Find an expression for the number of edges of $L(G)$ in terms of the degrees of the vertices of $G$.

Question

I have a hard time trying to understand this prove.

Find an expression for the number of edges of $L(G)$ in terms of the degrees of the vertices of $G$.

Let $\{v_1, v_2, . . . , v_n\}$ be the vertices of $G$ and let $d_i$ be the degree of the vertex $v_i$. An edge ${v_i, v_j}$ will be adjacent to $d_i − 1 + d_j − 1$ edges. Until this point I understand it. Since there are $d_i$ edges that contain $v_i$ in $G$, the sum of the degrees of the vertices in $L(G)$ will be $\sum_{i=1}^n d_i(d_i − 1)$ and so the number of edges in $L(G)$ is $\sum_{i=1}^n \frac{d_i(d_i − 1)}{2}$.

From: http://garsia.math.yorku.ca/~zabrocki/math3260w03/hw1sln.pdf

Answer 1

I think the given proof is a bit confusing.

Try this one:

An edge in $L(G)$ is a $2$-set $\{e_1,e_2\}$ of edges in $G$ with which are adjacent to a common vertex $v$. This vertex $v$ is uniquely determined by $\{e_1,e_2\}$. If a vertex $v$ has degree $d$, there are $\binom{d}{2}$ $2$-sets $\{e_1,e_2\}$ such that $e_1$ and $e_2$ are adjacent to $v$.

So the total number of edges in $L(G)$ is $$ \sum_{i = 1}^n \binom{d_i}{2} = \sum_{i = 1}^n \frac{d_i(d_i-1)}{2}\text{.} $$