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+33 votes

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
asked in Graph Theory by Veteran (406k points)
edited by | 3.9k views

3 Answers

+38 votes
Best answer

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.

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.

answered by (445 points)
edited by
In total there are 5 vertices.So 1*2+4*4=18. Where is ^2 coming from?Can you tell that?I am missing in this particular option:(.

Please elaborate a bit on the last calculation of getting 34 and 25?

for line graph we calculate in this way 

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.

i just follow this ... 

0 is this some property of line graph?
yes , and sachin metioned some method that could also used
GOt it.Thanks :)

V= 5 , e=9 in original graph now in line graph vertex=9 and edges will be ,given 1 vertex has 2 degree and other 4 has 4 degree so 2^2+4*(4^2)= 68 ,68/2= 34,

34- 9=25

Can somebody draw a graph for this ?  This selected answer should also tell how the the number of edges is 25 in the line graph along with the diagram. I am unable to get it.

nice explanation
Q is true? I have tried it and i am getting it true.

Can anyone verify if Q is also coming true?

Although ans is (A) only but i need to verify if i am doing all right.

@Shaik Masthan comment by sachin sir is very important to get intuition for this problem rather than rote learning which is given in second part of the answer. I think sachin sir comment should be included in answer for better understanding. 

the good reader, should always read all comments.... So, No problem, even it is not included in the answer.
+17 votes

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 


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

answered by Active (4.9k points)
edited by
perfect saket :)
@Saket, Sir how u have found number of edges in line graph? Which formula u r using?

Just small correction

Formula for finding no of edges in L(G):

If a vertex v has degree d, there are C(d,2) 2-sets {e1,e2} such that e1 and e2 are adjacent to v
So the total number of edges in L(G) is 

From i=1 to n ,∑C(di,2) =∑di(di−1) / 2 ,where i is vertex and di is degree of vertex i and n is the total no vertices in G.

last calculation is having some mistake,

no. of edges = $\frac{4*3}{2} + \frac{3*2}{2} + \frac{4*3}{2} + \frac{3*2}{2} + \frac{4*3}{2}$ =  6 + 3 +6 +3 +6 =24
+7 votes

Answer is A ...

answered by Boss (11.6k points)

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