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Let $G=(V, E)$ be a graph. Define $\overline{G}$ to be $(V, \overline{E})$, where for all $u, \: v \: \in V \: , (u, v) \in \overline{E}$ if and only if $(u, v) \notin E$. Then which of the following is true?

  1. $\overline{G}$ is always connected.
  2. $\overline{G}$ is connected if $G$ is not connected.
  3. At least one of $G$ and $\overline{G}$ connected.
  4. $G$ is not connected or $\overline{G}$ is not connected
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25 votes

A correct answer would be (C) At least one of G and G-bar is connected.

Option (A): It is straight forward wrong.

Option (B): This is a subset of Option (C).

Option (D): This also implies that G and G-bar are not connected at the same time, which is impossible

Here are the total possibilities:

$$\begin{array}{|l|l|l|} \hline \textbf{G} & \overline{\boldsymbol{G}} & \textbf{Possible/Not Possible}\\\hline \text{Connected} & \text{Connected}& \text{Possible}\\\hline \text{Connected} & \text{Disconnected}& \text{Possible}\\\hline \text{Disconnected} & \text{Connected}& \text{Possible}\\\hline \text{Disconnected} & \text{Disconnected}& \text{Not-Possible} \\\hline \end{array}$$                                                                 

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Option A - False. Counter Example - Complete Graph

Option B- $\overline{G}$ is connected if G is not connected. True

Proof -  Let G is disconnected. Suppose u and v are vertices. If (u.v) is not an edge in G, then it is an edge in $\overline{G}$, and so we have a path from u to v in $\overline{G}$. On the other hand, if (u,v) is an edge in G, then this means u and v are in the same component of G. Since G is disconnected, we can find a vertex w in a different component so that neither (u,w) nor (v,w) is edges of G. Then (u,w) and (v,w) are edges of $\overline{G}$ and hence there exist a path from u to v in $\overline{G}$.

This shows that any two vertices in $\overline{G}$ have a path (in fact a path of length one or two) between them in $\overline{G}$, so $\overline{G}$ is connected.

Option C - At least one of G and $\overline{G}$ connected.

Which means $\overline{G}$  is connected if G is not connected and vice versa. 

Proof:

If  $\overline{G}$ is not connected, then there exists two disjoint sets V1 and V2 such that V1 $\subset V$ and V2 $\subset V$ and for all v1∈V1 v2∈V2 we have (v1,v2) ∉ $\overline{E}$ .This means that $ \forall v1 \in V1 \ and \ v2 \in V2$ we have $(v1,v2)∈E$ Hence G is connected.

Option D - G is not connected or $\overline{G}$ is not connected. 
False. Both can be connected at the same time. 

Option B is a subset of option C.

So Option C is correct. At least one of G and $\overline{G}$ connected.

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A) If G is a complete graph then G' is a null graph which eliminates option A) 

D) Both G and G' can be connected. Example :

B) Suppose C1,C2,C3,...Cm be connected components in graph G (Note that these m components though are connected, there willnot be any edge between any 2 components). Now in G', all vertices of a component (say C1) will be connected to all vertices of all other components (C2,C3,C4,...). Similiarly, all vertices of a component (say C2) will be connected to all vertices of all other components (say C1,C3,C4,...).. Now G' is connected because 

                  1) Any vertex of a component Ci is connected to all other component's vertices. So there is a connection between all components.

                  2) Any 2 vertices say v1 and v2 inside a component (say C1) will also be connected because there is an edge from v1 of C1 to any vertex of C2 and there is an edge from that vertex of C2 to v2. 

C) If B) is true C) should also be TRUE... Hence both B) and C) are correct.

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Option B is the subset of option C. Hence option C is the correct answer.

Answer:

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