164 lines
12 KiB
TeX
164 lines
12 KiB
TeX
\subsection{\sdlog $\overset{\text{AGM}}{\Rightarrow}$ \igame}
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\label{sec:sdlog_imlies_igame}
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This section shows that \sdlog implies \igame using the algebraic group model. The section begins with an introduction to a special variant of the discrete logarithm problem, followed by an intuition of the proof, and finally a detailed security proof.
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\paragraph{\underline{Introducing \sdlog}}
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The \sdlog game is a variant of the discrete logarithm game that represents the clearing and setting of bits in the secret scalar during EdDSA key generation. The only difference to the normal discrete logarithm game is that the secret scalars are not randomly chosen from $\field{L}$, where $L$ is the order of the generator, but from the set $\{2^{n-1}, 2^{n-1} + 2^c, ..., 2^{n} - 2^c\}$. This set represents all valid private keys according to the key generation algorithm. The hardness of this version of the discrete logarithm problem is further analyzed in section \ref{sec:sdlog}. The \sdlog game is illustrated in figure \ref{fig:sdlog}.
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\begin{definition}[\sdlog]
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\label{def:sdlog}
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For an adversary $\adversary{A}$ we define its advantage in the \sdlog game as following:
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\[ \advantage{\group{G}, \adversary{A}}{\text{\sdlog}}(\secparamter) \assign | \Pr[\text{\sdlog}^{\adversary{A}} \Rightarrow 1] |. \]
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\end{definition}
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\begin{figure}[h]
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\hrule
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\vspace{1mm}
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\begin{algorithmic}
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\Statex \underline{\game \sdlog}
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\State $a \randomsample \{ 2^{n-1}, 2^{n-1} + 2^c, ..., 2^{n} - 2^c \}$
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\State $\groupelement{A} \assign a \groupelement{B}$
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\State $a' \randomassign \adversary{A}(\groupelement{A})$
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\State \Return $a \test a'$
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\end{algorithmic}
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\vspace{1mm}
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\hrule
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\caption{\sdlog}
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\label{fig:sdlog}
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\end{figure}
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\begin{theorem}
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\label{theorem:advgamez}
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Let $\adversary{A}$ be an adversary against \igame with $\group{G}$ being a cyclic group of prime order $L$, making at most $\oraclequeries$ oracle queries. Then
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\[ \advantage{\group{G},\adversary{A}}{\igame}(\secparamter) \leq \advantage{\group{G},\adversary{B}}{\sdlog}(\secparamter) + \frac{\oraclequeries \lceil \frac{2^{2b} - 1}{L} \rceil}{2^{2b}}. \]
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\end{theorem}
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\paragraph{\underline{Proof Overview}}
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The adversary must call the \ioracle oracle with a commitment $\groupelement{R}$ to get a challenge from the challenger. Due to the nature of the algebraic group model, the adversary must also provide a representation of the group element $\groupelement{R}$ as a linear combination of all known group elements. Since only the generator of the group and the public key are known to the adversary, the representation looks like this: $\groupelement{R} = r_1 \groupelement{B} + r_2 \groupelement{A}$. Together with a valid solution to the \igame game, this can be used to compute the discrete logarithm of the public key.
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\begin{figure}
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\hrule
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\begin{multicols}{2}
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\large
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\begin{algorithmic}
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\Statex \underline{\game $G_0$ / \textcolor{blue}{$G_1$} / \textcolor{red}{$G_2$}}
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\State $a \randomsample \{2^{n-1}, 2^{n-1} + 8, ..., 2^n - 8\}$
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\State $\groupelement{A} \assign a \groupelement{B}$
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\State $s^* \randomsample \adversary{A}^{\ioracle(\inp)}(\groupelement{A})$
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\State \Return $\exists (\groupelement{R}^*, \ch^*) \in \pset{Q}: \groupelement{R}^* = 2^c (s^* \groupelement{B} - \ch^* \groupelement{A})$
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\end{algorithmic}
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\columnbreak
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\begin{algorithmic}[1]
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\Statex \underline{\oracle \ioracle($\agmgroupelement{R_i}{r_i} \in \group{G}$)}
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\State Let $\groupelement{R}_i = r_1 \groupelement{B} + r_2 \groupelement{A}$
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\State $\ch_i \randomsample \{0,1\}^{2b}$
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\BeginBox[draw=blue]
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\State \textbf{If} $2^c \ch_i \equiv -r_2 \pmod L$ \textbf{then}
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\Comment{$G_1 - G_2$}
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\State \quad $bad \assign true$
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\BeginBox[draw=red,dashed]
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\State \quad $abort$
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\Comment{$G_2$}
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\EndBox
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\EndBox
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\State $\pset{Q} \assign \pset{Q} \cup \{ (\groupelement{R}_i, \ch_i) \}$
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\State \Return $\ch_i$
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\end{algorithmic}
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\end{multicols}
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\hrule
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\caption{Games $G_0 - G_2$}
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\label{fig:igamewithabort}
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\end{figure}
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\paragraph{\underline{Formal Proof}}
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\begin{proof}
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\item The proofs begin by showing that the only valid representation of a group element in the prime order subgroup is the one relative to all known elements in the subgroup and cannot include elements from outside the subgroup. This is followed by a discussion of the individual game-hops.
