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Francesco Romeo added section_Diffie_Hellman_Key_Exchange__.tex
about 8 years ago
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\section*{Diffie-Hellman Key Exchange}
The first use of the DLP is the creation of a a key that Alice and Bob could use later to set up a Symmetric Cipher. \\
Alice and Bob communicates in an insecure channel then they exchange their key, by using the difficulty of solving DLP.\\
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Alice and Bob choose, even communicating through the insecure channel, a prime $p$, a primitive root $g$ of $\mathbb{F}_{p}$.\\
Then Alice chooses a \textit{secret} exponent $a$ and computes $$A \equiv g ^{a} \ \ ( \ mod \ p \ ) $$
At the same time Bob chooses his exponent $b$ and computes $$B \equiv g ^{b} \ \ ( \ mod \ p \ ) $$
Then Alice sends $A$ to Bob and Bob sends $B$ to Alice. \\
Finally with their exponents, Alice computes $B^{a} \ \ ( \ mod \ p \ ) $, Bob computes $A^{b} \ \ ( \ mod \ p \ )$ and we have that
$$B^{a} \equiv_{p} (g^{b})^a = (g^{a})^{b} \equiv_{p} A^{b}$$ and $g^{ab}$ is the shared key.\\
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The attacker Eve that wants to decrypt Alice and Bob's mesages has to face the following problem: she knows $p$ and $g$, she knows $A$ and $B$ because they are exchanged on the insecure channel, but she couldn't determine $g^{ab}$. \\
So the \textbf{Diffie-Hellman Problem (DHP)} is that given $g^{a}$ and $g^{b}$ it's not possible to determine $g^{ab}$ without first solving the DLP. \\
It's indeed true that if Eve could find solution at one of the two equations:
$$g^{x} \equiv A \ \ ( \ mod \ p \ )$$
$$g^{x} \equiv B \ \ ( \ mod \ p \ )$$
finding e.g. $a$ she could compute $B^{a} \ \ ( \ mod \ p \ )$.
