first protocol part in math scratchpad

tex-stuff/math.tex

@@ -15,7 +15,7 @@ Alice and Bob know $v$ and $g$ with $|g| = n$, but only Alice knows $x$, so that
 	\item Bob checks that $rg = a + cv$.
 \end{enumerate}
 
-\subsection{Proof of equality of two EC DL}
+\subsubsection{Proof of equality of two EC DL}
 
 Alice and Bob know $v$, $w$, $g_1$ and $g_2$, but only Alice knows $x$, so that
 $v = xg_1$ and $w = xg_2$.
@@ -27,7 +27,7 @@ $v = xg_1$ and $w = xg_2$.
 	\item Bob checks that $rg_1 = a + cv$ and $rg_2 = b + cw$.
 \end{enumerate}
 
-\subsection{Proof that an encrypted value is one out of two values}
+\subsubsection{Proof that an encrypted value is one out of two values}
 
 Alice proves that an El Gamal encrypted value $(\alpha, \beta) = (m + ry, rg)$
 either decrypts to $0$ or to the fixed value $g$ without revealing which is the
@@ -55,6 +55,64 @@ Then regardless of the value of $m$:
 	\item Alice sends $(\alpha, \beta), a_1, b_1, a_2, b_2, c, d_1, d_2, r_1, r_2$ to Bob.
 	\item Bob checks that $c=d_1+d_2$ mod n, $a_1=r_1g+d_1\beta$, $b_1=r_1y+d_1(\alpha-g)$, $a_2=r_2g+d_2\beta$ and $b_2=r_2y+d_2\alpha$.
 \end{enumerate}
+
+\subsection{Protocol}
+
+\subsubsection{Generate public key}
+
+\begin{enumerate}
+	\item Choose $x_a$ and $m_{ij}, r_{aj}$ for each $i$ and $j$ at random.
+	\item Publish $y_a=g^{x_a}$ along with a zero-knowledge proof of knowledge of $y_a$'s EC DL.
+	\item Compute $y=\sum_{i=1}^ny_i$.
+\end{enumerate}
+
+\subsubsection{Round 1: Encrypt bid}
+
+\begin{enumerate}
+	\item Set $b_{aj}=\begin{cases}g & \mathrm{if}\quad j=b_a\\0 & \mathrm{else}\end{cases}$ and publish $\alpha_{aj}=b_{aj}+r_{aj}y$ and $\beta_{aj}=r_{aj}g$ for each j.
+\end{enumerate}
+
+\subsubsection{Round 2: Compute outcome}
+
+\begin{enumerate}
+	\item 
+\end{enumerate}
+
+\subsubsection{Round 3: Decrypt outcome}
+
+\begin{enumerate}
+	\item 
+\end{enumerate}
+
+\subsubsection{Epilogue: Outcome determination}
+
+\begin{enumerate}
+	\item 
+\end{enumerate}
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
 
 \section{first price auction with tie breaking and private outcome}
 \begin{align}