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\item \paragraph{\underline{AGM}} This proof is done in the algebraic group model. This means that the adversary has to provide a representation along each group element he provides to the reduction. The adversary must provide an element $\groupelement{R}$ which is an element in the prime order subgroup of the twisted Edwards curve. The question remains whether the representation should be defined relative to the prime order subgroup or the twisted Edwards curve. The answer to this question is that it is sufficient to define the representation relative to the prime order subgroup. The reason for this is given in the following paragraph.
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The twisted Edwards curve $\curve$ over the finite field $\mathbb{F}_{q}$ is a finite abelian group. Even though the group $\curve$ may not be cyclic, the Fundamental Theorem of Finitely Generated Abelian Groups tells us that every finite abelian group can be uniquely decomposed into the direct product of cyclic subgroups \cite{karpfinger_hauptsatz_2021}. This means that $\curve$ can be written as $\curve = \langle a_1 \rangle \bigotimes \langle a_2 \rangle \bigotimes ... \bigotimes \langle a_n \rangle$. The set of generators for each of the cyclic groups is called the generating set of $\curve$. Let us recall a well-known theorem of algebra:
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\item \begin{theorem}[Characterization of Inner Direct Products \cite{karpfinger_direkte_2021}]
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Let $N_1, ..., N_n$ be subgroups of a group $\group{G}$. Following statements are equivalent:
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\begin{enumerate}[label=(\arabic*)]
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\item $N_1, ..., N_n \trianglelefteq \group{G}$ and $\group{G} = N_1 \bigotimes ... \bigotimes N_n$.
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\item Each $x \in \group{G}$ can uniquely be represented in the following way:
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\[ x = a_i \cdot ... \cdot a_n, a_i \in N_i \]
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\end{enumerate}
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\end{theorem}
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Due to Sylow's theorems, the decomposition must include the large prime order subgroup $\group{G}$ used for EdDSA \cite{karpfinger_satze_2021}, and since twisted Edwards curves (like all elliptic curves) are abelian, each subgroup is also a normal subgroup. Together this means that the representation of each element $\groupelement{X} \in \curve$ is unique relative to the generating set. Since each element $\groupelement{Y} \in \group{G}$ can be represented as $\groupelement{Y} \assign y \groupelement{B}$, where $\groupelement{B}$ is the generator of the prime order subgroup, this must be the only representation with respect to the generating set. This means that an adversary, in the algebraic group model, must provide a representation in the prime order subgroup $\group{G}$.
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The only two group elements in $\group{G}$ provided to the adversary are the public key $\groupelement{A}$ and the generator $\groupelement{B}$. Therefore, the representation of the element $\groupelement{R}$, provided to the \ioracle oracle, looks like $\groupelement{R} = r_1 \groupelement{B} + r_2 \groupelement{A}$.
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\item \paragraph{\underline{$G_0$:}} Let $G_0$ be defined in figure \ref{fig:igamewithabort} by excluding all boxes. $G_0$ is the same as \igame. By definition,
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\[ \advantage{\group{G},\adversary{A}}{\igame}(\secparamter) = \Pr[\igame^{\adversary{A}} \Rightarrow 1] = \Pr[G_0^{\adversary{A}} \Rightarrow 1]. \]
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\item \paragraph{\underline{$G_1$:}} Game $G_1$ is exactly the same as $G_0$ with the only change being that the bad flag is set inside an if condition. The bad flag is set when $2^c \ch_i = -r_2$. This represents cases where not all solutions from the adversary $\adversary{A}$ can be used to calculate the discrete logarithm of $\groupelement{A}$. This is only a conceptual change, since the behavior of the game does not change whether the flag is set or not. Hence,
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\[ \Pr[G_0^{\adversary{A}} \Rightarrow 1] = \Pr[G_1^{\adversary{A}} \Rightarrow 1]. \]
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\item \paragraph{\underline{$G_2:$}} The game $G_2$ is aborted if the bad flag is set. For each individual \ioracle query, the $bad$ flag is set with probability at most $\frac{1}{2^{-\log_2(\lceil \frac{2^{2b} - 1}{L} \rceil 2^{-2b})}}$. $\ch_i$ is chosen by the game after the adversary has provided the representation of $\groupelement{R_i}$ and thus the value of $r_2$. This way the adversary has no way to choose $\ch_i$ after $r_2$ and therefore cannot influence the probability of the abort being triggered. $-\log_2(\lceil \frac{2^{2b} - 1}{L} \rceil 2^{-2b})$ is the min entropy of $\ch_i \pmod L$. $\ch_i$ is chosen uniformly at random from $\{0,1\}^{2b}$ and then reduced modulo $L$ during the if condition check. By the union bound over all oracle queries $\oraclequeries$ we obtain $\Pr[bad] \leq \frac{\oraclequeries}{2^{-\log_2(\lceil \frac{2^{2b} - 1}{L} \rceil 2^{-2b})}}$. Since $G_1$ and $G_2$ are identical-until-bad games, we have
