The Gateway to Computer Science Excellence
First time here? Checkout the FAQ!
x
+28 votes
2.8k views

An undirected graph $G(V,E)$ contains $n \: (n>2)$ nodes named $v_1,v_2, \dots, v_n$. Two nodes $v_i, v_j$  are connected if and only if $ 0 < \mid i-j\mid \leq 2$. Each edge $(v_i,v_j)$ is assigned a weight $i+j$. A sample graph with $n=4$ is shown below.

What will be the cost of the minimum spanning tree (MST) of such a graph with $n$ nodes?

  1. $\frac{1}{12} (11n^2 - 5 n)$
  2. $n^2-n+1$
  3. $6n-11$
  4. $2n+1$
asked in Algorithms by Veteran (98.4k points)
edited by | 2.8k views
0
Simple substitution .... For n=5 ... Option A will fail ... So option B ...
0
For those interested in recurrence relations, this would look like: $\color{red}{T(n) = T(n-1) + 2n-2}$  with  $\color{blue}{T(2) = 3}$ for base case.
+1
@Krish... how did you form the recurrence relation by seeing this question AND how did the idea hit your brain that you can form a recurrence relation and can solve the question like this? help me to learn also please :-)
+1
For n=5, Along with option A, option C and option D also fails

6 Answers

+21 votes
Best answer

Q 54.  Answer is B

$\text{ We observe a pattern in the weight of MST being formed }$

$\text{ For n=3 } (1+2+3)+(1)$
$\text{ For n=4 } (1+2+3+4)+(1+2)$
$\text{ For n=5 } (1+2+3+4+5)+(1+2+3)$
$\text{ These can be obtained by drawing graphs for these graphs. }$
$\therefore \text{ Total weight of MST is } \sum_{i=1}^{n}i+\sum_{i=1}^{n-2}i=n^2-n+1\\$

answered by Active (4.2k points)
edited by
0

But 55 is asking about V5 to V6 . right? and not Vto V6

+18

edge weights in mst:

=3 + 4 + 6 + 8 + 10 + 12 +...+*(2(n-1))

=1+(2+ 4+6+....till n-1 terms)

=1+ 2(1+2+3+4+...n-1) 

=1+(n-1)*n=n2-n+1

 

0
@papesh veteran, amazing approach
0

@papesh @abhishekmehta4u why in the end we have 2(n-1). I know, for n vertices in G we get (n-1) edges in MST, but where's the 2 coming from?

@Chhotu

Is it because every term is increasing by 2?

@jatin saini @srestha

+6 votes

Start from initial vertex and add each vertex at a time with its minimum weight edge.

answered by Active (4k points)
+6 votes

Caption

Q 54. Answer is B

answered by (81 points)
edited by
+1 vote

One more way to prove the same thing is as follow --> 

 

answered by Boss (10.4k points)
+2
Your answer is same as the one in the comments below the best answer. What's the point in adding the same thing ?
+1 vote

ANSWER is B:

The pattern that goes with the solution:.

 For n=3 (1+2+3)+(1) 
 For n=4 (1+2+3+4)+(1+2)
 For n=5 (1+2+3+4+5)+(1+2+3) 


For n=  {n(n+1)/2 } + {(n-2)(n-2+1)/2}

         ={(n2+n)/2} + {(n2-3n+2)/2}

         =(n2-3n+2+n2+n)/2

         =n2-n+1

answered by (21 points)
0 votes
I drew a graph for n=10 and found answer for Q54) as B and answer for Q55) as C

BUT that is a messy and time consuming procedure ... does anyone have a better approach to the given problem ?
answered by Active (3.8k points)
+2
Put the values 3,4,5,6...to check using which value the options are giving different anwers, then using the least such value, draw the required graph. In this case n=5 will suffice. You will see that the cost of MST is 21 which is satisfied only by option B. Having found the method you will know where to attach the new node, V6. Then from the new MST find the value of v5 to v6 which will be nothing but 21 + the minimum value to reach to v6 which will be (v4 to v6)=10, thus the answer will be 21+10=31.
Answer:

Related questions



Quick search syntax
tags tag:apple
author user:martin
title title:apple
content content:apple
exclude -tag:apple
force match +apple
views views:100
score score:10
answers answers:2
is accepted isaccepted:true
is closed isclosed:true

35,486 questions
42,746 answers
121,453 comments
42,138 users