diff --git a/tex-stuff/math.tex b/tex-stuff/math.tex index 2d2c1f7..6502f0b 100644 --- a/tex-stuff/math.tex +++ b/tex-stuff/math.tex @@ -5,58 +5,77 @@ \begin{document} \section{first price auction with tie breaking and private outcome (EC-Version)} \subsection{Zero Knowledge Proofs} -\subsubsection{Proof of Knowledge of a EC DL} +\subsubsection{Proof 1: Knowledge of an ECDL} -Alice and Bob know $v$ and $g$ with $|g| = n$, but only Alice knows $x$, so that $v = xg$. +Alice and Bob know $v$ and $g$ with $q = |g|$, but only Alice knows $x$, so that +$v = xg$. \begin{enumerate} \item Alice chooses $z$ at random and calculates $a = zg$. - \item Alice computes $c = HASH(g,v,a)$ mod n. - \item Alice sends $r = (z + cx)$ mod n and $a$ to Bob. + \item Alice computes $c = HASH(g,v,a)$ mod $q$. + \item Alice sends $r = (z + cx)$ mod $q$ and $a$ to Bob. \item Bob checks that $rg = a + cv$. \end{enumerate} -\subsubsection{Proof of equality of two EC DL} +\begin{tabular}{r l} + Prover only knowledge: & $x$ \\ + Common knowledge: & $v, g$ \\ + Proof: & $r, a$ +\end{tabular} + +\subsubsection{Proof 2: Equality of two ECDL} Alice and Bob know $v$, $w$, $g_1$ and $g_2$, but only Alice knows $x$, so that $v = xg_1$ and $w = xg_2$. \begin{enumerate} \item Alice chooses $z$ at random and calculates $a = zg_1$ and $b = zg_2$. - \item Alice computes $c = HASH(g_1,g_2,v,w,a,b)$ mod n. - \item Alice sends $r = (z + cx)$ mod n, $a$ and $b$ to Bob. + \item Alice computes $c = HASH(g_1,g_2,v,w,a,b)$ mod $q$. + \item Alice sends $r = (z + cx)$ mod $q$, $a$ and $b$ to Bob. \item Bob checks that $rg_1 = a + cv$ and $rg_2 = b + cw$. \end{enumerate} -\subsubsection{Proof that an encrypted value is one out of two values} +\begin{tabular}{r l} + Prover only knowledge: & $x$ \\ + Common knowledge: & $v, w, g_1, g_2$ \\ + Proof: & $r, a, b$ +\end{tabular} + +\subsubsection{Proof 3: 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 -case, in other words, it is shown that $m \in \{0, g\}$. +case, in other words, it is shown that $m \in \{0, g\}$. \\ -If $m = 0$: +\noindent If $m = 0$: \begin{enumerate} \item Alice chooses $r_1$, $d_1$, $w$ at random and calculates $a_1 = r_1g + d_1\beta$, $b_1 = r_1y + d_1(\alpha - g)$, $a_2=wg$ and $b_2=wy$. - \item Alice computes $c = HASH(g,\alpha,\beta,a_1,b_1,a_2,b_2)$ mod n. - \item Alice chooses $d_2=c-d_1$ mod n and $r_2=w-rd_2$ mod n. + \item Alice computes $c = HASH(g,\alpha,\beta,a_1,b_1,a_2,b_2)$ mod $q$. + \item Alice chooses $d_2=c-d_1$ mod $q$ and $r_2=w-rd_2$ mod $q$. \end{enumerate} -If $m = g$: +\noindent If $m = g$: \begin{enumerate} \item Alice chooses $r_2$, $d_2$, $w$ at random and calculates $a_1=wg$, $b_1=wy$, $a_2=r_2g + d_2\beta$ and $b_2=r_2y + d_2\alpha$. - \item Alice computes $c = HASH(g,\alpha,\beta,a_1,b_1,a_2,b_2)$ mod n. - \item Alice chooses $d_1=c-d_2$ mod n and $r_1=w-rd_1$ mod n. + \item Alice computes $c = HASH(g,\alpha,\beta,a_1,b_1,a_2,b_2)$ mod $q$. + \item Alice chooses $d_1=c-d_2$ mod $q$ and $r_1=w-rd_1$ mod $q$. \end{enumerate} -Then regardless of the value of $m$: +\noindent Then regardless of the value of $m$: \begin{enumerate} - \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 $b_2=r_2y+d_2\alpha$. + \item Alice sends $(\alpha, \beta), a_1, b_1, a_2, b_2, d_1, d_2, r_1, r_2$ to Bob. + \item