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| author | Markus Teich <markus.teich@stusta.mhn.de> | 2016-10-09 14:47:09 +0200 |
|---|---|---|
| committer | Markus Teich <markus.teich@stusta.mhn.de> | 2016-10-09 14:47:09 +0200 |
| commit | 9070e0fa9f58d2133443eacc4e05ec3b9ddfb93e (patch) | |
| tree | e4cf46c0489e60389d064a6997f68dd872176605 | |
| parent | ac5050919f2eaf9fa615d19ab3a638235d0b3054 (diff) | |
math.tex: fix M+1st price outcome determination formula
| -rw-r--r-- | tex-stuff/math.tex | 6 |
1 files changed, 3 insertions, 3 deletions
diff --git a/tex-stuff/math.tex b/tex-stuff/math.tex index 54a0fc8..04cc1dc 100644 --- a/tex-stuff/math.tex +++ b/tex-stuff/math.tex @@ -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} |