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minor protocol clarification

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Markus Teich 2016-06-20 01:38:16 +02:00
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tex-stuff/math.tex
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@ -77,14 +77,21 @@ the price $p_{b_a}$ bidder $a$ is willing to pay. $\forall j: p_j < p_{j+1}$.
\subsubsection{Round 1: Encrypt bid}
The message has $k$ parts, each consisting of $10$ Points plus an additional $3$
Points for the last proof. Therefore the message is $10k*32 + 3*32 = 320k + 96$
bytes large.
\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.
\item Prove that $\forall j:(\alpha_{aj}, \beta_{aj})$ decrypts to either $0$ or $g$.
\item Prove that $\forall j: ECDL_y\left(\left(\sum_{j=1}^k\alpha_{aj}\right) - g\right) = ECDL_g\left(\sum_{j=1}^k\beta_{aj}\right)$
\item $\forall j:$ 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.
\item $\forall j:$ Prove that $(\alpha_{aj}, \beta_{aj})$ decrypts to either $0$ or $g$.
\item Prove that $ ECDL_y\left(\left(\sum_{j=1}^k\alpha_{aj}\right) - g\right) = ECDL_g\left(\sum_{j=1}^k\beta_{aj}\right)$
\end{enumerate}
\subsubsection{Round 2: Compute outcome}
The message has $nk$ parts, each consisting of $5$ Points. Therefore the message
is $5nk*32 = 160nk$ bytes large.
\begin{enumerate}
\item Compute and publish for each $i$ and $j$: \\[2.0ex]
$\gamma_{ij}^{\times a} = m_{ij}^{+a}\displaystyle\left(\left(\sum_{h=1}^n\sum_{d=j+1}^k\alpha_{hd}\right)+\left(\sum_{d=1}^{j-1}\alpha_{id}\right)+\left(\sum_{h=1}^{i-1}\alpha_{hj}\right)\right)$ and \\[2.0ex]