math.tex: add m+1st price auction private outcome protocol

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Markus Teich 2016-10-08 16:59:01 +02:00
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\usepackage{amsmath} \usepackage{amsmath}
\usepackage{amsfonts} \usepackage{amsfonts}
\begin{document} \begin{document}
\section{first price auction with tie breaking and private outcome (EC-Version)}
\section{Ed25519 Elliptic Curve Based Algorithms And Protocols}
\subsection{Zero Knowledge Proofs} \subsection{Zero Knowledge Proofs}
\subsubsection{Proof 1: Knowledge of an ECDL} \subsubsection{Proof 1: Knowledge of an ECDL}
@ -76,7 +77,18 @@ case, in other words, it is shown that $M \in \{0, G\}$. \\
Proof: & $A_1, A_2, B_1, B_2, d_1, d_2, r_1, r_2$ Proof: & $A_1, A_2, B_1, B_2, d_1, d_2, r_1, r_2$
\end{tabular} \end{tabular}
\subsection{First Price Auction Protocol With Private Outcome} \subsection{public outcome auctions}
TODO: no need to unicast Round 3 to seller, implications
\subsection{M+1st price auctions}
TODO: explain blowing up $k$ to $nk$ to prevent ties and the additional check
needed in Round 1.
\subsection{Prologue}
These steps are the same for all following protocols in this section.
Let $n$ be the number of participating bidders/agents in the protocol and $k$ be Let $n$ be the number of participating bidders/agents in the protocol and $k$ be
the amount of possible valuations/prices for the sold good. Let $G$ be the the amount of possible valuations/prices for the sold good. Let $G$ be the
@ -106,6 +118,8 @@ bytes large.
\item Use Proof 2 to show that $ ECDL_Y\left(\left(\sum_{j=1}^k\alpha_{aj}\right) - G\right) = ECDL_G\left(\sum_{j=1}^k\beta_{aj}\right)$. \item Use Proof 2 to show 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} \end{enumerate}
\subsection{First Price Auction Protocol With Private Outcome}
\subsubsection{Round 2: Compute outcome} \subsubsection{Round 2: Compute outcome}
The message has $nk$ parts, each consisting of $5$ Points. Therefore the message The message has $nk$ parts, each consisting of $5$ Points. Therefore the message
@ -133,6 +147,41 @@ each $i, j$ and $h \neq i$ after having received all of them.
\subsection{First Price Auction Protocol With Public Outcome} \subsection{First Price Auction Protocol With Public Outcome}
TODO
\subsection{M+1st Price Auction Protocol With Private Outcome}
\subsubsection{Addition to Round 1: Encrypt bid}
The Bidders also have to use Proof 2 to show that $ ECDL_Y\left(\left(\sum_{j=1}^k\alpha_{a,jn+a}\right) - G\right) = ECDL_G\left(\sum_{j=1}^k\beta_{a,jn+a}\right)$.
This is to ensure bidders have only chosen valid bids for their bid index, since
in M+1st price auctions the amount of possible prices is multiplied by $n$ to
prevent ties. This increases the message size by $96$ bytes.
\subsubsection{Round 2: Compute outcome}
The message has $nk$ parts, each consisting of $5$ Points. Therefore the message
is $5nk*32 = 160nk$ bytes large.
$\forall i,j:$ Compute and publish \\[2.0ex]
$\gamma_{ij}^{\times a} = m_{ij}^{+a}\displaystyle\left(\left(2M+2\right)\left(\sum_{h=1}^n\left(\sum_{d=j}^k\alpha_{hd}+\sum_{d=j+1}^k\alpha_{hd}\right)+\sum_{d=1}^{j}\alpha_{id}\right) - \left(2M+1\right)Y \right)$ and \\[2.0ex]
$\delta_{ij}^{\times a} = m_{ij}^{+a}\displaystyle\left(\left(2M+2\right)\left(\sum_{h=1}^n\left(\sum_{d=j}^k\beta_{hd}+\sum_{d=j+1}^k\beta_{hd}\right)+\sum_{d=1}^{j}\beta_{id}\right)\right)$ \\[2.0ex]
with a corresponding Proof 2 for $ECDL(\gamma_{ij}^{\times a}) = ECDL(\delta_{ij}^{\times a})$.
\subsubsection{Round 3: Decrypt outcome}
$\forall i,j:$ Send $\varphi_{ij}^{\times a} =
x_{+a}\left(\sum_{h=1}^n\delta_{ij}^{\times h}\right)$ with a Proof 2 showing
$ECDL(\varphi_{ij}^{\times a}) = ECDL(Y_{\times a})$ to the seller who publishes
all $\varphi_{ij}^{\times h}$ and the corresponding proofs of correctness for
each $i, j$ and $h \neq i$ after having received all of them.
