61 lines
4.3 KiB
TeX
61 lines
4.3 KiB
TeX
\documentclass{article}
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\usepackage[a4paper, margin=2cm]{geometry}
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\usepackage{amsmath}
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\begin{document}
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\section{first price auction with tie breaking and private outcome (EC-Version)}
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\subsection{Zero Knowledge Proofs}
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\subsubsection{Proof of Knowledge of a EC DL}
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Alice and Bob know $v$ and $g$ with $|g| = n$, but only Alice knows $x$, so that $v = xg$.
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\begin{enumerate}
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\item Alice chooses $z$ at random and calculates $a = zg$.
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\item Alice computes $c = HASH(g,v,a)$ mod n.
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\item Alice sends $r = (z + cx)$ mod n and $a$ to Bob.
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\item Bob checks that $rg = a + cv$.
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\end{enumerate}
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\subsection{Proof of equality of tow EC DL}
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Alice and Bob know $v$, $w$, $g_1$ and $g_2$, but only Alice knows $x$, so that
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$v = xg_1$ and $w = xg_2$.
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\begin{enumerate}
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\item Alice chooses $z$ at random and calculates $a = zg_1$ and $b = zg_2$.
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\item Alice computes $c = HASH(g,v,w,a,b)$ mod n.
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\item Alice sends $r = (z + cx)$ mod n, $a$ and $b$ to Bob.
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\item Bob checks that $rg_1 = a + cv$ and $rg_2 = b + cw$.
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\end{enumerate}
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\subsection{Proof that an encrypted value is one out of two values}
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Alice proves that an El Gamal encrypted value $(\alpha, \beta) = (m + ry, rg)$
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either decrypts to $0$ or to a fixed value $z$ without revealing which is the
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case, in other words, it is shown that $m \epsilon \{0, z\}$.
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\section{first price auction with tie breaking and private outcome}
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\begin{align}
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v_{aj} & = \frac{\prod_{i=1}^n \gamma_{aj}^{\times i}}{\prod_{i=1}^n \varphi_{aj}^{\times i}} \\[2.0ex]
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& = \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]
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& = \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]
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& = \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]
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& = \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}}}
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\end{align}
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\subsection{outcome function}
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\begin{align}
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v_a & = \left((2U-I)\sum_{i=1}^n b_i-(2M+1)\mathbf{e}+(2M+2)Lb_a\right)R_a^* \\[2.0ex]
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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]
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& \text{switch from additive finite group to multiplicative finite group} \\[2.0ex]
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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]
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\end{align}
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\subsection{fixes to step 5 in (M+1)st Price auction from the 2003 paper pages 9 an 10}
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\begin{align}
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\gamma_{ij} = & \frac{\prod_{h=1}^n \prod_{d=j}^k (\alpha_{hd}\alpha_{h,d+1})\left(\prod_{d=1}^j \alpha_{id}\right)^{2M+2}}{(2M+1)Y} \\
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\text{changed to} & \frac{\prod_{h=1}^n \left(\prod_{d=j}^k \alpha_{hd} \cdot \prod_{d=j+1}^k \alpha_{hd}\right)\left(\prod_{d=1}^j \alpha_{id}\right)^{2M+2}}{Y^{2M+1}} \\[2.0ex]
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\delta_{ij} = & \prod_{h=1}^n \prod_{d=j}^k (\beta_{hd}\beta_{h,d+1})\left(\prod_{d=1}^j \beta_{id}\right)^{2M+2} \\
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\text{changed to} & \prod_{h=1}^n \left(\prod_{d=j}^k \beta_{hd} \prod_{d=j+1}^k \beta_{hd}\right)\left(\prod_{d=1}^j \beta_{id}\right)^{2M+2}
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\end{align}
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\end{document}
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