Add Dlog' ggm proof
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thesis/sections/edggm/sdlog.tex
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thesis/sections/edggm/sdlog.tex
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\subsection{Bounds on \sdlog} \label{sec:sdlog}
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This section concentrates on establishing a lower bound for the hardness of a modified version of the discrete logarithm problem in the generic group model. This variation is introduced in definition \ref{def:sdlog} and functions similarly to the original discrete logarithm problem, with the exception of the secret scalar generation, which is derived from the EdDSA signature scheme's key generation algorithm. The subsequent proof is carried out in the generic group model for twisted Edwards curves.
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\begin{theorem}
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\label{theorem:sdlog_ggm}
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Let $n$ and $c$ be positive integers. Consider a twisted Edwards curve $\curve$ with a cofactor of $2^c$ and a generating set consisting of $(\groupelement{B}, \groupelement{E_2}, ..., \groupelement{E_n})$. Among these, let $\groupelement{B}$ be the generator of the largest prime order subgroup with an order of $L$. Let $\adversary{A}$ be a generic adversary making at most $\oraclequeries$ group operations. Then,
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\[ \advantage{\curve, n, c, L, \adversary{A}}{\sdlog} \leq \frac{1}{2^{n-1-c}} + \frac{\oraclequeries^2}{2^{n-1-c}}. \]
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\end{theorem}
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\paragraph{\underline{Proof Overview}} This proof closely resembles the original proof on the lower bound for the discrete logarithm problem by Shoup \cite{EC:Shoup97}. The initial step involves working with the discrete logarithms of group elements rather than the actual group elements themselves. In the generic group model, this is equivalent as each group element can be uniquely represented by its discrete logarithms with respect to a generating set. For consistency the generating set is denoted as $(\groupelement{B}, \groupelement{E_1}, ..., \groupelement{E_n})$, with $\groupelement{B}$ being the generator of the prime order subgroup and $\groupelement{E_1}$ to $\groupelement{E_n}$ being the generators of the other subgroups. Subsequently, the discrete logarithm in the prime order subgroup is replaced by an indeterminate. By doing this, the discrete logarithm in the prime order subgroup can be chosen after the adversary has provided their solution. As a result, the generic adversary can only guess the discrete logarithm in the prime order subgroup, since it is generated only after the adversary has already submitted their solution. Figure \ref{fig:sdlog_ggm} shows the \sdlog game in the generic group model.
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\begin{figure}[h]
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\hrule
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\vspace{2mm}
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\begin{algorithmic}
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\Statex \underline{\game \sdlog}
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\vspace{1mm}
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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 $a^* \randomassign \adversary{A}^{GOp(\inp, \inp, \inp)}(Enc(\groupelement{B}), Enc(\groupelement{E_2}), ..., Enc(\groupelement{E_n}), Enc(\groupelement{A}))$
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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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\begin{algorithmic}
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\Statex \underline{\oracle GOp($x, y \in \mathbf{S}, b \in \{0,1\}$)}
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\State \quad \Return $Enc(\sum^{-1}[x] + (-1)^b \sum^{-1}[y])$
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\end{algorithmic}
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\vspace{1mm}
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\begin{algorithmic}
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\Statex \underline{\textbf{Procedure} Enc($\groupelement{X} \in \curve$)}
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\vspace{1mm}
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\State \textbf{If } $\sum[\groupelement{X}] = \bot$ \textbf{ then}
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\State \quad $\sum[\groupelement{X}] \randomsample \{0,1\}^{\lceil log_2(|\curve|) \rceil} \backslash \pset{S}$
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\State \quad $\mathbf{S} \assign \pset{S} \cup \{\sum[X]\}$
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\State \Return $\sum[\groupelement{X}]$
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\end{algorithmic}
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\hrule
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\caption{\sdlog in the generic group model}
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\label{fig:sdlog_ggm}
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\end{figure}
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\paragraph{\underline{Formal Proof}}
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\begin{figure}[h]
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\hrule
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\vspace{2mm}
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\begin{algorithmic}
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\Statex \underline{\game \textcolor{black}{$G_0$} / \textcolor{blue}{$G_1$} /\textcolor{red}{$G_2$} / \textcolor{green}{$G_3$} / \textcolor{orange}{$G_4$}}
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\vspace{1mm}
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\BeginBox[draw=black]