Bob checks that $c=d_1+d_2$ mod $q$, $a_1=r_1g+d_1\beta$, $b_1=r_1y+d_1(\alpha-g)$, $a_2=r_2g+d_2\beta$ and $b_2=r_2y+d_2\alpha$. \end{enumerate} +\begin{tabular}{r l} + Prover only knowledge: & $r, x$ \\ + Common knowledge: & $\alpha, \beta$ \\ + Proof: & $a_1, a_2, b_1, b_2, d_1, d_2, r_1, r_2$ +\end{tabular} + \subsection{Protocol} Let $n$ be the number of participating bidders/agents in the protocol and $k$ be @@ -70,8 +89,8 @@ the price $p_{b_a}$ bidder $a$ is willing to pay. $\forall j: p_j < p_{j+1}$. \subsubsection{Generate public key} \begin{enumerate} - \item Choose $x_{+a} \in \mathbb{Z}_q$ and $m_{ij}^{\times a}, r_{aj} \in \mathbb{Z}_q$ for each $i$ and $j$ at random. - \item Publish $y_{\times a}={x_{+a}}g$ along with a zero-knowledge proof of knowledge of $y_{\times a}$'s EC DL. + \item Choose $x_{+a} \in \mathbb{Z}_q$ and $\forall i,j: m_{ij}^{\times a}, r_{aj} \in \mathbb{Z}_q$ at random. + \item Publish $y_{\times a}={x_{+a}}g$ along with Proof 1 of $y_{\times a}$'s ECDL. \item Compute $y=\sum_{i=1}^ny_{\times i}$. \end{enumerate} @@ -82,9 +101,9 @@ Points for the last proof. Therefore the message is $10k*32 + 3*32 = 320k + 96$ bytes large. \begin{enumerate} - \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)$ + \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 $\beta_{aj}=r_{aj}g$. + \item $\forall j:$ Use Proof 3 to show that $(\alpha_{aj}, \beta_{aj})$ decrypts to either $0$ or $g$. + \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} \subsubsection{Round 2: Compute outcome} @@ -92,29 +111,24 @@ bytes large. 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] - $\delta_{ij}^{\times a} = m_{ij}^{+a}\displaystyle\left(\left(\sum_{h=1}^n\sum_{d=j+1}^k\beta_{hd}\right)+\left(\sum_{d=1}^{j-1}\beta_{id}\right)+\left(\sum_{h=1}^{i-1}\beta_{hj}\right)\right)$ \\[2.0ex] - with a proof of correctness. -\end{enumerate} +$\forall i,j:$ Compute and publish \\[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] +$\delta_{ij}^{\times a} = m_{ij}^{+a}\displaystyle\left(\left(\sum_{h=1}^n\sum_{d=j+1}^k\beta_{hd}\right)+\left(\sum_{d=1}^{j-1}\beta_{id}\right)+\left(\sum_{h=1}^{i-1}\beta_{hj}\right)\right)$ \\[2.0ex] +with a corresponding Proof 2 for $ECDL(\gamma_{ij}^{\times a}) = ECDL(\delta_{ij}^{\times a})$. \subsubsection{Round 3: Decrypt outcome} -\begin{enumerate} - \item Send $\varphi_{ij}^{\times a} = - x_{+a}\left(\sum_{h=1}^n\delta_{ij}^{\times h}\right)$ for each $i$ and - $j$ with a proof of correctness (ECDL is known \textbf{and} equal to the - ECDL used for $y_{\times a}$) 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. -\end{enumerate} +$\forall i,j:$ Send $\varphi_{ij}^{\times a} = +x_{+a}\left(\sum_{h=1}^n\delta_{ij}^{\times h}\right)$ with a Proof 2 +$ECDL(\varphi_{ij}^{\times a}) = ECDL(y_{\times a})$ 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 Compute $v_{aj}=\sum_{i=1}^n\gamma_{aj}^{\times i} - \sum_{i=1}^n\varphi_{aj}^{\times i}$ for each $j$. - \item If $v_{aw} = 1$ for any $w$, then bidder $a$ is the winner of the auction. $p_w$ is the selling price. + \item $\forall j:$ Compute $v_{aj}=\sum_{i=1}^n\gamma_{aj}^{\times i} - \sum_{i=1}^n\varphi_{aj}^{\times i}$. + \item If $\exists w: v_{aw} = 1$, then bidder $a$ is the winner of the auction. $p_w$ is the selling price. \end{enumerate}