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Consider a simple connected graph $G$ with $n$ vertices and $n$ edges $(n > 2)$. Then, which of the following statements are true?

  1. $G$ has no cycles
  2. The graph obtained by removing any edge from $G$ is not connected
  3. $G$ has at least one cycle
  4. The graph obtained by removing any two edges from $G$ is not connected
  5. None of the above 
asked in Graph Theory by Veteran (59.5k points)
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2 Answers

+17 votes
Best answer

This seems like multiple answer questions.

Here we have $n$ vertices & $n$ edges. So we must have cycle.

So (C) has at least one cycle is True & (A) is false.

(D) The graph obtained by removing any two edges from $G$ is not connected $\rightarrow$ This is true, for graph of $n$ vertices to be connected, we need at least $n-1$ edges. If we remove $2$ out of $n$, we get $n-2$ edges, which can connect at max $n-1$ vertices. $1$ Vertex at least will be disconnected. So D is true.

(B) is false as if graph is cyclic graph then removing any edge will not disconnect graph.

Answer $\rightarrow$ (C) & (D).

answered by Boss (42.8k points)
edited by
0

Consider case of a Cycle , each cycle(Cn) is going to have n vertices and n edges

now if u delete any 2  edges of cycle the graph will always be disconnected this proves (d) is  answer

(C) is the answer bcz max number of egdes required to connect n nodes in tree is n-1 hence adding 1 more edge leads to atleast one cyle

hence c and d

Thanks :)

+1
how a cycle will be connected if we remove 2 adjacent edges?
+1
I got it ... i did it by thinking vertices sryy :( i will edit that
+4

Option (C) might not be correct because they have used the word "at least one cycle" while a graph with n vertices and n edges, graph will contain exactly 1 cycle.

But 'at least' contains the possibility of 'exactly one', hence option (C) also can be considered as true statement.

0
Option (c) can be thought as

Average vertex degree will be $\frac{2e}{n} = \frac{2(n)}{n} = 2$

Yes, if we remove 2 edges, graph will be disconnected.
0

completely agree with @ Manu thakur here "Attest " won't be right it had "exactly one" then it would have been also the option so only "d" right here

+3 votes
if there are n vertices if you want to make it as a connected graph , there should be at least n-1 edges. hence if we remove two edges, it wont be a connected graph hence answer :d
answered by Boss (11.5k points)
0
(c) is also true rt?
0

No, C is not true as the graph will have exactly 1 cycle.

+2
But exactly one cycle doesn't mean at least one cycle is not true rt?
0
'exactly one' is a subset of 'atleast one'. Both are not equivalent.
+3
Yes. Both are not equivalent. But for exactly one, even at least one is true, not the other way.
0
yes..
0
@arjun sir

how c is correct ?

'exactly one' (1) is a subset of 'atleast one' (0,1) . Both are not equivalent.
+6
^Yes, but whenever exactly one is TRUE, at least one is also true. i.e.,

exactly one $\implies$ at least one

but

at least one need not imply exactly one.
0
got !! thanx sir
0
@Arjun

for given question i have doubt on option c, bcoz range of at least 1 = [1,.....infinity)

 

now {2,3,4,...infinity} subset of [1,2,...infinity) so it hav no of cycle is only possible is one , to being a statement true if it fails for any counterexample then it should not be true there are so many counterexamples for that at least one is false so c should not be the answer


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