $PrivK^{eav}_{\mathcal{A}}$ may output messages $m_0$ and $m_1$, that are NOT of the same length.

I just can (I think so) start this proof - if $Enc_k(\cdot)$ is some PPT algorithm, then there exists a polynomial $p(x) \in \mathbb{Z}[x]$ such that

$\forall n,k \leftarrow Gen(1^n), m \in \lbrace 0, 1 \rbrace^*: ||Enc_k(m)||<p(||k||+||m||)$.

Unfortunately I have no idea how to do the rest of the proof - by contradiction, or not? If you could help me with the rigorous proof, I'd be really grateful for your time.

P.S. Reminder (one-time computationally-secret).

An (efficient secret-key) encryption scheme $(Gen,Enc,Dec)$ is one-time computationally-secret if for any PPT adversary $\mathcal{A}$ it holds that $Pr[PrivK^{eav}_{\mathcal{A}}(n)=1]-\frac{1}{2}$ is negligible function, where $PrivK^{eav}_{\mathcal{A}}(n)$ denotes the output of the following experiment:

(a) The adversary $\mathcal{A}$ on input $1^n$ outputs a pair of messages $m_0,m_1$.

(b) Let $k \leftarrow Gen(1^n)$ and let $b \in \lbrace 0,1 \rbrace$ be chosen uniformly at random. Then a ciphertext $c \leftarrow Enc_k(m_b)$ is computed and given to $\mathcal{A}$.

(c) $\mathcal{A}$ on input $c$ outputs a bit $b'$.

(d) The output of the experiment is $1$ if $b'=b$ and $0$ otherwise.

share|improve this question
    
Hint: The adversary needs to construct two messages such that their ciphertexts are of different length. Use the efficiency condition to show that this is always possible. –  Paŭlo Ebermann Oct 28 '12 at 20:11
    
If this is not enough, have a look at our recent questions Why is a non fixed-length encryption scheme worse than a fixed-length one? and How to construct a variable length IND-CPA cipher from a fixed length one? (including the discussion in the comments). –  Paŭlo Ebermann Oct 28 '12 at 20:24
    
What is efficiency condition? Could you tell me a little bit more how should the formal proof be looking like? I think I know what is the idea, but I have big problem with writing it down rigorously. Thank you. –  BiggBen1989 Oct 30 '12 at 0:01
    
Look up the meaning of the word "efficient" in your definition. (It means something like "a short input will only take a short time", and as such it can also only have a short output. But look it up in your literature.) –  Paŭlo Ebermann Nov 1 '12 at 13:07

1 Answer 1

up vote 0 down vote accepted

This is already explained well, at a conceptual level, at a different question: Why is a non fixed-length encryption scheme worse than a fixed-length one?

To turn it into a formal proof, write down an adversary $A$ that breaks the security of $\Pi$. You may want to first work out the attack at a conceptual level to make sure you are clear on how to attack $\Pi$; then review the definition of what it means for an adversary to break $\Pi$; then turn your conceptual attack into a fully specified algorithm $A$, and check that it does indeed break $\Pi$.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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