# Is there any theory to measures and determine the strength of an encryption protocol?

To determine the strength of an encryption protocol, there are elements like key size, Strength of an algorithm, Performance of an algorithm or etc, I have studied these topics but I could not find the answer I am looking for.
Is there any mathematical explanation in this topic, like a "theory of everything" that makes it possible for every encryption protocol.

Is there any mathematical explanation in this topic, like a "theory of everything" that makes it possible for every encryption protocol.

Sure. It's called concrete security proofs.

With these, you define a model that seems reasonable for your attacker (usually someone already did this), then you go ahead and model your protocol as close as possible to reality as possible while staying compatible with the definitional framework. Finally you show that the protocol satisfies the definition. As this usually involves the kind of implication "if you can break the security of the protocol, you can break <insert assumption>" you get security results of the form

For any efficient adversary $$\mathcal A$$ against AKE security of Example, there exists adversaries $$\mathcal B_1,\mathcal B_2$$ such that

$$\mathbf{Adv}^{\text{AKE}}_{\text{Example}}(\mathcal A)\leq \mathbf{Adv}^{\text{CCA}}_{E}(\mathcal B_1)+3\mathbf{Adv}^{\text{EUF-CMA}}_{S}(\mathcal B_2)$$

for some made-up protocol "Example" that achieves "Authenticated Key Exchange" security if the used encryption scheme is CCA-secure and the used signature scheme is EUF-CMA secure.

You can then research the security bounds for the advantages of your favourite encryption scheme and your favourite signature scheme and insert the expressions to come up with a bound on the security of the protocol.

• The notion of "bit security" is also interesting in this context. – Occams_Trimmer Jun 3 '19 at 8:37
• Of course this will still require you first have to define your own model (and, possibly, definition). Above is a generic answer, but you'd still have to make it specific, in other words; it's not a magic wand. – Maarten Bodewes Jun 3 '19 at 9:38