diff --git a/ACNS/presentation/main.tex b/ACNS/presentation/main.tex index eca90406eb915b0fbfeb31b54e5068fea7c7383b..dabf6c180129f01a21af508f654e4a392513638d 100644 --- a/ACNS/presentation/main.tex +++ b/ACNS/presentation/main.tex @@ -8,12 +8,12 @@ \title{On Actively Secure Fine-Grained Access Structures from Isogeny Assumptions} \author{Fabio Campos\inst{1,2} \and \underline{Philipp Muth}\inst{3}} \institute{\relax - RheinMain University of Applied Sciences, Wiesbaden, Germany + $^1$RheinMain University of Applied Sciences, Wiesbaden, Germany \and - Radboud University, Nijmegen, The Netherlands %\\ + $^2$Radboud University, Nijmegen, The Netherlands %\\ %\email{campos@sopmac.de} \and - Technische Universität Darmstadt, Germany%\\ + $^3$Technische Universität Darmstadt, Germany%\\ %\email{philipp.muth@tu-darmstadt.de} } diff --git a/ACNS/presentation/motivation.tex b/ACNS/presentation/motivation.tex index 0eaf1219e439dc8a65495f1f20c1da71d79e002a..11cb711251338178c16b540ebde5bfa233b54bb8 100644 --- a/ACNS/presentation/motivation.tex +++ b/ACNS/presentation/motivation.tex @@ -27,10 +27,10 @@ \end{block} \begin{remark} For \(s,s'\in \Z_p\) and \(E\in\mathcal E\), we have - \[[s] \left(\left[s'\right] E\right) = \left[s+s'\right] E.\] + \[\left[s\right] \left(\left[s'\right] E\right) = \left[s+s'\right] E.\] \end{remark} - \begin{block}{The Group Action Inverse Problem} + \begin{block}{The Group Action Inverse Problem (GAIP)} Given two elements \(E,E' \in \mathcal E\), find \(g\in \mathcal G\) with \[g\ast E = E'.\] \end{block} @@ -108,10 +108,10 @@ \draw [->] (cipher) -- (sh1); \pause - \draw [->, >=Stealth, bend right] (sh1) edge node [midway, above] {$ E^1 = \left[L_{1} s_1\right] E_0$} (sh2) ; - \draw [->, >=Stealth, bend right] (sh2) edge node [midway, left] {$ E^2 = \left[L_{2} s_2\right] E_1$} (sh3) ; - \draw [->, >=Stealth, bend right] (sh3) edge node [midway, below] {$ E^3 = \left[L_{3} s_3\right] E_2$} (sh4) ; - \node [right = of sh4] (key) {$\key = \left[L_4 s_4\right] E^3$}; + \draw [->, >=Stealth, bend right] (sh1) edge node [midway, above] {$ E^1 = \left[L_{1,S'} s_1\right] E_0$} (sh2) ; + \draw [->, >=Stealth, bend right] (sh2) edge node [midway, left] {$ E^2 = \left[L_{2,S'} s_2\right] E_1$} (sh3) ; + \draw [->, >=Stealth, bend right] (sh3) edge node [midway, below] {$ E^3 = \left[L_{3,S'} s_3\right] E_2$} (sh4) ; + \node [right = of sh4] (key) {$\key = \left[L_{4,S'} s_4\right] E^3$}; \draw [->] (sh4) -- (key); \end{tikzpicture} @@ -154,10 +154,10 @@ \draw [->] (cipher) -- (sh1); %\pause - \draw [->, >=Stealth, bend right] (sh1) edge node [midway, above] {$ E^1 = \left[L_{1} s_1\right] E_0$} (sh2) ; - \draw [->, >=Stealth, bend right] (sh2) edge node [midway, left, color = red] {$ {E^2} \neq \left[L_{2} s_2\right] E_1$} (sh3) ; - \draw [->, >=Stealth, bend right] (sh3) edge node [midway, below] {$ E^3 = \left[L_{3} s_3\right] E_2$} (sh4) ; - \node [right = of sh4] (key) {$\key = \left[L_4 s_4\right] E^3$}; + \draw [->, >=Stealth, bend right] (sh1) edge node [midway, above] {$ E^1 = \left[L_{1,S'} s_1\right] E_0$} (sh2) ; + \draw [->, >=Stealth, bend right] (sh2) edge node [midway, left, color = red] {$ {E^2} \neq \left[L_{2,S'} s_2\right] E_1$} (sh3) ; + \draw [->, >=Stealth, bend right] (sh3) edge node [midway, below] {$ E^3 = \left[L_{3,S'} s_3\right] E_2$} (sh4) ; + \node [right = of sh4] (key) {$\key = \left[L_{4,S'} s_4\right] E^3$}; \draw [->] (sh4) -- (key); \end{tikzpicture}