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Consider a weighted complete graph $G$ on the vertex set $\{v_1,v_2,.....v_n\}$ such that the weight of the edge $(v_i, v_j)$ is  $2|i-j|$. The weight of a minimum spanning tree of $G$ is:

1. $n-1$
2. $2n-2$
3. $\begin{pmatrix} n \\ 2 \end{pmatrix}$
4. $n^2$

edited | 3.4k views
+11

for a weight of the edge (vi,vj) is 2|i-j|

The weight of MST will be 2(n-1)

for a weight of the edge (vi,vj) is |i-j|

The weight of MST will be (n-1)           // for this question, there is a typo. +5
Option B should be 2n-2...
+3
It seems this question needs small correction.
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what will be the case if weight of the edge v(i, j) is | i+j | ?
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I think it would be [(n+1)(n+2)/2] - 3.

$2(n-1)$ the spanning tree will traverse adjacent edges since they contain the least weight.
by Active (3.3k points)
edited by
+1
MST will be line graph.
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MST will be line graph.

Chhotu  pls explain how?

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@set2018

MST will add the edges with minimum weight, Weight of edges = 2|i−j|.

All successive number will have weights of 2. (like 1-2 weight is 2 , 2-3 weight is 2 so on) Hence it will be a line graph of weight 2 (n-1). i.e n-1 edges each of weight 2.

Take for example K4 an try.
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Suppose I take 4 vertices as an example. Then the minimum spanning tree is a line graph with 3 edges of weight 2 each.

now both options B and C gives me the correct minimum weight as 6.
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@Gupta731

To avoid that take a complete graph of $4$ vertices and then you will only get $\mathbf {B}$ as the answer.

None of the options match the Actual answer.

For connecting every i & i+1 node we have edge of weight 2 ,therefore We get 2*(n-1).

Correct Answer: $B$
by Boss (41.9k points)
edited
+1
which is equal to 2n-2
+1 vote

Ans is 2n-2 by Junior (737 points)
+1 vote
$\underline{\mathbf{Answer:B}}\Rightarrow$

$\underline{\mathbf{Explanation:}}\Rightarrow$

Make a square of $4$ vertices and make it as a complete graph, that is, having $6$ edges.

Now, make the spanning tree. It will be the $3$ adjacent edges and the weight will be $2+2+2=6$.

Now, check the options and substitute $\mathbf {n = 4}$, then only option $\mathbf B$ is satisfied.

$\therefore \mathbf B$ will be the answer.
by Boss (19.2k points)
edited by