\subsection{\sdlog $=>$ \igame (AGM)}

%TODO check if all c_i's are replaced by chall_i

This section shows that \sdlog implies \igame using the Algebraic Group Model. The section starts by introducing a special variant of the discrete logarithm problem followed by an intuition of the proof and at last giving a detailed security proof.

\paragraph{\underline{Introducing \sdlog}}

The \sdlog game is a variant of the discrete logarithm game which represents the clearing and setting of bits in the secret scalar during the EdDSA key generation. The only difference to the normal discrete logarithm game is that the secret scalars are not choosen uniformly random from $\field{L}$ with $L$ being the order of the generator but rather from the set $\{2^{n-1}, 2^{n-1} + 8, ..., 2^{n} - 8\}$. 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 depicted in figure \ref{fig:sdlog}. 

\begin{definition}[\sdlog]
	For an adversary $\adversary{A}$ we define its advantage in the \sdlog game as following:
	
	\[ \advantage{\adversary{A}}{\sdlog}(k) \assign | \Pr[\sdlog^{\adversary{A}} \Rightarrow 1] | \].
\end{definition}


\begin{figure}
	%TODO: include padding
	\hrule
	\begin{algorithmic}[1]
		\Statex \underline{\game \sdlog}
		\State $a \randomsample \{ 2^{n-1}, 2^{n-1} + 8, ..., 2^{n} - 8 \}$
		\State $\groupelement{A} \assign a \groupelement{B}$
		\State $a' \randomassign \adversary{A}(\groupelement{A})$
		\State \Return $a \test a'$
	\end{algorithmic}
	\hrule
	\caption{\sdlog}
	\label{fig:sdlog}
\end{figure}

\begin{theorem}
	\label{theorem:advgamez}
	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
	
	\[ \advantage{\group{G},\adversary{A}}{\igame}(k) \leq \advantage{\group{G},\adversary{B}}{\sdlog}(k) - \frac{\oraclequeries}{2^{-\log_2(\lceil \frac{2^{2b} - 1}{L} \rceil 2^{-2b})}} \].
\end{theorem}

\paragraph{\underline{Proof Overview}}

The adversary has to call the \ioracle oracle with a commitment $\groupelement{R}$ to get a challenge from the challenger. Due to the nature of Algebraic Group Model the adversary also has to provide a representation of the group element $\groupelement{R}$, as the 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 to the \igame game this can be used to calculate the discrete logarithm of the public key.

% TODO: clarify encoding of c
\begin{figure}
	\hrule
	\begin{multicols}{2}
		\large
		\begin{algorithmic}[1] 
			\Statex \underline{\game $G_0$ / \textcolor{blue}{$G_1$} / \textcolor{red}{$G_2$}}
			\State $a \randomsample \{2^{n-1}, 2^{n-1} + 8, ..., 2^n - 8\}$
			\State $\groupelement{A} \assign a \groupelement{B}$
			\State $s^* \randomsample \adversary{A}^{\ioracle(\inp)}(\groupelement{A})$	
			\State \Return $\exists \groupelement{R}^*, \ch^*: \groupelement{R}^* = 2^c (s^* \groupelement{B} - \ch^* \groupelement{A}) \wedge (\groupelement{R}^*, \ch^*) \in Q$		
		\end{algorithmic}
		\columnbreak
		\begin{algorithmic}[1] 
			\Statex \underline{\oracle \ioracle($\agmgroupelement{R_i}{r_i} \in \group{G}$)}
			\State Let $\groupelement{R}_i = r_1 \groupelement{B} + r_2 \groupelement{A}$
			\State $\ch_i \randomsample \{0,1\}^{2b}$
			\BeginBox[draw=blue]
			\State \textbf{If} $2^c \ch_i \equiv -r_2 \pmod L$ \textbf{then}
			\State \quad $bad \assign true$
			\BeginBox[draw=red,dashed]
			\State \quad $abort$
			\EndBox
			\EndBox
			\State $Q \assign Q \cup \{ (\groupelement{R}_i, \ch_i) \}$
			\State \Return $\ch_i$
		\end{algorithmic}
	\end{multicols}
	\hrule
	\caption{Games $G_0 - G_2$}
	\label{fig:igamewithabort}
\end{figure}

\paragraph{\underline{Formal Proof}}

\begin{proof}
	\item \paragraph{\underline{AGM}} This proof takes place in the algebraic group model. Meaning that the adversary has to provide a representation along each group element it provides to the reduction. The adversary has to provide an element $\groupelement{R}$, which is an element in the prime order subgroup of the Twisted Edwards curve. Leaving the question 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 enough to provide the representation relative to the prime order subgroup. The reason for that is shown in the following paragraph.
	
	The Twisted Edwards curve $\curve$ over the finite field $\field{q}$ is an finite abelian group. Even though the group $\curve$ might not be cyclic the fundamental theorem of finitely generated abelian groups tells us that each finite abelian groups can be uniquely decomposed into the direct product of cyclic subgroups \cite{karpfinger_hauptsatz_2021}. Meaning that $\curve$ can be represented 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 $\curve$. Lets recall a well known theorem of algebra:
	\item \begin{theorem}
		Let $N_1, ..., N_n$ be subgroups of an group $\group{G}$, following statements are equivalent:
		
		\begin{enumerate}[label=(\arabic*)]
			\item $N_1, ..., N_n \trianglelefteq \group{G}$ and $\group{G} = N_1 \bigotimes ... \bigotimes N_n$.
			\item Each $x \in \group{G}$ can uniquely be represented in the following way:
			
			\[ x = a_i \cdot ... \cdot a_n, a_i \in N_i \]
		\end{enumerate}
	\end{theorem}.\cite{karpfinger_direkte_2021}

	Due to Sylow theorems the decomposition has to include the large prime order subgroup $\group{G}$ used for EdDSA \cite{karpfinger_satze_2021} and since Twisted Edwards curve (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 $\groupelement{Y} \assign y \groupelement{B}$, with $\groupelement{B}$ being the generator of the prime order subgroup, this has to be the only representation regarding the generation set. Meaning that an adversary in the algebraic group model has to provide a representation in the prime order subgroup $\group{G}$.
	
