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Let $G$ be an arbitrary graph with $n$ nodes and $k$ components. If a vertex is removed from $G$, the number of components in the resultant graph must necessarily lie down between

  1. $k$ and $n$
  2. $k-1$ and $k+1$
  3. $k-1$ and $n-1$
  4. $k+1$ and $n-k$
asked in Graph Theory by Veteran (59.6k points)
edited by | 3.2k views
0

Nice explanation for this question :

3 Answers

+50 votes
Best answer

If a vertex is removed from the graph $G$,

Lower Bound: number of components decreased by one $= k - 1$ (remove an isolated vertex which was a component)

Upper Bound: number of components $= n-1$ (consider a vertex connected to all other vertices in a component as in a star and all other vertices outside this component being isolated. Now, removing the considered vertex makes all other $n-1$ vertices isolated making $n-1$ components)

Therefore (C).

answered by Active (3.9k points)
edited by
0

in a completely connected graph, there is only one component, so after removing one vertex, the number of connected components still stay 1, so should it not be k to n-1 ?

+4
We are talking about lower bound here.'k-1' is lesser than k and that's the lowest number of components possible.
0

is it true that lies between k-1 and n-1 include k-1 and [email protected] DanishArjun sir

0
@Arjun Sir lower bound will be 'K' if we remove one vertex here then component may not increase.
+3

This will help ...

+3 votes

Option C ...

answered by Boss (11.4k points)
–3 votes
Lower bound k-1 ,upper bound n-k.
answered by Active (3.3k points)
0
Because if u remove frm star graph it only has one compo so n-1 but the general case will be n-k for k compo graph.
+2
But consider the case where the other components are isolated vertices.
0
Hello anshu , he is asking about maximum possible number of components , which would be n-1 and n-k is supposed to be lesser or equal to n-1.
+1
yes when graph is star graph then components will be n-1 components will be come.
Answer:

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