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Let $G$ be an undirected connected graph with distinct edge weights. Let $e_{max}$ be the edge with maximum weight and $e_{min}$ the edge with minimum weight. Which of the following statements is false?

  1. Every minimum spanning tree of $G$ must contain $e_{min}$
  2. If $e_{max}$ is in a minimum spanning tree, then its removal must disconnect $G$
  3. No minimum spanning tree contains $e_{max}$
  4. $G$ has a unique minimum spanning tree
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3 Comments

Any example in support of c?
0

since all edge is distinct so emin must be tthere so true

option b is true.

..

remove of AE can give the disconnected G

Option C is false.

4
option B) is direct implication of option c) property of cut vertex
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4 Answers

31 votes
 
Best answer

C the case should be written as "may or may not", to be true.

D will always be true as per the question saying that the graph has distinct weights.

Correct Answer: $C$

edited by

6 Comments

if there is a bridge in the graph, that must be included in MST
16
If in a graph there is an edge which connects a pendant vertex to the graph, then however heavy that edge may be, that edge will always be included in all MST of the graph G.

So, option (c) is wrong.
9
option b is bidirectional na ?
0
do anyone have diagrammatic proof for option B. ??
0
this is a really good question.
0

@minal

NO, definitely not. pendant vertex can’t always be of maximum edge right?

0
15 votes

option a is true. emin should be there in all MST

option b is true - if emax there that means that is the only edge reachable to one of the incident vertices of it. Otherwise we will select lesser weight edge incident on that vertex, Hence its removal will disconnect G

we cannot infer whether c  and d are true always. sometimes they can be false

6 Comments

(d) is actually true. Since, edge weights are unique, minimum spanning tree can be only one.
13
you are right

actually i missed out the point distinct edge

hence option c is the answer to the question since from the given data we cannot infer that
1
Even if the edge weights are unique, Can't it have multiple spanning trees? Like if ST1 contains two edges with weights 4 and 3, and ST2 contains the same edges as ST1 but instead of those two edges it contains edges with weights 2 and 5 (both pairs summing up to 7). Both will have the same total edge weight sum.

I am not sure about this, is there anything wrong in my approach?
0

sir it is not always true that we get unique spanning tree

its counter example is in above image

here possible mst is 5+8+10=23

and 6+7+20=23

0

but in the given graph minimum spanning tree cost is not 23.

minimum spanning tree is 5+6+7 = 18

What you are commented is about there are two different spanning tree of same cost.

but they are not minimum spanning tree. MInimum spanning tree is a spanning tree with minimum cost.

Please note that option D is G has a unique minimum spanning tree

1
If G is the unique min spanning tree just bcz edge weights are unique then why not option 1 is true..
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3 votes
Option 1 :- Every minimum spanning tree of G must contain $e_{min}$.

Kruskal's algorithm always picks the edges in ascending order of their weights while constructing a MST of G. So yes , it is true.

 

Option 2 : -  If $e_{max}$ is in a minimum spanning tree, then its removal must disconnect G .

$e_{max}$ would be included in MST if and only if , $e_{max}$ is a bridge between two connected components , removal of which will surely disconnect the graph.

Option 3 :- No minimum spanning tree contains $e_{max}$.

Contradictory statement , already proved in option 2 that $e_{max}$ can be in MST. Thus option 3 is false.

Option 4 :- $G$ has a unique minimum spanning tree.

G has unique edge weights , so MST will be unique . In case if edge weights were repeating , there could've been a possibility of non-unique MSTs.

Thus it is true.
2 votes

Only option C is False.

2 Comments

but according to the question all  edges should be distinct
0
Yes it is... i have given all gate question possible into 1 page.
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Answer:

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