Graph Representations Examples
Adjacency Matrix Examples
Given the following graph:
The adjacency matrix is:
\[
\displaylines{
\hspace{0.7cm}\begin{array}{} a&b&c&d&h&i \end{array} \ \ \
\\
\begin{array}{} a\\b\\c\\d\\h\\i \end{array}
\begin{pmatrix}
0 & 1 & 0 & 1 & 0 & 0 \\
1 & 0 & 0 & 1 & 1 & 0 \\
1 & 0 & 0 & 0 & 0 & 1 \\
0 & 0 & 1 & 0 & 1 & 0 \\
0 & 0 & 0 & 0 & 0 & 1 \\
0 & 0 & 0 & 0 & 1 & 0 \\
\end{pmatrix}
}
\]
Given the following graph:
The adjacency matrix is:
\[
\displaylines{
\hspace{0.7cm}\begin{array}{} a&b&c&d&h&i \end{array} \ \ \
\\
\begin{array}{} a\\b\\c\\d\\h\\i \end{array}
\begin{pmatrix}
0 & 1 & 1 & 1 & 0 & 0 \\
1 & 0 & 0 & 1 & 1 & 0 \\
1 & 0 & 0 & 1 & 0 & 1 \\
1 & 1 & 1 & 0 & 1 & 0 \\
0 & 1 & 0 & 1 & 0 & 1 \\
0 & 0 & 1 & 0 & 1 & 0 \\
\end{pmatrix}
}
\]
Incidence Matrix Example
Given the following graph:
The incidence matrix is:
\[
\displaylines{
\hspace{0.7cm}\begin{array}{}\ a&\ \ \ b&\ \ \ c&\ \ \ d&\ \ \ h&i \end{array} \ \ \
\\
\begin{array}{} a\\b\\c\\d\\h\\i \end{array}
\begin{pmatrix}
0 & 1 & -1 & 1 & 0 & 0 \\
1 & 0 & 0 & 1 & 1 & 0 \\
1 & 0 & 0 & -1 & 0 & 1 \\
-1 & -1 & 1 & 0 & 1 & 0 \\
0 & -1 & 0 & -1 & 0 & 1 \\
0 & 0 & -1 & 0 & 1 & 0 \\
\end{pmatrix}
}
\]
Transitive closure Example
Example 1 (Roy-Warshall algorithm)
Given the following graph:
Apply the Roy-Warshall algorithm.
- picking A
- $p=C$, $s=B$, creating (C, B)? yes
- picking B
- $p=A$, $s=D$, creating (A, D)? yes
- $p=A$, $s=C$, creating (A, C)? yes
- $p=C$, $s=D$, creating (C, D)? yes
- picking C
- $p=B$, $s=A$, creating (B, A)? yes
- $p=A$, $s=B$, creating (A, B)? no
- picking D
- done
Example 2 (Roy-Warshall algorithm)
Given the following adjacency matrix with $V=\lbrace A, B, C, D\rbrace$:
\[
\begin{pmatrix}
0&1&1&0\\
1&0&0&1\\
0&0&0&1\\
0&1&0&0
\end{pmatrix}
\]
Apply the Roy-Warshall algorithm. π‘ Predecessors are in the column and successors are in the line of the vertex we picked.
- picking A
- $s=B$, $p=B$? creating (B, B)? no
- $s=C$, $p=B$? creating (B, C)? yes
- picking B
- $s=A$
- $p=A$? creating (A, A)? no
- $p=D$? creating (D, A)? yes
- $s=C$
- $p=A$? creating (A, C)? no
- $p=D$? creating (D, C)? yes
- $s=D$
- $p=A$? creating (A, D)? yes
- $p=D$? creating (D, D)? no
- picking C
- $s=D$
- $p=A$? creating (A, D)? no
- $p=B$? creating (B, D)? no
- $p=D$? creating (C, D)? no
- picking D
- $s=A$, $p=C$? creating (C, A)? yes
- $s=B$, $p=C$? creating (C, B)? yes
- $s=B$, $p=C$? creating (C, B)? yes