GATE CSE 2016 Set 1 | Question: 39

Question

Let $G$ be a complete undirected graph on $4$ vertices, having $6$ edges with weights being $1, 2, 3,  4, 5,$ and $6$. The maximum possible weight that a minimum weight spanning tree of $G$ can have is __________

Comments

It is not at all rocket science: Here is the solution

You pick up four min edges to get MST so you pick edge with weight 1 then 2 as no choice but when picking up 3 you can ask can I avoid this one so that I can pick 4 or above which has greater weight and answer is yes you simply make weight 3 edge as so that cycle is formed, and now will have to pick edge with weight 4 as you have no choice and no additional cycle can be formed as graph is complete 4

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@arjun sir graph is complete that is all distinct vertices are connected with unique edge so only one graph structure is possible ?
Square shape with diagonal vertices connected
No of edges = 4*(4-1)/2 = 6 edges as given in graph

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I think there are only 2 mst possible with min weight 6,7 so answer will be 7

this question gets to many view just due to  wrong interpretaion of question i.e. what actually asking in question is confusing means should we apply mst algo or not like this confusion .

Answer 1

Answer should be 1, 2 and 4…

Why?
Given that its complete graph so we will have cycle for sure as soon as we add 3^rd edge..
now there are many graphs possible….First draw 2 edges...which least weights(1 and 2 ) note we cannot have cycle with 2 edges but if we add 3^rd we will have cycle...now add the 3^rd edge such a way the next minimum which weight edge which is 3 in our case forms a cycle..next possible least weight edge would be 4...and includes all vertex..

Answer 2

The Best possible arrangement which can give minimum cost spanning tree with maximum weight is
 

Which gives weight as 1 + 2 + 4
= 7