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\State $a \randomsample \{2^{n-1}, 2^{n-1} + 2^c, ..., 2^{n} - 2^c\}$
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\Comment{$G_0 - G_4$}
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\EndBox
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\BeginBox[draw=black]
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\State $\groupelement{A} \assign a \groupelement{B}$
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\Comment{$G_0$}
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\EndBox
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\BeginBox[draw=blue]
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\State $\groupelement{A} \assign (a, 0, ..., 0)$
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\Comment{$G_1$}
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\EndBox
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\BeginBox[draw=red]
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\State $P \assign Z$
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\Comment{$G_2 - G_4$}
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\State $\groupelement{A} \assign (P, 0, ..., 0)$
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\EndBox
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\State $a^* \randomassign \adversary{A}^{GOp(\inp, \inp, \inp)}(Enc(\groupelement{B}), Enc(\groupelement{E_2}), ..., Enc(\groupelement{E_n}), Enc(\groupelement{A}))$
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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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\begin{algorithmic}
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\Statex \underline{\oracle GOp($x, y \in \mathbf{S}, b \in \{0,1\}$)}
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\State \quad \Return $Enc(\sum^{-1}[x] + (-1)^b \sum^{-1}[y])$
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\end{algorithmic}
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\vspace{1mm}
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\begin{algorithmic}
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\BeginBox[draw=black]
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\State \underline{\textbf{Procedure} Enc($\groupelement{X} \in \curve$)}
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\Comment{$G_0$}
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\EndBox
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\BeginBox[draw=blue]
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\State \underline{\textbf{Procedure} Enc($\groupelement{X} \in \field{L} \times \field{ord(E_1)} \times ... \times \field{ord(E_n)}$)}
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\Comment{$G_1$}
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\EndBox
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\BeginBox[draw=red]
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\State \underline{\textbf{Procedure} Enc($\groupelement{X} \in \field{L}[Z] \times \field{ord(E_1)} \times ... \times \field{ord(E_n)}$)}
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\Comment{$G_2 - G_4$}
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\vspace{1mm}
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\State Let $X = (P, x_2, ..., x_n)$
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\State $\pset{P} = \pset{P} \cup \{P\}$
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\EndBox
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\BeginBox[draw=green]
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\State \textbf{if } $\exists P_i \in \pset{P}: P_i(a) = P(a) \wedge P_i \neq P$
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\Comment{$G_3 - G_4$}
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\State \quad $bad \assign true$
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\BeginBox[draw=orange,dashed]
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\State \quad abort
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\Comment{$G_4$}
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\EndBox
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\EndBox
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\BeginBox[draw=red]
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\State $X \assign (P(a), x_2, ..., x_n)$
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\Comment{$G_2 - G_4$}
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\EndBox
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\State \textbf{If } $\sum[\groupelement{X}] = \bot$ \textbf{ then}
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\State \quad $\sum[\groupelement{X}] \randomsample \{0,1\}^{\lceil log_2(|\curve|) \rceil} \backslash \mathbf{S}$
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\State \quad $\mathbf{S} \assign \mathbf{S} \cup \{\sum[X]\}$
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\State \Return $\sum[\groupelement{X}]$
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\end{algorithmic}
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\hrule
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\caption{$G_0 - G_4$}
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\label{fig:sdlog_games_ggm_1}
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\end{figure}
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\begin{figure}[h]
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\hrule
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\vspace{2mm}
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\begin{algorithmic}
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\Statex \underline{\game \textcolor{black}{$G_4$} / \textcolor{blue}{$G_5$} /\textcolor{red}{$G_6$} / \textcolor{green}{$G_7$}}
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\vspace{1mm}
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\BeginBox[draw=black]
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\State $a \randomsample \{2^{n-1}, 2^{n-1} + 2^c, ..., 2^{n} - 2^c\}$
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\Comment{$G_4 - G_6$}
