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math.tex: fix M+1st price outcome determination formula

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Markus Teich 2016-10-09 14:47:09 +02:00
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@ -127,7 +127,7 @@ is $5nk*32 = 160nk$ bytes large.
$\forall i,j:$ 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]
$\delta_{ij}^{\times a} = m_{ij}^{+a}\displaystyle\left(\left(\sum_{h=1}^n\sum_{d=j+1}^k \beta_{hd}\right)+\left(\sum_{d=1}^{j-1} \beta_{id}\right)+\left(\sum_{h=1}^{i-1} \beta_{hj}\right)\right)$ \\[2.0ex]
with a corresponding Proof 2 for $ECDL(\gamma_{ij}^{\times a}) = ECDL(\delta_{ij}^{\times a})$.
\subsubsection{Round 3: Decrypt outcome}
@ -164,8 +164,8 @@ 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]
$\gamma_{ij}^{\times a} = m_{ij}^{+a}\displaystyle\left(\sum_{h=1}^n\left(\sum_{d=j}^k\alpha_{hd}+\sum_{d=j+1}^k\alpha_{hd}\right)+\left(2M+2\right)\sum_{d=1}^{j}\alpha_{id} - \left(2M+1\right)Y \right)$ and \\[2.0ex]
$\delta_{ij}^{\times a} = m_{ij}^{+a}\displaystyle\left(\sum_{h=1}^n\left(\sum_{d=j}^k \beta_{hd}+\sum_{d=j+1}^k \beta_{hd}\right)+\left(2M+2\right)\sum_{d=1}^{j} \beta_{id}\right)$ \\[2.0ex]
with a corresponding Proof 2 for $ECDL(\gamma_{ij}^{\times a}) = ECDL(\delta_{ij}^{\times a})$.
\subsubsection{Round 3: Decrypt outcome}