Done GA4
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2 changed files with 65 additions and 9 deletions
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GA4/ga4.pdf
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GA4/ga4.pdf
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GA4/ga4.tex
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GA4/ga4.tex
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@ -46,8 +46,8 @@ FUNCTION BEST-PATH(G=(V,E), v, w):
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prev_start[u] = NIL
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prev_start[u] = NIL
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prev_end[u] = NIL
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prev_end[u] = NIL
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prev_start[v] = START
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prev_start[v] = START $(\textit{non-NIL})$
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prev_end[w] = END
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prev_end[w] = END $(\textit{non-NIL})$
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HELP-SETUP(G, P, Adj[V(G)[0]], V(G)[0])
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HELP-SETUP(G, P, Adj[V(G)[0]], V(G)[0])
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@ -71,7 +71,12 @@ FUNCTION BEST-PATH(G=(V,E), v, w):
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s = prev_start[s]
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s = prev_start[s]
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prev_end[s] = P[s]
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prev_end[s] = P[s]
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return prev_end
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s = v
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while s != w:
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print(s)
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s = prev_end[s]
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if v != w:
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print(w)
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FUNCTION HELP-SETUP(G=(V,E), P, S, v):
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FUNCTION HELP-SETUP(G=(V,E), P, S, v):
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for each vertex u $\in$ S:
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for each vertex u $\in$ S:
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@ -85,9 +90,10 @@ arbitrary root (and consequent parent relation) on the tree.
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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
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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
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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
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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
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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
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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.
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$O(dist(v,w))$, since the number of traversed edges is at most two times the distance from $v$ to $w$, and the reversing operation
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The complexity of this step is $O(dist(v,w))$, since the number of traversed edges is at most two times the
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at the end requires at most $dist(v,w)$ steps.
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distance from $v$ to $w$, and the reversing operation
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at the end requires at most $dist(v,w)$ steps, as the printing operation.
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\section{Exercise 2}
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\section{Exercise 2}
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@ -134,7 +140,7 @@ first place, so $o$ does not have to be defined for this case.
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FUNCTION OBSERVATION-HOLDS(G=(V,E), o):
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FUNCTION OBSERVATION-HOLDS(G=(V,E), o):
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for each vertex u $\in$ V(G):
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for each vertex u $\in$ V(G):
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color[u] = WHITE
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color[u] = WHITE
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species[u] = Nil
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species[u] = NIL
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for each vertex u $\in$ V(G):
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for each vertex u $\in$ V(G):
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if color[u] $\neq$ WHITE:
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if color[u] $\neq$ WHITE:
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@ -177,5 +183,55 @@ Note that if the observation graph $G$ is composed by more than one connected co
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vertex encountered in each connected component does not compromise the solutions, since we are not asked to find the correct species
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vertex encountered in each connected component does not compromise the solutions, since we are not asked to find the correct species
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assignment but we are just asked to find inconsistencies.
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assignment but we are just asked to find inconsistencies.
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\section{Exercise 4}
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\subsection{Point 1}
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We assume the minimum spanning tree $T$ is represented as an adjacency-mapped graph. \textit{weight} is the weight mapping for every
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edge in the minimal spanning tree. The other parameters must be given as described in the assignment.
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\begin{lstlisting}[caption=Solution for exercise 4 point 1, label={lst:ex4p1}]
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FUNCTION IS-MST-MINIMAL(T=(V,E), weight, v, w, c):
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P[w] = NIL
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DEFINE-PARENT(T, P, Adj[w], w)
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edge1_w = weight($\textit{edge}$ (v, P[v]))
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s = v
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while P[s] $\neq$ w:
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s = P[s]
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edge2_w = weight($\textit{edge}$ (s, w))
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return c > edge1_w $\land$ c > edge2_w
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FUNCTION DEFINE-PARENT(G=(V,E), P, S, v):
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for each vertex u $\in$ S:
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P[u] = v
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DEFINE-PARENT(G, P, Adj[u] \ {v}, u)
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\end{lstlisting}
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The algorithm works by walking the entire tree with \texttt{DEFINE-PARENT} in order to define a parent relation considering $w$ as
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the root. Then, this relation is used to define the path between $v$ and $w$, and the weights of the
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first outgoing edge from $v$ and the final edge to $w$ are memorized in \texttt{edge1\_w} and \texttt{edge2\_w}.
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The complexity of this step is $O(|V|)$ since the total number of edges in a tree is linearly dependent to the number
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of vertices (i.e.: $|E_T| = |V| - 1$).
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Note that if $v$ and $w$ are adjacent then these two values are the same, but this does not compromise the algorithm.
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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
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case, then replacing \textit{edge1} or \textit{edge2} with $(v, w)$ would not give a spanning tree with minimum total weight, and
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thus we return \texttt{FALSE}. Otherwise, the opposite is true and we return \texttt{TRUE}. Note that this step is constant, so the
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total complexity is $O(|V|)$.
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\subsection{Point 2}
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\begin{lstlisting}[caption=Solution for exercise 4 point 2, label={lst:ex4p1}]
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FUNCTION MAKE-MST-MINIMAL(T=(V,E), weight, v, w, c):
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if not IS-MST-MINIMAL(T, weight, v, w, c):
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if edge1_w < edge2_w:
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$\textit{delete edge1 from T}$
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else:
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$\textit{delete edge2 from T}$
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$\textit{add (v, w) to T}$
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\end{lstlisting}
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For what said before, this algorithm updates $T$ to a valid MST and runs in $O(|V_T|)$ which is always $< O(|E|)$.
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\end{document}
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\end{document}
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