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Perfect security - is this definition correct?

br flag

I have this definition:

each ciphertext is equally probable for a given plaintext and key chosen at random

I know that perfect security can be defined as $$\forall c \in \mathcal{C} \ \forall m_1,m_2\in \mathcal{M} \ Pr[Enc_k(m_1)=c \ for \ k \ random]=Pr[Enc_k(m_2)=c \ for \ key \ random]$$

Are these equivalent?

The easiest thing to do is to somehow show that the first definition implies the second one and vice-versa. Although I have problems with understanding these conditions. If I rewrite the first definition in a more mathematical way: Let $m \in \mathcal{M}$ $$ \forall c_1, c_2\in \mathcal{C} \ Pr[Enc_k(m)=c_1]= Pr[Enc_k(m)=c_2] $$

However, this didn't help me much. My second thought was something along the lines: if the probability for every cipher is equal then it must be equal to $\frac{1}{|\mathcal{C}|}$ (I can't imagine it could have any other value). If this is the case for every message, then it also satisfies the second definition as every probability has the same value.

I'm not sure about this reasoning, is this ok or completely wrong?

Maarten Bodewes avatar
in flag
I don't see any particularly wrong with the reasoning and yes, I think the first sentence and formula are kind of the same, but the second formula probably fits it better.
meshcollider avatar
gb flag
You need to include that $k$ is drawn uniformly from the keyspace for the probabilities in the second equation to make sense. It looks like a fixed key $k$ at the moment.
Maarten Bodewes avatar
in flag
Yeah, true, because the definition reads "key chosen at random" while the message is fixed.
Awerde avatar
br flag
My reasoning is in one way (first -> second), how to prove it the other way?
Maarten Bodewes avatar
in flag
I'm not so good at rewriting these kind of proofs; for me this is kind of self-evident. Maybe somebody else can step in and help.
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