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

tex-stuff/math.tex

@@ -3,7 +3,8 @@
 \usepackage{amsmath}
 \usepackage{amsfonts}
 \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}
 \subsubsection{Proof 1: Knowledge of an ECDL}
 
@@ -76,7 +77,18 @@ case, in other words, it is shown that $M \in \{0, G\}$. \\
 	Proof: & $A_1, A_2, B_1, B_2, d_1, d_2, r_1, r_2$
 \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
 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)$.
 \end{enumerate}
 
+\subsection{First Price Auction Protocol With Private Outcome}
+
 \subsubsection{Round 2: Compute outcome}
 
 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}
 
+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}
 \begin{align}