	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 $\groupelement{R} = r_1 \groupelement{B} + r_2 \groupelement{A}$.
	
	\item \paragraph{\underline{$G_0$:}} Let $G_0$ be defined in figure \ref{fig:igamewithabort} by excluding all boxes and $G_0$ be \igame. By definition,
	
	\[ \advantage{\group{G},\adversary{A}}{\igame}(k) = \Pr[\igame^{\adversary{A}} \Rightarrow 1] = \Pr[G_0^{\adversary{A}} \Rightarrow 1] \].
	
	\item \paragraph{\underline{$G_1$:}} Game $G_1$ is exactly the same as $G_0$ with the only change being the bad flag being set inside an if condition. The bad flag is set if $2^c \ch_i = -r_2$. 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 just a conceptual change since the behavior of the game does not change whether the flag is set or not. Hence,
	
	\[ \Pr[G_0^{\adversary{A}} \Rightarrow 1] = \Pr[G_1^{\adversary{A}} \Rightarrow 1] \].
	
	\item \paragraph{\underline{$G_2:$}} Game $G_2$ aborts if the flag bad is set. For each individual \ioracle query the $bad$ flag is set with probability at most $\frac{1}{2^{-\log_2(\lceil \frac{2^{2b} - 1}{L} \rceil 2^{-2b})}}$. $c_i$ is chosen by the game after the adversary has provided the representation of $\groupelement{R_i}$ and therefore the value of $r_2$. This way the adversary has no way of choosing $\ch_i$ after $r_2$ and can not 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 check in the if condition. 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
	% TODO: Müsste das nicht floor statt ceil sein?
	
	\[ |\Pr[G_1^{\adversary{A}} \Rightarrow 1] - \Pr[G_2^{\adversary{A}} \Rightarrow 1]| \leq \Pr[bad] \leq \frac{\oraclequeries}{2^{-\log_2(\lceil \frac{2^{2b} - 1}{L} \rceil 2^{-2b})}} \].
	
	\item Finally, Game $G_2$ is well prepared to show that there exists an adversary $\adversary{B}$ satisfying
	
	\begin{align}
		\Pr[G_2^{\adversary{A}} \Rightarrow 1] = \advantage{\group{G},\adversary{B}}{\sdlog}(k) \label{eq:advbsdlog}
	\end{align}.
	
	\begin{figure}
		\hrule
		\begin{multicols}{2}
			\large
			\begin{algorithmic}[1]
				\Statex \underline{\textbf{Adversary} $\adversary{B}(\groupelement{A})$}
				\State $s^* \randomassign \adversary{A}^{\ioracle(\inp)}(\groupelement{A})$
				\State \textbf{If} $\nexists \agmgroupelement{R^*}{r^*}, \ch^*: \groupelement{R}^* = 2^c (s^* \groupelement{B} - \ch^* \groupelement{A}) \wedge (\agmgroupelement{R^*}{r^*}, \ch^*) \in Q$  \textbf{then}
				\State \quad $abort$
				\State Let $R^* = r_1 \groupelement{B} + r_2 \groupelement{A}$
				\State \Return $(2^c s^* - r_1)(r_2 + 2^c \ch^*)^{-1}$
			\end{algorithmic}
			\columnbreak
			\begin{algorithmic}[1] 
				\Statex \underline{\oracle \ioracle($\agmgroupelement{R_i}{r_i} \in \group{G}$)}
				\State Let $\groupelement{R}_i = r_1 \groupelement{B} + r_2 \groupelement{A}$
				\State $\ch_i \randomsample \{0,1\}^{2b}$
				\State \textbf{If} $2^c \ch_i \equiv -r_2 \pmod L$ \textbf{then}
				\State \quad $bad \assign true$
				\State \quad $abort$
				\State $Q \assign Q \cup \{ (\agmgroupelement{R_i}{r_i}, \ch_i) \}$
				\State \Return $\ch_i$
			\end{algorithmic}
		\end{multicols}
		\hrule
		\caption{Adversary $\adversary{B}$ breaking \sdlog}
		\label{fig:adversarybsdlog}
	\end{figure}
	
	To prove (\ref{eq:advbsdlog}), we define an adversary $\adversary{B}$ attacking \sdlog that simulates $\adversary{A}$'s view in $G_2$. Adversary $\adversary{B}$ formally defined in figure \ref{fig:adversarybsdlog} is run in the \sdlog game and adversary $\adversary{B}$ simulates \ioracle for adversary $\adversary{A}$. \ioracle is simulated perfectly.
	
	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 following equation:
	
	\begin{align*}
		r_1 \groupelement{B} + r_2 \groupelement{A} &= 2^c s^* \groupelement{B} - 2^c \ch^* \groupelement{A} \\
		(r_2 + 2^c \ch^*)A &= (2^c s^* - r_1)B \\
		A &= (2^c s^* - r_1)(r_2 + 2^c \ch^*)^{-1} B
	\end{align*}
	
	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 $\groupelement{A}$.
	
	\item This proves theorem \ref{theorem:advgamez}.
\end{proof}