Fixed security parameter

thesis/sections/security_of_eddsa/gamez_implies_uf-nma.tex

@@ -7,7 +7,7 @@ This section shows that \igame implies the UF-NMA security if the EdDSA signatur
 \begin{definition}[\igame]
 	For an adversary $\adversary{A}$ we define its advantage in the \igame game as following:
 	
-	\[ \advantage{\adversary{A}}{\igame}(k) \assign | \Pr[\igame^{\adversary{A}} \Rightarrow 1] | \].
+	\[ \advantage{\adversary{A}}{\igame}(\secparamter) \assign | \Pr[\igame^{\adversary{A}} \Rightarrow 1] | \].
 \end{definition}
 
 \begin{figure}
@@ -38,7 +38,7 @@ This section shows that \igame implies the UF-NMA security if the EdDSA signatur
 	\label{theorem:adv_igame}
 	Let $\adversary{A}$ be an adversary against $\text{UF-NMA}_{\text{EdDSA}}$. Then,
 	
-	\[ \advantage{\adversary{A}}{UF-NMA}(k) = \advantage{\adversary{B}}{\igame}(k) \].
+	\[ \advantage{\adversary{A}}{UF-NMA}(\secparamter) = \advantage{\adversary{B}}{\igame}(\secparamter) \].
 \end{theorem}
 
 \paragraph{\underline{Proof Overview}} The adversary has to query the random oracle to get the hash value $H(\encoded{R} | \encoded{A} | m)$. The programmability of the random oracle can be used to embed the challenge from the \ioracle oracle into the answer of the random oracle. This way a valid forgery of a signature also provides a valid solution for the \igame game.
@@ -70,12 +70,12 @@ This section shows that \igame implies the UF-NMA security if the EdDSA signatur
 \begin{proof}
 	\item \paragraph{\underline{$G_0$}} Let $G_0$ be defined in figure \ref{fig:igame_implies_uf-nma} and let $G_0$ be $\text{UF-NMA}_{\text{EdDSA}}$. By definition,
 	
-	\[ \advantage{\text{EdDSA}, \adversary{A}}{\text{UF-NMA}} = \Pr[\text{UF-NMA}_{\text{EdDSA}}^{\adversary{A}} \Rightarrow 1 ] = \Pr[G_0^{\adversary{A}} \Rightarrow 1] \].
+	\[ \advantage{\text{EdDSA}, \adversary{A}}{\text{UF-NMA}}(\secparamter) = \Pr[\text{UF-NMA}_{\text{EdDSA}}^{\adversary{A}} \Rightarrow 1 ] = \Pr[G_0^{\adversary{A}} \Rightarrow 1] \].
 	
 	\item $G_0$ is well prepared to show that there exists an adversary $\adversary{B}$ satisfying
 	
 	\begin{align}
-		\Pr[G_0^{\adversary{A}} \Rightarrow 1] = \advantage{\group{G}, \adversary{B}}{\igame}(k) \label{eq:adv_igame}
+		\Pr[G_0^{\adversary{A}} \Rightarrow 1] = \advantage{\group{G}, \adversary{B}}{\igame}(\secparamter) \label{eq:adv_igame}
 	\end{align}.
 
 	\begin{figure}