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<record version="1" id="8552">
 <title>line graph</title>
 <name>LineGraph</name>
 <created>2006-11-13 16:22:47</created>
 <modified>2006-11-13 16:22:47</modified>
 <type>Definition</type>
 <creator id="409" name="mps"/>
 <author id="409" name="mps"/>
 <classification>
	<category scheme="msc" code="05C75"/>
 </classification>
 <related>
	<object name="DeBruijnDigraph"/>
 </related>
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\newtheorem{example}{Example}</preamble>
 <content>\PMlinkescapeword{connected}
\PMlinkescapeword{right}
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Let $G$ be a graph.  The \emph{line graph} of $G$ is denoted by $L(G)$ and is defined as follows.  The vertices of $L(G)$ are the edges of $G$.  Two vertices of $L(G)$ are connected by an edge if and only if the corresponding edges in $G$ form a path (which must be a directed path if $G$ is a digraph).

\begin{example}
Let $G$ be the undirected graph on the vertex set $V(G)=\{a,b,c,d\}$ and edge set $E(G)=\{ab, bc, cd, bd\}$.  Then the line graph $L(G)$ has vertex set $V(L(G))=\{ab, bc, cd, bd\}$.  Since $ab$ and $bc$ are incident in $G$, they are connected by an edge in $L(G)$.  We label this edge by the unique walk from $a$ to $c$ determined by these edges, namely $abc$.  With this notation, the edge set is $E(L(G))=\{abc, abd, bcd, bdc, cbd\}$.
\[\xymatrix{
&amp; a\ar@{-}[d]^{ab}                     &amp; &amp;                          &amp; ab\ar@{-}[dl]_{abc}\ar@{-}[dr]^{abd} &amp; \\
                     &amp; b\ar@{-}[dl]_{bc}\ar@{-}[dr]^{bd} &amp;  &amp; bc\ar@{-}[rr]^{bcd}\ar@{-}[dr]_{cbd} &amp;                          &amp; bd\ar@{-}[dl]^{bdc} \\
c\ar@{-}[rr]_{cd} &amp; &amp; d &amp;                         &amp; cd
}\]
Above we display the graphs $G$ (left) and $L(G)$ (right).
\end{example}

Part of the reason for interest in line graphs is the following observation:

\begin{proposition*}
If a graph is Eulerian, then its line graph is Hamiltonian.
\end{proposition*}

This follows immediately from the fact that a sequence of incident edges in $G$ is a sequence of incident vertices in $L(G)$.

\begin{example}
Let $B_n$ be the de Bruijn digraph on binary strings of length $n$.  Then for $n\ge 2$, the next de Bruijn digraph can be obtained by taking the line graph of the previous one, that is, $B_{n+1}=L(B_n)$.
\end{example}</content>
</record>
