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author | Markus Teich <markus.teich@stusta.mhn.de> | 2016-10-09 14:47:09 +0200 |
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committer | Markus Teich <markus.teich@stusta.mhn.de> | 2016-10-09 14:47:09 +0200 |
commit | 9070e0fa9f58d2133443eacc4e05ec3b9ddfb93e (patch) | |
tree | e4cf46c0489e60389d064a6997f68dd872176605 | |
parent | ac5050919f2eaf9fa615d19ab3a638235d0b3054 (diff) | |
download | libbrandt-9070e0fa9f58d2133443eacc4e05ec3b9ddfb93e.tar.gz libbrandt-9070e0fa9f58d2133443eacc4e05ec3b9ddfb93e.zip |
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. | |||
127 | 127 | ||
128 | $\forall i,j:$ Compute and publish \\[2.0ex] | 128 | $\forall i,j:$ Compute and publish \\[2.0ex] |
129 | $\gamma_{ij}^{\times a} = m_{ij}^{+a}\displaystyle\left(\left(\sum_{h=1}^n\sum_{d=j+1}^k\alpha_{hd}\right)+\left(\sum_{d=1}^{j-1}\alpha_{id}\right)+\left(\sum_{h=1}^{i-1}\alpha_{hj}\right)\right)$ and \\[2.0ex] | 129 | $\gamma_{ij}^{\times a} = m_{ij}^{+a}\displaystyle\left(\left(\sum_{h=1}^n\sum_{d=j+1}^k\alpha_{hd}\right)+\left(\sum_{d=1}^{j-1}\alpha_{id}\right)+\left(\sum_{h=1}^{i-1}\alpha_{hj}\right)\right)$ and \\[2.0ex] |
130 | $\delta_{ij}^{\times a} = m_{ij}^{+a}\displaystyle\left(\left(\sum_{h=1}^n\sum_{d=j+1}^k\beta_{hd}\right)+\left(\sum_{d=1}^{j-1}\beta_{id}\right)+\left(\sum_{h=1}^{i-1}\beta_{hj}\right)\right)$ \\[2.0ex] | 130 | $\delta_{ij}^{\times a} = m_{ij}^{+a}\displaystyle\left(\left(\sum_{h=1}^n\sum_{d=j+1}^k \beta_{hd}\right)+\left(\sum_{d=1}^{j-1} \beta_{id}\right)+\left(\sum_{h=1}^{i-1} \beta_{hj}\right)\right)$ \\[2.0ex] |
131 | with a corresponding Proof 2 for $ECDL(\gamma_{ij}^{\times a}) = ECDL(\delta_{ij}^{\times a})$. | 131 | with a corresponding Proof 2 for $ECDL(\gamma_{ij}^{\times a}) = ECDL(\delta_{ij}^{\times a})$. |
132 | 132 | ||
133 | \subsubsection{Round 3: Decrypt outcome} | 133 | \subsubsection{Round 3: Decrypt outcome} |
@@ -164,8 +164,8 @@ The message has $nk$ parts, each consisting of $5$ Points. Therefore the message | |||
164 | is $5nk*32 = 160nk$ bytes large. | 164 | is $5nk*32 = 160nk$ bytes large. |
165 | 165 | ||
166 | $\forall i,j:$ Compute and publish \\[2.0ex] | 166 | $\forall i,j:$ Compute and publish \\[2.0ex] |
167 | $\gamma_{ij}^{\times a} = m_{ij}^{+a}\displaystyle\left(\left(2M+2\right)\left(\sum_{h=1}^n\left(\sum_{d=j}^k\alpha_{hd}+\sum_{d=j+1}^k\alpha_{hd}\right)+\sum_{d=1}^{j}\alpha_{id}\right) - \left(2M+1\right)Y \right)$ and \\[2.0ex] | 167 | $\gamma_{ij}^{\times a} = m_{ij}^{+a}\displaystyle\left(\sum_{h=1}^n\left(\sum_{d=j}^k\alpha_{hd}+\sum_{d=j+1}^k\alpha_{hd}\right)+\left(2M+2\right)\sum_{d=1}^{j}\alpha_{id} - \left(2M+1\right)Y \right)$ and \\[2.0ex] |
168 | $\delta_{ij}^{\times a} = m_{ij}^{+a}\displaystyle\left(\left(2M+2\right)\left(\sum_{h=1}^n\left(\sum_{d=j}^k\beta_{hd}+\sum_{d=j+1}^k\beta_{hd}\right)+\sum_{d=1}^{j}\beta_{id}\right)\right)$ \\[2.0ex] | 168 | $\delta_{ij}^{\times a} = m_{ij}^{+a}\displaystyle\left(\sum_{h=1}^n\left(\sum_{d=j}^k \beta_{hd}+\sum_{d=j+1}^k \beta_{hd}\right)+\left(2M+2\right)\sum_{d=1}^{j} \beta_{id}\right)$ \\[2.0ex] |
169 | with a corresponding Proof 2 for $ECDL(\gamma_{ij}^{\times a}) = ECDL(\delta_{ij}^{\times a})$. | 169 | with a corresponding Proof 2 for $ECDL(\gamma_{ij}^{\times a}) = ECDL(\delta_{ij}^{\times a})$. |
170 | 170 | ||
171 | \subsubsection{Round 3: Decrypt outcome} | 171 | \subsubsection{Round 3: Decrypt outcome} |