simplified equations
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@@ -40,7 +40,7 @@ This section shows that \somdl implies MU-\igame using the algebraic group model
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\label{theorem:adv_omdl'}
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Let $\adversary{A}$ be an adversary against \igame with $\group{G}$ being a cyclic group of prime order $L$, receiving $N$ public keys and making at most $\oraclequeries$ oracle queries. Then
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\[ \advantage{\group{G},\adversary{A}}{\text{MU-\igame}}(\secparamter) \leq \advantage{\group{G},\adversary{B}}{\somdl}(\secparamter) + \frac{\oraclequeries N}{2^{-\log_2(\lceil \frac{2^{2b} - 1}{L} \rceil 2^{-2b})}}. \]
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\[ \advantage{\group{G},\adversary{A}}{\text{MU-\igame}}(\secparamter) \leq \advantage{\group{G},\adversary{B}}{\somdl}(\secparamter) + \frac{\oraclequeries N \lceil \frac{2^{2b} - 1}{L} \rceil}{2^{2b}}. \]
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\end{theorem}
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\paragraph{\underline{Proof Overview}} In the multi-user setting the adversary gets access to not only the generator $\groupelement{B}$ and one public key $\groupelement{A}$ but rather a set of public keys $\groupelement{A_1}$ to $\groupelement{A_N}$. For this reason, the representation of a group element, the adversary has to provide, looks the following: $\groupelement{R} = r_1 \groupelement{B} + r_2 \groupelement{A_1} + ... + r_{N+1} \groupelement{A_N}$. Since there are multiple group elements with unknown discrete logarithms it is not possible to directly calculate the discrete logarithm of one of the public keys given a valid forgery of a signature. Upon receiving a valid solution the \textit{DL} oracle can be used to get the discrete logarithm of all the public keys except the one for which the solution is valid. This way it is again possible to construct a representation looking like $\groupelement{R} = r_1 \groupelement{B} + r_2 \groupelement{A_i}$. Then it is again possible to calculate the discrete logarithm of $\groupelement{A_i}$ and win the \somdl game.
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@@ -94,7 +94,7 @@ This section shows that \somdl implies MU-\igame using the algebraic group model
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\item \paragraph{\underline{$G_2:$}} $G_2$ also includes the abort instruction in the red box. The abort is triggered if the bad flag is set to true. For each individual \ioracle oracle query the bad flag is set with a probability of $\frac{N}{2^{-\log_2(\lceil \frac{2^{2b} - 1}{L} \rceil 2^{-2b})}}$. With $2^{-\log_2(\lceil \frac{2^{2b} - 1}{L} \rceil 2^{-2b})}$ being the min-entropy of $\ch$ and $N$ being the number of $r_i$ with which the equation $2^c \ch \equiv - r_i \pmod L$ could evaluate to true. By the Union bound over all $\oraclequeries$ oracle quries we obtain $\Pr[bad] \leq \frac{\oraclequeries N}{2^{-\log_2(\lceil \frac{2^{2b} - 1}{L} \rceil 2^{-2b})}}$. Since $G_1$ and $G_2$ are identical-until-bad games, we have
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\[ |\Pr[G_1^{\adversary{A}} \Rightarrow 1] - \Pr[G_2^{\adversary{A}} \Rightarrow 1]| \leq \Pr[bad] \leq \frac{\oraclequeries N}{2^{-\log_2(\lceil \frac{2^{2b} - 1}{L} \rceil 2^{-2b})}}. \]
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\[ |\Pr[G_1^{\adversary{A}} \Rightarrow 1] - \Pr[G_2^{\adversary{A}} \Rightarrow 1]| \leq \Pr[bad] \leq \frac{\oraclequeries N \lceil \frac{2^{2b} - 1}{L} \rceil}{2^{2b}}. \]
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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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