Minor changes

thesis/sections/security_of_eddsa/uf-nma_implies_suf-cma.tex

@@ -15,7 +15,7 @@ The EdDSA signature scheme is based on the Schnorr signature scheme, which is a
 
 To generate a signature without knowing the private key, the challenge and response are chosen randomly, and the commitment is calculated based on the chosen challenge and response. The random oracle is then programmed to output the challenge given the commitment and the message as input. In this way, the resulting tuple of challenge and response is a valid signature for the given message.
 
-For the reduction to be able to program the random oracle, the adversary must not have queried the hash function with this exact input before asking for the signature. Since the input to the hash query includes the commitment, which is unknown to the adversary prior to the \Osign query, and the result of a random function, the adversary can only guess it, which introduces the loss of advantage.
+For the reduction to be able to program the random oracle, the adversary must not have queried the hash function with this exact input before asking for the signature. Since the input to the hash query includes the commitment, which is the result of a random function and therefore unknown to the adversary prior to the \Osign query. For this reason, the adversary can only guess it.
 
 This method of simulating the \Osign oracle and the resulting loss of advantage was first introduced in \cite{SP:BCJZ21}.
 
@@ -112,7 +112,7 @@ This method of simulating the \Osign oracle and the resulting loss of advantage
 	
 	\[ \advantage{\group{G},\adversary{A}}{\cma}(\secparamter) = \Pr[\text{\cma}^{\adversary{A}} \Rightarrow 1] = \Pr[G_0^{\adversary{A}} \Rightarrow 1]. \]
 	
-	\item \paragraph{\underline{$G_1:$}} $G_1$ is now defined by replacing the black box with the blue one. This change inlines the call to the hash function and introduces a bad flag in the \Osign oracle, which is set in case the hash value for the challenge is already set before the \Osign oracle is called. In this cases the adversary already queried the challenge for that signature resulting in the challenger not being able to program the random oracle on that input. Without being able to program the random oracle the challenger is not able to generate a valid signature, without knowing the private key. This change is only conceptual, since it does not change the behavior of the oracle and only changes internal variables of the game. Therefore,
+	\item \paragraph{\underline{$G_1:$}} $G_1$ is now defined by replacing the black box with the blue one. This change inlines the call to the hash function and introduces a bad flag in the \Osign oracle, which is set in case the hash value for the challenge is already set before the \Osign oracle is called. In this cases the adversary already queried the challenge for that signature, resulting in the challenger not being able to program the random oracle on that input. Without being able to program the random oracle the challenger is not able to generate a valid signature, without knowing the private key. This change is only conceptual, since it does not change the behavior of the oracle and only changes internal variables of the game. Therefore,
 	
 	\[ \Pr[G_0^{\adversary{A}} \Rightarrow 1] = \Pr[G_1^{\adversary{A}} \Rightarrow 1]. \]
 	
@@ -120,7 +120,7 @@ This method of simulating the \Osign oracle and the resulting loss of advantage
 	
 	\[ |\Pr[G_1^{\adversary{A}} \Rightarrow 1] - \Pr[G_2^{\adversary{A}} \Rightarrow 1]| \leq \Pr[bad] \leq \frac{\oraclequeries \hashqueries}{2^{-\log_2(\lceil \frac{2^{2b} - 1}{L} \rceil 2^{-2b})}}. \]
 	
-	\item \paragraph{\underline{$G_3:$}} $G_3$ replaces the \sign oracle with the \sign oracle in the green box. Now the signature is not generated by using the secret key, but by using the \simalg procedure and manually setting the result of the hash function call. This change is conceptual only. \simalg returns a correctly distributed tuple $(R, \ch, S)$ with $2^c S \groupelement{B} = 2^c \groupelement{R} + 2^c \ch \groupelement{A}$ and it has been excluded that $H(\encoded{R} | \encoded{A} | \m)$ is set before calling the \sign oracle, so that the random oracle can be programmed to output $\ch$ when calling $H(\encoded{R} | \encoded{A} | m)$. This ensures that $2^c S \groupelement{B} = 2^c \groupelement{R} + 2^c H(\encoded{R} | \encoded{A} | \m) \groupelement{A}$, which means that $\signature \assign (\encoded{R}, S)$ is a valid signature for the message $\m$ and was generated without using the private key $s$. Therefore,
+	\item \paragraph{\underline{$G_3:$}} $G_3$ replaces the \sign oracle with the \sign oracle in the green box. Now the signature is not generated by using the secret key, but by using the \simalg procedure and manually setting the result of the hash function call. This change is conceptual only. \simalg returns a correctly distributed tuple $(R, \ch, S)$, with $2^c S \groupelement{B} = 2^c \groupelement{R} + 2^c \ch \groupelement{A}$, and it has been excluded that $H'(\encoded{R} | \encoded{A} | \m)$ is set before calling the \sign oracle, so that the random oracle can be programmed to output $\ch$ when calling $H'(\encoded{R} | \encoded{A} | m)$. This ensures that $2^c S \groupelement{B} = 2^c \groupelement{R} + 2^c H'(\encoded{R} | \encoded{A} | \m) \groupelement{A}$, which means that $\signature \assign (\encoded{R}, S)$ is a valid signature for the message $\m$ and was generated without using the private key $s$. Therefore,
 	
