part4 and part5

Lab04/content/part4/hashes.txt

@@ -3,3 +3,4 @@ jonas:$1$qPEx/5/4$4T0lDrIo5RnrDFH2rP7In/
 pony:$1$q1NwfAWz$5ulIYzhCjsLBsQjQ2CRRN1
 hannah:$1$ia1yKsCX$/Xdet.pCmIR3TdKoavIaP/
 root:$1$E2ST/4kf$6UnJsquPTK3S2lpLxShE/.
+florian:$1$AIi8cXMh$eiuFpI7YUjEvLaoq5mfss.

Lab04/documentation/main.tex

@@ -30,7 +30,7 @@
 \definecolor{backcolour}{rgb}{0.95,0.95,0.92}
 \definecolor{weborange}{rgb}{1,0.65,0}
 
-\usepackage{fancyvrb}
+\usepackage{fancyvrb,amsmath}
 \RecustomVerbatimCommand{\VerbatimInput}{VerbatimInput}%
 {fontsize=\footnotesize,
  commandchars=\|\(\), % escape character and argument delimiters for
@@ -113,6 +113,7 @@
     \include{part2/part2}
     \include{part3/part3}
     \include{part4/part4}
+    \include{part5/part5}
 
     \printbibliography[title=Literaturverzeichnis]
 

Lab04/documentation/part4/part4.tex

@@ -1,3 +1,25 @@
 \section{Part 4: Password Hashing and Cracking}
 
+\subsection{MD5 Hash}
 
+MD5 hash mit password aus Liste:
+
+\begin{verbatim}
+openssl passwd -1 !Florian
+\end{verbatim}
+
+\subsection{John}
+
+cracking it with john:
+
+\begin{verbatim}
+john --wordlist=crackstation.txt --format=md5crypt hashes.txt --fork=8
+\end{verbatim}
+
+\begin{figure}[H]
+    \begin{center}
+        \includegraphics[width=0.9\textwidth]{part4/john}
+        \caption{John}
+        \label{fig:John}
+    \end{center}
+\end{figure}

Lab04/documentation/part5/part5.tex

@@ -0,0 +1,45 @@
+\section{Part 5: Asymmetric Encryption}
+
+\subsection{Generierte Primzahlen}
+
+\begin{verbatim}
+openssl prime -generate -bits 8
+\end{verbatim}
+
+\begin{verbatim}
+prime1 = 211, prime2 = 223, prime3 = 227, e=11
+\end{verbatim}
+
+\subsection{Berechnungen}
+
+\begin{align}
+d = e^{-1} mod ((prime1-1)(prime2-1)) \\
+d = 11^{-1} mod ((211-1)(223-1)) = 21191
+\end{align}
+
+\begin{verbatim}
+g = 9, x = 2, y = 3
+\end{verbatim}
+
+\begin{align}
+    a = g^{x} (mod\ prime3) = 9^{2} (mod\ 227) = 81 \\
+    b = g^{y} (mod\ prime3) = 9^{3} (mod\ 227) = 48 \\
+    k_{1} = b^{x}(mod\ prime3) = 48^{2}(mod\ 227) = 34 \\
+    k_{2} = a^{y}(mod\ prime3) = 81^{3}(mod\ 227) = 34 \\
+    k = k_{1} = k_{2} = 34
+\end{align}
+
+\subsection{Fragen}
+
+1. What attack is the Diffie-Hellman key exchange vulnerable to?
+
+Man in the Middle attacks
+
+2. What measures can be taken to prevent this type of attack?
+
+RSA Encryption
+
+3. For the Diffie-Hellman, a generator g is used. Explain what a generator is and how can it be found
+
+4. Show why for the primes 61,23 and the public key e=60 no private key d can be found
+