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Fixed some spelling mistakes, thanks Henrik. again

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Aaron Kaiser
2023-06-14 16:23:56 +02:00
parent 497a789df3
commit 97e842892c
9 changed files with 28 additions and 28 deletions
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6
thesis/sections/mu_security_of_eddsa/omdl'_implies_mu-gamez.tex
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@@ -6,7 +6,7 @@ This section shows that \somdl implies MU-\igame using the algebraic group model
\begin{definition}[\somdl]
\label{def:somdl}
Let $n$ and $N$ be positive integer. For an adversary $\adversary{A}$, receiving $N$ challenge group elements, we define its advantage in the \somdl game as following:
Let $n$ and $N$ be positive integers. For an adversary $\adversary{A}$, receiving $N$ challenge group elements, we define its advantage in the \somdl game as following:
\[ \advantage{\adversary{A}}{\text{\somdl}}(\secparamter) \assign | \Pr[\text{\somdl}^{\adversary{A}} \Rightarrow 1] |. \]
\end{definition}
@@ -43,7 +43,7 @@ This section shows that \somdl implies MU-\igame using the algebraic group model
\[ \advantage{\group{G},\adversary{A}}{\text{MU-\igame}}(\secparamter) \leq \advantage{\group{G},\adversary{B}}{\somdl}(\secparamter) + \frac{\oraclequeries N}{2^{-\log_2(\lceil \frac{2^{2b} - 1}{L} \rceil 2^{-2b})}}. \]
\end{theorem}
\paragraph{\underline{Proof Overview}} In the multi-user setting the adversary gets access to not only the generator $\groupelement{B}$ and one public key $\groupelement{A}$ but rather a set of public keys $\groupelement{A_1}$ to $\groupelement{A_N}$. For this reason the representation of a group element, the adversary has to provide, looks the following: $\groupelement{R} = r_1 \groupelement{B} + r_2 \groupelement{A_1} + ... + r_{N+1} \groupelement{A_N}$. Since there are multiple group elements with unknown discrete logarithms it is not possible to directly calculate the discrete logarithm of one of the public keys given a valid forgery of a signature. Upon receiving a valid solution the \textit{DL} oracle can be used to get the discrete logarithm of all the public keys except the one for which the solution is valid. This way it again possible to construct a representation looking like $\groupelement{R} = r_1 \groupelement{B} + r_2 \groupelement{A_i}$. Then it is again possible to calculate the discrete logarithm of $\groupelement{A_i}$ and win the \somdl game.
\paragraph{\underline{Proof Overview}} In the multi-user setting the adversary gets access to not only the generator $\groupelement{B}$ and one public key $\groupelement{A}$ but rather a set of public keys $\groupelement{A_1}$ to $\groupelement{A_N}$. For this reason, the representation of a group element, the adversary has to provide, looks the following: $\groupelement{R} = r_1 \groupelement{B} + r_2 \groupelement{A_1} + ... + r_{N+1} \groupelement{A_N}$. Since there are multiple group elements with unknown discrete logarithms it is not possible to directly calculate the discrete logarithm of one of the public keys given a valid forgery of a signature. Upon receiving a valid solution the \textit{DL} oracle can be used to get the discrete logarithm of all the public keys except the one for which the solution is valid. This way it is again possible to construct a representation looking like $\groupelement{R} = r_1 \groupelement{B} + r_2 \groupelement{A_i}$. Then it is again possible to calculate the discrete logarithm of $\groupelement{A_i}$ and win the \somdl game.
\paragraph{\underline{Formal Proof}}
@@ -88,7 +88,7 @@ This section shows that \somdl implies MU-\igame using the algebraic group model
\[ \advantage{\group{G},\adversary{A}}{\text{MU-\igame}}(\secparamter) = \Pr[\text{MU-\igame}^{\adversary{A}} \Rightarrow 1] = \Pr[G_0^{\adversary{A}} \Rightarrow 1]. \]
\item \paragraph{\underline{$G_1$:}} $G_1$ is defined by including the if condition in the blue box, which sets a bad flag if the randomly chosen value $\ch$ fulfills $2^c \ch \equiv - r_i \pmod L$ for any $i \in \{2,3,...,N+1\}$. This represents challenges $\ch$ to which the solution might not be usable to break the discrete logarithm of one of the public keys, due to $(r_i + 2^c \ch)$ not being invertible in $\field{L}$. Since only the bad flag being introduced this change does not influence the behavior of the game and is therefore only conceptual.
\item \paragraph{\underline{$G_1$:}} $G_1$ is defined by including the if condition in the blue box, which sets a bad flag if the randomly chosen value $\ch$ fulfills $2^c \ch \equiv - r_i \pmod L$ for any $i \in \{2,3,...,N+1\}$. This represents challenges $\ch$ to which the solution might not be usable to break the discrete logarithm of one of the public keys, due to $(r_i + 2^c \ch)$ not being invertible in $\field{L}$. Since only the bad flag is introduced this change does not influence the behavior of the game and is therefore only conceptual.
\[ \Pr[G_0^{\adversary{A}} \Rightarrow 1] = \Pr[G_1^{\adversary{A}} \Rightarrow 1]. \]