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Let $T$ be a depth first search tree in an undirected graph $G$. Vertices $u$ and $ν$ are leaves of this tree $T$. The degrees of both $u$ and $ν$ in $G$ are at least $2$. which one of the following statements is true?

1. There must exist a vertex $w$ adjacent to both $u$ and $ν$ in $G$
2. There must exist a vertex $w$ whose removal disconnects $u$ and $ν$ in $G$
3. There must exist a cycle in $G$ containing $u$ and $ν$
4. There must exist a cycle in $G$ containing $u$ and all its neighbours in $G$
edited | 3.3k views
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Please someone explain why C is not answer?
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@Arjun sir, i m getting answer B), plz help...
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Is there a way to get the answer of this question without using counter examples?
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first one eliminate option C.

second one eliminate option B.

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Is there a way to get the answer of this question without using counter examples?

Yes possible. This question is straight forward.

It says Let T be a depth first search tree in an undirected graph G. Vertices u and ν are leaves of this tree T. The degrees of both u and ν in G are at least 2

That means :

G is original graph.

T is DFS Tree , where u and v are two leaves .

Both of u and v  have degree 2 only possible when G have a cycle with both u and v  .

Hence option D is only correct :)

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Before looking at answer try to find answer using proof via contradiction or counterexample.

One diagram, which is eliminating option A, B, C.
Hence D is the answer.

answered by Boss (10.7k points)
edited by
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@Bikram sir, Please verify.
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@ahwan

yes, correct.

one point i want to add , it says in question

Let T be a depth first search tree in an undirected graph G. Vertices u and ν are leaves of this tree T. The degrees of both u and ν in G are at least 2

G is original graph.

T is DFS Tree , where u and v are two leaves .

Both of these leaves have degree 2 only possible when G have a cycle with both u and v  .

Option D is straight forward from this given statement.

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@Bikram sir, but we have to prove G have a cycle wih u & all it's neighbours. So how that statement proves anything about all of it's neighbours. I did not get it.
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@Ahwan

see, " The degrees of both u and ν in G are at least 2" this only possible when they form a cycle in G . Bcoz in DFs tree T u and v have degree 1 as they are leaves  .

And when u ( that leaf in T) have cycle , it's neighbours also part of that cycle , is not it ? Think again :)
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Got it at last. Thank you sir @Bikram Sir.
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@ahwan

in ur diagram V and U are not having degree atleast 2

which is mentioned in the question
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@A_i_\$_h Check again. They have degree 2 each.

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@ahwan yes sorry i was looking at the DFS tree instead of the graph...

Consider following graph

Dfs is .

so D is answer.

answered by (89 points)
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I think it is correct as rest of the cases have counter examples.
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@ Nikunj Vinchhi  D is the right answer, as the degree of node "u" is atleast 2 and it is a leaf node in DFT

and this is possible only when node u forms a cycle and it is visited last in the cycle.

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@Marv can u plz gie counter examples for B and c
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according to me this may be the counter example for options a,b,c

start DFS traversal from b or f

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How option B) wrong??
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The Solution shown here contains Directed edge but question asked for undirected graph.

Also not provides counter for Option C as the given graph contains a cycle containing u(7) and v(5).

Correct Me @Anirudh
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Yes,it is a directed graph. Question is about undirected graph.
I took many examples and didnt understand the question for half an hour.

But now i realise, on reading clearly. its very easy and direct.

A LEAF. in a DFS TREE. has degree two.
Another LEAF. in a DFS TREE. has degree two.

Both can be non adjacent leaves also.

It implies that if a LEAF has degree two then it will form a cycle.

Since it is non adjacent leaves that have cycle. so Option C may be wrong.

Hence option D.
answered by Active (3.2k points)
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it will form a cycle, but how can you conclude it will include all it's neighbours from this statement ? please explain.
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draw a diagram and try to understand.. when u is in cycle then the neighbors also part of that cycle.
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@ahwan

Take that graph draw by Nikunj in his answer and see how u and it's neighbors form a cycle ..
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Thank u sir. got it.

By option Elimination :-

answered by Boss (22.7k points)
edited
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@rajesh pradhan in second graph G option A is satisfied..here vertex x is adjacent to both u and v in G..correct by adding one more node(say y) between x and v..
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@Nishant Thanks.Corrected. Now See the edited Version.

I also have doubt on this , may be helpful to remove some option .otherwise c,d is answer

answered by Veteran (60.4k points)
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@Anirudh I think instead of edge bu , edge uf should be there in the resultant Depth first Tree.
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I still have a doubt in option D why in a graph there must be a cycle .

In your example only where is the cycle , its not necessary to have a cycle at 'U' with all its neighbours ??

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1. A LEAF. in a DFS TREE. has degree two.
2. Another LEAF. in a DFS TREE. has degree two.
3. Both are non adjacent leaves also.

It is only possible that a LEAF has degree two  in original graph when it  form a cycle.

so option D is only correct :)

according to me this may be the counter example for options a,b,c

start DFS traversal from b or f

answered by Active (1.2k points)
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how option B is wrong?? removal of d or c can disconnect f and b