TIFR CSE 2017 | Part B | Question: 13

Question

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?

• A. the line graph for a complete graph is complete
• B. the line graph for a connected graph is connected
• C. the line graph for a bipartite graph is bipartite 
• D. the maximum degree of any vertex in the line graph is at most the maximum degree in the original graph
• E. each vertex in the line graph has degree one or two

Comments

you can check with K(2,2). The line graph obtained is not biparitite.

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@Chandrashis Mazumdar Line graph of K(2,2) is bipartite. It's K(2,2) only. So thats not a good example.

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For option C counter example is K(2,3) you you draw the line graph is produce a odd cycle(Not bipartile).

Answer 1

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))

Comments

The complete graph shown in the picture seems to be wrong. AC has multiple edges.

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Yes it is wrong.

Answer 2

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

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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