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59 votes
59 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
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4 Comments

This graph is the graph which is told in words in below answer

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Please someone explain @chotu link content. I am not able to understand how it can be used.
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For planar graph, this example which is easier to understand

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8 Answers

61 votes
61 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.
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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4 Comments

How is it possible to have graph G of 5  vertices with 4 vertices having degree 4 and only one vertex having degree 2.
At max, the possible case is 3 vertices with degree 4 and 2 vertices with degree 3.
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@ash_khola its possible only when we have multi edges or self loops on  vertices . Basically the graph @prashant singh 1 assumed in the answer is not a simple graph . 

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The line graph of any planar graph is planar:

Notation: Lines are edges in the line graph of G

An algorithm:

Consider the planar rep of G; Mark midpoints of every edge of G as a vertex in G1. Every line between two midpoints of 2 adjacent edges lie in a region. 2 lines in 2 different regions don't intersect. So, only lines in the same region might intersect. But the corresponding edges whose lines are in a particular region form a cycle. We know that the line graph of a cycle is a cycle (=> lines don’t intersect). Hence no pair of lines will intersect in this representation of L(G) and hence is a planar graph
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25 votes
25 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 

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

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

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4 Comments

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

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.

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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
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10 votes
10 votes

Answer is  option (A)

2 Comments

underrated answer .
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Although the answer is correct, the solution is not. Line graph of K4 is planar 

https://math.stackexchange.com/questions/3018581/is-lk4-graph-planar

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2 votes
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.

4 Comments

Oh u mean two vertex is coincide in the same point (ab coincide in a...) . R u sure it happens for any grah?

Planar graph are those graph in which no edge can intersect other. So how planar concept can be eliminated?
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take any complex example and you will see that the graph becomes non-planar,in my photo take an edge from a to c(diagonal) and you will see G2 becoming non-planar.

also, in any case cycle in G1 has to give cycle in G2.
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Yes. Taking examples is the better way to do, unless you know more about line graphs. 

http://mathworld.wolfram.com/LineGraph.html

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Answer:

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