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\[ |\Pr[G_1^{\adversary{A}} \Rightarrow 1] - \Pr[G_2^{\adversary{A}} \Rightarrow 1]| \leq \Pr[bad] \leq \frac{\oraclequeries \lceil \frac{2^{2b} - 1}{L} \rceil}{2^{2b}}. \]
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\item Finally, Game $G_2$ is well-prepared to show that there exists an adversary $\adversary{B}$ satisfying
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\begin{align}
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\Pr[G_2^{\adversary{A}} \Rightarrow 1] = \advantage{\group{G},\adversary{B}}{\sdlog}(\secparamter). \label{eq:advbsdlog}
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\end{align}
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\begin{figure}[h]
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\hrule
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\begin{multicols}{2}
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\large
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\begin{algorithmic}
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\Statex \underline{\textbf{Adversary} $\adversary{B}(\groupelement{A})$}
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\State $s^* \randomassign \adversary{A}^{\ioracle(\inp)}(\groupelement{A})$
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\State \textbf{If} $\nexists \agmgroupelement{R^*}{r^*}, \ch^*: \groupelement{R}^* = 2^c (s^* \groupelement{B} - \ch^* \groupelement{A}) \wedge (\agmgroupelement{R^*}{r^*}, \ch^*) \in \pset{Q}$ \textbf{then}
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\State \quad $abort$
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\State Let $R^* = r_1 \groupelement{B} + r_2 \groupelement{A}$
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\State \Return $(2^c s^* - r_1)(r_2 + 2^c \ch^*)^{-1}$
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\end{algorithmic}
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\columnbreak
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\begin{algorithmic}
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\Statex \underline{\oracle \ioracle($\agmgroupelement{R_i}{r_i} \in \group{G}$)}
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\State Let $\groupelement{R}_i = r_1 \groupelement{B} + r_2 \groupelement{A}$
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\State $\ch_i \randomsample \{0,1\}^{2b}$
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\State \textbf{If} $2^c \ch_i \equiv -r_2 \pmod L$ \textbf{then}
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\State \quad $bad \assign true$
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\State \quad $abort$
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\State $\pset{Q} \assign \pset{Q} \cup \{ (\agmgroupelement{R_i}{r_i}, \ch_i) \}$
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\State \Return $\ch_i$
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\end{algorithmic}
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\end{multicols}
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\hrule
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\caption{Adversary $\adversary{B}$ breaking \sdlog}
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\label{fig:adversarybsdlog}
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\end{figure}
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To prove (\ref{eq:advbsdlog}), we define an adversary $\adversary{B}$ attacking \sdlog, which simulates the view of $\adversary{A}$ in $G_2$. The adversary $\adversary{B}$ formally defined in figure \ref{fig:adversarybsdlog} is run in the \sdlog game and adversary $\adversary{B}$ simulates \ioracle for adversary $\adversary{A}$. The \ioracle oracle is simulated perfectly.
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Finally, consider $\adversary{A}$'s output $s^*$. We know that one $R^* = 2^c s^*B - 2^c \ch^*A$. We can use this together with the representation of $R^*$ to get the following equation:
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\begin{align*}
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r_1 \groupelement{B} + r_2 \groupelement{A} &= 2^c s^* \groupelement{B} - 2^c \ch^* \groupelement{A} \\
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\Leftrightarrow (r_2 + 2^c \ch^*)A &= (2^c s^* - r_1)B \\
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\Leftrightarrow A &= (2^c s^* - r_1)(r_2 + 2^c \ch^*)^{-1} B
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\end{align*}
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Assuming that $r_2 + 2^c \ch^*$ is invertible in $\field{L}$ (i.e. not equal to $0$), which is ensured due to the abort in $G_2$, both equations can be used to calculate the discrete logarithm of $\groupelement{A}$.
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\item Obviously, the runtime of $\adversary{B}$ is roughly the same as the runtime of $\adversary{A}$. The \ioracle just samples the challenge uniformly at random and returns it after checking the abort condition. After $\adversary{A}$ has provided its solution, adversary $\adversary{B}$ just does some additions, multiplications, and an inversion, which does not add much to its runtime.
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\item This proves theorem \ref{theorem:advgamez}.
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\end{proof} |