# GATE2004-81

5.3k views

Let $G_1=(V,E_1)$ and $G_2 =(V,E_2)$ be connected graphs on the same vertex set $V$ with more than two vertices. If $G_1 \cap G_2= (V,E_1\cap E_2)$  is not a connected graph, then the graph $G_1\cup G_2=(V,E_1\cup E_2)$

1. cannot have a cut vertex
2. must have a cycle
3. must have a cut-edge (bridge)
4. has chromatic number strictly greater than those of $G_1$ and $G_2$

edited
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Refere http://math.stackexchange.com/questions/1535029/union-of-two-graphs This question, if you are not able to interpret this image.
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I am not getting what he is trying to say. Can you explain a bit?
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https://www.gatementor.com/viewtopic.php?f=260&t=8394

This explains it in a simple way.....
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G1∩G2=(V,E1∩E2)  is not a connected graph,

wht does it mean

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Why we are not taking directed graph here? Will there also be a cycle then?

Take a tree for example

1. False. Every vertex of tree(other than leaves) is a cut vertex.
2. True.
3. False. Without E in $G1$ and $G2$, $G1 \cup G2$  has no bridge.
4. False. $G1 \cup G2$, $G1$, $G2$ three graphs have same chromatic number of $2$.

edited
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G1 and G2 are connected graph. As G1 $\cap$ G2 is disconnected graph, there must exist two vertices say V1 and V2 such that there is no path between them in G1 $\cap$ G2.

But as G1 is connected graph it must have some path P1 between vertices V1 and V2. Similarly, G2 must have some path P2 between V1 and V2.

Now, these P1 and P2 are not the same paths, if they are they will be present in G1 $\cap$ G2, which will contradict with our assumption.

In G1 $\cup$ G2, we have two paths P1 and P2 to between V1 and V2. That form the cycle. See the image.

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@srestha but G1 intersection G2 is connected but should be not connected as specified in question.

Correct me if m wrong.
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@Prateek

See the node D, if you try to take intersection of G1 and G2, D will be disconnected as the path is not common in G1 and G2.
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@srestha

G1 intersection G2 is disconnected or not?

I found that we have a common path in both G1 and G2 ,which is connected.

what actually G1 and G2 intersection is?

Say G1 is Black colored graph, and G2 is red colored graph.
In this figure i overlapped both graphs, to see them together.

Given that $G_1 \cap G_2$ is Disconnected, To make it happen G1 and G2 has to meet somewhere and then leave at some point, Once they leave each other and then have to meet somewhere again to have atleast two different disconnected component.
Now it is clear that this will form a cycle (See image has cycle.)

Note: If  $G_1 \cap G_2$ need not to be disconnected then $G_1\cup G_2$ need not to have a cycle.

(in this case when one of the Graph has cycles then only $G_1\cup G_2$ can have cycle.)

A very simple reasoning in this question may fetch the answer very fast. Since the two graphs G1 and G2 are connected, there will be path from a vertex to every other vertex in  both G1 and G2 but since G1 ∩ G2 is not connected , at least for one vertex pair v1,v2 the connecting paths must be different in both G1 and G2. Hence G1 union G2 will definitely have a cycle.

There would be "at least " one cycle there in G1 Union G2

Question is indirectly saying in $G_1$ U $G_2$ we have edge disjoin path between some vertices because  $G_1$ ∩  $G_2$ is disconnected (means at least two vertices could reach each other via edge disjoint path ). So option (B) is True in all the cases.
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As G1 and G2 are connected, let us say one vertex is V,now there is a way from V to all other vertices in both G1 and G2,but as their intersection  is disconnected,means there are some edges which are in G1 but not in G2 and vice versa.Or we can say there is atleast one from V to V' which is different in both G1 and G2 because if for all vertices it is same path from V to all other vertices then it will not give disconnected graph as intersection.

Now whn union will be done then their will be two paths from V to V' (one of G and other of G1) and then it will form a cycle.

Option B is correct.

In this example above there is cut vertex E is present so option A is false,

there is no cut edge so option C is false and

not necessarily chromatic number will be greater, it can be equal.  so option D is also false.

This can also be a approach to the solution.

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Just that handwriting is not good :p

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