 	\[ \Pr[G_2^{\adversary{A}} \Rightarrow 1] = \Pr[G_3^{\adversary{A}} \Rightarrow 1]. \]
 	
@@ -163,9 +163,9 @@ This method of simulating the \Osign oracle and the resulting loss of advantage
 		\label{fig:adversarybuf-nma}
 	\end{figure}
 
-	To prove (\ref{eq:adv_uf-nma}), an adversary $\adversary{B}$ is defined that attacks $\text{UF-NMA}$ simulating the view of $\adversary{A}$ in $G_3$. The adversary $\adversary{B}$, formally defined in figure \ref{fig:adversarybuf-nma}, is run in the $\text{UF-NMA}$ game, and the adversary $\adversary{B}$ simulates \Osign for the adversary $\adversary{A}$. \Osign is simulated perfectly. The hash queries of $\adversary{A}$ are forwarded to the UF-NMA challenger.
+	To prove (\ref{eq:adv_uf-nma}), an adversary $\adversary{B}$ is defined that attacks $\text{UF-NMA}$ simulating the view of $\adversary{A}$ in $G_3$. The adversary $\adversary{B}$, formally defined in figure \ref{fig:adversarybuf-nma}, is run in the $\text{UF-NMA}$ game, and the adversary $\adversary{B}$ simulates \Osign for the adversary $\adversary{A}$. \Osign is simulated perfectly. The hash queries of $\adversary{A}$ are forwarded to the UF-NMA challenger, when not set by the reduction itself.
 	
-	Finally, consider $\adversary{A}$'s output $(\m^*, \signature^* \assign (\encoded{R^*}, S^*))$. It is known that $2^c S^* \groupelement{B} = 2^c \groupelement{R^*} + 2^c H'(\encoded{R^*}|\encoded{A}|m^*) \groupelement{A}$. If strict parsing is used to decode $S$ from the signature, it is known that $0 \leq S < L$. Therefore, for every $R$, $\m$ pair, there is only one valid encoded $S$ that satisfies the equation. This means that a new and valid signature cannot be generated by simply changing the $S$ value of an already valid signature. Therefore, $R$ or $m$ must also be changed in order to create a new valid signature from another. Since $R$ and $m$ are inputs to the hash query used to generate the challenge, the result of this hash query is passed from the $H$ hash oracle provided to the adversary $\adversary{B}$ instead of being set by $\adversary{B}$ itself. Also, the existence of multiple encodings of the commitment $\groupelement{R}$ does not pose a problem, since if another representation of the same $\groupelement{R}$ is chosen, its hash value has not been set by $\adversary{B}$ and therefore must have been passed from the UF-NMA challenger. For this reason, $H'(\encoded{R^*}|\encoded{A}|m^*) = H(\encoded{R^*}|\encoded{A}|m^*)$. Therefore,
+	Finally, consider $\adversary{A}$'s output $(\m^*, \signature^* \assign (\encoded{R^*}, S^*))$. It is known that $2^c S^* \groupelement{B} = 2^c \groupelement{R^*} + 2^c H'(\encoded{R^*}|\encoded{A}|m^*) \groupelement{A}$. If strict parsing is used to decode $S$ from the signature, it is known that $0 \leq S < L$. Therefore, for every $R$, $\m$ pair, there is only one valid encoded $S$ that satisfies the equation. This means that a new and valid signature cannot be generated by simply changing the $S$ value of an already valid signature. Therefore, $R$ or $m$ must also be changed in order to create a new valid signature from another. Since $R$ and $m$ are inputs to the hash query used to generate the challenge, the result of this hash query is passed from the $H$ hash oracle provided to the adversary $\adversary{B}$ instead of being set by $\adversary{B}$ itself. Also, the existence of multiple encodings of the commitment $\groupelement{R}$ does not pose a problem, since if another representation of the same $\groupelement{R}$ is chosen, its hash value for the corresponding challenge has not been set by $\adversary{B}$ and therefore must have been passed from the UF-NMA challenger. For this reason, $H'(\encoded{R^*}|\encoded{A}|m^*) = H(\encoded{R^*}|\encoded{A}|m^*)$. Therefore,
 	
