Fixed exercise 4 GA4

GA4/ga4.tex

@@ -5,6 +5,7 @@
 \usepackage{xcolor}
 \usepackage{lmodern}
 \usepackage{listings}
+\usepackage{hyperref}
 
 \title{Graded Assignment 4 -- DSA}
 \author{Claudio Maggioni}
@@ -194,12 +195,14 @@ edge in the minimal spanning tree. The other parameters must be given as describ
 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:
+  while s $\neq$ w:
+    edge_w = weight($\textit{edge}$ (s, P[s]))
+    if edge_w > c:
+      return FALSE
     s = P[s]
-  edge2_w = weight($\textit{edge}$ (s, w))
+  
-  return c > edge1_w $\land$ c > edge2_w
+  return TRUE
     
 FUNCTION DEFINE-PARENT(G=(V,E), P, S, v):
   for each vertex u $\in$ S:
@@ -208,27 +211,19 @@ FUNCTION DEFINE-PARENT(G=(V,E), P, S, v):
 \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
+the root. Then, this relation is used to define the path between $v$ and $w$, and the weights of every edge in the path from $v$ to
-first outgoing edge from $v$ and the final edge to $w$ are memorized in \texttt{edge1\_w} and \texttt{edge2\_w}.
+$w$ are memorized in \texttt{edge\_w} and compared and checked against the weight of $(v, w)$. If we find an edge in the path with 
-The complexity of this step is $O(|V|)$ since the total number of edges in a tree is linearly dependent to the number
+weight bigger than $(v, w)$ we can then replace that edge with $(v, w)$, so we return \texttt{FLASE}\textit{ (sic)}
+\footnote{\url{https://medium.com/@DanielC7/dbf8773df767}} since we the old MST is not minimal anymore.
+The complexity of this 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:
+    $\textit{delete edge where IS-MST-MINIMAL stopped from T}$
-      $\textit{delete edge1 from T}$
-    else:
-      $\textit{delete edge2 from T}$
     $\textit{add (v, w) to T}$
 \end{lstlisting}