diff --git a/GA4/ga4.pdf b/GA4/ga4.pdf
index 25b41f7..0008bb8 100644
Binary files a/GA4/ga4.pdf and b/GA4/ga4.pdf differ
diff --git a/GA4/ga4.tex b/GA4/ga4.tex
index e9f1b67..c5f753f 100644
--- a/GA4/ga4.tex
+++ b/GA4/ga4.tex
@@ -21,7 +21,7 @@
   columns=fullflexible,
   commentstyle=\color{gray},
   keywordstyle=\bfseries,
-  keywords={,NIL,while,if,elif,else,FUNCTION,return,for,from,to,TRUE,FALSE},
+  keywords={,NIL,while,if,elif,else,continue,FUNCTION,return,for,from,to,TRUE,FALSE},
   mathescape=true,
   aboveskip=2em,
   captionpos=b,
@@ -76,7 +76,7 @@ FUNCTION BEST-PATH(G=(V,E), v, w):
 FUNCTION HELP-SETUP(G=(V,E), P, S, v):
   for each vertex u $\in$ S:
     P[u] = v
-    HELP-SETUP(G, P, Adj[u] \ v, u)
+    HELP-SETUP(G, P, Adj[u] \ {v}, u)
 	  
 \end{lstlisting}
 
@@ -84,4 +84,38 @@ The $O(n)$ setup happens between line 2 and line 14. This is mainly needed to in
 (and consequent parent relation) on the tree. 
 
 The rest of the algorithm walks the tree from the start to the root and from the end to the root concurrently, keeping track of the path taken and stopping when an edge was traversed by both walks. Then, the path memory to the start is reversed and inserted in the path memory for the end in order to obtain a mapping to the next node in the path from $v$ to $w$. The complexity of this step is $O(dist(v,w))$, since the number of traversed edges is at most two times the distance from $v$ to $w$, and the reversing operation at the end requires at most $dist(v,w)$ steps.
+
+\section{Exercise 2}
+
+\begin{lstlisting}[caption=Solution for exercise 2, label={lst:ex2}]	  
+FUNCTION CONNECTED-COMPONENTS(G=(V,E)):
+  for each vertex u $\in$ V(G):
+    color[u] = WHITE
+    
+  c = 0
+  
+  for each vertex s $\in$ V(G):
+    if color[u] $\neq$ WHITE:
+      continue
+      
+    color[s] = GRAY
+    Q = $\emptyset$
+    c = c + 1 
+    ENQUEUE(Q, s)
+    
+    while Q $\neq \emptyset$:
+      u = DEQUEUE(Q)
+      for each v $\in$ Adj[u]:
+        if color[v] == WHITE:
+          color[v] = GRAY
+          ENQUEUE(Q, v)
+        color[u] = BLACK
+        
+  return c
+\end{lstlisting}
+
+The algorithm is simply a modified color-only version of BFS with an extra iteration: using every vertex in the graph as a starting
+node. If the node was already visited, the color makes this iteration over all vertexes skip to the next vertex. The complexity of 
+this algorithm is $O(|V| + |E|)$ like BFS, since the iterative application to BFS over every connected component will cover every edge
+and node of the graph exactly once, and the extra check for nodes being which is just another $O(|V|)$ cost, which can be ignored. 
 \end{document}
\ No newline at end of file