 	\begin{align*}
 		2^c S^* \groupelement{B} &= 2^c \groupelement{R^*} + 2^c H'(\encoded{R^*}|\encoded{A}|m^*) \groupelement{A}\\
@@ -231,9 +231,9 @@ This section shows that the UF-NMA security of EdDSA implies the EUF-CMA securit
 		\label{fig:adversary_b_suf-nma}
 	\end{figure}
 
-	To prove (\ref{eq:adv2_uf-nma}), we define an adversary $\adversary{B}$ attacking $\text{UF-NMA}$ that simulates the view of $\adversary{A}$ in $G_3$. The adversary $\adversary{B}$ formally defined in figure \ref{fig:adversary_b_suf-nma} is run in the $\text{UF-NMA}$ game and the adversary $\adversary{B}$ simulates \Osign for the adversary $\adversary{A}$. \Osign is simulated perfectly. The hash queries of $\adversary{A}$ are forwarded to the UF-NMA challenger.
+	To prove (\ref{eq:adv2_uf-nma}), we define an adversary $\adversary{B}$ attacking $\text{UF-NMA}$ that simulates the view of $\adversary{A}$ in $G_3$. The adversary $\adversary{B}$ formally defined in figure \ref{fig:adversary_b_suf-nma} is run in the $\text{UF-NMA}$ game and the adversary $\adversary{B}$ simulates \Osign for the adversary $\adversary{A}$. \Osign is simulated perfectly. The hash queries of $\adversary{A}$ are forwarded to the UF-NMA challenger, when not set by the reduction itself.
 	
-	Finally, consider $\adversary{A}$ output $(\m^*, \signature^* \assign (\encoded{R^*}, S^*))$. It is known that $2^c S^* \groupelement{B} = 2^c \groupelement{R^*} + 2^c H'(\encoded{R^*}|\encoded{A}|m^*) \groupelement{A}$. The same argument as in the proof above cannot be used, since the lax parser could map multiple $2b$-bit bitstrings onto the same $S^* \pmod L$. Therefore, the adversary $\adversary{A}$ could generate a new valid signature from one generated by \Osign simply by choosing a different bitstring representation of the same $S^* \pmod L$. Since in this case $\sum[\encoded{R^*}|\encoded{A}|m^*]$ was set by the adversary $\adversary{B}$, this signature is not valid for the UF-NMA challenger. However, since we are in the EUF-CMA setting, we require the adversary $\adversary{A}$ to provide a signature for a message $m^*$ for which it has not requested a signature from the \Osign oracle. Since the signature for the message $m^*$ was not requested in the Sign oracle, the output of $H'(\encoded{R^*}|\encoded{A}|m^*)$ was not set by the adversary B, but was forwarded from the $H$ hash oracle. For this reason, $H'(\encoded{R^*}|\encoded{A}|m^*) = H(\encoded{R^*}|\encoded{A}|m^*)$. Therefore,
+	Finally, consider $\adversary{A}$ output $(\m^*, \signature^* \assign (\encoded{R^*}, S^*))$. It is known that $2^c S^* \groupelement{B} = 2^c \groupelement{R^*} + 2^c H'(\encoded{R^*}|\encoded{A}|m^*) \groupelement{A}$. Because we are in the EUF-CMA setting, the adversary $\adversary{A}$ is required to provide a signature for a message $m^*$ for which it has not requested a signature from the \Osign oracle. Since the signature for the message $m^*$ was not requested in the Sign oracle, the output of $H'(\encoded{R^*}|\encoded{A}|m^*)$ was not set by the adversary B, but must have been forwarded from the $H$ hash oracle. For this reason, $H'(\encoded{R^*}|\encoded{A}|m^*) = H(\encoded{R^*}|\encoded{A}|m^*)$. Therefore,
 	
 	\begin{align*}
 		2^c S^* \groupelement{B} &= 2^c \groupelement{R^*} + 2^c H'(\encoded{R^*}|\encoded{A}|m^*) \groupelement{A}\\