\subsubsection{Epilogue: Outcome determination}
\begin{enumerate}
\item $\forall j:$ Compute $V_{aj}=\sum_{i=1}^n\gamma_{aj}^{\times i} - \sum_{i=1}^n\varphi_{aj}^{\times i}$.
\item If $\exists w: V_{aw} = 0$, then bidder $a$ is the winner of the auction. $p_w$ is the selling price.
\end{enumerate}
@ -156,23 +205,6 @@ each $i, j$ and $h \neq i$ after having received all of them.
\section{first price auction with tie breaking and private outcome}
\begin{align}
v_{aj} & = \frac{\prod_{i=1}^n \gamma_{aj}^{\times i}}{\prod_{i=1}^n \varphi_{aj}^{\times i}} \\[2.0ex]
& = \frac{\prod_{i=1}^n \gamma_{aj}^{\times i}}{\prod_{i=1}^n \left(\prod_{h=1}^n \delta_{aj}^{\times h}\right)^{x_{+i}}} \\[2.0ex]
& = \frac{\prod_{i=1}^n \left(\left(\prod_{h=1}^n \prod_{d=j+1}^k \alpha_{hd}\right)\cdot\left(\prod_{d=1}^{j-1} \alpha_{ad}\right)\cdot\left(\prod_{h=1}^{a-1} \alpha_{hj}\right)\right)^{m_{aj}^{+i}}}{\prod_{i=1}^n \left(\prod_{h=1}^n \left(\left(\prod_{s=1}^n \prod_{d=j+1}^k \beta_{sd}\right)\cdot\left(\prod_{d=1}^{j-1} \beta_{ad}\right)\cdot\left(\prod_{s=1}^{a-1} \beta_{sj}\right)\right)^{m_{aj}^{+h}}\right)^{x_{+i}}} \\[2.0ex]
& = \frac{\prod_{i=1}^n \left(\left(\prod_{h=1}^n \prod_{d=j+1}^k b_{hd} y^{r_{hd}}\right)\cdot\left(\prod_{d=1}^{j-1} b_{ad} y^{r_{ad}}\right)\cdot\left(\prod_{h=1}^{a-1} b_{hj} y^{r_{hj}}\right)\right)^{m_{aj}^{+i}}}{\prod_{i=1}^n \left(\prod_{h=1}^n \left(\left(\prod_{s=1}^n \prod_{d=j+1}^k g^{r_{sd}}\right)\cdot\left(\prod_{d=1}^{j-1} g^{r_{ad}}\right)\cdot\left(\prod_{s=1}^{a-1} g^{r_{sj}}\right)\right)^{m_{aj}^{+h}}\right)^{x_{+i}}} \\[2.0ex]
& = \frac{\prod_{i=1}^n \left(\left(\prod_{h=1}^n \prod_{d=j+1}^k b_{hd} \left(\prod_{t=1}^n g^{x_{+t}}\right)^{r_{hd}}\right)\cdot\left(\prod_{d=1}^{j-1} b_{ad} \left(\prod_{t=1}^n g^{x_{+t}}\right)^{r_{ad}}\right)\cdot\left(\prod_{h=1}^{a-1} b_{hj} \left(\prod_{t=1}^n g^{x_{+t}}\right)^{r_{hj}}\right)\right)^{m_{aj}^{+i}}}{\prod_{i=1}^n \left(\prod_{h=1}^n \left(\left(\prod_{s=1}^n \prod_{d=j+1}^k g^{r_{sd}}\right)\cdot\left(\prod_{d=1}^{j-1} g^{r_{ad}}\right)\cdot\left(\prod_{s=1}^{a-1} g^{r_{sj}}\right)\right)^{m_{aj}^{+h}}\right)^{x_{+i}}}
\end{align}
\subsection{outcome function}
\begin{align}
v_a & = \left((2U-I)\sum_{i=1}^n b_i-(2M+1)\mathbf{e}+(2M+2)Lb_a\right)R_a^* \\[2.0ex]
v_{aj} & = \left(\sum_{i=1}^n \left(\sum_{d=j}^k b_{id} + \sum_{d=j+1}^k b_{id}\right)-(2M+1)+(2M+2)\sum_{d=1}^j b_{ad}\right)R_a^* \\[2.0ex]
& \text{switch from additive finite group to multiplicative finite group} \\[2.0ex]
v_{aj} & = \left(\frac{\displaystyle\prod_{i=1}^n \left(\prod_{d=j}^k b_{id} \cdot \prod_{d=j+1}^k b_{id}\right) \cdot \left(\prod_{d=1}^j b_{ad}\right)^{2M+2}}{(2M+1)g}\right)R_a^* \\[2.0ex]
\end{align}
\subsection{fixes to step 5 in (M+1)st Price auction from the 2003 paper pages 9 an 10} \subsection{fixes to step 5 in (M+1)st Price auction from the 2003 paper pages 9 an 10}
\begin{align} \begin{align}