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\title{Graded Assignment 4 -- DSA}
\author{Claudio Maggioni}
\setlength{\parindent}{0cm}

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\begin{document}

\maketitle
\tableofcontents
\lstlistoflistings
\newpage

\section{Exercise 1}

\begin{lstlisting}[caption=Solution for exercise 1, label={lst:ex1}]
FUNCTION BEST-PATH(G=(V,E), v, w):
  P[V(G)[0]] = NIL
  
  for each vertex u $\in$ V(G):
    prev_start[u] = NIL
    prev_end[u] = NIL
  
  prev_start[v] = START $(\textit{non-NIL})$
  prev_end[w] = END $(\textit{non-NIL})$
  
  HELP-SETUP(G, P, Adj[V(G)[0]], V(G)[0])
  
  s = v
  e = w
  
  while prev_end[s] is NIL and prev_start[e] is NIL:
    if P[s] is not NIL:
      prev_start[P[s]] = s
      s = P[s]
    if P[e] is not NIL:
      prev_end[P[e]] = e
      e = P[e]
  
  if prev_end[s] is not NIL:
    n = s
  else:
    n = e
  
  while s is not v:
    s = prev_start[s]
    prev_end[s] = P[s]
  
  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:
    P[u] = v
    HELP-SETUP(G, P, Adj[u] \ {v}, u)
	  
\end{lstlisting}

The $O(n)$ setup happens between line 2 and line 14. This is mainly needed to initialize some help arrays and define an
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 $w$. This mapping is then printed.
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, as the printing operation.

\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.

\section{Exercise 3}

Assume data is provided in Graph-like form $G=(V,E)$ where $V$ is the set of butterflies, $E$ is the set of edges, and a relation
$o : E \to \textsc{true, false}$ where \textsc{false} means ``same'' and \textsc{true} means ``different'' to determine the the type
of observation. Ambiguous observations are not included in $E(G)$ in the
first place, so $o$ does not have to be defined for this case.

\begin{lstlisting}[caption=Solution for exercise 3, label={lst:ex3}]	  
FUNCTION OBSERVATION-HOLDS(G=(V,E), o):
  for each vertex u $\in$ V(G):
    color[u] = WHITE
    species[u] = NIL   
     
  for each vertex u $\in$ V(G):
    if color[u] $\neq$ WHITE:
      continue
    species[u] = FALSE  
    if not DFS-CHECK-OBSERVATION(species, o, v):
      return FALSE  
       
  return TRUE
    
FUNCTION DFS-CHECK-OBSERVATION(species, o, v):
  color[u] = GREY
  
  for each vertex v $\in$ Adj[u]:
    if color[v] == WHITE:
      species[v] = species[u] XOR o($\textit{edge}$ (u, v))
      if not DFS-CHECK-OBSERVATION(species, o, v):
        return FALSE
    else:
      if species[u] $\neq$ species[v] XOR o($\textit{edge}$ (v, u)) $\lor$
         species[v] $\neq$ species[u] XOR o($\textit{edge}$ (u, v)):
         return FALSE
         
    color[u] = BLACK
    return TRUE
\end{lstlisting}

The code given traverses $G$ using a modified DFS by first assigning an arbitrary species (\texttt{FALSE}) to the first
vertex encountered, then by computing the species of the subsequent nodes encountered using the observation mapping $o : E \to
\textsc{true, false}$. A XOR is used to assign the opposite species (flip the species[\ldots] bit) to the newly discovered node
if $o(\textsc{current node, new node})$ is ``different'', and to assign the same species if the observation is ``same''
\footnote{\textsc{true, false} mean ``different'' and ``same'' when they are result of $o(\ldots)$. When assigning to
\texttt{species[\ldots]}, they represent an arbitrary assignment of the two species of butterfly.}.

Paths between already visited vertexes (non-white vertexes) are checked in order to find inconsistencies. Both edge traversal
directions are checked in order to handle cases where observation are directed (i.e. $o(\textit{edge} (v, u)) \neq 
o(\textit{edge} (v, u)))$. If an inconsistency is found, we return \texttt{FALSE}, otherwise we return true.

Note that if the observation graph $G$ is composed by more than one connected component assigning an arbitrary species to the first
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)
  s = v
  while s $\neq$ w:
    edge_w = weight($\textit{edge}$ (s, P[s]))
    if edge_w > c:
      return FALSE
    s = P[s]
  
  return TRUE
    
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 every edge in the path from $v$ to
$w$ 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 
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.: $|E_T| = |V| - 1$). 

\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):
    $\textit{delete edge where IS-MST-MINIMAL stopped from T}$
    $\textit{add (v, w) to T}$
\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}