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\EndBox
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\State $P \assign Z$
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\State $\groupelement{A} \assign (P, 0, ..., 0)$
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\State $a^* \randomassign \adversary{A}^{GOp(\inp, \inp, \inp)}(Enc(\groupelement{B}), Enc(\groupelement{E_2}), ..., Enc(\groupelement{E_n}), Enc(\groupelement{A}))$
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\BeginBox[draw=green]
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\State $a \randomsample \{2^{n-1}, 2^{n-1} + 8, ..., 2^{n} - 8\}$
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\Comment{$G_7$}
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\EndBox
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\BeginBox[draw=red]
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\State \textbf{if } $\exists P_i, P_j \in \pset{P}: P_i(a) = P_j(a) \wedge P_i \neq P$
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\Comment{$G_6 - G_7$}
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\State \quad $bad \assign true$
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\State \quad abort
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\EndBox
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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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\begin{algorithmic}
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\Statex \underline{\oracle GOp($x, y \in \mathbf{S}, b \in \{0,1\}$)}
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\State \quad \Return $Enc(\sum^{-1}[x] + (-1)^b \sum^{-1}[y])$
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\end{algorithmic}
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\vspace{1mm}
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\begin{algorithmic}
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\State \underline{\textbf{Procedure} Enc($\groupelement{X} \in \field{L}[Z] \times \field{ord(E_1)} \times ... \times \field{ord(E_n)}$)}
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\vspace{1mm}
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\State Let $X = (P, x_2, ..., x_n)$
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\State $\pset{P} = \pset{P} \cup \{P\}$
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\BeginBox[draw=black]
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\State \textbf{if } $\exists P_i \in \pset{P}: P_i(a) = P(a) \wedge P_i \neq P$
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\Comment{$G_4 - G_5$}
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\State \quad $bad \assign true$
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\State \quad abort
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\EndBox
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\BeginBox[draw=black]
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\State $X \assign (P(a), x_2, ..., x_n)$
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\Comment{$G_4$}
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\EndBox
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\State \textbf{If } $\sum[\groupelement{X}] = \bot$ \textbf{ then}
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\State \quad $\sum[\groupelement{X}] \randomsample \{0,1\}^{\lceil log_2(|\curve|) \rceil} \backslash \mathbf{S}$
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\State \quad $\mathbf{S} \assign \mathbf{S} \cup \{\sum[X]\}$
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\State \Return $\sum[\groupelement{X}]$
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\end{algorithmic}
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\hrule
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\caption{$G_4 - G_7$}
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\label{fig:sdlog_games_ggm_2}
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\end{figure}
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\begin{proof}
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\item Let $G_0$ represent the \sdlog game in the generic group model. In this proof, the discrete logarithm within the prime order subgroup of the group element $\groupelement{A}$ will be substituted with an indeterminate. Following that, it will be demonstrated that the challenger, by working with polynomials rather than actual discrete logarithms, makes errors in the simulation with negligible probability. Finally, it will be established that the discrete logarithm of the group element $\groupelement{A}$ can be selected after the adversary has submitted its solution for the game.
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\item This proof utilizes the Schwartz-Zippel lemma. The Schwarz-Zippel lemma is defined as following:
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\begin{definition}[Schwatz-Zippel lemma]
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Let $L$ be a prime number and $P \in \mathbb{F}_{L}[X]$ be a non-zero polynomial of total degree $d \geq 0$ over a field $\mathbb{F}_{L}$. Let $S$ be a finite subset of $\mathbb{F}_{L}$ and let $x$ be selected uniformly at random from $S$. Then
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\[ \Pr[P(x) = 0] \leq \frac{d}{|S|}. \]
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A proof for this lemma can be found in \cite{schwartz_fast_1980}.
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\end{definition}
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\item \paragraph{\underline{$G_0:$}} Let $G_0$ be defined in figure \ref{fig:sdlog_games_ggm_1} by excluding all boxes except the black ones. This is equivalent to the \sdlog in the generic group model. By definition,
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\[ \advantage{\curve, \adversary{A}}{\sdlog} = \prone{\sdlog^{\adversary{A}}}. \]
