| Version 2 |
Version 1 |
| Let $G$ be a finite graph with $n$ vertices, $\{v_1, \ldots, v_n\}$ |
Let $G$ be a finite graph with $n$ vertices, $\{v_1, \ldots, v_n\}$ |
| and $m$ edges, $\{e_1, \ldots, e_m\}$. |
and $m$ edges, $\{e_1, \ldots, e_m\}$. |
| For each edge $e = (v_i,v_j)$ of $G$ choose one vertex |
For each edge $e = (v_i,v_j)$ of $G$ choose one vertex |
| to be the positive end and the other to be the negative end. In this way, |
to be the positive end and the other to be the negative end. In this way, |
|
we assign an \emph{orientation} to $G$. The \emph{\PMlinkescapetext{incidence matrix}}
|
we assign an \emph{orientation} to $G$. The \emph{incidence matrix}
|
| of $G$ with respect an orientation is an $n \times m$ matrix |
of $G$ with respect an orientation is an $n \times m$ matrix |
| $D=(d_{i,j})$ |
$D=(d_{i,j})$ |
| where |
where |
| \begin{displaymath} |
\begin{displaymath} |
| d_{i,j} = \left\{ \begin{array}{ll} |
d_{i,j} = \left\{ \begin{array}{ll} |
| +1 & \textrm{if $v_i$ is the positive end of $e_j$} \\ |
+1 & \textrm{if $v_i$ is the positive end of $e_j$} \\ |
| -1 & \textrm{if $v_i$ is the negative end of $e_j$} \\ |
-1 & \textrm{if $v_i$ is the negative end of $e_j$} \\ |
| 0 & \textrm{otherwise}. |
0 & \textrm{otherwise}. |
| \end{array} \right. |
\end{array} \right. |
| \end{displaymath} |
\end{displaymath} |
