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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
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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.

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 a line graph, no. 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
No. of vertices in line graph $= 6$
So, no. of edges to make complete graph with $6$ vertices $= \frac{6\times 5}{2} = 3\times 5=15$
But for given line graph from complete graph of $4$ vertices we have only $12$ edges.

Option (B)   True

1. Smallest line graph for original graph one edge

which is also connected graph

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

Option (C)    False

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

Option (D)    False

Because line graph degree of vertex depends on the attribute

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

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

So, this is wrong.

Option (E)    False (wrong as proved in above option (D))

edited

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.

edited
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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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