Done GA4

GA4/ga4.tex

@@ -46,8 +46,8 @@ FUNCTION BEST-PATH(G=(V,E), v, w):
     prev_start[u] = NIL
     prev_end[u] = NIL
   
-  prev_start[v] = START
+  prev_start[v] = START $(\textit{non-NIL})$
-  prev_end[w] = END
+  prev_end[w] = END $(\textit{non-NIL})$
   
   HELP-SETUP(G, P, Adj[V(G)[0]], V(G)[0])
   
@@ -70,8 +70,13 @@ FUNCTION BEST-PATH(G=(V,E), v, w):
   while s is not v:
     s = prev_start[s]
     prev_end[s] = P[s]
-    
+  
-  return prev_end
+  s = v
+  while s != w:
+    print(s)
+    s = prev_end[s]
+  if v != w:
+    print(w) 
 
 FUNCTION HELP-SETUP(G=(V,E), P, S, v):
   for each vertex u $\in$ S:
@@ -85,9 +90,10 @@ arbitrary root (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
+path memory for the end in order to obtain a mapping to the next node in the path from $v$ to $w$. This mapping is then printed.
-$O(dist(v,w))$, since the number of traversed edges is at most two times the distance from $v$ to $w$, and the reversing operation
+The complexity of this step is $O(dist(v,w))$, since the number of traversed edges is at most two times the 
-at the end requires at most $dist(v,w)$ steps.
+distance from $v$ to $w$, and the reversing operation
+at the end requires at most $dist(v,w)$ steps, as the printing operation.
 
 \section{Exercise 2}
 
@@ -134,7 +140,7 @@ first place, so $o$ does not have to be defined for this case.
 FUNCTION OBSERVATION-HOLDS(G=(V,E), o):
   for each vertex u $\in$ V(G):
     color[u] = WHITE
-    species[u] = Nil   
+    species[u] = NIL   
      
   for each vertex u $\in$ V(G):
     if color[u] $\neq$ WHITE:
@@ -177,5 +183,55 @@ Note that if the observation graph $G$ is composed by more than one connected co
 vertex encountered in each connected component does not compromise the solutions, since we are not asked to find the correct species
 assignment but we are just asked to find inconsistencies.
 
- 
+\section{Exercise 4}
+
+\subsection{Point 1}
+
+We assume the minimum spanning tree $T$ is represented as an adjacency-mapped graph. \textit{weight} is the weight mapping for every
+edge in the minimal spanning tree. The other parameters must be given as described in the assignment.
+
+\begin{lstlisting}[caption=Solution for exercise 4 point 1, label={lst:ex4p1}]	  
+FUNCTION IS-MST-MINIMAL(T=(V,E), weight, v, w, c):
+  P[w] = NIL
+  DEFINE-PARENT(T, P, Adj[w], w)
+  edge1_w = weight($\textit{edge}$ (v, P[v]))
+  s = v
+  while P[s] $\neq$ w:
+    s = P[s]
+  edge2_w = weight($\textit{edge}$ (s, w))
+  return c > edge1_w $\land$ c > edge2_w
+    
+FUNCTION DEFINE-PARENT(G=(V,E), P, S, v):
+  for each vertex u $\in$ S:
+    P[u] = v
+    DEFINE-PARENT(G, P, Adj[u] \ {v}, u)
+\end{lstlisting}
+
+The algorithm works by walking the entire tree with \texttt{DEFINE-PARENT} in order to define a parent relation considering $w$ as 
+the root. Then, this relation is used to define the path between $v$ and $w$, and the weights of the
+first outgoing edge from $v$ and the final edge to $w$ are memorized in \texttt{edge1\_w} and \texttt{edge2\_w}.
+The complexity of this step is $O(|V|)$ since the total number of edges in a tree is linearly dependent to the number
+of vertices (i.e.: $|E_T| = |V| - 1$). 
+
+Note that if $v$ and $w$ are adjacent then these two values are the same, but this does not compromise the algorithm.
+
+Finally, we check if $(v, w)$ is the the edge with highest weight with respect to \textit{edge1} and \textit{edge2}. If this is the
+case, then replacing \textit{edge1} or \textit{edge2} with $(v, w)$ would not give a spanning tree with minimum total weight, and 
+thus we return \texttt{FALSE}. Otherwise, the opposite is true and we return \texttt{TRUE}. Note that this step is constant, so the
+total complexity is $O(|V|)$.
+
+\subsection{Point 2}
+
+\begin{lstlisting}[caption=Solution for exercise 4 point 2, label={lst:ex4p1}]	  
+FUNCTION MAKE-MST-MINIMAL(T=(V,E), weight, v, w, c):
+  if not IS-MST-MINIMAL(T, weight, v, w, c):
+    if edge1_w < edge2_w:
+      $\textit{delete edge1 from T}$
+    else:
+      $\textit{delete edge2 from T}$
+    $\textit{add (v, w) to T}$
+\end{lstlisting}
+
+For what said before, this algorithm updates $T$ to a valid MST and runs in $O(|V_T|)$ which is always $< O(|E|)$.
+
 \end{document}