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\item \paragraph{\underline{$G_1$}} $G_1$ is defined by substituting some black boxes with blue ones, causing the challenger to work with discrete logarithms rather than group elements. This modification remains undetectable to the adversary, as it only deals with labels representing group elements, and each group element can be uniquely represented by its discrete logarithms. These discrete logarithms are denoted by an integer vector, where each element corresponds to the discrete logarithm concerning the generator in the generating set. The addition of these vectors is carried out component-wise. Therefore,
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\[ \prone{G_0^{\adversary{A}}} = \prone{G_1^{\adversary{A}}}. \]
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\item \paragraph{\underline{$G_2:$}} $G_2$ replaces the blue boxes with the red ones. This change replaces the discrete logarithm of the prime order subgroup with a polynomial. This change is only conceptual since the polynomial is evaluated in the Enc procedure. Hence,
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\[ \prone{G_1^{\adversary{A}}} = \prone{G_2^{\adversary{A}}}. \]
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\item \paragraph{\underline{$G_3:$}} $G_3$ introduces the "if" condition within the green box. This modification is purely conceptual, as it only impacts internal variables that do not influence the game's behavior. Therefore,
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\[ \prone{G_1^{\adversary{A}}} = \prone{G_2^{\adversary{A}}}. \]
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\item \paragraph{\underline{$G_4:$}} $G_4$ terminates if the bad flag is activated. This bad flag signifies situations where collisions of group elements would not be identified by merely comparing polynomials without evaluating them. The likelihood of the bad flag being activated can be determined using the Schwartz-Zippel lemma. For a fixed $P_i \in \pset{P}$ we define $P^* = P_i - P$. If and only if $P_i(a) = P(a) \wedge P_i \neq P$ then $P^*(a) = 0$. Since $P^* \neq 0$, the degree of $P^*$ being less or equal to $1$ and $a$ being chosen uniformly at random from $\{2^{n-1}, 2^{n-1} + 2^c, ..., 2^{n} - 2^c\}$ the Schwarz-Zippel lemma can be used to calculate the probability that $P^*(a) = 0$, which is $\Pr[P^*(a) = 0] \leq \frac{1}{2^{n-1-c}}$. Since the set $\pset{P}$ can hold at most $\oraclequeries$ many polynomials by the Union bound over all polynomials in $\pset{P}$ the probability of bad being set, for each individual oracle query, is less or equal to $\frac{\oraclequeries}{2^{n-1-c}}$. By the Union bound over all oracle queries the probability of bad being set to true is $\Pr[bad] \leq \frac{\oraclequeries^2}{2^{n-1-c}}$. For this reason,
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\[ |\prone{G_2^{\adversary{A}}} - \prone{G_3^{\adversary{A}}}| \leq \Pr[bad] \leq \frac{\oraclequeries^2}{2^{n-1-c}}. \]
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For improved readability, $G_4$ is also depicted in \ref{fig:sdlog_games_ggm_2} by including only the black boxes and excluding all others.
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\item \paragraph{\underline{$G_5:$}} $G_5$ removes the evaluation of the polynomial in the Enc procedure. This alteration is purely conceptual, as the previous abort condition ensured that no two distinct polynomials would yield the same value upon evaluation. Consequently, it is feasible to work directly with the polynomials rather than evaluating them.
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\[ \prone{G_4^{\adversary{A}}} = \prone{G_5^{\adversary{A}}}. \]
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\item \paragraph{\underline{$G_6:$}} The difference in $G_6$ is that the abort condition was moved into the main game after the adversary provided its solution. To demonstrate that this alteration is solely conceptual, it will be proven that $G_6$ aborts if and only if $G_5$ would do the same.
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$G_5$ aborts $\Rightarrow G_6$ aborts: If $G_5$ aborts, it means that a polynomial $P_i$ has been added to the set $\pset{P}$ during the call to the Enc procedure, which satisfies the abort condition. In $G_6$, the polynomials in the set $\pset{S}$ remain the same, since the instruction for adding polynomials to the set during the Enc procedure has not been altered between the games. After the adversary provides its solution, the challenger checks for any pair of polynomials in the set that meet the abort condition. Thus, $G_6$ will abort if $G_5$ would have aborted.
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$G_6$ aborts $\Rightarrow G_5$ aborts: If $G_6$ were to abort, the set $\pset{P}$ would contain a pair of polynomials that satisfy the abort condition. The distinction between $G_6$ and $G_5$ is that $G_5$ checks for the existence of such a pair immediately after inserting a new polynomial. Consequently, if $G_6$ were to abort, $G_5$ would also abort.
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This proofs that this change is only conceptual. Hence,
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\[ \prone{G_5^{\adversary{A}}} = \prone{G_6^{\adversary{A}}}. \]
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\item \paragraph{\underline{$G_7:$}} The generation of the secret scalar $a$ in $G_7$ occurs after the adversary has provided its solution. This modification is purely conceptual, as the secret scalar is not utilized prior to this point. As a result, the adversary has no improved likelihood of computing its solution $a^*$ other than guessing, given that the challenger does not select $a$ until the adversary has submitted its solution. Thus,
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\[ \prone{G_6^{\adversary{A}}} = \prone{G_7^{\adversary{A}}} = \frac{1}{2^{n-1-c}}. \]
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\item This proofs theorem \ref{theorem:sdlog_ggm}.
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\end{proof}
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