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?

- Every minimum spanning tree of $G$ must contain $e_{min}$
- If $e_{max}$ is in a minimum spanning tree, then its removal must disconnect $G$
- No minimum spanning tree contains $e_{max}$
- $G$ has a unique minimum spanning tree

## 4 Answers

### 6 Comments

option a is true. e_{min} should be there in all MST

option b is true - if e_{max }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

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

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**

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.