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Let $G$ be a weighted graph with edge weights greater than one and $G'$ be the graph constructed by squaring the weights of edges in $G$. Let $T$ and $T'$ be the minimum spanning trees of $G$ and $G'$, respectively, with total weights $t$ and $t'$. Which of the following statements is TRUE?

  1. $T' = T$ with total weight $t' = t^2 $
  2. $T' = T$ with total weight $t' < t^2$
  3. $T' \neq T$ but total weight $t' = t^2$
  4. None of the above
asked in Algorithms by Veteran (14.6k points)
edited by | 2.1k views
Please remove the tag calculus. It is misleading.
(a+b)^2  >= a^2 + b^2.  Hehe. :)
what is the final answer now?

3 Answers

+32 votes
Best answer
When the edge weights are squared the minimum spanning tree won't change.

$t'$ < $t^2$, because sum of squares is always less than the square of the sums except for a single element case.

Hence, B is the general answer and A is also true for a single edge graph. Hence, in GATE 2012, marks were given to all.
answered by Veteran (14.6k points)
edited by
A is true for single edge graph

B is true for most graphs

C is true when G has multiple MSTs.

Since option D says None of the above, shouldn't D be the answer?

Are you sure that marks were awarded to all for this question? (Please cite some source.)

None of the above would mean all others are false rt? Or the question should have asked "Always TRUE". Here is the answer key same file as published by IIT-K. 

http://gatecse.in/w/images/b/b5/Key_CS_2012.pdf

This official site supposedly still has the key. But as usual no words for IIT IT department. 

http://gate.iitk.ac.in/gate2012/anskeys.php

 

Thanks :)
pragy can u plz provide example for mst where this is true........
None of the above means None is True- not Multiple Choices are TRUE.
For D to be correct, A, B and C should be FALSE. Here, A and B can be TRUE. But now they stopped giving marks to all and so should go for the most general case- B here.
Then the question should have been "ALWAYS TRUE", though you are correct- D can also be chosen. May be due to this ambiguity, GATE gave marks to All.

in case of  multiple min spanning tree also. t > t2 and T=T'.

I don't understand your 3rd point. Could u explain? @pragya

Sir ,will you plz explain how single edge graph is satifying option a.?Like if i take a graph T with two vertices A n B with edge weight 2 and so  T' will have edge weight 4 ,so MST will have T=T' but  not t=t'.
"Let G be a weighted graph with edge weights greater than one and G′." It will always greater than G' so it will not affect any how. @neal
Hope yu got right answer now and it's clear. Thank You.
so it can be "Less than or equal to" for all cases, right?
Here (D) is the possible answer. Can anyone tell me why were marks given to all?

Pragy Agarwal Sir, How option C can be true.. Please give an example... Even in case of multiple MSTs, although T and T' may be different but still t' will not be equal to t2 in any case. Please clarify

@anchitjindal07 C can never be true as for single edge graphs multiple spanning trees are not possible
sir here by saying T' = T ... do they mean they are isomorphic ...? How can we say 2 graphs (or trees) are equal ?
@Arjun sir , marks to all when we have attempted ?
+5 votes

we can make different as well as same MST also so option (d) is correct option

answered by Active (2.1k points)
D IS MORE CORRECT THAN B
+2 votes
d will be the answer. t' may or may not b equal to t . if distict weight it will be equal and if same weight then diffrent structure may be obtained . plus the weight of t <= t'
answered by Veteran (15.6k points)
Can you Explain why T and T' are not equal and t'<t^2.

so why not ans B.


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