\\[ A^4 = \left(
\begin{array}{c | cc}
s \rightarrow t & x & y \\\\
\hline
x & \lbrace f;f;f;f, \; g;h;f;f, \; f;g;h;f \; f;f;g;h, \; f;h;g;h \rbrace & \lbrace f;f;f;g, \; g;h;f;g, \; f;g;h;g \rbrace \\\\
y & \lbrace h;f;f;f \; h;f;g;h \; h;g;h;f \rbrace & \lbrace h;f;f;g \; h;g;h;g \rbrace
\end{array}
\right) \\]

\[ A^1 = \left( \begin{array}{c | cc} s \rightarrow t & x & y \\\\ \hline x & \lbrace f \rbrace & \lbrace g \rbrace \\\\ y & \lbrace h \rbrace & \emptyset \end{array} \right) \]

Oh, I see! ^:)^

This actually does not require a pointed graph.
The matrix gives all the paths between every pair of nodes.