diff --git a/ea/ubA/eaA.pdf b/ea/ubA/eaA.pdf
index 463adac..1ee9b4d 100644
--- a/ea/ubA/eaA.pdf
+++ b/ea/ubA/eaA.pdf
Binary files differ
diff --git a/ea/ubA/eaA.tex b/ea/ubA/eaA.tex
index 49ab2f5..81d5eb9 100644
--- a/ea/ubA/eaA.tex
+++ b/ea/ubA/eaA.tex
@@ -94,7 +94,7 @@
                 \Leftrightarrow \sin\left(\pi\cdot w_e\right)=\frac{s}{2r}\\
                 \Leftrightarrow r=\frac{s}{2\sin\left(w_e\pi\right)}\\
                 \Rightarrow r=\frac{3cm}{2\sin\left(0.2\pi\right)}\approx 2.552cm=\sigma_{opt}$
-        \end{enumerate}\pagebreak\\
+        \end{enumerate}\pagebreak
     \Aufgabe{1/5-Erfolgsregel (Praxis)}{11}\\
     \foreach \x in {0.15,0.2,0.25,0.3,0.5,0.6}{
         \includegraphics[width=0.33\textwidth]{figures/sigma\x .png}
@@ -168,7 +168,7 @@
     j=j+1
 np.save("stats.npy",stats)
 
-\end{lstlisting}\\
+\end{lstlisting}
     \Aufgabe{Rekombination in ES}{4}
         \begin{enumerate}[(a)]
             \item $a_1'=(0.23,-0.92,-0.23,-0.85,0.21)$
diff --git a/mr/ub10/A1a.png b/mr/ub10/A1a.png
new file mode 100644
index 0000000..3f57430
--- /dev/null
+++ b/mr/ub10/A1a.png
Binary files differ
diff --git a/mr/ub10/A1b.png b/mr/ub10/A1b.png
new file mode 100644
index 0000000..85a7f96
--- /dev/null
+++ b/mr/ub10/A1b.png
Binary files differ
diff --git a/mr/ub10/A2/A2.m b/mr/ub10/A2/A2.m
index 3c4f4de..3f72f35 100644
--- a/mr/ub10/A2/A2.m
+++ b/mr/ub10/A2/A2.m
@@ -109,6 +109,7 @@
 cp= goal;
 path =goal;
 
+
 while  ~isequal(cp,start)
        neighbours=[];
        for x = [cp(1)-1:cp(1)+1]
@@ -130,4 +131,4 @@
 end    
 path
 latexmat(path, '%g')
-
+ 
diff --git a/mr/ub10/mr10.pdf b/mr/ub10/mr10.pdf
index f2a3a02..6b2a763 100644
--- a/mr/ub10/mr10.pdf
+++ b/mr/ub10/mr10.pdf
Binary files differ
diff --git a/mr/ub10/mr10.tex b/mr/ub10/mr10.tex
index 53adf0c..76cc542 100644
--- a/mr/ub10/mr10.tex
+++ b/mr/ub10/mr10.tex
@@ -89,10 +89,24 @@
 
     \Aufgabe{Convex Hull Algorithms}{10}
         \begin{enumerate}[(a)]
-            \item done?
-            \item done?
-            \item ~
-
+\item \includegraphics[width=0.7\linewidth]{A1a}
+            \item \includegraphics[width=0.7\linewidth]{A1b}
+             ~\\$ \begin{tabular}{|c|c|c|c|c|c|c|c|c|c|}
+            	1&2&3&4&5&6&7&8&9&10\\ 
+            	\hline
+            	2&3&4&5&6&7&8&9&10&9\\
+            	1&2&2&4&4&6&7&8&9&8\\
+            	& 1&1&2&2&4&6&7&8&7\\
+            	&&   &1&1&2&4&6&7&6\\
+            	&&&&     &1&2&4&6&4\\
+            	&&&&&&      1&2&4&2\\
+            	&&&&&&&       1&2&1\\
+            	&&&&&&&&        1&\\
+            	&&&&&&&&&\\
+            \end{tabular}$\\
+            S pushes the new point if it is on the left of the line through the two highest elements of the stack.\\
+            S pops elements until the new point is on the right of the resultin line. 
+            \item \pagebreak~
 \includegraphics[width=0.48\textwidth]{A1c/step1}
 \includegraphics[width=0.48\textwidth]{A1c/step2}
 \includegraphics[width=0.48\textwidth]{A1c/step3}
@@ -104,6 +118,7 @@
             \item Sklansky's algorithm runs in linear time, i.e. $\mathcal{O}(n)$
             \item QuickHull has a worst case runningtime of $\mathcal{O}(n^2)$, when there are no points that can be ignored. A geometric example of the worst case is a cylindric obstacle because a circle doesn't allow any point to be ignored because the drawn line is always nearer to the circles center then a point on the circle.
         \end{enumerate}
+        \pagebreak
      \Aufgabe{Path Planning by dynamic programming}{10}   
 	\begin{enumerate}[(a)]
 		\item  $\left[ \begin{array}{cccccccccc}