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The line graph $L(G)$ of a simple graph $G$ is defined as follows:

  • There is exactly one vertex $v(e)$ in $L(G)$ for each edge $e$ in $G$.
  • For any two edges $e$ and $e'$ in $G$, $L(G)$ has an edge between $v(e)$ and $v(e')$, if and only if $e$ and $e'$ are incident with the same vertex in $G$.

Which of the following statements is/are TRUE?

  • (P) The line graph of a cycle is a cycle.
  • (Q) The line graph of a clique is a clique.
  • (R) The line graph of a planar graph is planar.
  • (S) The line graph of a tree is a tree.
     
  1. $P$ only
  2. $P$ and $R$ only
  3. $R$ only
  4. $P, Q$ and $S$ only
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61 votes

P) True. Because every edge in cycle graph will become a vertex in new graph $L(G)$ and every vertex of cycle graph will become an edge in new graph.

R) False. We can give counter example. Let $G$ has $5$ vertices and $9$ edges which is a planar graph. Assume degree of one vertex is $2$ and of all others are $4$. Now,  $L(G)$ has $9$ vertices (because $G$ has $9$ edges ) and $25$ edges. (See below). But for a graph to be planar $|E| <= 3|V| - 6$.

For $9$ vertices $|E| \leq 3*9 - 6$

$⇒|E| \leq 27 - 6$

$⇒|E| \leq 21$. But $L(G)$ has $25$ edges and so is not planar.

As R) is False option (B), (C) are eliminated.
http://www.personal.kent.edu/~rmuhamma/GraphTheory/MyGraphTheory/planarity.htm


S) False. By counter example. Try drawing a simple tree which has a Root node ,Root node has one child A, node A has two child B and C. Draw its Line graph acc. to given rules in question you will get a cycle graph of $3$ vertices.

So (D) also not correct.

$\therefore$  option (A) is correct.


For a graph G with n vertices and m edges, the number of vertices of the line graph L(G) is m, and the number of edges of L(G) is half the sum of the squares of the degrees of the vertices in G, minus m.

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L(G) is a 2-set {e1,e2} of edges in G with which are adjacent to a common vertex v. This vertex v is uniquely determined by {e1,e2}. 

If a vertex vv has degree d, there are (d2) 2-sets{e1,e2} such that e1 and e2 are adjacent to v

So the total number of edges in L(G) is 

$\sum_{i=0}^{n}\binom{n}{2}=n(n-1)/2$

There is very simple and easy proof to drive expression for number of edges in lines graph- 

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2 votes

i think

(A) P only

because by taking a simple graph of 4 vertices(i.e planar graph)(take a 2 regular graph) and we can some to know it's planar    :    option C eliminated ,

but tree for tree is not possible as it seems to be a cycle : option D eliminated.

also planar doesn't seem possible ,

hence possible answer is option A.

Answer:

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