in Graph Theory retagged by
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17 votes
17 votes

For an undirected graph $G=(V, E)$, the line graph $G'=(V', E')$ is obtained by replacing each edge in $E$ by a vertex, and adding an edge between two vertices in $V'$ if the corresponding edges in $G$ are incident on the same vertex. Which of the following is TRUE of line graphs?

  1. the line graph for a complete graph is complete
  2. the line graph for a connected graph is connected
  3. the line graph for a bipartite graph is bipartite 
  4. the maximum degree of any vertex in the line graph is at most the maximum degree in the original graph
  5. each vertex in the line graph has degree one or two
in Graph Theory retagged by
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4 Comments

Option B). is true.
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yes, Agreed. (y)
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why not option C?
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you can check with K(2,2). The line graph obtained is not biparitite.
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2 Answers

15 votes
15 votes
Best answer

Option B is the right answer

We can solve this question by eliminating options

Option (A) False

Let us take a complete graph of  $4$ vertices:

In the corresponding line graph, number of edges is

$\sum_{i=0}^{n}$  ${}^{d_{i}}C_{{2}},$     $d_{i}$is degree of each vertex

$=\frac{3\times 2}{2}+\frac{3\times 2}{2}+\frac{3\times 2}{2}+\frac{3\times 2}{2} = 3\times 4=12$

No. of vertices in line graph $=$ No. of edges in original graph $=6.$
So, no. of edges to make complete graph with $6$ vertices $= \frac{6\times 5}{2} = 3\times 5=15$
But we got only $12$ edges for the line graph.
Contradiction.

Option (B) True

Smallest line graph for the original graph with just one edge    

which is also connected graph

If a graph is connected with more then one edge, it's line graph will never be disconnected 

Option (C) False

 

This cannot be 2-colorable and hence is not bipartite. 

Option (D) False

Because in the line graph degree of a vertex depends on the number of neighbors its corresponding edge has in the original graph.

e.g., $\left [ A, B \right ]$ as a point in the line graph, then the degree of this vertex depends on the degree of $A$ and degree of $B$ in the original graph.

 

I'm drawing a degree for a point $[AB]$ in the line graph.

 

So, this is wrong.

Option (E) is false (wrong as proved in the above option (D))

edited by

2 Comments

The complete graph shown in the picture seems to be wrong. AC has multiple edges.
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Yes it is wrong.
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3 votes
3 votes

The line graph of a connected graph is connected. If G is connected, it contains a path connecting any two of its edges, which translates into a path in L(G) containing any two of the vertices of L(G). Therefore, option B is correct.

We can also do this question using elimination of options. 

You can view the following google drive link for the example - https://drive.google.com/open?id=0B1OKeqz0MEWwNTh4czgzalVDNGM

edited by

4 Comments

but your L(G2): e1---e2 it is bipertite right?e1 may be in one partition and e2 on another...it is bipertite i guess.
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Hi,

I have corrected the example, and updated the file.
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great example you have updated...proving line of tree is not a tree too :p
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Link is not working
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Answer:

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